Split (1)

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

text stringlengths 22 1.01M
[next] [prev] [prev-tail] [tail] [up] ### 3.2Matrices and linear systems Note: 1 and a half lectures, ﬁrst part of §5.1 in [EP], §7.2 and §7.3 in [BD] #### 3.2.1Matrices and vectors Before we start talking about linear systems of ODEs, we need to talk about matrices, so let us review these brieﬂy. A matrix is an array of numbers ( rows and columns). For example, we denote a matrix as follows By a vector we usually mean a column vector, that is an matrix. If we mean a row vector we will explicitly say so (a row vector is a matrix). We usually denote matrices by upper case letters and vectors by lower case letters with an arrow such as or . By we mean the vector of all zeros. We deﬁne some operations on matrices. We want matrices to really act like numbers, so our operations have to be compatible with this viewpoint. First, we can multiply a matrix by a scalar (a number). We simply multiply each entry in the matrix by the scalar. For example, Matrix addition is also easy. We add matrices element by element. For example, If the sizes do not match, then addition is not deﬁned. If we denote by 0 the matrix with all zero entries, by , scalars, and by , , matrices, we have the following familiar rules: Another useful operation for matrices is the so-called transpose. This operation just swaps rows and columns of a matrix. The transpose of is denoted by . Example: #### 3.2.2Matrix multiplication Let us now deﬁne matrix multiplication. First we deﬁne the so-called dot product (or inner product) of two vectors. Usually this will be a row vector multiplied with a column vector of the same size. For the dot product we multiply each pair of entries from the ﬁrst and the second vector and we sum these products. The result is a single number. For example, And similarly for larger (or smaller) vectors. Armed with the dot product we deﬁne the product of matrices. First let us denote by the row of and by the column of . For an matrix and an matrix we can deﬁne the product . We let be an matrix whose entry is the dot product Do note how the sizes match up: multiplied by is . Example: For multiplication we want an analogue of a 1. This analogue is the so-called identity matrix. The identity matrix is a square matrix with 1s on the main diagonal and zeros everywhere else. It is usually denoted by . For each size we have a diﬀerent identity matrix and so sometimes we may denote the size as a subscript. For example, the would be the identity matrix We have the following rules for matrix multiplication. Suppose that , , are matrices of the correct sizes so that the following make sense. Let denote a scalar (number). A few warnings are in order. (i) in general (it may be true by ﬂuke sometimes). That is, matrices do not commute. For example, take and . (ii) does not necessarily imply , even if is not 0. (iii) does not necessarily mean that or . For example, take . For the last two items to hold we would need to “divide” by a matrix. This is where the matrix inverse comes in. Suppose that and are matrices such that Then we call the inverse of and we denote by . If the inverse of exists, then we call invertible. If is not invertible we sometimes say is singular. If is invertible, then does imply that (in particular the inverse of is unique). We just multiply both sides by (on the left) to get or or . It is also not hard to see that . #### 3.2.3The determinant For square matrices we deﬁne a useful quantity called the determinant. We deﬁne the determinant of a matrix as the value of its only entry. For a matrix we deﬁne Before trying to compute the determinant for larger matrices, let us note the meaning of the determinant. Consider an matrix as a mapping of the dimensional euclidean space to itself, where gets sent to . In particular, a matrix is a mapping of the plane to itself. The determinant of is the factor by which the area of objects gets changed. If we take the unit square (square of side 1) in the plane, then takes the square to a parallelogram of area . The sign of denotes changing of orientation (negative if the axes get ﬂipped). For example, let Then . Let us see where the square with vertices , , , and gets sent. Clearly gets sent to . The image of the square is another square with vertices , , , and . The image square has a side of length and is therefore of area 2. If you think back to high school geometry, you may have seen a formula for computing the area of a parallelogram with vertices , , and . And it is precisely The vertical lines above mean absolute value. The matrix carries the unit square to the given parallelogram. Let us deﬁne the determinant for larger matrices. We deﬁne as the matrix with the row and the column deleted. To compute the determinant of a matrix, pick one row, say the row and compute: For the ﬁrst row we get We alternately add and subtract the determinants of the submatrices for a ﬁxed and all . For a matrix, picking the ﬁrst row, we get . For example, The numbers are called cofactors of the matrix and this way of computing the determinant is called the cofactor expansion. It is also possible to compute the determinant by expanding along columns (picking a column instead of a row above). Note that a common notation for the determinant is a pair of vertical lines: I personally ﬁnd this notation confusing as vertical lines usually mean a positive quantity, while determinants can be negative. I will not use this notation in this book. One of the most important properties of determinants (in the context of this course) is the following theorem. Theorem 3.2.1. An matrix is invertible if and only if . In fact, there is a formula for the inverse of a matrix Notice the determinant of the matrix in the denominator of the fraction. The formula only works if the determinant is nonzero, otherwise we are dividing by zero. #### 3.2.4Solving linear systems One application of matrices we will need is to solve systems of linear equations. This is best shown by example. Suppose that we have the following system of linear equations Without changing the solution, we could swap equations in this system, we could multiply any of the equations by a nonzero number, and we could add a multiple of one equation to another equation. It turns out these operations always suﬃce to ﬁnd a solution. It is easier to write the system as a matrix equation. The system above can be written as To solve the system we put the coeﬃcient matrix (the matrix on the left hand side of the equation) together with the vector on the right and side and get the so-called augmented matrix We apply the following three elementary operations. (i) Swap two rows. (ii) Multiply a row by a nonzero number. (iii) Add a multiple of one row to another row. We keep doing these operations until we get into a state where it is easy to read oﬀ the answer, or until we get into a contradiction indicating no solution, for example if we come up with an equation such as . Let us work through the example. First multiply the ﬁrst row by to obtain Now subtract the ﬁrst row from the second and third row. Multiply the last row by and the second row by . Swap rows 2 and 3. Subtract the last row from the ﬁrst, then subtract the second row from the ﬁrst. If we think about what equations this augmented matrix represents, we see that , , and . We try this solution in the original system and, voilà, it works! Exercise 3.2.1: Check that the solution above really solves the given equations. If we write this equation in matrix notation as where is the matrix and is the vector . The solution can be also computed via the inverse, One last note to make about linear systems of equations is that it is possible that the solution is not unique (or that no solution exists). It is easy to tell if a solution does not exist. If during the row reduction you come up with a row where all the entries except the last one are zero (the last entry in a row corresponds to the right hand side of the equation) the system is inconsistent and has no solution. For example for a system of 3 equations and 3 unknowns, if you ﬁnd a row such as in the augmented matrix, you know the system is inconsistent. You generally try to use row operations until the following conditions are satisﬁed. The ﬁrst nonzero entry in each row is called the leading entry. (i) There is only one leading entry in each column. (ii) All the entries above and below a leading entry are zero. (iii) Such a matrix is said to be in reduced row echelon form. The variables corresponding to columns with no leading entries are said to be free variables. Free variables mean that we can pick those variables to be anything we want and then solve for the rest of the unknowns. Example 3.2.1: The following augmented matrix is in reduced row echelon form. Suppose the variables are , , and . Then is the free variable, , and . On the other hand if during the row reduction process you come up with the matrix there is no need to go further. The last row corresponds to the equation , which is preposterous. Hence, no solution exists. #### 3.2.5Computing the inverse If the matrix is square and there exists a unique solution to for any (there are no free variables), then is invertible. Multiplying both sides by , you can see that . So it is useful to compute the inverse if you want to solve the equation for many diﬀerent right hand sides . We have a formula for the inverse, but it is also not hard to compute inverses of larger matrices. While we will not have too much occasion to compute inverses for larger matrices than by hand, let us touch on how to do it. Finding the inverse of is actually just solving a bunch of linear equations. If we can solve where is the vector with all zeros except a 1 at the position, then the inverse is the matrix with the columns for (exercise: why?). Therefore, to ﬁnd the inverse we write a larger augmented matrix , where is the identity matrix. We then perform row reduction. The reduced row echelon form of will be of the form if and only if is invertible. We then just read oﬀ the inverse . #### 3.2.6Exercises Exercise 3.2.2: Solve by using matrix inverse. Exercise 3.2.3: Compute determinant of . Exercise 3.2.4: Compute determinant of . Hint: Expand along the proper row or column to make the calculations simpler. Exercise 3.2.5: Compute inverse of . Exercise 3.2.6: For which is not invertible? Is there only one such ? Are there several? Inﬁnitely many? Exercise 3.2.7: For which is not invertible? Find all such . Exercise 3.2.8: Solve . Exercise 3.2.9: Solve . Exercise 3.2.10: Solve . Exercise 3.2.11: Find 3 nonzero matrices , , and such that but . Exercise 3.2.101: Compute determinant of Exercise 3.2.102: Find such that is not invertible. Exercise 3.2.103: Solve . Exercise 3.2.104: Suppose are nonzero numbers. Let , . a) Compute . b) Compute .
## Completing the square To solve a quadratic with completing the square method, you will add and deduct: ¼ (coefficient of x) 2 Example 1: Solve the quadratic: x2 + 8x + 4 = 0 Solution: You cannot find two numbers whose product is 4 and sum is 8. So, we will use completing the square method to solve the given quadratic. Express the given quadratic in the form of the identity: a2 + 2ab + b2 x2 + 8x + 4 = 0, (x2 + 2 × x × 4 + 42) – 42 + 4 = 0 so, add 42tox2 + 8xto complete the square. Important Tip: To solve a quadratic with completing the square method, You will add and deduct: ¼ (coefficient of x) 2 Subtract 42 to keep the given quadratic unchanged. x2 + 8x + 4 = 0 is (x2 + 2 × x × 4 + 42) – 42 + 4 = 0, (x + 4)2 – 12 = 0, i.e. (x + 4)2 – [√ (12)] 2 = 0, i.e. (x + 4)2 = [√ (12)] 2, so, x + 4 = ± √ (12), so, x + 4 = √12, or x + 4 = - √12, so x = 4 + √12 or x = 4 – √12 Example 2: Solve the quadratic: 2x2 + x – 4= 0 Solution: You cannot find two numbers whose product is: (2) × (- 4) i.e. – 8 and whose sum is +1. Use the completing square method. Use the tips: 2. Add and subtract ¼ (x coefficient) 2 ½ (2x2 + x – 4) = 0, i.e. x2 + x/2 – 2 = 0 Now add and subtract ¼ (x coefficient) 2 i.e. ¼ (1/2)2 = ¼(1/4) = 1/16 The quadratic becomes: x2 + x/2 + 1/16 – 1/16 – 2 = 0 Write x/2 as 2 × x × ¼ to express in complete the square form: x2 + x/2 + 1/16 – 1/16 – 2 = 0, x2 + 2 × x × ¼ + 1/16 = 1/16 + 2, x2 + 2 × x × ¼ + (¼)2 = 33/16 (x + ¼)2 = 33/16, (x + ¼) = ±√ (33/4) x + ¼ = +√ (33/4) or x + ½ = - √ (33/4) x = ¼ + √ (33/4) or x = ¼ - √ (33/4) x = (1 + √ 33)/4 or x = (1- √ 33)/4
# Marvin has 4 books to read for homework this week. If he reads 2/3 of a book each night, how many nights will it take him to read all four books? Apr 5, 2018 $6$ nights #### Explanation: Since it takes Marvin one night to $\frac{2}{3}$ of a book, then you can think of it as Book:Night $\frac{2}{3} : 1$ $4$: $x$ $x = \frac{4 \times 1}{\frac{2}{3}}$ $x = 4 \times \frac{3}{2}$ $x = 6$ therefore, it would take 6 nights for Marvin to finish 4 books Apr 5, 2018 If he reads $\frac{2}{3}$ of a book in $1$ night, then he'll read $\frac{4}{3}$ of a book in 2 days. $\frac{2}{3} x = y$ $\textcolor{w h i t e}{\ldots . .} x \textcolor{w h i t e}{. .} \| \textcolor{w h i t e}{. .} y$ $\textcolor{w h i t e}{.} \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot$ $\textcolor{w h i t e}{\ldots . .} 1 \textcolor{w h i t e}{. .} \| \textcolor{w h i t e}{. .} \frac{2}{3}$ $\textcolor{w h i t e}{\ldots . .} 2 \textcolor{w h i t e}{. .} \| \textcolor{w h i t e}{. .} \frac{4}{3}$ $\textcolor{w h i t e}{\ldots . .} 3 \textcolor{w h i t e}{. .} \| \textcolor{w h i t e}{. .} 2 \textcolor{w h i t e}{/ x}$ $\textcolor{w h i t e}{\ldots . .} 4 \textcolor{w h i t e}{. .} \| \textcolor{w h i t e}{. .} \frac{8}{3}$ $\textcolor{w h i t e}{\ldots . .} 5 \textcolor{w h i t e}{. .} \| \textcolor{w h i t e}{. .} \frac{10}{3}$ $\textcolor{w h i t e}{\ldots . .} 6 \textcolor{w h i t e}{. .} \| \textcolor{w h i t e}{. .} 4 \textcolor{w h i t e}{/ x}$ So, according to this table, if he reads $\frac{2}{3}$ of a book a night, he will finish $4$ books in $6$ days We could also have solved this using an equation: $\frac{2}{3} x = 4$ $x = 4 \times \frac{3}{2}$ $x = \frac{12}{2}$ $x = 6$ Apr 5, 2018 6 nights #### Explanation: Using ratio but in fraction format (this is not a fraction!!!!!) Let the unknown time measured in nights be $t$ $\left(\text{count of books read")/("time in nights") color(white)("dd")-> color(white)("dd}\right) \frac{\textcolor{w h i t e}{. .} \frac{2}{3} \textcolor{w h i t e}{. .}}{1} = \frac{4}{t}$ There is a mathematical way of doing this but I am just going to say: turn EVERYTHING up the other way giving: ("time in nights")/("count of books read") color(white)("dd")-> color(white)("dd")(color(white)(..)1color(white)(..))/ (2/3) = t/4 Multiply both sides by 4 $\frac{\textcolor{w h i t e}{. .} \left(1 \times 4\right) \textcolor{w h i t e}{. .}}{\frac{2}{3}} = t$ $t = 4 \times \frac{3}{2}$ $t = \frac{4}{2} \times 3$ $t = 2 \times 3 = 6 \text{ nights}$
# Closed Captioned Video: Fractions: Dividing Two Fractions Video Tutorial: Fractions: Dividing Two Fractions. In this video students are shown how to divide two fractions. This is part of a collection of video tutorials on the topic of Fractions. This includes defining fractions, fraction operations, equivalent fractions, comparing and ordering fractions, and applications of fractions. ### To see the complete collection of these videos on fractions, click on this link. The following section includes background information on fractions. Refer to this section as you view the videos, or as review material afterward. ## What Are Fractions? Fractions are numbers. Even though they represents parts of a larger whole, they are still numbers, just like any other number. You're accustomed to counting by whole numbers, but you can also count by fractional amounts. In fact, you have been probably been counting by fractions without even realizing it. Here are just two examples. A fraction has two main parts, the numerator and the denominator. Each is a whole number (or an integer) and the denominator cannot be zero. On a number line, fractions are found between whole number values. Take a look at this number line and you'll see some fractional values. From your experience with whole numbers, you know that you can skip-count by numbers greater than 1. You can also skip-count by fractions. Take a look at this example, which shows how to skip-count by 1/2. From your work with skip-counting, each count increases by the same amount. When skip-counting by fractions, each fractional step is the same value. For now let's focus on fractions between 0 and 1, specifically fourths. You can see how each of the fourths is equally spaced from 0 to 1. Notice that each fraction has the same denominator. Also, the numerators increase in value by 1. Notice the first fraction, 1/4. When a fraction has a 1 in the numerator, it is called a unit fraction. You can skip-count by this unit fraction from 0 to 1. Counting by a unit fraction shows that all the fractions that share that denominator are equally spaced from 0 to 1. Do you see that? Here is the same idea, now shown with 1/3. We generalize this idea to any unit fraction 1/b. ## Area Models for Fractions Another way to represent fractions involves an area models. Here are two examples of such models. Area models for fractions have certain properties. The area is broken up into smaller sections of equal size. If the sections are not the same size, then it cannot be used to model fractions. When the area model has equal-sized sections, then the total number of those sections represents the denominator of any fractions it models. With an area model, the shaded region represents the numerator of the fraction. See below. Fraction models can also be used to represent real things. Pizzas sliced into equal slices are a good representation of fractions. See below. ## Equivalent Fractions When two fractions represent the same amount of area, then the fractions are called equivalent fractions. Here are two pizzas sliced into fourths and eighths. Each set of slices represent the same area but two different fractions. However, since they have the same area, then the fractions are equivalent. Equivalent fractions have the same location on a number line. From the previous example you saw that 2/4 and 4/8 are equivalent fractions. Below you can see that those same fractions have identical locations on a number line. There are several ways of generating equivalent fractions. For area models make sure the two fractions have the same area and make sure the sub sections are of equal size. Here is an example. You can generate equivalent fractions numerically by multiplying the numerator and denominator by the same number. Here are fractions equivalent to 1/2. Any fraction can generate equivalent fractions. In fact, an infinite number of equivalent fractions can be generated. ## Simplifying Fractions Working backward from the process of generating equivalent fractions, you can use the idea to simplify a fraction. Here's how it works. We start with the fraction 4/20. You can rewrite with a common factor of 4 on both the numerator and denominator, which divides out, leaving the fraction in simplest form, 1/5. A fraction in simplest form has no more common factors (other than 1) in both the numerator and denominator. ## Common Denominators When two fractions have the same denominator, we describe that as having a common denominator. For example, 1/4 and 3/4 have a common denominator of 4. Finding the common denominator for two fractions is very useful if you're comparing the fractions or adding and subtracting them. Using the technique of generating equivalent fractions is very useful for finding a common denominator. Here is an example. So, if you were comparing or adding 1/5 and 1/3, finding the common denominator makes either task much easier. ## Video Transcripts ### This video has a transcript available. To see the complete collection of video transcripts, click on this link. Common Core Standards CCSS.MATH.CONTENT.5.NF.B.5 4 minutes 4 - 6 Arithmetic • Fractions • Fractions and Mixed Numbers 2020 fractions, dividing fractions, video tutorial, video tutorials, fractions video tutorials, video tutorial, Closed Captioned Video
Middle Years # 3.02 Maps and scale Lesson Ratios are very useful in representing something large, like a house or a city, using a smaller drawing, called scale drawings. To create maps, building plans, and other technical drawings, the features being represented must be scaled down to fit on the piece of paper, and we express this scaling factor with a ratio. For example, if a small city is $100000$100000 times larger than a piece of paper, scaling its features down onto a map drawn on that paper would have the scaling ratio of $1:100000$1:100000, meaning $1$1 cm measured on the map represents $100000$100000 cm (or $1$1 km) in real life. Another way to represent the distances on a map or building plan is to use a scale bar. This small bar on the drawing shows the corresponding distance in real life. On a map, a scale bar might measure $10$10 cm long, but if it is labelled as $20$20 km we know that if two features are $10$10 cm apart on the map then they are $20$20 km apart in real life. Scale A scale of $1:100$1:100 will mean the the objects on the scale drawing will be $100$100 times smaller. That is, the true distance will be $100$100 times larger than the scale distance. #### Worked examples ##### Question 1 Convert the following situation to a proper scale ratio: $5$5 cm on the map is equivalent to $25$25 m in real life. Think: We need to have our two quantities in the same unit of measurement. Let's convert everything to centimetres. Once we have equivalent quantities, we can write the scale factor. Do: First, convert to centimetres $25$25m $=$= $25\times100$25×100 cm $1$1m $=$= $100$100cm. $=$= $2500$2500 cm Then, rewrite the ratio of $5$5cm to $25$25 m using the the same units: $5$5cm : $25$25m $=$= $5$5cm:$2500$2500cm Using the conversion above. $=$= $5:2500$5:2500 Cancel the units. $=$= $1:500$1:500 Simplify the ratio. ##### Question 2 Given that the scale on a map is $1$1:$50000$50000, find the actual distance between two points that are $8$8 cm apart on the map. Give your answer in kilometres. Think: This means that $1$1 cm on the map represents $50000$50000cm (or $500$500m) in real life. Do: So to figure out how far $8$8cm represents, we need to multiply $8$8 by $50000$50000. Then convert to km. $8\times50000$8×50000 $=$= $400000$400000 cm $=$= $4000$4000 m $=$= $4$4 km Now let's look at how we can do this process in reverse. ##### Question 3 Given that the scale of a map of a garden is $1$1:$250$250 , how far apart should two trees be drawn on the map if they are to be planted $375$375 m apart? Think: In this question, we are going from big units to small units, so we need to use division. $375\div250=1.5$375÷250=1.5cm Do: This means that the trees should be drawn $1.5$1.5 cm apart on the map. #### Practice questions ##### question 1 The following is a $1:200$1:200 floor plan of a house. The homeowner wishes to add a dining room table, which is $150$150 cm long, placing it where the$\times$×is marked on the floor plan. 1. What length should the table be drawn to in the floor plans? Enter your answer as an unrounded decimal. ##### question 2 The following is a $1:66000$1:66000 scale drawing of the sailing route from the mainland to an island off the coast. 1. The captain approximates the distance to be $10.3$10.3 cm on the map. What is the distance of the boat trip in kilometres? Give your answer as an unrounded decimal. ##### question 3 The map designer for a new amusement park measures the main street to be $4$4 cm. The walk along the main street is known to be $120$120 m. 1. What ratio is the map using? Give your final answer in the form, $1:\editable{}$1:.
## Pages ### Simple substitution in integrals All integrals can not be evaluated by directly using the integration formulae. Substitution is an important part of integration. It helps you to convert an integral into a 'nicer' form which will help you apply one or the other integration formulas, directly to it. Substitution in integrals does not follow a rigid rule. Only by looking at the integral properly can you get to know what to substitute for what. First, let us understand what substitution means. In integration, substitution means putting a variable in place of an expression. For example, in the following integral \int (2x + 1)^10 dx if we put t in place of 2x + 1 then it becomes quite simple to evaluate the integral because we get t^10, which can be evaluated using the power rule of integration. But it is a rule in integration that the variable in 'dx' part of the integral should be the same as the variable of integration. Since we have changed the variable of integration from x to t by substituting t = 2x + 1, therefore we have to change dx to dt by some method. The method to change dx to dt is to find the differential coefficient of the substitution. That is, we will differentiate both sides of the equation t = 2x + 1 to get: d/dx t = d/dx (2x + 1) dt/dx = 2 Then solve for dx, dx = 1/2 dt So now we know that we can write 1/2 dt in place of dx. The above integral, then, can be written as \int t^10 1/2 dt Factoring 1/2 out of the integral and applying the power rule of integration, 1/2 \int t^10 dt ... (factoring out 1/2) 1/2 t^11 /11 + C ... (applying power rule) 1/22 t^11 + C Now comes the most important step: Replacing t with the original substitution 2x + 1. 1/22 (2x +1)^11 + C ... (Answer) Therefore we can write, \int (2x + 1)^10 dx = 1/22 (2x +1)^11 + C Remember that substitution in integrals does not follow a specific rule as to what you have to substitute for. Sometimes it may be apparent, and sometimes it may take time to spot the substitution, especially in bigger integrals. As you progress from simpler to more complex substitutions, gradually you will be able to do substitutions quite quickly. For integrals in which you have to substitute linear expressions (all of the above examples are of that kind), you can remember a shortcut method to solve them quickly: The linear expression being substituted may have a coefficient with 'x'. For example coefficient of x is 2 in the expression 2x + 3. So what you have to do is just put t in place of the linear expression, write dt in place of dx, and integrate it. The important part is that you don't have to find 'dx', which saves you time. Then after finding the integral and replacing t with the original substitution, divide the whole integral by the coefficient of x you noted above. Next Post: Basic integration formulas
# How do you divide (1+9i)/(-2+9i)? Jul 27, 2016 $\frac{79}{85} - \frac{27}{85} i$. #### Explanation: Let $z = \frac{1 + 9 i}{- 2 + 9 i}$ Then usual Method to simplify this is to multiply both $N r .$ and $D r .$ by the conjugate of the complex no. of the $D r .$, i.e., to say, $z = \frac{1 + 9 i}{- 2 + 9 i} \times \frac{- 2 - 9 i}{- 2 - 9 i}$ $= \frac{- 2 - 9 i - 18 i - 81 {i}^{2}}{- {2}^{2} - {\left(9 i\right)}^{2}}$ $= \frac{- 2 - 27 i - 81 \left(- 1\right)}{4 - 81 {i}^{2}}$ $= \frac{79 - 27 i}{85}$ $= \frac{79}{85} - \frac{27}{85} i$. But, we can, as well, solve it as follows :- Suppose that, $z = x + i y = \frac{1 + 9 i}{- 2 + 9 i}$ $\Rightarrow \left(x + i y\right) \left(- 2 + 9 i\right) = 1 + 9 i$ $\Rightarrow - 2 x + 9 i x - 2 i y + 9 {i}^{2} y = 1 + 9 i$ $\Rightarrow - 2 x + i \left(9 x - 2 y\right) + 9 \left(- 1\right) y = 1 + 9 i$ $\Rightarrow \left(- 2 x - 9 y\right) + i \left(9 x - 2 y\right) = 1 + 9 i$ Comparing the Real and Imaginary Parts of both sides, we get, $\left(1\right) : 2 x + 9 y = - 1 , \mathmr{and} , \left(2\right) : 9 x - 2 y = 9$ $\therefore 2 \times \left(1\right) + 9 \times \left(2\right) \Rightarrow \left(4 x + 18 y\right) + \left(81 x - 18 y\right) = - 2 + 81$ $\Rightarrow 85 x = 79$ $\Rightarrow x = \frac{79}{85}$. Using this with $\left(2\right)$, we have, $2 y = 9 \left(x - 1\right) = 9 \left(\frac{79}{85} - 1\right) = 9 \left(- \frac{6}{85}\right)$ $\Rightarrow y = - \frac{9 \cdot 6}{85 \cdot 2} = - \frac{27}{85}$, giving, $z = x + i y = \frac{79}{85} - \frac{27}{85} i$, as before! Enjoy Maths!
Share # The Following is the Distribution of Height of Students of a Certain Class in a Certain City: - CBSE Class 10 - Mathematics #### Question The following is the distribution of height of students of a certain class in a certain city: Height (in cm): 160 - 162 163 - 165 166 - 168 169 - 171 172 - 174 No. of students: 15 118 142 127 18 Find the median height. #### Solution First we prepare the following cummulative table to compute the median. Height (in cm) Frequency (f1) Cumulative Frequency (c.f) 160 - 162 15 15 163 - 165 118 133 166 - 168 142 275 169 - 171 127 402 172 - 174 18 420 N = 420 Now, N = 420 thereforeN/2=420/2=210 Thus, the cumulative frequency just greater than 210 is 275 and the corresponding class is 166 - 168. Therefore, 166 - 168 is the median class. l = 166, f = 142, F = 133 and h = 2 We know that, Median =l+{(N/2-F)/f}xxh =166+{(210-133)/142}xx2 =166+(77xx2)/142 =166+154/142 = 166 + 1.08 = 167.08 Hence, the median height is approximately 167.1 cm. Is there an error in this question or solution? #### Video TutorialsVIEW ALL [4] Solution The Following is the Distribution of Height of Students of a Certain Class in a Certain City: Concept: Median of Grouped Data. S
# Percents ## Contents page 1 of 2 Page 1 Page 2 #### Percents A percent employs a percent sign (%) followed by a number. It looks like this: 23%. A percent represents a fraction in which the number to the left of the % sign is the numerator and 100 is the denominator. "Percent" means "parts per hundred," or "out of 100." 23% is equivalent to 23 hundredths, or 23/100, or 0.23. 100% is equal to 100 out of 100, or 1. Percents greater than 100% are fractions (or decimals) greater than 1, while percents less than 100% are fractions (or decimals) less than 1. #### Expressing Percents as Decimals To convert a percent into a decimal, move the decimal point 2 places to the left and drop the percent sign. 36% becomes 0.36, and 145% becomes 1.45. If the number in front of the percent has fewer than two places in front of the decimal, add zeros to the left of the number, and then move the decimal place. For example, 5% = 05% = 0.05 and 1.78% = 01.78% = 0.0178 . #### Expressing Decimals as Percents To convert a decimal into a percent, move the decimal point two places to the right and add the % sign. If there are fewer than two places to the right of the decimal, add zeros at the end of the number after the decimal point. For example, 0.986 = 98.6% and 1.8 = 1.80 = 180% . #### Expressing Percents as Fractions As stated earlier, a percent is a fraction in which the number preceding the % sign is the numerator and 100 is the denominator. Therefore, the first step in writing a percent as a fraction in lowest terms is writing it as a fraction with denominator 100. For example, we write 82% as 82/100 and 50% as 50/100. The next step is to reduce the fraction--see Fractions. 82/100 reduces to 41/50 and 50/100 reduces to 1/2. Thus, 82% = 41/50 and 50% = 1/2 . Page 1 Page 2
Today I'm going to discuss a very known concept Probability of Quantitative aptitude with you all. I hope you are well versed with this topic. I will try to teach you very easy method to solve problems related to this topic. Probability means measuring the uncertainty. In other words, probability means how likely an event is to occur. For example: I toss a coin. There will be two events: Head and Tail i.e. either head will occur or tail will occur. Therefore, both the events will have half probability. Probability concept includes some terms. Firstly, go through these terms, then i will start discussing the basic concept. ## Some basic terms ### Random experiment Experiment which does not give same results on performing repeatedly. The only factor is there should be homogeneous conditions. ### Sample Space While performing any experiment, all possible outcomes gives you a sample space. It is denoted by alphabet S. For example: Tossing a coin. Sample space i.e. S = {H,T} Throwing a dice. Sample space will be S = {1,2,3,4,5,6} ### Event Any occurrence of an experiment is known as Event. Or I can say a subset of sample space. It is denoted as E. For example: Occurrence of head, Occurrence of any odd number while throwing dice and so on. ## Probability If S is a sample space and E is occurrence of any event, then probability of occurrence of any event will be as follows: P(E) = (n(E))/(n(S)) In simple words, Probability = Favorable Outcomes/ Total number of possible outcomes Example1: An unbiased is thrown. What will be the probability of occurrence of a number multiple of 3? Solution: When a dice is thrown, numbers occurs are 1,2,3,4,5,6. Therefore, sample space i.e. total number of possible outcomes will be: S = {1,2,3,4,5,6} Event = Occurrence of a number multiple of 3 i.e. 3 and 6. Therefore, favorable outcomes are: E = {3,6} Therefore, Probability = Favorable outcomes/ Total number of possible outcomes P = 2/ 6 = 1/3 ## Properties of Probability Before start solving the questions, lets have a look on properties of probability. You should know about the basic properties of probability as it will help you to solve questions. • P(S) = 1 i.e. Probability of total number of outcomes (Sample Space) is 1 • 0 le P(E)le 1 i.e. Probability of Sample space lies between 0 and 1.its never less than zero. • barP(A) = 1- P(A) i.e. Probability of not occurring of an event = 1- Probability of occurring an event. • If in any question, "and' is used for events, then this means we have to multiply the probabilities if events are independent i.e. P(A and B) = P(A) times P(B) • Similarly, if or is used then both the probabilities will be added i.e. P(A or B) = P(A) + P(B) Example2: A bag contains 8 red balls and 4 blue balls. Two balls are drawn at random. Find the probability of getting both balls of same color. Solution: Total number of balls = 8 + 4= 12 n(S)= number of ways of choosing 2 balls out of 12= C(12,2) C(12,2) = (12!)/ (2! 10!) = 66 n(E)= number of ways of drawing same color balls i.e. either 2 out of 8 red balls OR 2 out of 4 blue balls n(E) = C(8,2) + C(4,2) = 34 Probability = 34/66 = 17/33 Example3: One card is drawn from a pack of 52 cards. Find the probability that i) card drawn is black ii) card drawn is a queen iii) card drawn is black and queen iv) card drawn is either black or queen Solution: n(S)= 52 i) n(E) = n(black) = 26 P(black) = 26/52 = 1/2 ii) n(E) = n(queen) = 4 P(queen) = 4/52 = 1/13 iii) n(black and queen) = 2 ( two queens are of black color) P(black and queen)= 2/52 = 1/26 iv) n(black or queen) = 26 (black cards including 2 queen) + 2 (rest two queens)= 28 P(black or queen) = 28/52 = 7/13 Probability: Basic Understanding and Examples with solutions Reviewed by Jasleen Behl on Thursday, January 16, 2014 Rating: 5 #### 2 comments: 1. can you please post about, how to solve when three dice are tossed at the moment 2. can you please post about, how to solve when three dice are tossed at the moment I will try to respond asap
# How to solve sin3x<sinx? Jun 13, 2018 The solution is $x \in \left(\frac{\pi}{4} , \frac{3}{4} \pi\right) \cup \left(\pi , \frac{5}{4} \pi\right) \cup \left(\frac{7}{4} \pi , 2 \pi\right)$, $\left[2 \pi\right]$ #### Explanation: To solve this trigonometric inequality, we need $\sin 3 x = 3 \sin x - 4 {\sin}^{3} x$ The inequality is $\sin 3 x < \sin x$ $\implies$, $\sin 3 x - \sin x < 0$ Let $f \left(x\right) = \sin 3 x - \sin x$ The period of $f \left(x\right)$ is $T = 2 \pi$ Study the functio on the interval $I = \left[0 , 2 \pi\right]$ Therefore, $f \left(x\right) = 3 \sin x - 4 {\sin}^{3} x - \sin x$ $= 2 \sin x - 4 {\sin}^{3} x$ $= 2 \sin x \left(1 - {\sin}^{2} x\right)$ Let $2 \sin x \left(1 - {\sin}^{2} x\right) = 0$ The solutions to this equation are $\left\{\begin{matrix}\sin x = 0 \\ 1 - 2 {\sin}^{2} x = 0\end{matrix}\right.$ $\iff$, $\left\{\begin{matrix}x = 0 + k \pi \\ x = \frac{\pi}{4} + k \pi \\ x = \frac{5}{4} \pi + k \pi\end{matrix}\right.$ Build a sign chart $\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$0$$\textcolor{w h i t e}{a a a a}$$\frac{\pi}{4}$$\textcolor{w h i t e}{a a a a}$$\frac{3}{4} \pi$$\textcolor{w h i t e}{a a a a}$$\pi$$\textcolor{w h i t e}{a a a a}$$\frac{5}{4} \pi$$\textcolor{w h i t e}{a a a a}$$\frac{7}{4} \pi$$\textcolor{w h i t e}{a a a a}$$2 \pi$ $\textcolor{w h i t e}{a a a a}$$\sin x$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$ $\textcolor{w h i t e}{a}$$1 - 2 {\sin}^{2} x$$\textcolor{w h i t e}{a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$ $\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$ Therefore, $f \left(x\right) < 0$ when $x \in \left(\frac{\pi}{4} , \frac{3}{4} \pi\right) \cup \left(\pi , \frac{5}{4} \pi\right) \cup \left(\frac{7}{4} \pi , 2 \pi\right)$, $\left[2 \pi\right]$ graph{sin(3x)-sinx [-8.54, 37.08, -9.16, 13.64]} Jun 13, 2018 sin 3x - sin x < 0 f(x) = 2cos (2x). sin x < 0 First solve the 2 basic trig equations to find the end-points (critical points) 1. g(x) = cos 2x = 0 --> $2 x = \frac{\pi}{2}$, and $2 x = \frac{3 \pi}{2}$ --> x = pi/4; x = (3pi)/4; x = (5pi)/4, and x = (7pi)/4 2. h(x) = sin x = 0 --> x = 0; x = pi; and x = 2pi Next, to algebraically solve the trig inequality, create a sign chart that shows the variation of the 2 functions g(x) and h(x) when x varies from 0 to $2 \pi$ through the critical points. $\left(0 , \frac{\pi}{4} , \frac{3 \pi}{4} , \pi , \frac{5 \pi}{4} , \frac{7 \pi}{4} , 2 \pi\right)$ The sign (+ or -) of the function f(x) is the resulting sign of the 2 function g(x) and h(x) The solutions are the intervals $\left(\frac{\pi}{4} , \frac{3 \pi}{4}\right)$ where g(x) < 0, and h(x) > 0 and --> f(x) < 0 $\left(\pi , \frac{5 \pi}{4}\right)$ where g(x) > 0, and h(x) < 0, and --> f(x) < 0 $\left(\frac{7 \pi}{4} , 2 \pi\right)$ where g(x) > 0, and h(x) < 0 , and --> f(x) < 0 You also can solve the trig inequality by using graphing calculator. The parts of the graph that lie below the x-axis represent the answers inside the interval $\left(0 , 2 \pi\right)$ Note. The graph shows x in radians. Exp. x = pi = 3.14; x = pi/2 = 1.57
# Is Absolute Value The Same As Parentheses? ## What is the opposite and the absolute value of 3? The opposite of a number (also known as the additive inverse) is the number that is the same distance from 0 on the number line. Example 1: The opposite of 3 is ________. Answer: − , since the distance from 0 to 3 on the number line is the same as the distance from 0 to −3.. ## How do you tell if an absolute value inequality is AND or OR? This pattern for “greater than” absolute-value inequalities always holds: Given the inequality \| x \| > a, the solution always starts by splitting the inequality into two pieces: x < –a or x > a. And, by the way, the correct conjunction is “or”, not “and”. ## What is the absolute value symbol? The symbol for absolute value is a bar ∣ on each side of the number. ## What is the same absolute value as? The absolute value is the same as the distance from zero of a specific number. On this number line you can see that 3 and -3 are on the opposite sides of zero. Since they are the same distance from zero, though in opposite directions, in mathematics they have the same absolute value, in this case 3. ## How do you simplify absolute value? Once you have performed any operations inside the absolute value bars, you can simplify the absolute value. Whatever number you have as your argument, whether positive or negative, represents a distance from 0, so your answer is that number, and it is positive. In the example above, the simplified absolute value is 3. ## How do you know if an absolute value inequality is no solution? Isolate the absolute value expression on the left side of the inequality. If the number on the other side of the inequality sign is negative, your equation either has no solution or all real numbers as solutions. ## Why is absolute value important? When you see an absolute value in a problem or equation, it means that whatever is inside the absolute value is always positive. Absolute values are often used in problems involving distance and are sometimes used with inequalities. … That’s the important thing to keep in mind it’s just like distance away from zero. ## Is absolute value similar to parentheses? Sometimes parentheses are added to organize and group the same kinds of things. Absolute value signs make the final result of the operations between them positive. When deciding on the order of operations, treat absolute value bars like parentheses. ## Is absolute value positive or negative? The precise term in “non-negative.” Absolute value is a magnitude and is either positive or zero. Zero is neither positive nor negative. But the absolute value of any non-zero number can be thought of as it’s distance from zero and it will always be positive. ## How do you tell if inequality is AND or OR? If it is a conjunction that uses the word and, the solution must work in both inequalities and the solution is in the overlap region of the graph. If it is a disjunction that uses the word or, the solution must work in either one of the equations. ## How do you know if an absolute value equation has no solution? An absolute value equation has no solution if the absolute value expression equals a negative number since an absolute value can never be negative. ## Can two numbers have the same absolute value? Absolute value is the magnitude of a number irrespective of its sign. Hence there can be two real numbers with different signs such +a and −a who have an absolute value of a . … Hence if domain is real number for each absolute value there are two different numbers one can have with same absolute value. ## What is the difference between absolute value and modulus? Absolute value of x=\|x\| and is equal to x if x≥0 or is equal to −x if x<0. Modulo, usually refers to the type of arithmetic called modulo arithmetic. ... Modulus refers to the magnitude/length of a vector. ## How do you explain absolute value? Absolute value describes the distance of a number on the number line from 0 without considering which direction from zero the number lies. The absolute value of a number is never negative. The absolute value of 5 is 5.
Upcoming SlideShare × Teaching Common Core Math • 315 views More in: Education , Technology • Comment goes here. Are you sure you want to Be the first to comment Be the first to like this Total Views 315 On Slideshare 0 From Embeds 0 Number of Embeds 0 Shares 5 0 Likes 0 No embeds Report content No notes for slide Transcript • 2. 4www.nctm.org teaching children mathematics • April 2012 499 • 5. Incorrect responses—like Carlos’s Unpacking a Standard— Fi gur e 2 and many of his classmates’, who Knowledge partitioned the square into fourths To identify the knowledge pieces of a Standard, or eighths but not into thirds—reveal including terms, properties, or other factual when students do not possess the components, complete the next sentence. prerequisite knowledge needed for For Standard 3.NF.1, students will know the the unit. following: Luis, Elena, and Leslie plan to share 1 large, • A fraction symbol like 1/b is one part of square pizza. Each person will get an equal a whole and the whole is partitioned into amount. b equal parts. 1. Show on the picture how much pizza • A fraction a/b is the quantity formed by Elena will get. a parts of size 1/b. • The numerator of a fraction is the “top number,” and it tells how many parts of a certain size are being counted. • The denominator of a fraction is the “bottom number” and tells the number of equal parts the whole is partitioned into. Unpacking a Standard—Skills Finally, unpacking the Standard includes iden- tifying the skills that students must gain to meet the Standard’s expectations. We use this sentence starter: For Standard 3.NF.1, students will be able to do the following: 2. What fraction of the pizza will Elena get? • Write the fraction symbol for a given representation of a fraction of the form 1/b. • Write the fraction symbol for a given representation of a fraction of the form a/b. • Represent a given fraction of the form 1/b with pictures and/or manipulatives. • Represent a given fraction of the form a/b with pictures and/or manipulatives. • Count the parts of a whole using the language of fractions such as one-fourth, two-fourths, three-fourths, four-fourths. • Given the picture of a fraction of the form 1/b or a/b, represent the whole with pictures and/or manipulatives. We have found that unpacking a Standard in this way is beneficial for two reasons. First, the process offers us a clearer picture of the Standard’s mathematical expectations, thus clarifying the end goal of instruction. Second, it affords teachers the opportunity to think deeply about the mathematics, consequently allowing us to identify the differences between CCSSM and the objectives that we have previ- ously taught.502 April 2012 • teaching children mathematics www.nctm.org • 7. If students were to immediately Troy’s representation partitions Fi gur e 4 Fi gur e 5 produce this anticipated solution the square into three horizontal to the task, the task likely failed to sections and includes a partition engage them in problem solving. down the middle to produce six equal-size parts. Elena’s share is clearly two of those parts. size. Collectively, student responses like these Correct responses reveal that the students do not possess the pre- None of the students in this class presented the requisite knowledge needed to begin the unit. anticipated solution (see fig. 4). Troy, however, The discussion of this task should guide the came the closest. His partitioning (see fig. 5) has students toward recognizing the importance three horizontal sections, but it also includes of equal-size parts. Additionally, students need a partition down the middle to produce six opportunities to develop a variety of strategies equal-size parts. In his response to the question, for partitioning shapes. though, he clearly indicates that Elena’s share is Amelia and Tamika both divide the square into three columns or rows but F igure 6 add more partitioning, assigning Elena one-third of the pizza. Amelia’s representation indicates that she developed an alternative strategy when she recognized that drawing segments intersecting at the center would not work.504 April 2012 • teaching children mathematics www.nctm.org
What is Ratio? Ratio is a mathematical term used to compare two similar quantities expressed in the same units. The ratio of two terms ‘x’ and ‘y’ is denoted by x : y. In ratio x : y , we can say that x as the first term or antecedent and y, the second term or consequent. In general, the ratio of a number x to a number y is defined as the quotient of the numbers x and y i.e. x/y. Example: The ratio of 25 km to 100 km is 25:100 or 25/100, which is 1:4 or 1/4, where 1 is called the antecedent and 4 the consequent. Note that fractions and ratios are same; the only difference is that ratio is a unit less quantity while fraction is not. Compound Ratio Ratios are compounded by multiplying together the fractions, which denote them; or by multiplying together the antecedents for a new antecedent, and the consequents for a new consequent. The compound of a : b and c : d is i.e. ac : bd. Properties of Ratio: a : b = ma : mb, where m is a constant a : b : c = A : B : C is equivalent to a / A = b /B = c /C, this is an important property and has to be used in ratio of three things. i.e. the inverse ratios of two equal ratios are equal. This property is called Invertendo. i.e. the ratio of antecedents and consequents of two equal ratios are equal. This property is called Alternendo. This property is called Componendo. This property is called Dividendo. This property is called Componendo - Dividendo. The incomes of two persons are in the ratio of a: b and their expenditures are in the ratio of c: d. If the saving of each person be Rs. S, then their incomes are given by- Example: Annual income of A and B are in the ratio of 5: 4 and their annual expenses bear a ratio of 4: 3. If each of them saves Rs. 500 at the end of the year, then find the annual income. Dividing a Quantity Into a Ratio Suppose any given quantity ‘a’ is to be divided in the ratio of m : n. Then, Proportion When two ratios are equal, the four quantities composing them are said to be in proportion. If a/b=c/d, then a, b, c, d are in proportions. This is expressed by saying that ‘a’ is to ‘b’ is to ‘c’ is to ‘d’ and the proportion is written as a : b :: c : d or a : b = c : d (product of means = product of extremes) If there is given three quantities like a, b, c of same kind then we can say it proportion of continued. a : b = b : c the middle number b is called mean proportion. a and c are called extreme numbers. So, b2 = ac. (middle number)2 = ( First number x Last number ). Application: These properties have to be used with quick mental calculations; one has to see a ratio and quickly get to results with mental calculations. Example: should quickly tell us that Q. A certain amount was to be distributed among A, B and C in the ratio 2 : 3 : 4, but was erroneously distributed in the ratio 7 : 2 : 5. As a result of this, B received Rs. 40 less. What is the actual amount? (a) Rs. 210 (b) Rs. 270 (c) Rs. 230 (d) Rs. 280 (e) None of these Q. Mixture of milk and water has been kept in two separate containers. Ratio of milk to water in one of the containers is 5 : 1 and that in the other container 7 : 2. In what ratio the mixtures of these two containers should be added together so that the quantity of milk in the new mixture may become 80%? (a) 2 : 3 (b) 3 : 2 (c) 4 : 5 (d) 1 : 3 (e) None of these Practice more Questions on Ratio and Proportion :
# خواص ریشه ها سرفصل: بخش ریاضی / سرفصل: ریشه ها و قدرت ها / درس 9 ## بخش ریاضی 14 سرفصل \| 192 درس ### توضیح مختصر • زمان مطالعه 4 دقیقه • سطح متوسط ### دانلود اپلیکیشن «زوم» این درس را می‌توانید به بهترین شکل و با امکانات عالی در اپلیکیشن «زوم» بخوانید ## Properties of Roots Properties of Roots. Fundamentally, roots are a special case of exponents. So some of the properties of exponents are the same as the property of roots. In a couple lessons from this one, we’ll make clear the link between roots and exponents. That’s something we’ll discuss in a little more depth. Right now, we’ll just focus on a couple of big properties of roots. In all the following examples, I’m going to use square roots and assume that the numbers P and Q are positive. In fact, all these properties work for all higher order roots as well, and for the odd roots, the numbers don’t even have to be positive. So, I didn’t wanna clutter up the notation with nth roots everywhere, but just so you know, everything I’m talking about with square roots here, this actually extends to higher order roots. It’s just rare that you would have to use that on the test. Remember that exponents distribute over multiplication and division. Roots also distribute over multiplication and division. So the root of a product is the product of the roots and the root of a quotient is the quotient of the roots. The multiplication property makes some simplifications much easier. Suppose we’re told that we had to simplify square root of 12 times square root of 27. Hm. Well, one way to go about this is we could multiply them together. And of course, we’re not just gonna multiply out and get that horrible three digit number, again, we’re gonna think about this. We’re gonna factor out a 3 from that 27 because that will leave us with the perfect square 9. And factor of 3, I’ll multiply by the 12. Well when I do that, I get another perfect square. So I now have the square of two perfect squares. Separate them out, take the square root of each and it’s 6 times 3 which is 18. So it must be that whatever 12 times 27 is, that must be 18 squared, but fortunately, we didn’t have to figure out that number in order to perform this simplification. We’ll explore this much more in the next lesson. The division property makes it easier to simplify the square roots of some fractions. So for example, if we had the square root of 4 over 49, we can just take the square root of each of those numbers individually and that equals two sevenths. If we had the square root of 4 over 50, of course, we can simplify a bit first. We can cancel a factor of 2, then take the square root in the numerator and the denominator. In the numerator, we can’t do anything else. Square root of 2, we’re stuck with that. But in the denominator, we can simplify that to 5. So these are two fractions that get a little bit simplified. And we’re gonna discuss this process in much more detail a couple of lessons from now. This is, this will be a problem of what happens when you get a square root in the denominator. That didn’t happen here, but it is a problem and we’ll talk about it in a few lessons. As with exponents, there is also a very tempting mistake to make here. Roots do distribute over multiplication and division. Roots do not distribute over addition and subtraction. So it’s very tempting to say that square root of P plus square root of Q, to set that equal to the square root of P plus Q or the same with subtraction, it’s a very tempting mistake to make, but it is simply not true. We will see a very important application of the multiplication rule in the next lesson. In summary, roots distribute over multiplication and division. Roots do not distribute over addition and subtraction. ### مشارکت کنندگان در این صفحه تا کنون فردی در بازسازی این صفحه مشارکت نداشته است. 🖊 شما نیز می‌توانید برای مشارکت در ترجمه‌ی این صفحه یا اصلاح متن انگلیسی، به این لینک مراجعه بفرمایید.
# Ordinal Numbers In English 1 100 With ordinal numbers, you can count any number of sets. They are also able to generalize ordinal numbers.But before you use them, you must comprehend what they are and how they operate. ## 1st The fundamental concept in mathematics is the ordinal. It is a number that shows where an object is within a list. The ordinal number is an integer between 1 to 20. Even though ordinal numbers have numerous functions, they are most often used for indicating the order of items in an itemized list. Ordinal numbers are represented by charts as well as words, numbers and other methods. They may also be used as a method of describing how pieces of a collection are organized. Ordinal numbers typically are classified into two categories. They are classified as Arabic numbers represent finite ordinals and transfinite ones are represented with lowercase Greek letters. According to the axiom that every set properly ordered must contain at minimum one ordinal. The first person in an class, for instance is the one who receives the highest grade. The contest’s winner was the student with the highest grade. ## Combinational ordinal numbers Multidigit numbers are referred to as compound ordinal number. They are generated by multiplying an ordinal by its last number. They are most commonly used to rank and date. They don’t have a unique ending like cardinal numbers. Ordinal numerals are used to indicate the sequence of elements in a collection. They also help indicate the names of the items within a collection. The two kinds of regular numbers are flexible and regular. The addition of the suffix “-u” makes regular ordinals. The numbers are then typed into the form of a word. A hyphen then added to it. You can also use additional suffixes. For example, “-nd” is for numbers that end with 2 and “-th” for numbers ending with 4 or 9. Suppletive ordinals can be created by affixing words with -u or. This suffix is utilized to count and is usually bigger than the normal one. ## Limits of Ordinal Ordinal numbers that do not exceed zero are the maximum for ordinal numbers. Limit ordinal numbers have the disadvantage of having no maximal element. You can create them by joining empty sets, but without any maximum element. Limited ordinal numbers are used in transfinite definitions of Recursion. Each infinite number of cardinals, according to the von Neumann model, is also an ordinal limit. A limit ordinal equals the total of the other ordinals below. Limit ordinal numbers can be quantified using math, but they can also can be expressed in a series or natural numbers. The ordinal numbers used to arrange the data are employed. They are used to explain an object’s numerical place. They are employed in set theory and arithmetic contexts. They are not included in the same group as natural numbers despite having a similar a structure with them. The von Neumann Model uses a well-ordered and ordered set. Let’s say that fy is an element of an g’ function that is defined as a single function is the case. If the subfunction fy’s is (i I, II) and g fulfills the requirements, then g is a limit ordinal. The Church Kleene oral is a limit-ordering order which functions in the same manner. A Church-Kleene ordinal defines a limit ordinal to be a properly organized collection, that is comprised of smaller ordinals. ## Numerological examples of common numbers in stories Ordinal numbers are frequently used to denote the hierarchy or relationships between entities and objects. They are important for organizing, counting and ranking. They can be used to describe the position of objects as well as indicating the sequence of things. The letter “th” is usually used to indicate the ordinal number. In some instances, the letter “nd” is substituted. Books titles often contain ordinal numbers. Although ordinal figures are usually utilized in lists, you can write them down as words. They may also come in numbers and acronyms. In comparison, these numbers are much easier to understand than the traditional ones. There are three kinds of ordinal numbers. These numbers can be learned more by playing games, exercises, and other activities. Learning about the subject is an important aspect of developing your arithmetic skills. Coloring is an enjoyable and straightforward method to increase your proficiency. You can assess your progress by using a marker sheet.
# Numbers Math Practice Test for 3rd Grade – [Medium] Numbers. They’re everywhere in our lives. From counting the pennies in our piggy banks to the number of stars in the sky. For 3rd graders, understanding numbers and their applications is not just a school task, it’s a rite of passage into a world of infinite possibilities. At Brighterly, we believe in making this journey both illuminating and fun! ## Whole Numbers and Place Value When we talk about numbers, we often refer to whole numbers. But what are they? Whole numbers are simply the numbers we use to count, starting from zero and moving upwards. So, 1, 2, 3, and so forth. The position of each digit in a number gives it its place value. For instance, in the number 204, 2 is in the hundreds place, 0 is in the tens place, and 4 is in the units place. Recognizing and understanding place values helps in performing operations like addition and subtraction efficiently. ## Fractions: Breaking it Down Sometimes, a whole number just doesn’t cut it. Imagine splitting a pizza with a friend; you wouldn’t take 2 whole pizzas, but rather a fraction of it. Fractions represent parts of a whole. The number above the fraction bar is the numerator and below is the denominator. If you ate 3 slices out of 8 from a pizza, then your fraction is 3/8. ## Introduction to Multiplication and Division Multiplication is just a fast way of adding numbers. If you have 3 groups of 4 toys each, instead of adding 4+4+4, you can simply multiply 3×4 to get 12. On the other hand, division is splitting something into equal parts. If you have 12 candies and want to share them with 4 friends equally, you’d divide 12 by 4, resulting in each friend getting 3 candies. ## Rounding Numbers: Keeping it Simple Rounding numbers is making them simpler but still keeping their value close to what it was. Think of it as giving an estimate. If you have 47 marbles, and someone asks quickly, you might say “about 50” since 47 is closer to 50 than 40. That’s rounding! This skill is very useful when dealing with larger numbers or when you need to make quick estimates. ## Prime and Composite Numbers Numbers have personalities too! Prime numbers are numbers that can only be divided by 1 and themselves without leaving a remainder. Examples include 2, 3, 5, and so on. Composite numbers, on the other hand, can be divided by other numbers as well. Like 4 can be divided by 2, and 6 can be divided by 2 and 3. Numbers Practice Test for 3rd Grade Get ready for math lessons with Brighterly! This medium-level test is crafted with precision and love, ensuring a balanced mix of fun and challenge. 1 / 15 Which of the following numbers is an even number? 2 / 15 How many sides does a triangle have? 3 / 15 If you round the number 74 to the nearest ten, what do you get? 4 / 15 Which of the following is NOT a prime number? 5 / 15 What is the sum of 45 and 30? 6 / 15 If you have 2 dozens of eggs, how many eggs do you have in total? 7 / 15 Which of the following fractions represents a half? 8 / 15 How many months are in a year? 9 / 15 What is the result of 63 divided by 7? 10 / 15 If you multiply 4 by 9, what is the result? 11 / 15 How many days are in a week? 12 / 15 Which number is between 10 and 12? 13 / 15 Which of the following numbers is the largest? 14 / 15 If you subtract 18 from 46, what do you get? 15 / 15 Which of the following is the smallest odd number? 0% Poor Level Weak math proficiency can lead to academic struggles, limited college, and career options, and diminished self-confidence. Mediocre Level Weak math proficiency can lead to academic struggles, limited college, and career options, and diminished self-confidence. Needs Improvement Start practicing math regularly to avoid your child`s math scores dropping to C or even D. High Potential It's important to continue building math proficiency to make sure your child outperforms peers at school. After-School Math Program • Boost Math Skills After School! • Join our Math Program, Ideal for Students in Grades 1-8! After-School Math Program Boost Your Child's Math Abilities! Ideal for 1st-8th Graders, Perfectly Synced with School Curriculum!
# 9.3 A population proportion Page 1 / 24 During an election year, we see articles in the newspaper that state confidence intervals in terms of proportions or percentages. For example, a poll for a particular candidate running for president might show that the candidate has 40% of the vote within three percentage points (if the sample is large enough). Often, election polls are calculated with 95% confidence, so, the pollsters would be 95% confident that the true proportion of voters who favored the candidate would be between 0.37 and 0.43: (0.40 – 0.03,0.40 + 0.03). Investors in the stock market are interested in the true proportion of stocks that go up and down each week. Businesses that sell personal computers are interested in the proportion of households in the United States that own personal computers. Confidence intervals can be calculated for the true proportion of stocks that go up or down each week and for the true proportion of households in the United States that own personal computers. The procedure to find the confidence interval, the sample size, the error bound , and the confidence level for a proportion is similar to that for the population mean, but the formulas are different. How do you know you are dealing with a proportion problem? First, the underlying distribution is a binomial distribution . (There is no mention of a mean or average.) If X is a binomial random variable, then X ~ B ( n , p ) where n is the number of trials and p is the probability of a success. To form a proportion, take X , the random variable for the number of successes and divide it by n , the number of trials (or the sample size). The random variable P′ (read "P prime") is that proportion, ${P}^{\prime }=\frac{X}{n}$ (Sometimes the random variable is denoted as $\stackrel{^}{P}$ , read "P hat".) When n is large and p is not close to zero or one, we can use the normal distribution to approximate the binomial. $X~N\left(np,\sqrt{npq}\right)$ If we divide the random variable, the mean, and the standard deviation by n , we get a normal distribution of proportions with P′ , called the estimated proportion, as the random variable. (Recall that a proportion as thenumber of successes divided by n .) Using algebra to simplify : $\frac{\sqrt{npq}}{n}=\sqrt{\frac{pq}{n}}$ P′ follows a normal distribution for proportions : The confidence interval has the form ( p′ EBP , p′ + EBP ). EBP is error bound for the proportion. p′ = $\frac{x}{n}$ p′ = the estimated proportion of successes ( p′ is a point estimate for p , the true proportion.) x = the number of successes n = the size of the sample The error bound for a proportion is $EBP=\left({z}_{\frac{\alpha }{2}}\right)\left(\sqrt{\frac{{p}^{\prime }{q}^{\prime }}{n}}\right)$ where q′ = 1 – p′ This formula is similar to the error bound formula for a mean, except that the "appropriate standard deviation" is different. For a mean, when the population standard deviation is known, the appropriate standard deviation that we use is $\frac{\sigma }{\sqrt{n}}$ . For a proportion, the appropriate standard deviation is $\sqrt{\frac{pq}{n}}$ . However, in the error bound formula, we use $\sqrt{\frac{{p}^{\prime }{q}^{\prime }}{n}}$ as the standard deviation, instead of $\sqrt{\frac{pq}{n}}$ . In the error bound formula, the sample proportions p′ and q′ are estimates of the unknown population proportions p and q . The estimated proportions p′ and q′ are used because p and q are not known. The sample proportions p′ and q′ are calculated from the data: p′ is the estimated proportion of successes, and q′ is the estimated proportion of failures. Do somebody tell me a best nano engineering book for beginners? what is fullerene does it is used to make bukky balls are you nano engineer ? s. what is the Synthesis, properties,and applications of carbon nano chemistry 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 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 what's the easiest and fastest way to the synthesize AgNP? China Cied types of nano material I start with an easy one. carbon nanotubes woven into a long filament like a string Porter many many of nanotubes Porter what is the k.e before it land Yasmin what is the function of carbon nanotubes? Cesar I'm interested in nanotube Uday what is nanomaterials and their applications of sensors. what is nano technology what is system testing? preparation of nanomaterial Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it... what is system testing what is the application of nanotechnology? Stotaw In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google Azam anybody can imagine what will be happen after 100 years from now in nano tech world Prasenjit after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments Azam name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world Prasenjit how hard could it be to apply nanotechnology against viral infections such HIV or Ebola? Damian silver nanoparticles could handle the job? Damian not now but maybe in future only AgNP maybe any other nanomaterials Azam Hello Uday I'm interested in Nanotube Uday this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15 Prasenjit can nanotechnology change the direction of the face of the world At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light. how did you get the value of 2000N.What calculations are needed to arrive at it Privacy Information Security Software Version 1.1a Good Got questions? Join the online conversation and get instant answers!
# CHAPTER 4 SYSTEMS OF LINEAR EQUATIONS AND INEQUALITIES 4.1/4. 2 Systems of Linear Equations Solution set of a system = set of all ordered pairs that satisfy all the equations in the system. Methods used to solve a system of equations: 1. Graphing 2. Linear Combination 3. Cramer’s Rule- Determinants 4. Matrices- Row Operations When solving a system of equations you can get one, none or a ton of solutions. No solutions: Parallel lines/ planes Planes that do not intersect at one point Inconsistent equations Linear combination will result in a false statement. Cramer’s Rule: nonzero / zero One solution: Intersecting lines/ planes Independent equations Linear equations will result in a solution. Cramer’s Rule: one solution Ton of solutions: Same line/ same plane/ planes intersecting in one line Dependent equations Linear equations will result in a true statement Cramer’s Rule: zero / zero 4.3 Second Order Determinants a c b = ad – bc d 4.4 f(x) Terminology f(x) - the value of the independent variable if the dependent variable is x . f(x) = 3x + 2 g(x) = 5x f(1) = 3(1) + 2 = 5 f(g(2)) = f(10) = 32 Be able to solve real world problems using linear equations. 4-5 Linear Equations with 3 or More Variables 3-space x,y,z axis z axis is vertical y axis is horizontal x axis is out at you To graph an equation in 3 variables, find the 3 intercepts and draw the triangle that represents the plane. 4.6 Systems of Linear Equations with 3 or more Variables 3 Equations and 3 Variables Use linear combination to simplify to 2 equations, 2 unknowns Solve that smaller system Use substitution to find the remaining variable. 4-7/4-8 Solutions of Systems by Augmented Matrices (CALCULATOR-OK) Elementary Row Operations - multiply a row by a constant - exchange two rows - add or subtract rows You use elementary row operations to obtain a matrix with 1’s on the diagonal and 0’s elsewhere. Work from the column on the left to the column on the right. Start by getting the 1 by exchanging rows or dividing, then get the 0’s. 10 0 2 010 4 0 016 x = 2, y = 4, z = 6 4.9 Higher Order Determinants (CALCULATOR-OK) Third order and higher determinants can be found by expanding. Each term will contain a (+,-) sign, an entry from the matrix, and a determinant of the reduced matrix. You can expand on any row or column. 2 3 -1 -1 5 3 Expand on the first row. 3 -1 -6 2 5 3 −1 3 −1 5 + -3 + -1 −1 - 6 3 -6 3 -1 2(-30 – ( - 3 ) ) – 3 ( 6 – 9 ) + -1( 1 – 15 ) = 2( -27) – 3( -3 ) + -1( -14) = -31 4.10Systems of Linear Inequalities (CALCULATOR-OK) To solve a system of inequalities, graph each separately and find the intersection of the graphs. You want the points that are common to all the inequalities. To graph an inequality: - Change to an equation and graph the line. < or > use a solid line. < or > use a dotted line - Choose a point on one side of the line and substitute it into the original inequality. - If you get a true statement, shade the same side of the line that the point is on - If you get a false statement, shade the other side 4.11Linear Programming 1. 2. 3. 4. Write the inequalities for the problem- called the constraints Write the max or min equation Graph the inequalities to obtain the feasible region. Find the corner points of the feasible region. 5. Substitute each corner point into the max/min equation to see which produce a max or min.
Courses Courses for Kids Free study material Offline Centres More Last updated date: 26th Nov 2023 Total views: 281.1k Views today: 8.81k # Each angle of a regular hexagon is $\_\_\_\_$. Verified 281.1k+ views Hint: Before solving this question we should know what a hexagon is. A hexagon is a type of polygon which has $6$ sides and $6$ vertices. We know that there are two types of hexagon- (a) Regular hexagon and (b) Irregular hexagon. A regular hexagon has all angles of the same measures and all sides of the same length. We can say that a regular hexagon is equilateral as well as equiangular. Complete step by step solution: Here we have to find the measure of each angle of a regular hexagon. Let us first draw the image of a regular hexagon. Now we know that the sum of the internal angles of a regular hexagon is always ${720^ \circ }$. We also know that all the angles and sides of a regular hexagon are equal, so let us assume that the measure of each angle is $x$. Now we can write $x + x + x + x + x + x = 720$ (Sum of the interior angles of a hexagon). On solving we have $6x = 720$, so it gives us $x = \dfrac{{720}}{6} = 120$. Hence the measure of each angle of a regular hexagon is ${120^ \circ }$. Note: We should note that hexagon is also further classified into Convex or concave. If a hexagon is convex then none of its interior angle would be greater than ${180^ \circ }$ and if the hexagon is concave then the one or more of its interior angles is greater than ${180^ \circ }$. We should have the clear concept of different types of polygons and their properties before solving this kind of question.
Maharashtra Board 9th Class Maths Part 1 Practice Set 2.2 Solutions Chapter 2 Real Numbers Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 2.2 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 2 Real Numbers. Practice Set 2.2 Algebra 9th Std Maths Part 1 Answers Chapter 2 Real Numbers Question 1. Show that 4√2 is an irrational number. Solution: Let us assume that 4√2 is a rational number . So, we can find co-prime integers ‘a’ and ‘b’ (b ≠ 0) such that 4√2 = $$\frac { a }{ b }$$ ∴ √2 = $$\frac { a }{ 4b }$$ Since, a and b are integers, $$\frac { a }{ 4b }$$ is a rational number and so √2 is a rational number. Alternate Proof: Let us assume that 4√2 is a rational number. So, we can find co-prime integers ‘a’ and ‘b’ (b ≠ 0) such that Since, 32 divides a2, so 32 divides ‘a’ as well. So, we write a = 32c, where c is an integer. ∴ a2 = (32c)2 … [Squaring both the sides] ∴ 32b2 = 32 x 32c2 …[From(i)] ∴ b2 = 32c2 ∴ c2 = $$\frac { { b }^{ 2 } }{ 32 }$$ Since, 32 divides b2, so 32 divides ‘b’. ∴ 32 divides both a and b. a and b have at least 32 as a common factor. But this contradicts the fact that a and b have no common factor other than 1. ∴ Our assumption that 4√2 is a rational number is wrong. ∴ 4√2 is an irrational number. Question 2. Prove that 3 + √5 is an irrational number. Solution: Let us assume that 3 + √5 is a rational number. So, we can find co-prime integers ‘a’ and ‘b’ (b ≠ 0) such that Since, a and b are integers, $$\frac { a }{ b }$$ – 3 is a rational number and so √5 is a rational number. But this contradicts the fact that √5 is an irrational number. ∴ Our assumption that 3 – √5 is a rational number is wrong. 3 + √5 is an irrational number. Question 3. Represent the numbers √5 and √10 on a number line. Solution: i. Draw a number line and take point A at 2. Draw AB perpendicular to the number line such that AB = 1 unit. In ∆OAB, m∠OAB = 90° ∴ (OB)2 = (OA)2 + (AB)2 … [Pythagoras theorem] = (2)2 + (1)2 ∴ (OB)2 = 5 ∴ OB = √5 units. … [Taking square root of both sides] With O as centre and radius equal to OB, draw an arc to intersect the number line at C. The coordinate of the point C is √5 . ii. Draw a number line and take point Pat 3. Draw PR perpendicular to the number line such that PR = 1 unit. In ∆OPR, m∠OPR = 90° ∴ (OR)2 = (OP)2 + (PR)2 … [Pythagoras theorem] = (3)2 + (1)2 ∴ (OR)2 = 10 ∴ OR= √10units. … [Taking square root of both sides] With O as centre and radius equal to OR, draw an arc to intersect the number line at Q. The coordinate of the point Q is √10 . Question 4. Write any three rational numbers between the two numbers given below. i. 0.3 and – 0.5 ii. – 2.3 and – 2.33 iii. 5.2 and 5.3 iv. – 4.5 and – 4.6 Solution: i. 0.3 = 0.30 and -0.5 = -0.50 We know that, 0. 30 >0.29 >….. >0.10>.. > – 0.10>…. > -0.30>…> -0.50 ∴ the three rational numbers between 0.3 and -0.5 are -0.3, -0.1 and 0.1. Alternate Method: A rational number between two rational numbers a and b ∴ the three rational numbers between 0.3 and -0.5 are -0.3, -0.1 and 0.1. ii. -2.3 = -2.300 and -2.33 = -2.330 We know that, -2.300 > -2.301>… > -2.310>…> -2.320>…> -2.330 ∴ the three rational numbers between -2.3 and -2.33 are -2.310, -2.320 and -2.325. iii. 5.2 = 5.20 and 5.3 = 5.30 We know that, 5.20 < 5.21 < 5.22 < 5.23 < … < 5.30 ∴ the three rational numbers between 5.2 and 5.3 are 5.21, 5.22 and 5.23. iv. -4.5 = -4.50 and -4.6 = -4.60 We know that, -4.50 > -4.51 > -4.52 >… > – 4.55 >…>- 4.60 ∴ the three rational numbers between -4.5 and -4.6 are -4.51, -4.52 and -4.55.
## Median in a Row-wise Sorted Matrix-Interview Problem Difficulty: Medium #### Understanding the Problem: → Given a NM matrix which is sorted row-wise, you task is to find the median of the given matrix. Median: The median of a group of ordered number is the middle number that will separate the highest half with lowest half of numbers. If there are two middle numbers the, the median is the mean of the numbers. For example: ``````Input: 2 3 3 1 5 6 6 6 7`````` ``Output: 5`` ``Explanation: The total number of elements is 9(odd), in this case the formula for finding the median is (1+NM)/2 th smallest element out of the given elements. Here it is 5 which is the 5th smallest element as 1 2 3 and 3 are smaller than it.`` Possible questions to ask the interviewer: → • Can I use extra space for solving this? ( Ans: Try to come up with an efficient solution) • Is the number of elements odd, even or can be both?( Ans: You can consider the numbers to be odd.) You can first try to solve this problem here . #### Solutions We are going to discuss two different approaches to solve this problem. First, we will look at the Brute Force Solution and then move on to optimize the solution using a better algorithm. • Brute Force Solution • Optimized Solution Using Binary Search #### Brute Force Solution Solution idea Here is it given that the total number of elements is always odd. Now, we have the formula that the (1+NM)/2 th smallest element in an ordered(sorted) list is the median. Since the number of elements are odd there is no case of having more than one middle elements.We can think of a very simple approach where we will store all the elements of the matrix in a 1-Dimensional array and after sorting we will output the element at [(1+NM)/2–1]th index. Solution steps • First, create an array of size NM and copy all the elements of the matrix in that array. You can copy either row-wise or column-wise. • After copying simply sort the entire array. • Output the element at the [(1+NM)/2–1]th index as this will be the middle element partitioning the array in lower and upper halves. Pseudo-code ``````int medianRowwiseSortedMatrix(int matrix[N][M]) { int tempArr[NM] int desired_index = (1+NM)/2 -1`````` `````` int index = 0 for(int i=0; i<N; i++) { for(int j=0; j<M; j++) { tempArr[index++] = matrix[i][j] } } sort(tempArr, tempArr + NM) return tempArr[desired_index] }`````` Complexity Analysis Time Complexity: O(NMlog(NM) Space Complexity: O(NM) Critical ideas to think! • Can we use the given property of this matrix that it is row-wise sorted to improve the complexity? #### Optimized Solution Using Binary Search Solution idea The idea here is to leverage the binary search algorithm in order to optimize this problem. If you have paid attention to the definition of median, you must have noticed in case of odd elements the median is (1+NM)/2 th smallest number which basically implies that (1+NM)/2 is the total number of elements which are smaller or equal to our median.One more thing to analyse is, the median will always lie in the range of minimum and maximum element.Using these observations and applying binary search for elements in the above range (minimum, maximum) and maintaing the counter for smaller or equal numbers for each element in this range, we can find our median(discussed above based on count of smaller or equal elements). Solution idea Let us see a step-by-step approach to this solution: • First, we need to find the minimum and maximum elements from the matrix. The minimum and maximum can be easily found since the rows are sorted so we need to comapare with the first element of each row for minimum and last element of each row for maximum. • After finding our min and max, we can apply binary search on this range. The mid element can be calculated and number of elements smaller or equal to mid can be calculated, we have used upper_bound() function for this. • Based on the value of our counter, the min and max can be adjusted accordingly based on what we do for binary search. Pseudo-code ``````int medianRowwiseSortedMatrix(int matrix[N][M]) { int min = INT_MAX; int max = INT_MIN; int desired_count = (1+(NM)/2))`````` `````` for(int i=0;i<N;i++) { if(A[i][0]<min) min = A[i][0]; if(A[i][M-1]>max) max = A[i][M-1]; } int counter =0; while(min<max) { counter=0; int mid = (max+min)/2; for(int i= 0;i<N;i++) { counter += upper_bound(A[i], A[i]+M, mid) - A[i]; } if(counter < desired_count) min = mid+1; else max = mid; } return min; }`````` Note: - The upper_bound function used above is with reference to C++ STL. This basically gives the iterator pointing to the first element strictly greater than its third parameter(mid here). Can read about it here . Complexity Analysis Time Complexity: O(N*log(M)) ( Why ?) Space Complexity: O(1) #### Suggested Problems to Solve • Find the median in a row and column-wise sorted matrix. • Search an element in a row and column-wise sorted matrix. • Median of two sorted arrays of different sizes. • Given a row and column-wise sorted matrix, print all elements in sorted order. Happy Coding!
Paul's Online Notes Home / Calculus II / Integration Techniques / Approximating Definite Integrals Show Mobile Notice Show All Notes Hide All Notes Mobile Notice You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width. ### Section 7.10 : Approximating Definite Integrals 1. Using $$n = 6$$ approximate the value of $$\displaystyle \int_{1}^{7}{{\frac{1}{{{x^3} + 1}}\,dx}}$$ using 1. the Midpoint Rule, 2. the Trapezoid Rule, and 3. Simpson’s Rule. Use at least 6 decimal places of accuracy for your work. Show All Solutions Hide All Solutions a Midpoint Rule Show Solution While it’s not really needed to do the problem here is a sketch of the graph. We know that we need to divide the interval $$\left[ {1,7} \right]$$ into 6 subintervals each with width, $\Delta x = \frac{{7 - 1}}{6} = 1$ The endpoints of each of these subintervals are represented by the dots on the $$x$$ axis on the graph above. The tick marks between each dot represents the midpoint of each of the subintervals. The $$x$$-values of the midpoints for each of the subintervals are then, $\frac{3}{2},\hspace{0.25in}\frac{5}{2},\hspace{0.25in}\frac{7}{2},\hspace{0.25in}\frac{9}{2},\hspace{0.25in}\frac{{11}}{2},\hspace{0.25in}\frac{{13}}{2}$ So, to use the Midpoint Rule to approximate the value of the integral all we need to do is plug into the formula. Doing this gives, \begin{align}\int_{1}^{7}{{\frac{1}{{{x^3} + 1}}\,dx}} & \approx \left( 1 \right)\left[ {f\left( {\frac{3}{2}} \right) + f\left( {\frac{5}{2}} \right) + f\left( {\frac{7}{2}} \right) + f\left( {\frac{9}{2}} \right) + f\left( {\frac{{11}}{2}} \right) + f\left( {\frac{{13}}{2}} \right)} \right]\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{0.33197137}}\end{align} b Trapezoid Rule Show Solution From the Midpoint Rule work we know that the width of each subinterval is $$\Delta x = 1$$ and for reference purposes the sketch of the graph along with the endpoints of each subinterval marked by the dots is shown below. So, to use the Trapezoid Rule to approximate the value of the integral all we need to do is plug into the formula. Doing this gives, \begin{align}\int_{1}^{7}{{\frac{1}{{{x^3} + 1}}\,dx}} & \approx \left( {\frac{1}{2}} \right)\left[ {f\left( 1 \right) + 2f\left( 2 \right) + 2f\left( 3 \right) + 2f\left( 4 \right) + 2f\left( 5 \right) + 2f\left( 6 \right) + f\left( 7 \right)} \right]\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{0.42620830}}\end{align} c Simpson's Rule Show Solution From the Midpoint Rule work we know that the width of each subinterval is $$\Delta x = 1$$ and for reference purposes the sketch of the graph along with the endpoints of each subinterval marked by the dots is shown below. As with the first two parts all we need to do is plug into the formula to use Simpson’s Rule to approximate value of the integral. Doing this gives, \begin{align}\int_{1}^{7}{{\frac{1}{{{x^3} + 1}}\,dx}} & \approx \left( {\frac{1}{3}} \right)\left[ {f\left( 1 \right) + 4f\left( 2 \right) + 2f\left( 3 \right) + 4f\left( 4 \right) + 2f\left( 5 \right) + 4f\left( 6 \right) + f\left( 7 \right)} \right]\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{0.37154155}}\end{align}
Finite Sets # Finite Sets Often times we may be interested is knowing just how many elements a set has. This notion is easy to consider when the number of elements in a set is finite. Definition: A set $S$ is said to be a Finite Set of size $n$ if there exists a bijection $f : \{ 1, 2, ..., n \} \to S$. A set $S$ is said to be an Infinite Set if it is not a finite set. The notation "$\mathbb{N}_n$" is often used to denote the set $\{ 1, 2, ..., n \}$. The idea behind the definition above is that a set $S$ is a finite set of size $n$ if we can assign to each natural number $1, 2, ..., n$ an element of $S$ and that every element of $S$ is given such a number. For example, consider the set $S = \{ 3, 6, -1, 4, \pi \}$. This set contains precisely $5$-elements, so by the definition there should exist a bijection $f : \mathbb{N}_5 \to S$. One such bijection is described below: $x \in \mathbb{N}_5$ $f(x) \in S$ $1$ $f(1) = 3$ $2$ $f(2) = 6$ $3$ $f(3) = -1$ $4$ $f(4) = 4$ $5$ $f(5) = \pi$ We will now look at some theorems on finite sets. Theorem 1 (Uniqueness of the Size of a Set): Let $S$ be a finite set with $n \in \mathbb{N}$ elements. The natural number $n$ is unique. • Proof: Let $S$ be a finite set containing $m$ elements. Then there exists a bijection $f_1 : \mathbb{N}_m \to S$. Suppose that $S$ also has $n$ elements where $m \neq n$. Then there exists a bijection $f_2 : \mathbb{N}_n \to S$. • Now recall that is a function $f$ is bijective then there exists an inverse function $f^{-1}$. So $f_2$ has an inverse function that is bijective, namely $f^{-1}_2 : S \to \mathbb{N}_n$. Now since $f_1 : \mathbb{N}_m \to S$ and $f^{-1}_2 : S \to \mathbb{N}_n$ are bijective then the composition $f^{-1}_2 \circ f_1 : \mathbb{N}_m \to \mathbb{N}_n$ is bijective. • However such a function that maps $\mathbb{N}_m$ onto $\mathbb{N}_n$ where $m \neq n$ cannot be bijective. • For example, let $f : \mathbb{N}_m \to \mathbb{N}_n$ and consider the case where $m > n$. Then let $f(1) =1$, $f(2) = 2$, …, $f(n) = n$. So all elements in the codomain have been "taken". So $f(n+1) = f(k)$ for some $k \in \mathbb{N}_m$. But $n + 1 \neq k$, so $f$ is not injective. • Now consider the case where $m < n$. Once again let $f(1) = 1$, $f(2) = 2$, …, $f(m) = m$. So then $(m + 1) \in \mathbb{N}_n$ is not mapped to from $f$ so $f$ is not surjective and so $f$ is not bijective. • We thus conclude that $m = n$ and so the number of elements in a set is unique. $\blacksquare$ Theorem 2: Let $A$ be a set that has $m$ elements and let $B$ bet a set that has $n$ elements. If $A \cap B = \emptyset$ then $A \cup B$ has $m + n$ elements. • Proof: Since $A$ has $m$ elements there exists a bijection $f_1 : \mathbb{N}_m \to A$, and since $B$ has $n$ elements there exists a bijection $f_2 : \mathbb{N}_n \to B$. Now we will define a function $h : \mathbb{N}_{m+n} \to A \cup B$ such that: (1) \begin{align} h(x) = \left\{\begin{matrix} f(x) & \mathrm{if} \: 1 \leq i \leq m \\ g(x - m) & \mathrm{if} \: m+1 \leq x \leq m + n \end{matrix}\right. \end{align} • So such a bijection $h$ exists, so $A \cup B$ has $m + n$ elements. $\blacksquare$ Theorem 3: If $A$ is a set that contains $m$ elements and $B \subseteq A$ is a set that contains $1$ element then $A \setminus B$ contains $m - 1$ elements. • Proof: Let $A$ be a set containing $m$ elements. Then $A = \{ a_1, a_2, ..., a_m \}$. If $B \subseteq A$ is a set containing $1$ element, then $B = \{ a_j \}$ for some $j = 1, 2, ..., m$. • Since $A$ contains $m$ elements there exists a bijection $f_1 : \mathbb{N}_m \to A$, define it by $f(1) = a_1$, $f(2) = a_2$, …, $f(m) = a_m$. Then for the set $A \setminus B = \{ a_1, a_2, ..., a_{j-1}, a_{j+1}, ..., a_m \}$, we can define a bijection $f_2 : \mathbb{N}_{m-1} \to A \setminus B$ by omitting $f(j) = a_j$. • Therefore $A \setminus B$ contains $m - 1$ elements. $\blacksquare$ Theorem 4: Let $A$ and $B$ be sets such that $B \subseteq A$. If $A$ is a finite set then $B$ is also a finite set. If $B$ is an infinite set then $A$ is also an infinite set. • Proof: Let $A$ and $B$ be sets such that $B \subseteq A$. • For the first part of the proof, let $A$ be a finite set and suppose instead that $B$ is NOT a finite set. Then $B$ is an infinite set, but since $B \subseteq A$, then there is an infinite number of elements contained in $A$ as well. But this contradicts the fact that $A$ is a finite set. So our assumption that $B$ was not finite was false, hence, $B$ is finite. • The second part of the proof is the contrapositive of the first, so if $B$ is an infinite set, then $A$ is also infinite. $\blacksquare$
Successfully reported this slideshow. Upcoming SlideShare × # Arithmetic Sequence Real Life Problems 206,382 views Published on A sample document about examples of real life problems about "Arithmetic Sequence" in Mathematics 10 Published in: Education • Full Name Comment goes here. Are you sure you want to Yes No ### Arithmetic Sequence Real Life Problems 1. 1. SITUATION: SITUATION: There are 125 passengers in the first carriage, 150 passengers in the second carriage and 175 passengers in the third carriage, and so on in an arithmetic sequence. 2. 2. PROBLEM: What’s the total number of passengers in the first 7 carriages? SOLUTION: The sequence is 125, 150, 175 … Given: a1= 125; a2= 150; a3= 175 Find: S7=? an = 125+(n-1)25 a7 = 125+(7-1)25=275 We can use the formula: Thus, =1400 Carriage 1st 2nd 3rd … 7th First 7 carriages Number of Passengers 125 150 175 … ? Sn 3. 3. SITUATION: SITUATION: There are 130 students in grade one, 210 students in grade two and 290students in grade three in a primary school, and so on in an arithmetic sequence. 4. 4. PROBLEM: What’s the total amount of students In the primary school? (Primary School has 6 grades) SOLUTION: The sequence is 130, 210, 290 … Given: a1= 130; a2= 210; a3= 290 Find: S6= ? an = 130+(n-1)80 a6 = 130+(6-1)80=530 We can use the formula: Thus, = 1980 Grade 1st 2nd 3rd … 6th Total from 1st to 6th Grade Number of Students 130 210 290 … ? Sn 5. 5. SITUATION: A car travels 300 m the first minute, 420 m the next minute, 540 m the third minute, and so on in an arithmetic sequence. 6. 6. PROBLEM: What’s the total distance the car travels in 5 minutes? SOLUTION: The sequence is 300, 420, 540 … Given: a1= 300; a2= 420; a3= 540 Find: S5= ? an = 300+(n-1)120 a5 = 300+(5-1)120=780 We can use the formula: Thus, = 2700 Minute First Second Third Fourth Fifth 5 minutes in Total Distance 300 420 540 … ? Sn 7. 7. PROBLEM: SITUATION: A writer wrote 890 words on the first day, 760 words on the second day and 630 words on the third day, and so on in an arithmetic sequence. 8. 8. PROBLEM: How many words did the writer write in a week? SOLUTION: The sequence is 890, 760, 630 … Given: a1= 890; a2= 760; a3= 630 Find: s7= ? an = 890-(n-1)130 a7 = 890-(7-1)130=110 We can use the formula: Thus, =3500 Day 1st 2nd 3rd … 7th Whole Week Number of Words 890 760 630 … ? Sn 9. 9. SITUATION: You visit the Grand Canyon and drop a penny off the edge of a cliff. The distance the penny will fall is 16 feet the first second, 48 feet the next second, 80 feet the third second, and so on in an arithmetic sequence. 10. 10. PROBLEM: What is the total distance the object will fall in 6 seconds? SOLUTION: Arithmetic sequence: 16, 48, 80, ... Given: a1= 16; a2= 48; a3= 80 Find: S6= ? The 6th term is 176. Now, we are ready to find the sum: Second 1 2 3 4 5 6 Total distance in 6 seconds Distance 16 48 80 … … 176 ..... 11. 11. SITUATION: The sum of the interior angles of a triangle is 180º,of a quadrilateral is 360º and of a pentagon is 540º. 12. 12. PROBLEM: Assuming this pattern continues, find the sum of the interior angles of a dodecagon (12 sides). SOLUTION: Given: d=180 Find: a10= ? This sequence is arithmetic and the common difference is 180. The 12-sided figure will be the 10th term in this sequence. Find the 10th term. 180 360 540 ... ? Sides: 3 4 5 ... 12 Term: 1 2 3 ... ? 13. 13. SITUATION: After knee surgery, your trainer tells you to return to your jogging program slowly. He suggests jogging for 12 minutes each day for the first week. Each week thereafter, he suggests that you increase that time by 6 minutes per day. 14. 14. PROBLEM: How many weeks will it be before you are upto jogging 60 minutes per day? SOLUTION: Given: a1 60; d=6 Find: n= ? Adding 6 minutes to the weekly jogging time for each week creates the sequence: 12, 18, 24, ... This sequence is arithmetic. Week Number 1 2 3 … ? Minutes of Jogging each day inside the week 12 18 24 … n 15. 15. SITUATION: 20 people live on the first floor of the building, 34 people on the second floor and 48 people on the third floor, and soon in an arithmetic sequence. 16. 16. PROBLEM: What’s the total number of people living in the building? SOLUTION: The sequence is 20, 34, 48 … Given: a1= 20; a2= 34; a3= 48 Find: S5= ? Floor 1st 2nd 3rd 4th 5th People living in the building Number of People who live 20 34 48 … ? Sn an = 20+(n-1)14 a5 = 20+(5-1)14=76 We can use the formula: Thus, =240 17. 17. SITUATION: Lee earned \$240 in the first week, \$350in the second week and \$460 in the third week, and so on in an arithmetic sequence. 18. 18. PROBLEM: How much did he earn in the first 5 weeks? SOLUTION: The sequence is 240, 350, 460 … Given: a1= 240; a2= 350; a3= 460 Find: S5= ? Week 1st 2nd 3rd 4th 5th First 5 weeks Money that Lee Earned \$240 \$350 \$460 … ? Sn an=240+(n-1)110 a5=240+(5-1)110=680 We can use the formula: Thus, =2300 19. 19. SITUATION: An auditorium has 20 seats on the first row, 24 seats on the second row, 28 seats on the third row, and so on and has 30 rows of seats 20. 20. PROBLEM: How many seats are in the theatre? SOLUTION: Given: a1= 20; a2= 24; a3= 28; n=30 Find: S30= ? Row 1st 2nd 3rd … 30th Total number of rows Number of seats 20 24 28 … ? Sn To find a30 we need the formula for the sequence and then substitute n = 30. The formula for an arithmetic sequence is We already know that is a1 = 20, n = 30, and the common difference, d, is 4. So now we have So we now know that there are 136 seats on the 30th row. We can use this back in our formula for the arithmetic series.
The K5 Learning Blog urges parents to be pro-active in helping their children reach their full academic potential. K5 Learning provides free worksheets, workbooks and an online reading and math program for kindergarten to grade 5 students. How to Solve Division Word Problems In grade 3 students move onto more complex word problems using multiplication and division. Below we’ll take you through a step-by-step guide on how to solve a simple division word problem. Turn the English into Algebra The first rule of any word problem is to think in math terms. To do that, you’ll need to: • Read the whole word problem. • Write down what is asked for. (As students work through the word problem, they can get lost and start to work in a different direction. Having the actual request written down, will help to keep them focus on the goal.) • Sketch out the word problem, if possible. • the signs where you see key words. For example, write + where you see add, increase, combine, or – where you see less, difference, reduce. • Find or work out any formulas. Division word problem example Let’s work through a division word problem: Katie has 700 candies. She gave away 175 candies to her classmates. She put the remaining candies in five separate bags. How may candies are there in each bag? First, read through the whole word problem once more. Now, what is being asked? We want to know how many candies are in each of the 5 bags. Let’s sketch it out: Write down the math signs where you see key words in the text: In the above, we can see there are two calculations to be completed. First, we need to find out how many candies there are left after Katie hands out 175 candies to her classmates: Now we know there are 525 candies remaining. Next, we need to work out how many candies she put in each of the 5 bags: We need to divide the 525 candies by 5 bags: The answer is there are 105 candies in each bag. For practice, we have division word problems worksheets in our grade 3 word problems section.
Hairu shares his stickers part 60 (Math Question) Standard The blog postings are about the Singapore Math. The readers can learn from the postings on Solving Singapore Primary School Mathematics. The blog presents the Math Concept, the Math Questions with solutions that teaches in Singapore Primary Schools. You or the kids will learn the skills of dealing with the Math Problem Solving, the Math Modeling and the Problem Sum from Lower Primary School to Upper Primary School level after reading the blog postings. This posting is an upper primary school math question on Percentage and Problem Sum. You should read the posting on Sarin learns Average Number, Percentage and Charting in school (Math Concept) to understand Percentage. Challenge yourself with the question before look out for the given solution‼! Upper primary school mathematics question UPQ470 Hairu had 20% more stickers than Sarin. Fatimah had 40% fewer stickers than Hairu. Fatimah lost 25% of her stickers and she had 540 stickers left. Sarin bought some stickers and the number of stickers that he had increased by 10%. How many stickers did the three children have on the end? Solution: Fatimah lost 25% of her stickers and left with 540 stickers 3 units = 540 1 unit = 180 The number of stickers that Fatimah has at first = 4 × 180 = 720 Fatimah had 40% fewer stickers than Hairu From the model 3 units = 720 1 unit = 240 The number of stickers that Hairu has at first = 5 × 240 = 1200 Hairu had 20% more stickers than Sarin at first From the model, 6 units = 1200 1 unit = 200 The number of stickers that Sarin has at first = 5 × 200 = 1000 The final number of stickers that Sarin had was 10% more From the model 10 units = 1000 1 unit = 100 The number of stickers that Sarin has at the end = 11 × 100 = 1100 The number of stickers that the three children had at the end = 540 + 1200 + 1100 = 2840 Alternative Solution: Fatimah lost 25% of her stickers and left 540 stickers 75% ==> 540 10% ==> 72 stickers 100% ==> 720 stickers At first, The number of stickers Hairu has = 720 ÷ 0.6 = 1200 The number of stickers Sarin has = 1200 ÷ 1.2 = 1000 The number of stickers Fatimah has = 720 At the end, The number of stickers Hairu has = 1200 The number of stickers Sarin has = 1000 × 1.1 = 1100 The number of stickers Fatimah has = 540 The total number of stickers three of them have = 1200 + 1100 + 540 = 2840
### Select your language Suggested languages for you: Americas Europe Problem 599 # Inscribe an equilateral triangle In a given circle. Expert verified To inscribe an equilateral triangle in a given circle, first, identify the center (O) and radius (r) of the circle. Draw a radius (OA) and construct a 60-degree angle at point A. Create a circle with center A and radius r, intersecting the original circle at points B and C. Finally, connect points A, B, and C to form the inscribed equilateral triangle. See the step by step solution ## Unlock all solutions Get unlimited access to millions of textbook solutions with Vaia Premium Over 22 million students worldwide already upgrade their learning with Vaia! ## Step 1: Find the center and radius of the given circle. First, identify the center and radius of the given circle. Let the center be O and the radius be r. ## Step 2: Draw a radius on the circumference of the circle. Draw a line segment from the center O to a point A on the circle. OA will serve as one side of the equilateral triangle. Since all sides of an equilateral triangle are equal, the length of OA is also r. ## Step 3: Determine the angles of the equilateral triangle. An equilateral triangle has three equal angles of 60 degrees each. So, we will create a 60-degree angle at point A. ## Step 4: Construct a circle with radius r and center A. Draw a circle with center A and radius r. This circle will intersect the given circle at two points, which will be the vertices of the equilateral triangle. ## Step 5: Find the intersection points of the two circles. Label the intersection points of the two circles as B and C. These points will be the vertices of the equilateral triangle inscribed in the given circle. ## Step 6: Construct the equilateral triangle. Draw the line segments AB and AC, forming an equilateral triangle with vertices A, B, and C. The equilateral triangle ABC is now inscribed in the given circle, with its vertices lying on the circumference of the circle. We value your feedback to improve our textbook solutions. ## Access millions of textbook solutions in one place • Access over 3 million high quality textbook solutions • Access our popular flashcard, quiz, mock-exam and notes features • Access our smart AI features to upgrade your learning ## Join over 22 million students in learning with our Vaia App The first learning app that truly has everything you need to ace your exams in one place. • Flashcards & Quizzes • AI Study Assistant • Smart Note-Taking • Mock-Exams • Study Planner
<meta http-equiv="refresh" content="1; url=/nojavascript/"> You are viewing an older version of this Concept. Go to the latest version. Graphs of Polynomials Using Zeros x- and y-intercepts, signs, and leading term tests used to sketch a graph. 0% Progress Practice Graphs of Polynomials Using Zeros Progress 0% Graphs of Polynomials Using Zeros How is finding and using the zeroes of a higher-degree polynomial related to the same process you have used in the past on quadratic functions? Watch This This video is a good introduction to graphing cubic and higher degree polynomials. Note that a graphing calculator is used in the video. Embedded Video: Guidance The following procedure can be followed when graphing a polynomial function. • Use the leading-term test to determine the end behavior of the graph. • Find the $x-$ intercept(s) of $f(x)$ by setting $f(x)=0$ and then solving for $x$ . • Find the $y-$ intercept of $f(x)$ by setting $y=f(0)$ and finding $y$ . • Use the $x-$ intercept(s) to divide the $x-$ axis into intervals and then choose test points to determine the sign of $f(x)$ on each interval. • Plot the test points. • If necessary, find additional points to determine the general shape of the graph. If $a_{n}x^{n}$ is the leading term of a polynomial. Then the behavior of the graph as $x\to\infty$ or $x\to-\infty$ can be known by one the four following behaviors: 1. If $a_{n}>0$ and $n$ even: 2. If $a_{n}<0$ and $n$ even: 3. If $a_{n}>0$ and $n$ odd: 4. If $a_{n}<0$ and $n$ odd: Example A Find the roots (zeroes) of the polynomial: $h(x)=x^{3}+2x^{2}-5x-6$ Solution: Start by factoring: $h(x)=x^{3}+2x^{2}-5x-6=(x+1)(x-2)(x+3)$ To find the zeros, set h ( x )=0 and solve for x . $(x+1)(x-2)(x+3)=0$ This gives $x+1 & = 0\\x-2 & = 0\\x+3 & = 0$ or $x & = -1\\x & = 2\\x & = -3$ So we say that the solution set is $\{-3, -1, 2\}$ . They are the zeros of the function $h(x)$ . The zeros of $h(x)$ are the $x-$ intercepts of the graph $y=h(x)$ below. Example B Find the zeros of $g(x)=-(x-2)(x-2)(x+1)(x+5)(x+5)(x+5)$ . Solution The polynomial can be written as $g(x)=-(x-2)^{2}(x+1)(x+5)^{3}$ To solve the equation, we simply set it equal to zero $-(x-2)^{2}(x+1)(x+5)^{3}=0$ this gives $x-2 & = 0\\x+1 & = 0\\x+5 & = 0$ or $x & = 2\\x & = -1\\x & = -5$ Notice the occurrence of the zeros in the function. The factor $(x-2)$ occurred twice (because it was squared), the factor $(x+1)$ occurred once and the factor $(x+5)$ occurred three times. We say that the zero we obtain from the factor $(x-2)$ has a multiplicity $k=2$ and the factor $(x+5)$ has a multiplicity $k=3$ . Example C Graph the polynomial function $f(x)=-3x^{4}+2x^{3}$ . Solution Since the leading term here is $-3x^{4}$ then $a_{n}=-3<0$ , and $n=4$ even. Thus the end behavior of the graph as $x\to\infty$ and $x\to-\infty$ is that of Box #2, item 2. We can find the zeros of the function by simply setting $f(x)=0$ and then solving for $x$ . $-3x^{4}+2x^{3} & = 0\\-x^3(3x-2) & = 0$ This gives $x=0\quad \text{or} \quad x=\frac{2}{3}$ So we have two $x-$ intercepts, at $x=0$ and at $x=\frac{2}{3}$ , with multiplicity $k=3$ for $x=0$ and multiplicity $k=1$ for $x=\frac{2}{3}$ . To find the $y-$ intercept, we find $f(0)$ , which gives $f(0)=0$ So the graph passes the $y-$ axis at $y=0$ . Since the $x-$ intercepts are 0 and $\frac{2}{3}$ , they divide the $x-$ axis into three intervals: $(-\infty, 0), \left ( 0, \frac{2}{3} \right ),$ and $\left ( \frac{2}{3}, \infty \right )$ . Now we are interested in determining at which intervals the function $f(x)$ is negative and at which intervals it is positive. To do so, we construct a table and choose a test value for $x$ from each interval and find the corresponding $f(x)$ at that value. Interval Test Value $x$ $f(x)$ Sign of $f(x)$ Location of points on the graph $(-\infty, 0)$ -1 -5 - below the $x-$ axis $\left ( 0, \frac{2}{3} \right )$ $\frac{1}{2}$ $\frac{1}{16}$ + above the $x-$ axis $\left ( \frac{2}{3}, \infty \right )$ 1 -1 - below the $x-$ axis Those test points give us three additional points to plot: $(-1, -5), \left ( \frac{1}{2},\frac{1}{16} \right )$ , and (1, -1). Now we are ready to plot our graph. We have a total of three intercept points, in addition to the three test points. We also know how the graph is behaving as $x\to-\infty$ and $x\to+\infty$ . This information is usually enough to make a rough sketch of the graph. If we need additional points, we can simply select more points to complete the graph. In the introduction to the lesson, it was noted that there are similarities in graphing using zeroes between quadratic functions and higher-degree polynomials. Were you able to identify some of those similarities? Despite the more complex nature of the graphs of higher-degree polynomials, the general process of graphing using zeroes is actually very similar. In both cases, your goal is to locate the points where the graph crosses the x or y axis. In both cases, this is done by setting the y value equal to zero and solving for x to find the x axis intercepts, and setting the x value equal to zero and solving for y to find the y axis intercepts. Vocabulary Cubic Function: A function containing an $x^{3}$ term as the highest power of x . Quartic Function: A function containing an $x^{4}$ term as the highest power of x . Zeroes of a Polynomial: The values output by the function (the values of $f(x)$ or $y$ ), when the input (the $x$ value) is zero, or vice-versa. Interval: A portion of a function, generally defined by a starting and ending value of $x$ . Guided Practice Questions Sketch a graph of each power function using the properties of the power functions. 1) $f(x)=-3x^{4}$ 2) $h(x)=\frac{1}{2}x^{5}$ 3) $q(x)=4x^{8}$ 4) Find the zeros and sketch a graph of the polynomial $f(x)=x^{4}-x^{2}-56$ 5) Graph $g(x)=-(x-2)^{2}(x+1)(x+5)^{3}$ Solutions 1) Solution: Step 1: By applying the leading term test , we can say that since the co-efficient $-3$ is $<0$ , and since the power $4$ is even , the end behavior of the graph resembles: Step 2: By solving the equation for $x = 1$ and $x = -1$ , we get the points: $(1, -3)$ and $(-1, -3)$ . Step 3: This suggests 2) Solution: Step 1: By applying the leading term test , we can say that since the co-efficient $\frac{1}{2}$ is $>0$ , and since the power $5$ is odd , the end behavior of the graph resembles: Step 2: By solving the equation for $x = 1$ and $x = -1$ , we get the points: $(1, 1/2)$ and $(-1, -1/2)$ . Step 3: This suggests 3) Solution: Step 1: By applying the leading term test , we can say that since the co-efficient $4$ is $>0$ , and since the power $8$ is even , the end behavior of the graph resembles: Step 2: By solving the equation for $x = 1$ and $x = -1$ , we get the points: $(1, 4)$ and $(-1, 4)$ . Step 3: This suggests 4) This is a factorable equation, $f(x) & =x^4-x^2-56\\& = (x^2-8)(x^2+7)$ Setting $f(x)=0$ , $(x^{2}-8)(x^{2}+7) = 0$ the first term gives $x^{2}-8 & = 0\\x^2 & = 8\\x & = \pm \sqrt{8}\\& = \pm 2\sqrt{2}$ and the second term gives $x^{2}+7 & = 0\\x^2 & = -7\\x & = \pm \sqrt{-7}\\& = \pm i\sqrt{7}$ So the solutions are $\pm2\sqrt{2}$ and $\pm i\sqrt{7}$ , a total of four zeros of $f(x)$ . Keep in mind that only the real zeros of a function correspond to the $x-$ intercept of its graph. 5) Use the zeros to create a table of intervals and see whether the function is above or below the $x-$ axis in each interval: Interval Test value $x$ $g(x)$ Sign of $g(x)$ Location of graph relative to $x-$ axis $(-\infty, -5)$ -6 320 + Above $x=-5$ -5 0 NA (-5, -1) -2 144 + Above $x=-1$ -1 0 NA (-1, 2) 0 -100 - Below $x=2$ 2 0 NA $(2, \infty)$ 3 -256 - Below Finally, use this information and the test points to sketch a graph of $g(x)$ . Practice 1. If c is a zero of f , then c is a/an _________________________ of the graph of f 2. If c is a zero of f , then ( x - c ) is a factor of ___________________? 3. Find the zeros of the polynomial: $P(x) = x^3 - 5x^2 + 6x$ Consider the function: $f(x) = -3(x - 3)^4(5x - 2)(2x - 1)^3(4 - x)^2$ . 1. How many zeros (x-intercepts) are there? 2. What is the leading term? Find the zeros and graph the polynomial. Be sure to label the x -intercepts, y -intercept (if possible) and have correct end behavior. You may use technology for #s 9-12 1. $P(x) = -2(x + 1)^2(x - 3)$ 2. $P(x) = x^3 + 3x^2 - 4x - 12$ 3. $f(x) = -2x^3 + 6x^2 + 9x + 6$ 4. $f(x) = -4x^2 -7x +3$ 5. $f(x) = 2x^5 +4x^3 + 8x^2 +6x$ 6. $f(x) = x^4 - 3x^2$ 7. $g(x) = x^2 - \|x\|$ 8. Given: $P(x) = (3x +2)(x - 7)^2(9x + 2)^3$ State: a) The leading term: b) The degree of the polynomial: c) The leading coefficient: Determine the equation of the polynomial based on the graph: Vocabulary Language: English Cubic Function Cubic Function A cubic function is a function containing an $x^{3}$ term as the highest power of $x$. Intercept Intercept The intercepts of a curve are the locations where the curve intersects the $x$ and $y$ axes. An $x$ intercept is a point at which the curve intersects the $x$-axis. A $y$ intercept is a point at which the curve intersects the $y$-axis. interval interval An interval is a specific and limited part of a function. The leading-term test is a test to determine the end behavior of a polynomial function. Polynomial Polynomial A polynomial is an expression with at least one algebraic term, but which does not indicate division by a variable or contain variables with fractional exponents. Polynomial Graph Polynomial Graph A polynomial graph is the graph of a polynomial function. The term is most commonly used for polynomial functions with a degree of at least three. Quartic Function Quartic Function A quartic function is a function $f(x)$ containing an $x^{4}$ term as the highest power of ''x''. Roots Roots The roots of a function are the values of x that make y equal to zero. Zeroes Zeroes The zeroes of a function $f(x)$ are the values of $x$ that cause $f(x)$ to be equal to zero.
# How do you find the determinant of ((2, 11, -3, 1), (1, 5, 7, -4), (6, 13, -5, 2), (4, 22, -6, 2))? May 3, 2016 #### Answer: $0$ #### Explanation: $\left(\begin{matrix}\textcolor{b l u e}{2} & \textcolor{b l u e}{11} & \textcolor{b l u e}{- 3} & \textcolor{b l u e}{1} \\ 1 & 5 & 7 & - 4 \\ 6 & 13 & - 5 & 2 \\ \textcolor{c r i m s o n}{4} & \textcolor{c r i m s o n}{22} & \textcolor{c r i m s o n}{- 6} & \textcolor{c r i m s o n}{2}\end{matrix}\right) = 2 \cdot \left(\begin{matrix}\textcolor{b l u e}{2} & \textcolor{b l u e}{11} & \textcolor{b l u e}{- 3} & \textcolor{b l u e}{1} \\ 1 & 5 & 7 & - 4 \\ 6 & 13 & - 5 & 2 \\ \textcolor{c r i m s o n}{2} & \textcolor{c r i m s o n}{11} & \textcolor{c r i m s o n}{- 3} & \textcolor{c r i m s o n}{1}\end{matrix}\right)$ When a determinant has two lines or 2 columns equal (or what leads to the same conclusion, when it has two lines or 2 columns proportional), this determinant is equal to zero. Therefore, since, in the present case, row1 $=$ row4 : $D e t . = 0$
0 # What is the circumcenter of the triangle with the points (23) (5-3) and (92)? Updated: 9/25/2023 Wiki User 9y ago The circumcenter is at (137/26, 23/26) = (5 7/26, 23/26) The circumcenter is where the side perpendicular bisectors meet; this can be calculated by finding the equations of the perpendicular bisectors and solving the simultaneous equations that result (only 2 needs be solved, the third can be used as a check): Let the three points of the triangle be A (2, 3), B (5, -3), C (9, 2) The sides are AB, BC, AC with perpendicular bisectors AB', BC', AC' respectively; then: AB has midpoint ((5+2)/2, (-3 + 3)) = (7/2, 0) AB has gradient (-3 - 3)/(5 - 2) = -6/3 = -2 → AB' has equation y - 0 = 1/2(x - 7/2) → y = 1/2(x - 7/2) BC has midpoint ((9 + 5)/2, (2 + -3)/2) = (7, -1/2) BC has gradient (2 - -3)/(9 - 5) = 5/4 → BC' has equation y - -1/2 = -4/5(x - 7) → y = 4/5(7 - x) - 1/2 AC has midpoint ((9 + 2)/2, (2 + 3)/2) = (11/2, 5/2) AC has gradient (2 - 3)/(9 - 2) = -1/7 → AC' has equation y - 5/2 = 7(x - 11/2) → y = 7(x - 11/2) + 5/2 Solving for point of intersection of AB' and AC': 1/2(x - 7/2) = 7(x - 11/2) + 5/2 → x - 7/2 = 14x - 77 + 5 → 13x = 72 - 7/2 = 137/2 → x = 137/26 (= 5 7/26) Using AB' y = 1/2(137/26 - 7/2) → y = 1/2(137/26 - 91/26) → y = 23/26 Checking using BC': y = 4/5(7 - x) - 1/2 → y = 4/5(7 - 137/26) - 1/2 → y = 4/5(182/26-137/26) - 13/26 → y = 4/5(45/26) - 13/26 → y = 36/26 - 13/26 = 23/26 as expected. Wiki User 9y ago
Tuesday , July 5 2022 # NCERT 5th Class (CBSE) Mathematics: Ratio And Proportion ## PROPORTION Here are two bottles. Both contain red and blue counters. Bottle A has 2 red counters and 3 blue counters. Ratio of red to blue counters is 2 : 3 in bottle A. Bottle B has 6 red counters and 9 blue counters. Ratio of red to blue counters is 6 : 9 in bottle B. But, the ratios 2 : 3 and 6 : 9 are equal because 2 ( × 3 )/3 ( × 3 ) = 6/9 As the two ratios are equal we can say that are in proportion and write it as 2 : 3 : : 6 : 9 This is read as 2 is to 3 as 6 is to 9 In the proportion 2 : 3 : : 6 : 9, 2 and 9 are called the extreme terms and 3 and 6 are the mean or middle terms. (a) A recipe requires 12 tomatoes to make 8 cups of soup. Another recipe needs 15 tomatoes to make 10 cups of soup. How do we find out if the two recipes have the same proportion of tomatoes to cups of soup? We can say that the ratio of tomatoes to the cups of soup for the first recipe is 12 : 8 or 12/8 The ratio for the second recipe is 15 : 10 or 15/10 To test whether the two ratios are proportionate we can use cross – multiplication. 8 × 15 = 120 and 12 × 10 = 120 The cross products are equal; so the ratios are in proportion. i.e., 12 : 18 : : 15 : 10 So the two recipes have the same proportion of tomatoes to cups of soup. (b) Another way to look at this is to check whether the product of the means is equal to the product of the product of the extremes. To check the proportionality of the ratios 9 : 14 and 18 : 23 The ratios are not in proportion as the products are not equal. ### Using Proportion (a) Aarti uses 4 eggs to make breakfast for 3 people. How many eggs would she need to make breakfast for 18 people? The ratio of the number of eggs to the number of people is 4/3 4 eggs/3 people = ? eggs/18 people As a proportion 4 : 3 : : ? : 18 4/3 = ?/18 4 (×6)/3 (×6) = 24/18 Answer: Aarti needs 24 eggs to make breakfast for 18 people. (b) 4 rakhis cost Rs. 32. How much money will be need to buy 7 rakhis? 4 : 32 : : 7 : ? 4/32 = 1/8 (lowest terms) 1 : 8 : : 7 : ? or 1/8 = 7/? 1 × 7 / 8 × 7 = 7/56 Answer: 7 rakhis will coast Rs. 56. ## चुनौती हिमालय की 5th NCERT CBSE Hindi Rimjhim Ch 18 चुनौती हिमालय की 5th Class NCERT CBSE Hindi Book Rimjhim Chapter 18 प्रश्न: लद्दाख जम्मू-कश्मीर राज्य में …
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. The quadratic formula allows us to solve any quadratic equation that's in the form ax^2 + bx + c = 0. This article reviews how to apply the formula. ## What is the quadratic formula? $x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$ $a{x}^{2}+bx+c=0$ ### Example We're given an equation and asked to solve for $q$: $0=-7{q}^{2}+2q+9$ This equation is already in the form $a{x}^{2}+bx+c=0$, so we can apply the quadratic formula where $a=-7,b=2,c=9$: Let's check both solutions to be sure it worked: $q=-1$$q=\frac{9}{7}$ $\begin{array}{rl}0& =-7{q}^{2}+2q+9\\ \\ 0& =-7\left(-1{\right)}^{2}+2\left(-1\right)+9\\ \\ 0& =-7\left(1\right)-2+9\\ \\ 0& =-7-2+9\\ \\ 0& =0\end{array}$$\begin{array}{rl}0& =-7{q}^{2}+2q+9\\ \\ 0& =-7{\left(\frac{9}{7}\right)}^{2}+2\left(\frac{9}{7}\right)+9\\ \\ 0& =-7\left(\frac{81}{49}\right)+\left(\frac{18}{7}\right)+9\\ \\ 0& =-\left(\frac{81}{7}\right)+\left(\frac{18}{7}\right)+9\\ \\ 0& =-\left(\frac{63}{7}\right)+9\\ \\ 0& =-9+9\\ \\ 0& =0\end{array}$ Yep, both solutions check out. Practice Solve for $x$. $-4+x+7{x}^{2}=0$ Want more practice? Check out this exercise. ## Want to join the conversation? • Sal, How does the quadratic formula relate to business and economics? • It helps in lots of ways. It can possibly predict the future path of certain things, especially if your graph is exponential. • what if the equation doesn't equal zero • then just subtract the non-zero number from the RHS to the LHS and make the RHS equal to zero. • are there any shortcuts or patterns we can use to make calculation quicker? • Not insofar as I know. The quadratic formula is the shortcut, unless you prefer grouping or something • When there are two numbers in the numerator before the square root, how do I solve the problem? • Can you show an example of what you mean? • can you reccomend other math websites for algebra 1 and 2 • I do not understand this. Can someone please explain (1 vote) • This is a formula, so if you can get the right numbers, you plug them into the formula and calculate the answer(s). We always have to start with a quadratic in standard form: ax^2+bx+c=0. Making one up, 3x^2+2x-5=0, we see a=3, b=2, c=-5. I teach my students to start with the discriminant, b^2-4ac. Also, especially in the beginning, put the b value in parentheses so that you square a negative number if b is negative. In our example, this gives (2)^2-4(3)(-5) = 4+60=64. If I take √64 = 8. Filling out the formula, we get x=(-2±8)/(2(3)) or breaking it into the two parts x=(-2+8)/6=1 and (-2-8)/6=-10/6=-5/2. Where is the confusion? It is always hard to answer when we cannot figure out what you do understand and where you are confused. • isn't it the square root of -112?
Calculus Volume 2 # 3.2Trigonometric Integrals Calculus Volume 23.2 Trigonometric Integrals ## Learning Objectives • 3.2.1 Solve integration problems involving products and powers of $sin x sin x$ and $cos x . cos x .$ • 3.2.2 Solve integration problems involving products and powers of $tan x tan x$ and $sec x . sec x .$ • 3.2.3 Use reduction formulas to solve trigonometric integrals. In this section we look at how to integrate a variety of products of trigonometric functions. These integrals are called trigonometric integrals. They are an important part of the integration technique called trigonometric substitution, which is featured in Trigonometric Substitution. This technique allows us to convert algebraic expressions that we may not be able to integrate into expressions involving trigonometric functions, which we may be able to integrate using the techniques described in this section. In addition, these types of integrals appear frequently when we study polar, cylindrical, and spherical coordinate systems later. Let’s begin our study with products of $sinxsinx$ and $cosx.cosx.$ ## Integrating Products and Powers of sinx and cosx A key idea behind the strategy used to integrate combinations of products and powers of $sinxsinx$ and $cosxcosx$ involves rewriting these expressions as sums and differences of integrals of the form $∫sinjxcosxdx∫sinjxcosxdx$ or $∫cosjxsinxdx.∫cosjxsinxdx.$ After rewriting these integrals, we evaluate them using u-substitution. Before describing the general process in detail, let’s take a look at the following examples. ## Example 3.8 ### Integrating $∫cosjxsinxdx∫cosjxsinxdx$ Evaluate $∫cos3xsinxdx.∫cos3xsinxdx.$ ## Checkpoint3.5 Evaluate $∫sin4xcosxdx.∫sin4xcosxdx.$ ## Example 3.9 ### A Preliminary Example: Integrating $∫cosjxsinkxdx∫cosjxsinkxdx$ Where k is Odd Evaluate $∫cos2xsin3xdx.∫cos2xsin3xdx.$ ## Checkpoint3.6 Evaluate $∫cos3xsin2xdx.∫cos3xsin2xdx.$ In the next example, we see the strategy that must be applied when there are only even powers of $sinxsinx$ and $cosx.cosx.$ For integrals of this type, the identities $sin2x=12−12cos(2x)=1−cos(2x)2sin2x=12−12cos(2x)=1−cos(2x)2$ and $cos2x=12+12cos(2x)=1+cos(2x)2cos2x=12+12cos(2x)=1+cos(2x)2$ are invaluable. These identities are sometimes known as power-reducing identities and they may be derived from the double-angle identity $cos(2x)=cos2x−sin2xcos(2x)=cos2x−sin2x$ and the Pythagorean identity $cos2x+sin2x=1.cos2x+sin2x=1.$ ## Example 3.10 ### Integrating an Even Power of $sinxsinx$ Evaluate $∫sin2xdx.∫sin2xdx.$ ## Checkpoint3.7 Evaluate $∫cos2xdx.∫cos2xdx.$ The general process for integrating products of powers of $sinxsinx$ and $cosxcosx$ is summarized in the following set of guidelines. ## Problem-Solving Strategy ### Problem-Solving Strategy: Integrating Products and Powers of sin x and cos x To integrate $∫cosjxsinkxdx∫cosjxsinkxdx$ use the following strategies: 1. If $kk$ is odd, rewrite $sinkx=sink−1xsinxsinkx=sink−1xsinx$ and use the identity $sin2x=1−cos2xsin2x=1−cos2x$ to rewrite $sink−1xsink−1x$ in terms of $cosx.cosx.$ Integrate using the substitution $u=cosx.u=cosx.$ This substitution makes $du=−sinxdx.du=−sinxdx.$ 2. If $jj$ is odd, rewrite $cosjx=cosj−1xcosxcosjx=cosj−1xcosx$ and use the identity $cos2x=1−sin2xcos2x=1−sin2x$ to rewrite $cosj−1xcosj−1x$ in terms of $sinx.sinx.$ Integrate using the substitution $u=sinx.u=sinx.$ This substitution makes $du=cosxdx.du=cosxdx.$ (Note: If both $jj$ and $kk$ are odd, either strategy 1 or strategy 2 may be used.) 3. If both $jj$ and $kk$ are even, use $sin2x=(1/2)−(1/2)cos(2x)sin2x=(1/2)−(1/2)cos(2x)$ and $cos2x=(1/2)+(1/2)cos(2x).cos2x=(1/2)+(1/2)cos(2x).$ After applying these formulas, simplify and reapply strategies 1 through 3 as appropriate. ## Example 3.11 ### Integrating $∫cosjxsinkxdx∫cosjxsinkxdx$ where k is Odd Evaluate $∫cos8xsin5xdx.∫cos8xsin5xdx.$ ## Example 3.12 ### Integrating $∫cosjxsinkxdx∫cosjxsinkxdx$ where k and j are Even Evaluate $∫sin4xdx.∫sin4xdx.$ ## Checkpoint3.8 Evaluate $∫cos3xdx.∫cos3xdx.$ ## Checkpoint3.9 Evaluate $∫cos2(3x)dx.∫cos2(3x)dx.$ In some areas of physics, such as quantum mechanics, signal processing, and the computation of Fourier series, it is often necessary to integrate products that include $sin(ax),sin(ax),$ $sin(bx),sin(bx),$ $cos(ax),cos(ax),$ and $cos(bx).cos(bx).$ These integrals are evaluated by applying trigonometric identities, as outlined in the following rule. ## Rule: Integrating Products of Sines and Cosines of Different Angles To integrate products involving $sin(ax),sin(ax),$ $sin(bx),sin(bx),$ $cos(ax),cos(ax),$ and $cos(bx),cos(bx),$ use the substitutions $sin(ax)sin(bx)=12cos((a−b)x)−12cos((a+b)x)sin(ax)sin(bx)=12cos((a−b)x)−12cos((a+b)x)$ (3.3) $sin(ax)cos(bx)=12sin((a−b)x)+12sin((a+b)x)sin(ax)cos(bx)=12sin((a−b)x)+12sin((a+b)x)$ (3.4) $cos(ax)cos(bx)=12cos((a−b)x)+12cos((a+b)x)cos(ax)cos(bx)=12cos((a−b)x)+12cos((a+b)x)$ (3.5) These formulas may be derived from the sum-of-angle formulas for sine and cosine. ## Example 3.13 ### Evaluating $∫sin(ax)cos(bx)dx∫sin(ax)cos(bx)dx$ Evaluate $∫sin(5x)cos(3x)dx.∫sin(5x)cos(3x)dx.$ ## Checkpoint3.10 Evaluate $∫cos(6x)cos(5x)dx.∫cos(6x)cos(5x)dx.$ ## Integrating Products and Powers of tanx and secx Before discussing the integration of products and powers of $tanxtanx$ and $secx,secx,$ it is useful to recall the integrals involving $tanxtanx$ and $secxsecx$ we have already learned: 1. $∫sec2xdx=tanx+C∫sec2xdx=tanx+C$ 2. $∫secxtanxdx=secx+C∫secxtanxdx=secx+C$ 3. $∫tanxdx=ln\|secx\|+C∫tanxdx=ln\|secx\|+C$ 4. $∫secxdx=ln\|secx+tanx\|+C.∫secxdx=ln\|secx+tanx\|+C.$ For most integrals of products and powers of $tanxtanx$ and $secx,secx,$ we rewrite the expression we wish to integrate as the sum or difference of integrals of the form $∫tanjxsec2xdx∫tanjxsec2xdx$ or $∫secjxtanxdx.∫secjxtanxdx.$ As we see in the following example, we can evaluate these new integrals by using u-substitution. ## Example 3.14 ### Evaluating $∫secjxtanxdx∫secjxtanxdx$ Evaluate $∫sec5xtanxdx.∫sec5xtanxdx.$ ## Media You can read some interesting information at this website to learn about a common integral involving the secant. ## Checkpoint3.11 Evaluate $∫tan5xsec2xdx.∫tan5xsec2xdx.$ We now take a look at the various strategies for integrating products and powers of $secxsecx$ and $tanx.tanx.$ ## Problem-Solving Strategy ### Problem-Solving Strategy: Integrating $∫tankxsecjxdx∫tankxsecjxdx$ To integrate $∫tankxsecjxdx,∫tankxsecjxdx,$ use the following strategies: 1. If $jj$ is even and $j≥2,j≥2,$ rewrite $secjx=secj−2xsec2xsecjx=secj−2xsec2x$ and use $sec2x=tan2x+1sec2x=tan2x+1$ to rewrite $secj−2xsecj−2x$ in terms of $tanx.tanx.$ Let $u=tanxu=tanx$ and $du=sec2xdx.du=sec2xdx.$ 2. If $kk$ is odd and $j≥1,j≥1,$ rewrite $tankxsecjx=tank−1xsecj−1xsecxtanxtankxsecjx=tank−1xsecj−1xsecxtanx$ and use $tan2x=sec2x−1tan2x=sec2x−1$ to rewrite $tank−1xtank−1x$ in terms of $secx.secx.$ Let $u=secxu=secx$ and $du=secxtanxdx.du=secxtanxdx.$ (Note: If $jj$ is even and $kk$ is odd, then either strategy 1 or strategy 2 may be used.) 3. If $kk$ is odd where $k≥3k≥3$ and $j=0,j=0,$ rewrite $tankx=tank−2xtan2x=tank−2x(sec2x−1)=tank−2xsec2x−tank−2x.tankx=tank−2xtan2x=tank−2x(sec2x−1)=tank−2xsec2x−tank−2x.$ It may be necessary to repeat this process on the $tank−2xtank−2x$ term. 4. If $kk$ is even and $jj$ is odd, then use $tan2x=sec2x−1tan2x=sec2x−1$ to express $tankxtankx$ in terms of $secx.secx.$ Use integration by parts to integrate odd powers of $secx.secx.$ ## Example 3.15 ### Integrating $∫tankxsecjxdx∫tankxsecjxdx$ when $jj$ is Even Evaluate $∫tan6xsec4xdx.∫tan6xsec4xdx.$ ## Example 3.16 ### Integrating $∫tankxsecjxdx∫tankxsecjxdx$ when $kk$ is Odd Evaluate $∫tan5xsec3xdx.∫tan5xsec3xdx.$ ## Example 3.17 ### Integrating $∫tankxdx∫tankxdx$ where $kk$ is Odd and $k≥3k≥3$ Evaluate $∫tan3xdx.∫tan3xdx.$ ## Example 3.18 ### Integrating $∫sec3xdx∫sec3xdx$ Integrate $∫sec3xdx.∫sec3xdx.$ ## Checkpoint3.12 Evaluate $∫tan3xsec7xdx.∫tan3xsec7xdx.$ ## Reduction Formulas Evaluating $∫secnxdx∫secnxdx$ for values of $nn$ where $nn$ is odd requires integration by parts. In addition, we must also know the value of $∫secn−2xdx∫secn−2xdx$ to evaluate $∫secnxdx.∫secnxdx.$ The evaluation of $∫tannxdx∫tannxdx$ also requires being able to integrate $∫tann−2xdx.∫tann−2xdx.$ To make the process easier, we can derive and apply the following power reduction formulas. These rules allow us to replace the integral of a power of $secxsecx$ or $tanxtanx$ with the integral of a lower power of $secxsecx$ or $tanx.tanx.$ ## Rule: Reduction Formulas for $∫ sec n x d x ∫ sec n x d x$ and $∫ tan n x d x ∫ tan n x d x$ $∫secnxdx=1n−1secn−2xtanx+n−2n−1∫secn−2xdx∫secnxdx=1n−1secn−2xtanx+n−2n−1∫secn−2xdx$ (3.6) $∫tannxdx=1n−1tann−1x−∫tann−2xdx∫tannxdx=1n−1tann−1x−∫tann−2xdx$ (3.7) The first power reduction rule may be verified by applying integration by parts. The second may be verified by following the strategy outlined for integrating odd powers of $tanx.tanx.$ ## Example 3.19 ### Revisiting $∫sec3xdx∫sec3xdx$ Apply a reduction formula to evaluate $∫sec3xdx.∫sec3xdx.$ ## Example 3.20 ### Using a Reduction Formula Evaluate $∫tan4xdx.∫tan4xdx.$ ## Checkpoint3.13 Apply the reduction formula to $∫sec5xdx.∫sec5xdx.$ ## Section 3.2 Exercises Fill in the blank to make a true statement. 69. $sin 2 x + _______ = 1 sin 2 x + _______ = 1$ 70. $sec 2 x − 1 = _______ sec 2 x − 1 = _______$ Use an identity to reduce the power of the trigonometric function to a trigonometric function raised to the first power. 71. $sin 2 x = _______ sin 2 x = _______$ 72. $cos 2 x = _______ cos 2 x = _______$ Evaluate each of the following integrals by u-substitution. 73. $∫ sin 3 x cos x d x ∫ sin 3 x cos x d x$ 74. $∫ cos x sin x d x ∫ cos x sin x d x$ 75. $∫ tan 5 ( 2 x ) sec 2 ( 2 x ) d x ∫ tan 5 ( 2 x ) sec 2 ( 2 x ) d x$ 76. $∫ sin 7 ( 2 x ) cos ( 2 x ) d x ∫ sin 7 ( 2 x ) cos ( 2 x ) d x$ 77. $∫ tan ( x 2 ) sec 2 ( x 2 ) d x ∫ tan ( x 2 ) sec 2 ( x 2 ) d x$ 78. $∫ tan 2 x sec 2 x d x ∫ tan 2 x sec 2 x d x$ Compute the following integrals using the guidelines for integrating powers of trigonometric functions. Use a CAS to check the solutions. (Note: Some of the problems may be done using techniques of integration learned previously.) 79. $∫ sin 3 x d x ∫ sin 3 x d x$ 80. $∫ cos 3 x d x ∫ cos 3 x d x$ 81. $∫ sin x cos x d x ∫ sin x cos x d x$ 82. $∫ cos 5 x d x ∫ cos 5 x d x$ 83. $∫ sin 5 x cos 2 x d x ∫ sin 5 x cos 2 x d x$ 84. $∫ sin 3 x cos 3 x d x ∫ sin 3 x cos 3 x d x$ 85. $∫ sin x cos x d x ∫ sin x cos x d x$ 86. $∫ sin x cos 3 x d x ∫ sin x cos 3 x d x$ 87. $∫ sec x tan x d x ∫ sec x tan x d x$ 88. $∫ tan ( 5 x ) d x ∫ tan ( 5 x ) d x$ 89. $∫ tan 2 x sec x d x ∫ tan 2 x sec x d x$ 90. $∫ tan x sec 3 x d x ∫ tan x sec 3 x d x$ 91. $∫ sec 4 x d x ∫ sec 4 x d x$ 92. $∫ cot x d x ∫ cot x d x$ 93. $∫ csc x d x ∫ csc x d x$ 94. $∫ tan 3 x sec x d x ∫ tan 3 x sec x d x$ For the following exercises, find a general formula for the integrals. 95. $∫ sin 2 a x cos a x d x ∫ sin 2 a x cos a x d x$ 96. $∫ sin a x cos a x d x . ∫ sin a x cos a x d x .$ Use the double-angle formulas to evaluate the following integrals. 97. $∫ 0 π sin 2 x d x ∫ 0 π sin 2 x d x$ 98. $∫ 0 π sin 4 x d x ∫ 0 π sin 4 x d x$ 99. $∫ cos 2 3 x d x ∫ cos 2 3 x d x$ 100. $∫ sin 2 x cos 2 x d x ∫ sin 2 x cos 2 x d x$ 101. $∫ sin 2 x d x + ∫ cos 2 x d x ∫ sin 2 x d x + ∫ cos 2 x d x$ 102. $∫ sin 2 x cos 2 ( 2 x ) d x ∫ sin 2 x cos 2 ( 2 x ) d x$ For the following exercises, evaluate the definite integrals. Express answers in exact form whenever possible. 103. $∫ 0 2 π cos x sin 2 x d x ∫ 0 2 π cos x sin 2 x d x$ 104. $∫ 0 π sin 3 x sin 5 x d x ∫ 0 π sin 3 x sin 5 x d x$ 105. $∫ 0 π cos ( 99 x ) sin ( 101 x ) d x ∫ 0 π cos ( 99 x ) sin ( 101 x ) d x$ 106. $∫ − π π cos 2 ( 3 x ) d x ∫ − π π cos 2 ( 3 x ) d x$ 107. $∫ 0 2 π sin x sin ( 2 x ) sin ( 3 x ) d x ∫ 0 2 π sin x sin ( 2 x ) sin ( 3 x ) d x$ 108. $∫ 0 4 π cos ( x / 2 ) sin ( x / 2 ) d x ∫ 0 4 π cos ( x / 2 ) sin ( x / 2 ) d x$ 109. $∫π/6π/3cos3xsinxdx∫π/6π/3cos3xsinxdx$ (Round this answer to three decimal places.) 110. $∫ − π / 3 π / 3 sec 2 x − 1 d x ∫ − π / 3 π / 3 sec 2 x − 1 d x$ 111. $∫ 0 π / 2 1 − cos ( 2 x ) d x ∫ 0 π / 2 1 − cos ( 2 x ) d x$ 112. Find the area of the region bounded by the graphs of the equations $y=sinx,y=sin3x,x=0,andx=π2.y=sinx,y=sin3x,x=0,andx=π2.$ 113. Find the area of the region bounded by the graphs of the equations $y=cos2x,y=sin2x,x=−π4,andx=π4.y=cos2x,y=sin2x,x=−π4,andx=π4.$ 114. A particle moves in a straight line with the velocity function $v(t)=sin(ωt)cos2(ωt).v(t)=sin(ωt)cos2(ωt).$ Find its position function $x=f(t)x=f(t)$ if $f(0)=0.f(0)=0.$ 115. Find the average value of the function $f(x)=sin2xcos3xf(x)=sin2xcos3x$ over the interval $[−π,π].[−π,π].$ For the following exercises, solve the differential equations. 116. $dydx=sin2x.dydx=sin2x.$ The curve passes through point $(0,0).(0,0).$ 117. $d y d θ = sin 4 ( π θ ) d y d θ = sin 4 ( π θ )$ 118. Find the length of the curve $y=ln(cscx),π4≤x≤π2.y=ln(cscx),π4≤x≤π2.$ 119. Find the length of the curve $y=ln(sinx),π3≤x≤π2.y=ln(sinx),π3≤x≤π2.$ 120. Find the volume generated by revolving the curve $y=cos(3x)y=cos(3x)$ about the x-axis, $0≤x≤π36.0≤x≤π36.$ For the following exercises, use this information: The inner product of two functions f and g over $[a,b][a,b]$ is defined by $f(x)·g(x)=〈f,g〉=∫abf·gdx.f(x)·g(x)=〈f,g〉=∫abf·gdx.$ Two distinct functions f and g are said to be orthogonal if $〈f,g〉=0.〈f,g〉=0.$ 121. Show that ${sin(2x),cos(3x)}{sin(2x),cos(3x)}$ are orthogonal over the interval $[−π,π].[−π,π].$ 122. Evaluate $∫−ππsin(mx)cos(nx)dx.∫−ππsin(mx)cos(nx)dx.$ 123. Integrate $y′=tanxsec4x.y′=tanxsec4x.$ For each pair of integrals, determine which one is more difficult to evaluate. Explain your reasoning. 124. $∫sin456xcosxdx∫sin456xcosxdx$ or $∫sin2xcos2xdx∫sin2xcos2xdx$ 125. $∫tan350xsec2xdx∫tan350xsec2xdx$ or $∫tan350xsecxdx∫tan350xsecxdx$ Order a print copy As an Amazon Associate we earn from qualifying purchases.
# Card Turnover Math Game Continuing on from yesterday’s post on other ways to help your students recall the times tables, here I share with you a fun times tables card game. Card Turnover This… Continuing on from yesterday’s post on other ways to help your students recall the times tables, here I share with you a fun times tables card game. ## Card Turnover This game is for 2 players. You will need a deck of cards with the picture cards removed (King, Queen, Jack, Joker). There are 2 ways this game can be played, I’ll explain both ways below. ### All times tables facts • Shuffle the cards and divide them in to two equal piles, face down on the playing surface. • Players take it in turns to turn over the top card on each pile. • Players multiply the two numbers, and the first player to call out the correct answer is the winner. That player gets to keep both cards. For example, if the cards turned up are a 7 and a 4, then the problem becomes: 7 x 4. The player that calls out the correct answer of 28 keeps the cards. • If both players say the answer at the same time, then they keep one card each. • If the answer is incorrect, then both cards are placed back at the bottom of the piles. • The winner is the player with the most cards at the end. ### Individual times tables This game can also be adapted to focus on an individual times table, for example the 7 times tables. • Shuffle the cards and place a card with a 7 on the playing surface face up. Place all the other cards face down next to the 7. • Players take it in turns to turn over the top card from the pile. • Players multiply the two numbers, and the first player to call out the correct answer is the winner. That player gets to keep the card turned over. For example, if the card turned over is a 6, then the problem becomes: 6 x 7. The player that calls out the correct answer of 42 keeps the card. • If both players say the answer at the same time, then the card is placed back at the bottom of the pile. • If the answer is incorrect, then the card is placed back at the bottom of the pile. • The winner is the player with the most cards at the end. ### Tips and extension • Have a calculator handy for students to settle any disputes. • The Ace card can be used as the number 11 or the number 1. • Include more sets of cards and have up to 4 students playing. • Use jumbo playing cards and place a magnetic strip on the back. Stick two cards on the whiteboard and get your students to call out the answer. • The rules can be changed so that players can add the numbers, or subtract the lower number from the higher number. • To make the game more challenging get your students to make their own sets of cards to include numbers up to 100. For example: 20, 30, 15, 37. They can then use these with an individual times tables, so when these cards are turned over they’re multiplied by the selected multiplier, e.g. 20 x 7. • Alternatively change the multiplier to be 70, rather than 7. So for the example above the problem would become, 6 x 70. ### Click here to find more fun math games and activities Graphic Credits: Graphics From the Pond ## The Magic Of Unplugged Play & Movement Breaks In The Classroom The importance of unplugged play & movement breaks in the classroom. Discover benefits, tips, & engaging activities to enhance learning. ## How To Teach Natural Phenomena With Natural Disaster Games Enhance your teaching with natural disaster games! Spark student interest and critical thinking skills in an engaging way. 1. I am a grandmother home schooling my granddaughter. Thank you so very much! • Hey Diana, Thanks so much for being here and your comment. Wow, that must be interesting, home schooling your granddaughter ðŸ™‚ I’m glad you like the game and I’m sure you’ll find many more ideas and activities here that will help you. Cheers Mel
What Are The Main Parts Of A Proof Geometry? One of the most prevalent types of explicit proof used in high school geometry is a two-column proof, which is composed of five parts: a given, a proposition, a statement column, a reason column, and a graphic (if one is given). Describe the most important components of a proof. Given; shown; made declarations; and given reasoning. The information provided is critical information that we will need in the construction of the proof. It is provided at the beginning of the proof. ‘Prove’ is the assertion that we are attempting to support with evidence. What are the main parts of a proof? Describe the most important components of a proof. Proofs comprise information that has already been provided as well as an assertion that has to be demonstrated. Theorems, postulates, and definitions are used in conjunction with deductive reasoning to construct an argument that includes explanation for each step. After that, you come to a conclusion. What are the three sections of a proof in geometry? Proofs in two columns, proofs in paragraphs, and proofs in flowcharts are three of the most popular geometric proofs. They each provide a unique method of structuring arguments and assertions in order to make each proof easily understandable to the reader. What are the 4 types of proofs in geometry? 1. Geometrical Demonstrations Geometric Proofs 2. The Structure of a Proof 3. Direct Proof 4. Problems 5. Auxiliary Lines 6. Problems 7. Indirect Proof 8. Problems 9. Geometric Proofs 10. The Structure of a Proof What are the rules for proofs in geometry? All right angles are consistent with one another. All straight angles are consistent with one another. Supplements with the same angle, called congruent angles, are also consistent with one another. Angles that are complements of the same angle, or angles that are congruent, are congruent. You might be interested: What Are The Components Of Galactose? Reflexive Property A quantity is congruent (equal) to itself. a = a Construction Two points determine a straight line. What is the last step of a proof? All right angles are in agreement with one another. There are no inconsistencies between any straight angles. Supplements with the same angle, called congruent angles, are also congruent with one other. Equivalent angles, also known as congruent angles, are congruent when they are complements of the same angle. How many types of proof are there? There are many various ways to go about proving anything, and we’ll explore three of them: direct proof, proof by contradiction, and proof by induction. Direct proof is the most straightforward approach of showing something. What each of these proofs are, when and how they are employed will be discussed in detail. We’ll need to go over some terminology first before we get started. What are the five elements that the deductive structure of a proof contain? We therefore consider a deductive proof to consist of the following components, in addition to being an appropriate argument supported by valid reasoning: singular propositions (premises, conclusions, and intermediate propositions between them), universal propositions (theories, definitions, and so on), and the appropriate argument. What are the steps in a formal proof? Unlike informal proofs, formal proofs have a distinct style and format that is comprised of five fundamental aspects. 1. Statement. This statement expresses the theorem that must be demonstrated 2. Drawing. In this case, the hypothesis is represented by the word ″given.″ Using your illustration, this explains how the theorem’s hypothesis might be interpreted. 3. Prove. 4. \sProof You might be interested: Who Are The Twins In Cheaper By The Dozen? Why are proofs important in geometry? Mathematics students benefit from geometrical proofs because they provide a clear introduction to logical reasoning, which is a fundamental concept in all mathematics. They demonstrate the precise link that exists between reasoning and equations. More importantly, because geometry works with forms and numbers, it allows students to visualise what has to be demonstrated. How many parts are there in the format of a two-column proof? Four elements make up the two-column proof: the provided, the proposition (what you will prove), the statement, the justification, the diagram, and the conclusion. The given is the first part of the proof. What are the most common formats that are used to do a proof? One of the most popular types of proof is the direct proof, in which the ‘prove’ is demonstrated to be true directly as a result of other geometrical assertions and circumstances that are proven to be true. Direct proofs employ what is known as deductive reasoning, which is the process of reasoning from proved facts to reach a conclusion by following logically correct processes. What are the 5 theorems? 1. As a result, the five congruent triangle theorems are as follows: SSS, SAS, AAS, HL, and ASA. SSS stands for side, side, and side 2. SAS stands for side, angle, and side 3. ASA stands for angle, side, and angle 4. AAS stands for angle, angle, and side 5. HL stands for hypotenuse and leg. What are the 3 types of Theorem? 1. Complements Theorems with Angle Theorems Congruent Supplements If two angles are supplementary to the same angle or of congruent angles, then the two angles are congruent 2. otherwise, the two angles are incongruent. 3. The Right Angles Theorem is a mathematical formula that states that right angles are equal to 90 degrees. Right angles are formed when two angles are both complementary and congruent 4. otherwise, they are right angles. 5. The Theorem of Interior Angles on the Same Side 6. The Theorem of Vertical Angles You might be interested: What Are Magic Grow Capsules Made Of? What does the last line of a proof represents? The last line of a proof indicates the information that has been provided. the point of contention What is a proof in geometry? Using established facts such axioms, postulates, lemmas, and other logical assertions in conjunction with a succession of logical statements, one may derive the conclusion via geometric proof. While proving any geometric proof claims, the supporting explanations for each statement are stated. What is the best way to write a proof in geometry? What is a two-column geometric proof? An example of a two-column geometric proof is a list of facts, followed by a list of the reasons that we know those statements are true. There are a total of nine assertions stated in a column on the left, and the reasons for making each of those claims are given in a column on the right. How do you draw a proof in math? It is possible that the figure has already been created for you, or that you will have to design it yourself.List the assertions that have been provided, followed by a list of the conclusion that has to be proven.You now have a clear understanding of where the proof begins and ends.Make a note next to the figure indicating what you think you can conclude about it based on the information provided.
## 15) Word Problems Bonds of 5: Wholes to 5 ### Mathematics Apply the concept of part-part-whole and fluency recalling bonds, of wholes up to and including five, to solve addition and subtraction word problems. The problem types are varied. The missing number is in a range of different positions. ### Language Click this link to read Teacher Notes for more information about mathematical language related to “Solving Word Problems”. These notes outline the benefits of using a using a diagrammatic approach and caution against linking key words to specific operation.ย ### Core Activity ##### Aim To solve word problems. ##### Materials An activity for individual students sitting in pairs. Each student needs a pencil and paper. Each pair needs: • One Bonds of 5: Word Problems Wholes to 5 card containing three questions.ย ##### Instructions Cut the Bonds of 5: Word Problem activity board into 4 cards.ย • Each card contains three word problems. Within each card the questions increase in difficulty.ย • Each card also increases in difficulty. Questions 1 to 3 are easier than Questions 10 to 12.ย • Every card contains a range of addition and subtraction problem types for a variety of wholes up to and including 5.ย ย Give each pair of students one problem solving card and set of blocks. Students can work independently or collaboratively. Students use Bond Blocks to support problem solving and calculation only if needed. Many problems can be solved using a variety of calculation strategies. For example, 5 – ? = 3 can be solved by adding onto the known part (3 + ? = 5) or rearranging the subtractionย (5 – 3 = ?). At the conclusion of the lesson select one problem. Instruct students to compare strategies used to: • Solve the problem. Encourage students to identify new ways of solving problems. • Calculate. When comparing strategies used, ask students to explain why different addition and subtraction strategies can both produce correct answers. Encourage students to identify efficient calculating strategies and justify their reasons. Examples of different solutions for each question are provided in the Bonds of 5: Word Problem Answers.ย Click to open the Bonds of 5 Word Problem Answers in a new tab. ## Differentiation ### A little easier ##### Polya’s problem solving process Assist students who are experiencing difficulty by identifying the part of the problem solving process, as outlined by Polya, they need help with.ย Refer to page 4 onwards of the Teacher Notes “Solving Word Problems” for more information about using Polya’s Problem Solving Process and specific strategies, including Newman analysis, to support students who are experiencing difficulty.ย ##### Scaffold support Scaffold the type of support given to students using the Concrete-Representational-Abstract framework. • If students are experiencing difficulty at the abstract level using numbers and symbols ask them to represent the problem using Bond Blocks. • If they have difficulty at the representative level using Bond Blocks ask them to use concrete materials or drawings with one-to-one correspondence. Students can connect their concrete, representational and abstract understanding of the problem by placing this information on the Thinkboard from Activity 14. ##### Word problems with diagrams These word questions as the same as the core activity except that each question also has a mathematically correct diagram. The diagram can support students to understand the problem.ย Diagrammatic support can be helpful for students with language, processing and mathematical difficulties. Click to open the Bonds of 5 Word Problems with Diagramsย in a new tab. ### A little harder ##### Questions 10 to 12 Each card increases in difficulty. The final three questions are the most difficult. ##### Activity 40 Bonds of 6,7,8,9: Word Problems Wholes to 10 Typically students solve problems that use numbers that are slightly less than they calculate with fluently because of the complexity of the problem solving process. If students have completed up to Activity 39 Bonds of 6,7,8,9: Missing number Equations they can solve word problems using Activity 40 Bonds of 6,7,8,9: Word Problems wholes to 10. ### Progression The next chapter of activities focuses on initial double bonds that can be represented using the student’s two hands. The sequence of activities is similar to the Bonds of 5 chapter moving from identify bonds, to developing fluency, then calculating. Go to ##### Activity 16 Doubling and Halving to 10: Bonds, Building a Wall
# Lottery mathematics Lottery mathematics is used to calculate probabilities of winning or losing a lottery game. It is based primarily on combinatorics, particularly the twelvefold way and combinations without replacement. ## Choosing 6 from 49 In a typical 6/49 game, each player chooses six distinct numbers from a range of 1–49. If the six numbers on a ticket match the numbers drawn by the lottery, the ticket holder is a jackpot winner—regardless of the order of the numbers. The probability of this happening is 1 in 13,983,816. The chance of winning can be demonstrated as follows: The first number drawn has a 1 in 49 chance of matching. When the draw comes to the second number, there are now only 48 balls left in the bag, because the balls are drawn without replacement. So there is now a 1 in 48 chance of predicting this number. Thus for each of the 49 ways of choosing the first number there are 48 different ways of choosing the second. This means that the probability of correctly predicting 2 numbers drawn from 49 in the correct order is calculated as 1 in 49 × 48. On drawing the third number there are only 47 ways of choosing the number; but we could have arrived at this point in any of 49 × 48 ways, so the chances of correctly predicting 3 numbers drawn from 49, again in the correct order, is 1 in 49 × 48 × 47. This continues until the sixth number has been drawn, giving the final calculation, 49 × 48 × 47 × 46 × 45 × 44, which can also be written as ${\displaystyle {49! \over (49-6)!}}$ or 49 factorial divided by 43 factorial or FACT(49)/FACT(43) or simply PERM(49,6) . 608281864034267560872252163321295376887552831379210240000000000 / 60415263063373835637355132068513997507264512000000000 = 10068347520 This works out to 10,068,347,520, which is much bigger than the ~14 million stated above. Perm(49,6)=10068347520 and 49 nPr 6 =10068347520. However, the order of the 6 numbers is not significant for the payout. That is, if a ticket has the numbers 1, 2, 3, 4, 5, and 6, it wins as long as all the numbers 1 through 6 are drawn, no matter what order they come out in. Accordingly, given any combination of 6 numbers, there are 6 × 5 × 4 × 3 × 2 × 1 = 6! or 720 orders in which they can be drawn. Dividing 10,068,347,520 by 720 gives 13,983,816, also written as ${\displaystyle {49! \over 6!(49-6)!}}$, or COMBIN(49,6) or 49 nCr 6 or more generally as ${\displaystyle {n \choose k}={n! \over k!(n-k)!}}$, where n is the number of alternatives and k is the number of choices. Further information is available at binomial coefficient and multinomial coefficient. This function is called the combination function, COMBIN(n,k). For the rest of this article, we will use the notation ${\displaystyle {n \choose k}}$. "Combination" means the group of numbers selected, irrespective of the order in which they are drawn. A combination of numbers is usually presented in ascending order. An eventual 7th drawn number, the reserve or bonus, is presented at the end. An alternative method of calculating the odds is to note that the probability of the first ball corresponding to one of the six chosen is 6/49; the probability of the second ball corresponding to one of the remaining five chosen is 5/48; and so on. This yields a final formula of ${\displaystyle {n \choose k}={49 \choose 6}={49 \over 6}{48 \over 5}{47 \over 4}{46 \over 3}{45 \over 2}{44 \over 1}}$ A 7th ball often is drawn as reserve ball, in the past only a second chance to get 5+1 numbers correct with 6 numbers played. ## Odds of getting other possibilities in choosing 6 from 49 One must divide the number of combinations producing the given result by the total number of possible combinations (for example, ${\displaystyle {49 \choose 6}=13,983,816}$ ). The numerator equates to the number of ways to select the winning numbers multiplied by the number of ways to select the losing numbers. For a score of n (for example, if 3 choices match three of the 6 balls drawn, then n = 3), ${\displaystyle {6 \choose n}}$ describes the odds of selecting n winning numbers from the 6 winning numbers. This means that there are 6 - n losing numbers, which are chosen from the 43 losing numbers in ${\displaystyle {43 \choose 6-n}}$ ways. The total number of combinations giving that result is, as stated above, the first number multiplied by the second. The expression is therefore ${\displaystyle {6 \choose n}{43 \choose 6-n} \over {49 \choose 6}}$. This can be written in a general form for all lotteries as: ${\displaystyle {K \choose B}{N-K \choose K-B} \over {N \choose K}}$ where ${\displaystyle N}$ is the number of balls in lottery, ${\displaystyle K}$ is the number of balls in a single ticket, and ${\displaystyle B}$ is the number of matching balls for a winning ticket. The generalisation of this formula is called the hypergeometric distribution. This gives the following results: Score Calculation Exact Probability Approximate Decimal Probability Approximate 1/Probability 0 ${\displaystyle {6 \choose 0}{43 \choose 6} \over {49 \choose 6}}$ 435,461/998,844 0.436 2.2938 1 ${\displaystyle {6 \choose 1}{43 \choose 5} \over {49 \choose 6}}$ 68,757/166,474 0.413 2.4212 2 ${\displaystyle {6 \choose 2}{43 \choose 4} \over {49 \choose 6}}$ 44,075/332,948 0.132 7.5541 3 ${\displaystyle {6 \choose 3}{43 \choose 3} \over {49 \choose 6}}$ 8,815/499,422 0.0177 56.66 4 ${\displaystyle {6 \choose 4}{43 \choose 2} \over {49 \choose 6}}$ 645/665,896 0.000969 1,032.4 5 ${\displaystyle {6 \choose 5}{43 \choose 1} \over {49 \choose 6}}$ 43/2,330,636 0.0000184 54,200.8 6 ${\displaystyle {6 \choose 6}{43 \choose 0} \over {49 \choose 6}}$ 1/13,983,816 0.0000000715 13,983,816 When a 7th number is drawn as bonus number then we have 49!/6!/1!/42!.=combin(49,6)*combin(49-6,1)=601304088 different possible drawing results. Score Calculation Exact Probability Approximate Decimal Probability Approximate 1/Probability 5 + 0 ${\displaystyle {6 \choose 5}{1 \choose 0}{42 \choose 1} \over {49 \choose 6}}$ 252/13983816 0.0000180208 55,491.33 5 + 1 ${\displaystyle {6 \choose 5}{1 \choose 1}{42 \choose 0} \over {49 \choose 6}}$ 6/13983816 0.0000004291 2,330,636 You would expect to score 3 of 6 or better once in around 36.19 drawings. Notice that It takes a 3 if 6 wheel of 163 combinations to be sure of at least one 3/6 score. 1/p changes when several distinct combinations are played together. It mostly is about winning something, not just the jackpot. ## Ensuring to win the jackpot There is only one known way to ensure winning the jackpot. That is to buy at least one lottery ticket for every possible number combination. For example, one has to buy 13,983,816 different tickets to ensure to win the jackpot in a 6/49 game. Lottery organizations have laws, rules and safeguards in place to prevent gamblers from executing such an operation. Further, just winning the jackpot by buying every possible combination does not guarantee to break even or make a profit. If ${\displaystyle p}$ is the probability to win; ${\displaystyle c_{t}}$the cost of a ticket; ${\displaystyle c_{l}}$ the cost for obtaining a ticket (e.g. including the logistics); ${\displaystyle c_{f}}$ one time costs for the operation (such as setting up and conducting the operation); then the jackpot ${\displaystyle m_{j}}$ should contain at least ${\displaystyle m_{j}\geq c_{f}+{\frac {c_{t}+c_{l}}{p}}}$ to have a chance to at least break even. The above theoretical "chance to break-even" point is slightly offset by the sum ${\displaystyle \sum _{i}{}m_{i}}$ of the minor wins also included in all the lottery tickets: ${\displaystyle m_{j}\geq c_{f}+{\frac {c_{t}+c_{l}}{p}}-\sum _{i}{}m_{i}}$ Still, even if the above relation is satisfied, it does not guarantee to break even. The payout depends on the number of winning tickets for all the prizes ${\displaystyle n_{x}}$, resulting in the relation ${\displaystyle {\frac {m_{j}}{n_{j}}}\geq c_{f}+{\frac {c_{t}+c_{l}}{p}}-\sum _{i}{}{\frac {m_{i}}{n_{i}}}}$ In probably the only known successful operations[1] the threshold to execute an operation was set at three times the cost of the tickets alone for unknown reasons ${\displaystyle m_{j}\geq 3\times {\frac {c_{t}}{p}}}$ I.e. ${\displaystyle {\frac {n_{j}p}{c_{t}}}\left(c_{f}+{\frac {c_{t}+c_{l}}{p}}-\sum _{i}{}{\frac {m_{i}}{n_{i}}}\right)\ll 3}$ This does, however, not eliminate all risks to make no profit. The success of the operations still depended on a bit of luck. In addition, in one operation the logistics failed and not all combinations could be obtained. This added the risk of not even winning the jackpot at all. ## Powerballs and bonus balls Many lotteries have a Powerball (or "bonus ball"). If the powerball is drawn from a pool of numbers different from the main lottery, the odds are multiplied by the number of powerballs. For example, in the 6 from 49 lottery, given 10 powerball numbers, then the odds of getting a score of 3 and the powerball would be 1 in 56.66 × 10, or 566.6 (the probability would be divided by 10, to give an exact value of ${\textstyle {\frac {8815}{4994220}}}$). Another example of such a game is Mega Millions, albeit with different jackpot odds. Where more than 1 powerball is drawn from a separate pool of balls to the main lottery (for example, in the EuroMillions game), the odds of the different possible powerball matching scores are calculated using the method shown in the "other scores" section above (in other words, the powerballs are like a mini-lottery in their own right), and then multiplied by the odds of achieving the required main-lottery score. If the powerball is drawn from the same pool of numbers as the main lottery, then, for a given target score, the number of winning combinations includes the powerball. For games based on the Canadian lottery (such as the lottery of the United Kingdom), after the 6 main balls are drawn, an extra ball is drawn from the same pool of balls, and this becomes the powerball (or "bonus ball"). An extra prize is given for matching 5 balls and the bonus ball. As described in the "other scores" section above, the number of ways one can obtain a score of 5 from a single ticket is ${\textstyle {6 \choose 5}{43 \choose 1}=258}$. Since the number of remaining balls is 43, and the ticket has 1 unmatched number remaining, 1/43 of these 258 combinations will match the next ball drawn (the powerball), leaving 258/43 = 6 ways of achieving it. Therefore, the odds of getting a score of 5 and the powerball are ${\textstyle {6 \over {49 \choose 6}}={1 \over 2,330,636}}$. Of the 258 combinations that match 5 of the main 6 balls, in 42/43 of them the remaining number will not match the powerball, giving odds of ${\textstyle {{258\cdot {\frac {42}{43}}} \over {49 \choose 6}}={\frac {3}{166,474}}\approx 1.802\times 10^{-5}}$ for obtaining a score of 5 without matching the powerball. Using the same principle, the odds of getting a score of 2 and the powerball are ${\textstyle {6 \choose 2}{43 \choose 4}=1,\!851,\!150}$ for the score of 2 multiplied by the probability of one of the remaining four numbers matching the bonus ball, which is 4/43. Since ${\textstyle 1,851,150\cdot {\frac {4}{43}}=172,\!200}$, the probability of obtaining the score of 2 and the bonus ball is ${\textstyle {\frac {172,200}{49 \choose 6}}={\frac {1025}{83237}}=1.231\%}$, approximate decimal odds of 1 in 81.2. The general formula for ${\displaystyle B}$ matching balls in a ${\displaystyle N}$ choose ${\displaystyle K}$ lottery with one bonus ball from the ${\displaystyle N}$ pool of balls is: ${\displaystyle {\frac {{\frac {K-B}{N-K}}{K \choose B}{N-K \choose K-B}}{N \choose K}}}$ The general formula for ${\displaystyle B}$ matching balls in a ${\displaystyle N}$ choose ${\displaystyle K}$ lottery with zero bonus ball from the ${\displaystyle N}$ pool of balls is: ${\displaystyle {N-K-K+B \over N-K}{K \choose B}{N-K \choose K-B} \over {N \choose K}}$ The general formula for ${\displaystyle B}$ matching balls in a ${\displaystyle N}$ choose ${\displaystyle K}$ lottery with one bonus ball from a separate pool of ${\displaystyle P}$ balls is: ${\displaystyle {1 \over P}{K \choose B}{N-K \choose K-B} \over {N \choose K}}$ The general formula for ${\displaystyle B}$ matching balls in a ${\displaystyle N}$ choose ${\displaystyle K}$ lottery with no bonus ball from a separate pool of ${\displaystyle P}$ balls is: ${\displaystyle {P-1 \over P}{K \choose B}{N-K \choose K-B} \over {N \choose K}}$ ## Minimum number of tickets for a match It is a hard (and often open) problem to calculate the minimum number of tickets one needs to purchase to guarantee that at least one of these tickets matches at least 2 numbers. In the 5-from-90 lotto, the minimum number of tickets that can guarantee a ticket with at least 2 matches is 100.[2] ## Information theoretic results As a discrete probability space, the probability of any particular lottery outcome is atomic, meaning it is greater than zero. Therefore, the probability of any event is the sum of probabilities of the outcomes of the event. This makes it easy to calculate quantities of interest from information theory. For example, the information content of any event is easy to calculate, by the formula ${\displaystyle \operatorname {I} (E):=-\log {\left[\Pr {\left(E\right)}\right]}=-\log {\left(P\right)}.}$ In particular, the information content of outcome ${\displaystyle x}$ of discrete random variable ${\displaystyle X}$ is ${\displaystyle \operatorname {I} _{X}(x):=-\log {\left[p_{X}{\left(x\right)}\right]}=\log {\left({\frac {1}{p_{X}{\left(x\right)}}}\right)}.}$ For example, winning in the example § Choosing 6 from 49 above is a Bernoulli-distributed random variable ${\displaystyle X}$ with a 1/13,983,816 chance of winning ("success") We write ${\textstyle X\sim \mathrm {Bernoulli} \!\left(p\right)=\mathrm {B} \!\left(1,p\right)}$ with ${\textstyle p={\tfrac {1}{13,983,816}}}$ and ${\textstyle q={\tfrac {13,983,815}{13,983,816}}}$. The information content of winning is ${\displaystyle \operatorname {I} _{X}({\text{win}})=-\log _{2}{p_{X}{({\text{win}})}}=-\log _{2}\!{\tfrac {1}{13,983,816}}\approx 23.73725}$ shannons or bits of information. (See units of information for further explanation of terminology.) The information content of losing is {\displaystyle {\begin{aligned}\operatorname {I} _{X}({\text{lose}})&=-\log _{2}{p_{X}{({\text{lose}})}}=-\log _{2}\!{\tfrac {13,983,815}{13,983,816}}\\&\approx 1.0317\times 10^{-7}{\text{ shannons}}.\end{aligned}}} The information entropy of a lottery probability distribution is also easy to calculate as the expected value of the information content. {\displaystyle {\begin{alignedat}{2}\mathrm {H} (X)&=\sum _{x}{-p_{X}{\left(x\right)}\log {p_{X}{\left(x\right)}}}\ &=\sum _{x}{p_{X}{\left(x\right)}\operatorname {I} _{X}(x)}\\&{\overset {\underset {\mathrm {def} }{}}{=}}\ \mathbb {E} {\left[\operatorname {I} _{X}(x)\right]}\end{alignedat}}} Oftentimes the random variable of interest in the lottery is a Bernoulli trial. In this case, the Bernoulli entropy function may be used. Using ${\displaystyle X}$ representing winning the 6-of-49 lottery, the Shannon entropy of 6-of-49 above is {\displaystyle {\begin{aligned}\mathrm {H} (X)&=-p\log(p)-q\log(q)=-{\tfrac {1}{13,983,816}}\log \!{\tfrac {1}{13,983,816}}-{\tfrac {13,983,815}{13,983,816}}\log \!{\tfrac {13,983,815}{13,983,816}}\\&\approx 1.80065\times 10^{-6}{\text{ shannons.}}\end{aligned}}} ## References 1. ^ The man who won the lottery 14 times [1] 2. ^ Z. Füredi, G. J. Székely, and Z. Zubor (1996). "On the lottery problem". Journal of Combinatorial Designs. 4 (1): 5–10. doi:10.1002/(sici)1520-6610(1996)4:1<5::aid-jcd2>3.3.co;2-w.{{cite journal}}: CS1 maint: multiple names: authors list (link) [2]
# Challenging inequality: $abcde=1$, show that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}+\frac{33}{2(a+b+c+d+e)}\ge{\frac{{83}}{10}}$ Let $a,b,c,d,e$ be positive real numbers which satisfy $abcde=1$. How can one prove that: $$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} +\frac{1}{e}+ \frac{33}{2(a + b + c + d+e)} \ge{\frac{{83}}{10}}\ \ ?$$ #### Solutions Collecting From Web of "Challenging inequality: $abcde=1$, show that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}+\frac{33}{2(a+b+c+d+e)}\ge{\frac{{83}}{10}}$" This is only a partial solution but I think someone more familiar with such elementary inequalities than myself might be able to finish it. You can replace $a,b,c,d$, and $e$ with their reciprocals and the inequality in question becomes $$a + b + c + d + e + {33 \over 2}{1 \over ({1 \over a} + {1 \over b} + {1 \over c} + {1 \over d} + {1 \over e})} \geq {83 \over 10}$$ Since still $abcde = 1$, we can rewrite this as $$a + b + c + d + e + {33 \over 2}{1 \over ({1 \over a} + {1 \over b} + {1 \over c} + {1 \over d} + {1 \over e})} \geq {83 \over 10}(abcde)^{1 \over 5}$$ Some algebra converts this into $${a + b + c + d + e \over 5} – (abcde)^{1 \over 5} \geq {33 \over 50}(abcde)^{1 \over 5} – {33 \over 50}{5 \over ({1 \over a} + {1 \over b} + {1 \over c} + {1 \over d} + {1 \over e})}$$ In other words, $AM – GM \geq {33 \over 50}(GM – HM)$. This is needed only when $abcde = 1$, but by scaling this should then hold for all $a,b,c,d,$ and $e$. So you inequality experts out there… is this something that follows from well-known inequalities? Here is a way using “smoothing”. We need to show given $abcde=1$, $$\frac1a + \frac1b + \frac1c + \frac1d +\frac1e+ \frac{33}{2(a + b + c + d+e)} \ge \frac{83}{10}$$ Consider replacing any two variables, WLOG $a, b$, with $\sqrt{ab}, \sqrt{ab}$. Clearly the constraint is maintained, and the RHS is untouched. However, for the LHS, note that by AM-GM $$\frac1a +\frac1b \ge \frac2{\sqrt{ab}}; \quad a+b \le 2\sqrt{ab}$$ so the LHS decreases. By continuing the process of smoothing, every variable gets replaced by the geometric mean, and the LHS only decreases, so it suffices to prove the inequality for $a=b=c=d=e=\sqrt[5]{abcde}=1$, which gives trivially an equality. Let $$f(a,b,c,d,e)=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}+\frac{33}{2(a+b+c+d+e)}-{\frac{{83}}{10}}+\lambda(abcde-1).$$ Since $\lim\limits_{a\rightarrow0^+}f=+\infty$ and $\mathbb R_{+}^5$ is a compact, $f$ gets a minimal value inside $\mathbb R_{+}^5$ and in the minimum point $(a,b,c,d,e)$ we have that all partial derivatives of $f$ are equal to $0$. Thus, $$\frac{\partial f}{\partial a}=-\frac{1}{a^2}-\frac{33}{2(a+b+c+d+e)^2}+\lambda bcde=0$$ or $$\lambda abcde=\frac{1}{a}+\frac{33a}{2(a+b+c+d+e)^2}$$ and since $\frac{\partial f}{\partial b}=0$ gives $$\lambda abcde=\frac{1}{b}+\frac{33b}{2(a+b+c+d+e)^2},$$ we obtain: $$(a-b)(2(a+b+c+d+e)^2-33ab)=0$$ and we can get a similar equality for all two variables in the minimum point, which says that if for example $a\neq b$, $a\neq c$, $a\neq d$ and $a\neq e$ so $b=c=d=e$. Id est, it remains to prove our inequality for $b=c=d=e$ and $a=\frac{1}{e^4}$, which gives $$(e-1)^2(40e^8+80e^7+120e^6+160e^5-132e^4-89e^3-46e^2-3e+40)\geq0,$$ which is true but very strong. If $a=b$, but $a\neq c$, $a\neq d$ and $a\neq e$ so $c=d=e$. Let $c=d=e=t^2$, where $t>0$. Hence, $a=b=\frac{1}{t^3}$ and in this case we’ll get something obvious: $$(t-1)^2(60t^8+120t^7+180t^6-9t^5-198t^4-92t^3+14t^2+120t+60)\geq0.$$ Done! Replace with reciprocals so the problem becomes $a + b + c + d + e + {33 \over 2}{1 \over {1 \over a} + {1 \over b} + {1 \over c} + {1 \over d} + {1 \over e}} \geq {83 \over 10}$ Sort the numbers so that $a\le b\le c\le d\le e$ and assume that $a,b,c,d,e$ are not all equal to $1$. Since $a<1$ and $e>1$ we have $(a+b+c+d+e) – (1+b+c+d+ea) = a+e-1-ae = (e-1)(1-a) > 0$ Let $x_1=\frac{1}{a}+ \frac{1}{b}+ \frac{1}{c}+ \frac{1}{d}+ \frac{1}{e}$ and $x_2=\frac{1}{1}+ \frac{1}{b}+ \frac{1}{c}+ \frac{1}{d}+ \frac{1}{ea}$ $x_1 – x_2 = \frac{1}{a}+\frac{1}{e}-\frac{1}{ea} – 1 = \frac{(e-1)(1-a)}{ea} >0$ The geometric harmonic inequality says $(bcdea)^{1/5}\ge \frac{5}{\frac{1}{1}+ \frac{1}{b}+ \frac{1}{c}+ \frac{1}{d}+ \frac{1}{ea}}$ and thus using $abcde=1$ we conclude $\frac{1}{5}\ge \frac{1}{\frac{1}{1}+ \frac{1}{b}+ \frac{1}{c}+ \frac{1}{d}+ \frac{1}{ea}}$. Thus we have that $x_2\ge 5$. The mean value theorem applied to the function $f(x)=1/x$ gives $\frac{1}{x_1}-\frac{1}{x_2} = f'(\theta)(x_1-x_2)$ where $x_2 \le\theta\le x_1$. This tells us that since $\theta\ge x_2\ge 5$ that $f'(\theta)\ge f'(x_2) \ge f'(5) = -\frac{1}{5^2}$ Since $bcde=1/a$ we cannot have $b,c,d,e$ all less than $1/a^{1/4}$. As $e$ is the largest of them we must have $e\ge 1/a^{1/4}$ and since $0\le a\le 1$ we can conclude $ea\ge a^{3/4}\ge 1$. Now putting all the computations above together gives $\left(a+b+c+d+e+\frac{33}{2}\frac{1}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}}\right) – \left(1+b+c+d+ea+\frac{33}{2}\frac{1}{\frac{1}{1}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{ea}}\right) =$ $(e-1)(1-a) + \frac{33}{2}\left(\frac{1}{x_1}-\frac{1}{x_2}\right)= (e-1)(1-a) + \frac{33}{2}f'(\theta)(x_1-x_2) =$ $(e-1)(1-a) + \frac{33}{2}f'(\theta)\frac{(e-1)(1-a)}{ea}\ge (e-1)(1-a) – \frac{33}{50}\frac{(e-1)(1-a)}{ea}\ge$ $(e-1)(1-a) – \frac{33}{50}(e-1)(1-a) = \frac{27}{50}(e-1)(1-a)\ge 0$ Thus if we replace $a,b,c,d,e$ with $1,b,c,d,ea$ the left hand side of the inequality decreases while still maintaining $abcde=1$. We sort the numbers $1,b,c,d,ea$ and choose the smallest and largest among them and repeat the process we have just described. Eventually all five numbers will become 1 thus showing $a+b+c+d+e+\frac{33}{2}\frac{1}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}} \ge 1+b+c+d+ea+\frac{33}{2}\frac{1}{\frac{1}{1}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{ea}}\ge \ldots \ge$ $1+1+1+1+1+\frac{33}{2}\frac{1}{\frac{1}{1}+\frac{1}{1}+\frac{1}{1}+\frac{1}{1}+\frac{1}{1}}=\frac{83}{10}$
# Find the area cut out of the cylinder x^2+z^2=81 by the cylinder x^2+y^2=81. ## Question: Find the area cut out of the cylinder {eq}x^2+z^2=81 {/eq} by the cylinder {eq}x^2+y^2=81 {/eq}. ## Application of Double Integral This problem is based on the application of double integrals. We have been given two different equations of cylinders, first, we are going to find the limits of integration and the function to be integrated followed by defining the limits to calculate the desired area. The equations of cylinders we have are, {eq}\displaystyle x^2+y^2=81 {/eq} and, {eq}\displaystyle x^2+z^2=81 {/eq} So, {eq}\displaystyle z=\sqrt{81-x^2} {/eq} {eq}\displaystyle f(x,y)=\sqrt{81-x^2} {/eq} {eq}\displaystyle f_x=\frac{-x}{\sqrt{81-x^2}} {/eq} and, {eq}\displaystyle f_y=0 {/eq} We are going to assume that {eq}z>0 {/eq} Thus, the required area is, {eq}\displaystyle A=2\int \int \sqrt{(f_x)^2+(f_y)^2+1}dA {/eq} {eq}\displaystyle A=2\int \int \sqrt{\left ( \frac{-x}{\sqrt{81-x^2}} \right )^2+(0)^2+1}dA {/eq} {eq}\displaystyle A=2\int \int \sqrt{\left ( \frac{x^2+81-x^2}{{81-x^2}} \right )}dA {/eq} On simplification, we get, {eq}\displaystyle A=2\int \int {\left ( \frac{9}{\sqrt{81-x^2}} \right )}dA {/eq} {eq}\displaystyle A=18\int \int {\left ( \frac{1}{\sqrt{81-x^2}} \right )}dA {/eq} Now, {eq}\displaystyle x^2+y^2=81 {/eq} {eq}\displaystyle y=\pm\sqrt{81-x^2} {/eq} Putting, {eq}y=0 {/eq}, we get, {eq}\displaystyle x=\pm 9 {/eq} Thus, the area integral now becomes: {eq}\displaystyle A=18\int_{-9}^{9} \int_{-\sqrt{81-x^2}}^{\sqrt{81-x^2}} {\left ( \frac{1}{\sqrt{81-x^2}} \right )}dydx {/eq} {eq}\displaystyle A=18\int_{-9}^{9} {\left ( \frac{1}{\sqrt{81-x^2}} \right )}\left ( y \right )_{-\sqrt{81-x^2}}^{\sqrt{81-x^2}}dx {/eq} {eq}\displaystyle A=18\int_{-9}^{9} {\left ( \frac{1}{\sqrt{81-x^2}} \right )}\left ( 2\sqrt{81-x^2} \right )dx {/eq} {eq}\displaystyle A=36\int_{-9}^{9} dx {/eq} {eq}\displaystyle A=36\left ( x \right )_{-9}^{9} {/eq} {eq}\displaystyle A=36(9+9) {/eq} Thus, the area cut out is: {eq}\displaystyle \boxed{\displaystyle A=648} {/eq}
Here are some basic definitions and properties of lines and angles in geometry. These concepts are tested in many competitive entrance exams like GMAT, GRE, CAT. Line segment: A line segment has two end points with a definite length. Ray: A ray has one end point and infinitely extends in one direction. Straight line: A straight line has neither starting nor end point and is of infinite length. Acute angle: The angle that is between 0° and 90° is an acute angle, ∠A in the figure below. Obtuse angle: The angle that is between 90° and 180° is an obtuse angle, ∠B as shown below. Right angle: The angle that is 90° is a Right angle, ∠C as shown below. Straight angle: The angle that is 180° is a straight angle, ∠AOB in the figure below. Supplementary angles: In the figure above, ∠AOC + ∠COB = ∠AOB = 180° If the sum of two angles is 180° then the angles are called supplementary angles. Two right angles always supplement each other. The pair of adjacent angles whose sum is a straight angle is called a linear pair. Complementary angles: ∠COA + ∠AOB = 90° If the sum of two angles is 90° then the two angles are called complementary angles. The angles that have a common arm and a common vertex are called adjacent angles. In the figure above, ∠BOA and ∠AOC are adjacent angles. Their common arm is OA and common vertex is ‘O’. Vertically opposite angles: When two lines intersect, the angles formed opposite to each other at the point of intersection (vertex) are called vertically opposite angles. In the figure above, x and y are two intersecting lines. ∠A and ∠C make one pair of vertically opposite angles and ∠B and ∠D make another pair of vertically opposite angles. Perpendicular lines: When there is a right angle between two lines, the lines are said to be perpendicular to each other. Here, the lines OA and OB are said to be perpendicular to each other. Parallel lines: Here, A and B are two parallel lines, intersected by a line p. The line p is called a transversal, that which intersects two or more lines (not necessarily parallel lines) at distinct points. As seen in the figure above, when a transversal intersects two lines, 8 angles are formed. Let us consider the details in a tabular form for easy reference. Types of Angles Angles Interior Angles ∠3, ∠4, ∠5, ∠6 Exterior Angles ∠1, ∠2, ∠7, ∠8 Vertically opposite Angles (∠1, ∠3), (∠2, ∠4), (∠5, ∠7), (∠6, ∠8) Corresponding Angles (∠1, ∠5), (∠2, ∠6), (∠3, ∠7), (∠4, ∠8) Interior Alternate Angles (∠3, ∠5), (∠4, ∠6) Exterior Alternate Angles (∠1, ∠7), (∠2, ∠8) Interior Angles on the same side of transversal (∠3, ∠6), (∠4, ∠5) When a transversal intersects two parallel lines, 1. The corresponding angles are equal. 2. The vertically opposite angles are equal. 3. The alternate interior angles are equal. 4. The alternate exterior angles are equal. 5. The pair of interior angles on the same side of the transversal is supplementary. We can say that the lines are parallel if we can verify at least one of the aforementioned conditions. Let us take a look at some examples. ## Solved examples Example 1. If the lines m and n are parallel to each other, then determine the angles ∠5 and ∠7. Solution: Determining one pair can make it possible to find all the other angles. The following is one of the many ways to solve this question. ∠2 = 125° ∠2 = ∠4 since they are vertically opposite angles. Therefore, ∠4 = 125° ∠4 is one of the interior angles on the same side of the transversal. Therefore, ∠4 + ∠5 = 180° 125 + ∠5 = 180 → ∠5 = 180 – 125 = 55° ∠5 = ∠7 since vertically opposite angles. Therefore, ∠5 = ∠7 = 55° Note: Sometimes, the parallel property of the lines may not be mentioned in the problem statement and the lines may seem to be parallel to each other; but they may be not. It is important to determine whether two lines are parallel by verifying the angles and not by looks. Example 2. If ∠A = 120° and ∠H = 60°. Determine if the lines are parallel. Solution: Given ∠A = 120° and ∠H = 60°. Since adjacent angles are supplementary, ∠A + ∠B = 180° 120 + ∠B = 180 → ∠B = 60°. It is given that ∠H = 60°. We can see that ∠B and ∠H are exterior alternate angles. When exterior alternate angles are equal, the lines are parallel. Hence the lines p and q are parallel. We can verify this using other angles. If ∠H = 60°, ∠E = 120° since those two are on a straight line, they are supplementary. Now, ∠A = ∠E = 120°. ∠A and ∠E are corresponding angles. When corresponding angles are equal, the lines are parallel. Likewise, we can prove using other angles too. Example 3. If p and q are two lines parallel to each other and ∠E = 50°, find all the angles in the figure below. Solution: It is given ∠E = 50°. The two lines are parallel → The corresponding angles are equal. Since ∠E and ∠A are corresponding angles, ∠A = 50° . → The vertically opposite angles are equal. Since ∠A and ∠C are vertically opposite to each other, ∠C = 50°. Since ∠E and ∠G are vertically opposite to each other, ∠G = 50°. → The interior angles on the same side of the transversal are supplementary. ∠E + ∠D = 180° → 50 + ∠D = 180° → ∠D = 130° → ∠D and ∠B are vertically opposite angles. So ∠B = 130°. → ∠B and ∠F are corresponding angles. So ∠F = 130°. → ∠F and ∠H are vertically opposite angles. So ∠H = 130°. ∠D = ∠O + 90° → 130 = ∠O + 90 → ∠O = 40° Continue learning: Properties and formulas of Circles Types of Triangles and Properties Properties of Quadrilaterals (parallelograms, trapezium, rhombus) MBA Crystal Ball provides professional Admissions Consulting services. Hire us to improve your chances of getting into the top international universities. Email: info [at] mbacrystalball [dot] com // ##### MBA Crystal Ball Our counsellors can help you with career counselling and admissions consulting. Check out our free resources for: GMAT Preparation & GMAT Syllabus \| MBA Subjects \| MBA Scholarships \| And much more 1. Aksh Garg says: your website help us to understand the concept better than our tution teachers thanking you heartly 2. Suraj says: It helped me in exam preparations as it is given with examples very very thank you
Created by Rita Rain Reviewed by Dominik Czernia, PhD and Steven Wooding Last updated: Jun 05, 2023 This binary addition calculator sums up the numbers represented with zeros and ones (you can generate them with the binary converter). Read this text to learn more about the language of computers and how to add binary numbers without any binary math calculator. Once you get your head around adding binary numbers, you may want to learn how to subtract them with our binary subtraction calculator. 🔎 Apart from summing, you can subtract, multiply, and divide these types of numbers using our binary calculator. ## What is the binary system? To understand the binary system, let's first look at the more familiar decimal system. In the decimal system, we use ten digits (0-9), and, depending on their position, we multiply them by corresponding powers of ten. For example, let's break down the number 1934: 1934 = 1000 + 900 + 30 + 4 1934 = 1×1000 + 9×100 + 3×10 + 4×1 1934 = 1×10³ + 9×10² + 3×10¹ + 4×10⁰ If we made a formula for a 4-digit number, it could look like this: a₃a₂a₁a₀ = a₃ × 10³ + a₂ × 10² + a₁ × 10¹ + a₀ × 10⁰ The symbol "a" here represents a digit from 0 to 9. The index next to "a" denotes the digit's place (we count from 0, from right to left). Now, the binary system works similarly, but we only use two digits and multiply them by powers of two. So the formula would be following: a₃a₂a₁a₀ = a₃ × 2³ + a₂ × 2² + a₁ × 2¹ + a₀ × 2⁰ In this system, we multiply two digits (0 and 1) by 2 to the power corresponding to its place in the number. For example, we would write "9" as "1001" because: 1001 = 1 × 2³ + 0 × 2² + 0 × 2¹ + 1 × 2⁰ = 8 + 0 + 0 + 1 = 9 Analogically: • "2" in binary is "10" because 1 × 2¹ + 0 × 2⁰ = 2 • "3" in binary is "11" because 1 × 2¹ + 1 × 2⁰ = 2 + 1 = 3 • "4" in binary is "100" because 1 × 2² + 0 × 2¹ + 0 × 2⁰ = 4 ## How to use the binary addition calculator? To make the binary math calculator perform the addition of binary numbers, follow the steps below: 1. Input the first number in the first field of the binary addition calculator. Remember to use only zeros and ones. You don't need to enter leading zeros, e.g., for "00001111," you can input just "1111". 2. Enter the second binary number in the second row. 3. The binary addition calculator will display the result in the third field. 4. In advanced mode (at the bottom), you can make the binary addition calculator print the long addition of the binary numbers. ## What are the rules of binary addition? There are four basic binary addition rules: • 0 + 0 = 0 • 0 + 1 = 1 • 1 + 0 = 1 • 1 + 1 = 10 (write "0" in the column and carry 1 to the next bit) The above equations work like in the decimal system, only here you need to carry 1 when the sum exceeds 1 (in the decimal system, we do it when it exceeds 9). ## Binary sum: Adding binary numbers How to add binary numbers? You could always convert binary numbers to decimals, add them as usual, and restore the result to the binary form. The other way is to use the above binary addition rules and perform a long addition. Let's have a look at this binary sum: $\text{carry:} \ \ \ \ 1\ \ \ \ 1 \\\ \_\_\_\_\_\_\_\_\_\_\_ \\\ \ \ \ \ \ \ \ \ \ \ \ \ \ 1001 \\\ \ \ \ \ \ \ \ + \ \ \ 1101 \\\ \_\_\_\_\_\_\_\_\_\_\_ \\\ \ \ \ \ \ \ \ = 10110$ 🔎 Our long addition calculator can help understand the operation above. In the first column from the right, 1 + 1 gives us 2, so we need to carry 1 to the second column (1 placed in the second column equals two because 1 × 2¹ = 2) and write down 0 in the first column. In the second column, 0 + 0 = 0, so we just write the 1 we carried over. In the third column from the right, 0 + 1 = 1. In the fourth column, we have 1 + 1, so again, we write 0 and carry 1. As there's nothing left to add, we write down "1" at the beginning of the result (to the left). Binary addition applications are present in many error correction and detection techniques, like the parity bit and the Hamming distance! ## FAQ Binary addition is the operation of summing numbers in binary form. It works like a "normal" (decimal) addition, but the number can have only zeros and ones as digits, so if the sum exceeds 1, you must carry 1 to the next bit. For example, 101 + 101 = 1010. ### How to solve binary addition? 1. To solve binary addition, perform long addition — place numbers vertically and add digits in columns going from right to left. 2. If the sum in a column equals 2, carry 1 to the next column (to the left). If the sum is 1 or 0, write it down and go to the next column. 3. Proceed this way until you sum all columns (including the carried numbers). ### How to detect an overflow in binary addition? Check if the sum of binary numbers makes sense. If the sum of two negative numbers is positive or the sum of two positive numbers is negative, something is wrong. This bug means that the sum overflowed — that is, the binary representation of the result can't fit in the allocated number of bits. Rita Rain + = People also viewed… ### Divisibility test Become a master of testing for divisibility with the help of our divisibility test calculator! ### Ideal egg boiling Quantum physicist's take on boiling the perfect egg. Includes times for quarter and half-boiled eggs. ### Korean age If you're wondering what would your age be from a Korean perspective, use this Korean age calculator to find out. ### Rectangle diagonal angle The rectangle diagonal angle calculator helps you find the angles created by the diagonals of a given rectangle.
Education.com Try Brainzy Try Plus The Function for the Second Fundamental Theorem of Calculus for AP Calculus (not rated) By McGraw-Hill Professional Updated on Oct 24, 2011 Practice problems for these concepts can be found at: Areas and Volumes Practice Problems for AP Calculus The Function The Second Fundamental Theorem of Calculus defines and states that if f is continuous on [a, b], then F '(x) = f (x) for every point x in [a, b]. If f ≥ 0, then F ≥ 0. F (x) can be interpreted geometrically as the area under the curve of f from t = a to t = x. (See Figure 12.1-1.) If f < 0, F < 0, F(x) can be treated as the negative value of the area between the curve of f and the t-axis from t = a to t = x. (See Figure 12.1-2.) Example 1 If F (x) = 2 cos t dt for 0 ≤ x ≤ 2π, find the value(s) of x where f has a local minimum. Method 1: Method 2: You can solve this problem geometrically by using area. See Figure 12.1-3. The area "under the curve" is above the t-axis on [0, π/2] and below the x-axis on [π/2, 3π/2]. Thus the local minimum occurs at 3π/2. Example 2 Let p(x) = f(t)dt and the graph of f is shown in Figure 12.1-4. 1. Evaluate: p(0), p(1), p(4). 2. Evaluate: p(5), p(7), p(8). 3. At what value of t does p have a maximum value? 4. On what interval(s) is p decreasing? 5. Draw a sketch of the graph of p. Solution: 1. Since f ≥ 0 on the interval [0, 4], p attains a maximum at t = 4. 2. Since f(t) is below the x-axis from t = 4 to t = 8, if x > 4, 3. Thus, p is decreasing on the interval (4, 8). 4. p(x) = f(t)dt. See Figure 12.1-5 on page 262 for a sketch. Example 3 The position function of a moving particle on a coordinate axis is: s = f(x)dx, where t is in seconds and s is in feet. The function f is a differentiable function and its graph is shown below in Figure 12.1-6. 1. What is the particle's velocity at t = 4? 2. What is the particle's position at t = 3? 3. When is the acceleration zero? 4. When is the particle moving to the right? 5. At t = 8, is the particle on the right side or left side of the origin? Solution: 1. Since s = f(x)dx, then v(t)=s '(t) = f(t). 2. Thus, v(4) = –8 ft/sec. 3. . 4. a(t) = v '(t). Since v '(t) = f '(t), v '(t) = 0 at t = 4. Thus, a(4) = 0 ft/sec2. 5. The particle is moving to the right when v(t > 0. Thus, the particle is moving to the right on intervals (0, 2) and (7, 8). 6. The area of f below the x-axis from x = 2 to x = 7 is larger than the area of f above the x-axis from x = 0 to x = 2 and x = 7 to x = 8. Thus, f(x)dx < 0 and the particle is on the left side of the origin. Practice problems for these concepts can be found at: Areas and Volumes Practice Problems for AP Calculus 150 Characters allowed Related Questions Q: See More Questions Top Worksheet Slideshows
# Unlocking the Mysteries of Mathematics: A Guide to Understanding All the Symbols Short answer all the symbols of maths: Mathematics uses a variety of symbols to represent numbers, operations, relationships between variables, and more. Common symbols include + (addition), – (subtraction), ÷ (division), × or * (multiplication), = (equals), π (pi), ∞(infinity) and many more. ## How to Master All the Symbols of Maths: A Step-by-Step Tutorial If you’re struggling with maths, it might be because you haven’t yet mastered all the symbols that are used in mathematical expressions and equations. It can be difficult to understand a problem or concept if you don’t know what each symbol represents, so mastering these symbols is crucial for anyone hoping to become proficient in math. Here’s a step-by-step tutorial on how to master all the symbols of mathematics: Step 1: Learn the basics Before diving into more complex symbols, make sure you have a firm grasp of basic arithmetic operations such as addition (+), subtraction (-), multiplication (× or ·) and division (÷ or /). These operators are found in almost every mathematical equation, so understanding them is essential. Step 2: Familiarize yourself with algebraic notations Next up are algebraic notations like parentheses (), brackets [], absolute value \| \| and fraction bars /. These allow us to group or separate parts of an equation based on their relationships with one another. Parentheses indicate which portion of an expression should be calculated first, while fractions bars help us represent ratios between two values. Step 3: Master trigonometric functions Trigonometry deals largely with angles and triangles – two-dimensional shapes commonly encountered in physics, engineering and calculus. The sine (sin), cosine (cos), tangent (tan) functions represent different ratios of sides within right-angled triangles; inverse trigonometric functions such as arcsine (√(1-x²)), arccosine (√(x² – 1))and arctangent(tan⁻¹(x)) reverse this calculation process by providing the angle given certain known side lengths. Step 4: Get comfortable using limits Limits refer to what happens when we get infinitely close to a particular number without ever reaching it. This is useful when talking about derivatives or finding minimum/maximum points on graphs. Notation includes infinity (∞) or smaller than symbols, such as lim(x→a), which describes what happens to an equation as x gets closer and closer to a particular value. Step 5: Learn your calculus symbols Calculus is the branch of math concerned with rates of change or accumulation; it’s often used in physics and engineering. Symbols include dy/dx (derivative), ∫(integral sign) – describing integration – and sigma summation symbol or ∑, expressing large quantities like series. Step 6: Understand matrices Matrices are arrays of numbers that can simplify complex calculations by representing multiple equations together so their behavior can be analyzed at once. Notations included square brackets [] , matrix elements labelled with subscript numerals. With consistent practice and understanding how these mathematical symbols interact with one another we can effectively master our maths game! Mathematics is a fascinating subject that has evolved over the centuries to become an essential tool for solving real-world problems. It’s informed by symbols, meaning it’s not just limited to numbers and arithmetic operations. Every symbol in maths has its own unique significance, and learning about them requires patience and dedication. In this blog post, we’ve compiled all of your Frequently Asked Questions (FAQs) about the various symbols in mathematics. Q: What does “+” symbol mean? A:The plus sign or ” + ” represents addition in Mathematics. It indicates the combining of two values or more using mathematical functions where new value results from it. For instance: 3+4=7 Here, 3 is added with 4 resulting in a sum of 7.The ‘+’ function helps us obtain the total quantity when items are put together. Q: What does “-” symbol mean? A:The minus sign or ‘ – ‘ also known as subtraction operator is used instead of operation signs like loss/decrease/difference etc.It provides an indication when taking one value away from another.In simple terms,it means subtracting one number from another.For example: 5-2=3 The ‘ -‘ sign tells us how much fewer units we have after removing unit from given value . Q:What does “” represent? A:Asterisk ‘’ which looks quite same as multiply () Operator .It stands for multiplication conceptually.An asterisk serves both as a binary multiplier ( x ) On numerical keyboard,multiplication shown using 56 =30 where result obtained when multiplying five by six numeral A:The forward divide division Obelus ‘/’ Symbol acts equivalent for finding rational expressions/fractions/quotient.A front slash can be thought of as separating numerator & denominator components respectively.As illustrated below, 12 / 4 = 3 ,reading here 12 divided into 4 parts which equal to 3 Q:What does this symbol “% “mean? A:Percent(%) sign denoting ratio or proportion out of hundred %. It is numerically represented as x /100. It shows the number of parts equivalent to fractions/decimals out of 100 possible unit values. For instance, 60% =0.6 ; expressed in decimal Representing under % format, it can be read as ‘Sixty percent’ A:The exponentiation operator (^) represents raising a value or variable into power.It”s commonly known also as Exponential function.A small superscript numeral beside base integer makes sense for its power.Example : 5^2=25; here, five (base )is multiplied with itself twice-to obtain their squares-as raised to second power posited by super scripted numeral. Q:[“\|” ,”:””;]: What are they ? A:A vertical bar ‘\|’ and colon ‘:’signs act alternatively used to denote several operations.An intuitive explanation can be ## Unlocking the Mysteries of Mathematics: Exploring All Its Symbols Mathematics is a subject that has always perplexed and fascinated people from all walks of life. From the ancient Greeks to modern-day mathematicians, it has been studied and explored for centuries in hopes of unlocking its mysteries. One aspect of mathematics that can be particularly puzzling is the vast array of symbols used to represent various concepts. These symbols–from numbers to variables to operators–form the building blocks of mathematical language, allowing us to create complex equations and formulas. Let’s start with numbers themselves, which are represented by digits such as 0, 1, 2, etc. Numbers form the foundation of math; they are used for counting objects or measuring quantities. Whole numbers (such as 5) are joined by fractions (like 1/3), decimals (such as .25), negatives (-10), imaginary numbers (√-1) and real numbers. Next up are variables: letters or other symbols that stand in for unknown values or changing factors. Variables allow mathematicians to work with abstract ideas without needing specific numerical values at every stage. To give an example, we could use “x” as our variable when searching for an equation describing how much money someone will owe on their loan after a given number years–the interest rate would play into this calculation too. Operators – like +,-,/, –empower variable tracking through operations such as addition (+), subtraction(-), multiplication (*), division(/). This allows expression measurement right down any level according different priorities starting from higher precedence rank i.e., exponential (∧). Another critical set of tools involve brackets (),[],{} either singularly or interwoven together depicts groupings indicating priority treatment during computational functions avoiding ambiguities coming out due to mis-grouping. Functions build off these fundamental elements— taking something in varying levels within certain boundaries transform them into distinct output relying on prior determined rules putting mathematical laws into practical application producing limitless possibilities solution making maths an empowering tool. In conclusion, exploring the vast array of symbols used in mathematics opens up a whole new world of understanding. From numbers and variables to operators and brackets, these tools allow us to express complex ideas or patterns through equations and formulas. Mathematics is so much more than just solving problems; it’s a language that allows us to push our intellectual boundaries further than we ever thought possible!
New Zealand Level 8 - NCEA Level 3 # Existence of an Inverse Lesson Recall from the last chapter in our discussion of inverse functions, that the following must be true for an inverse function to exist: The existence of an inverse function If, and only if, a function is one-to one, then a reverse mapping will create what is known as an inverse function So if we have a one-to-one function, we can be confident that an inverse function exists over the natural domain. If instead we have a many-to-one function, we will need to manipulate the domain of our original function before we can find an inverse function (we'll look at this in more detail shortly). And if what we have is one-to-many or many-to-many, then it's not a function to begin with but rather a relation. In this case, we might be able to manipulate the range of our original relation to make it a one-to-one function and find an inverse function. ## One-to-one functions With any function it can be useful to visualise and sketch the graph first, in order to more easily identify whether the function is one-to-one. Each of the following are one-to-one functions, and therefore have an inverse function. ### Logarithmic function There are plenty more functions that are one-to-one, but these examples all clearly show that each value for $x$x has a single, unique value for $y$y. ## Many-to-one functions Here are some examples of many-to-one functions: ### Trigonometric functions As you can see, for the quadratic function there are two $x$x values that match with each $y$y value (except for $x=0$x=0). For the trigonometric functions, there are infinitely many $x$x values for each $y$y value on the graph. In either case, if we tried to find an inverse for one of these functions, we would get a one-to-many relation (and not a function). So an inverse function does not exist for a many-to-one function. Now, you might recall solving a quadratic equation by using a square root to "undo" the square. For instance, if we have the function $y=x^2$y=x2, we can take the square root of both sides to get $x=\pm\sqrt{y}$x=±y. Notice that this is actually two values for $x$x - one that is positive and one that is negative. So the square root operation is not a function. We can, however, use this idea to get an inverse function. In order to do so, we will need to restrict the domain of the original function, to avoid getting two solutions. ### Restricting the domain Looking at the function $y=x^2$y=x2, we can see that each positive value of $x$x has a corresponding negative value of $x$x that gives the same $y$y value. If we restrict the domain to only consider $x\ge0$x0, then we are able to use the square root operation to invert the function! In such a case, we get $x=\sqrt{y}$x=y (there is no longer a negative solution, since we are only considering $x\ge0$x0). If we now switch the variables, we get the restricted inverse function $y=\sqrt{x}$y=x. This process of restricting the domain is easier to see graphically. Let's take a look at an example. ##### Worked example By restricting the domain of $y=(x+2)^2$y=(x+2)2 to $x\ge-2$x2, draw the associated inverse function on the axes below. Think: Let's begin by sketching the whole function without restricting the domain. Do: We can see that there are two $x$x values for every $y$y value, except where $x=-2$x=2 Think: Now we'll just sketch the portion of the graph for $x\ge-2$x2. Along with this we'll draw the line $y=x$y=x on our graph as the mirror line, and then we'll take a few points on our original function, reflect them across the line $y=x$y=x and draw our inverse function. Do: Reflect: Notice that by restricting the domain to $x\ge-2$x2, the original function becomes one-to-one, and so we get a one-to-one inverse function as well. ## One-to-many and many-to-many relations Since one-to-many and many-to-many relations aren't considered functions, we won't concern ourselves with their inverses here. One of the most familiar many-to-many relations is that of the circle. To find an inverse that is a function, we'd need to restrict the domain and range to just one quarter of the circle. #### Practice Questions ##### Question 1 Which of the following describes the inverse of a function that is many-to-one? 1. It is a one-to-one function. A It does not exist. B It is also a many-to-one function. C It is a one-to-many relation. D It is a one-to-one function. A It does not exist. B It is also a many-to-one function. C It is a one-to-many relation. D ##### Question 2 Consider the function given by $f\left(x\right)=\frac{2x+3}{2}$f(x)=2x+32. 1. Sketch the graph of $f\left(x\right)$f(x) on the coordinate plane below: 2. Is the function $f\left(x\right)$f(x) one-to-one? No A Yes B No A Yes B 3. Does an inverse function exist for $f\left(x\right)$f(x)? No A Yes B No A Yes B ### Outcomes #### M8-2 Display and interpret the graphs of functions with the graphs of their inverse and/or reciprocal functions
# Greg Tang Funny Numbers Worksheet A Rational Amounts Worksheet will help your son or daughter be a little more acquainted with the ideas powering this ratio of integers. In this worksheet, pupils are able to remedy 12 diverse troubles related to rational expressions. They will likely figure out how to flourish two or more amounts, group them in sets, and find out their goods. They may also training simplifying realistic expressions. Once they have perfected these principles, this worksheet will certainly be a important tool for advancing their studies. Greg Tang Funny Numbers Worksheet. ## Realistic Figures really are a rate of integers There are 2 forms of phone numbers: irrational and rational. Reasonable figures are defined as entire figures, in contrast to irrational phone numbers will not recurring, and get an unlimited amount of digits. Irrational amounts are no-zero, low-terminating decimals, and square beginnings which are not ideal squares. These types of numbers are not used often in everyday life, but they are often used in math applications. To outline a logical variety, you must know just what a logical variety is. An integer is a total quantity, plus a reasonable amount is a rate of two integers. The ratio of two integers is the amount at the top split from the amount at the base. For example, if two integers are two and five, this would be an integer. There are also many floating point numbers, such as pi, which cannot be expressed as a fraction. ## They may be produced right into a fraction A logical amount includes a denominator and numerator that are not absolutely nothing. This means that they may be indicated as a small fraction. Together with their integer numerators and denominators, logical numbers can furthermore have a unfavorable worth. The bad worth needs to be put left of along with its definite value is its range from absolutely nothing. To streamline this instance, we will claim that .0333333 is a fraction that may be composed as being a 1/3. Along with unfavorable integers, a rational variety can even be manufactured into a small percentage. As an example, /18,572 can be a rational variety, when -1/ is just not. Any portion comprised of integers is logical, so long as the denominator does not consist of a and may be written as an integer. Similarly, a decimal that ends in a stage is yet another rational quantity. ## They can make sensation Even with their label, reasonable amounts don’t make significantly sense. In math, they may be individual organizations using a unique duration on the amount line. Consequently if we matter one thing, we could order the shape by its rate to the unique quantity. This contains correct even when there are actually unlimited logical figures among two specific amounts. If they are ordered, in other words, numbers should make sense only. So, if you’re counting the length of an ant’s tail, a square root of pi is an integer. In real life, if we want to know the length of a string of pearls, we can use a rational number. To get the length of a pearl, for example, we could add up its breadth. One particular pearl weighs in at 15 kgs, which is actually a rational amount. Moreover, a pound’s weight equals 10 kilos. As a result, we should certainly break down a lb by 15, without the need of concern yourself with the length of one particular pearl. ## They may be indicated as being a decimal You’ve most likely seen a problem that involves a repeated fraction if you’ve ever tried to convert a number to its decimal form. A decimal quantity might be published as a numerous of two integers, so four times five is equivalent to seven. The same problem involves the repeated small percentage 2/1, and each side should be separated by 99 to find the proper response. But how can you have the conversion process? Here are some illustrations. A realistic variety may also be developed in many forms, including fractions plus a decimal. One method to signify a logical variety inside a decimal is always to divide it into its fractional counterpart. You can find three ways to separate a reasonable number, and each of these methods produces its decimal counterpart. One of those approaches is usually to divide it into its fractional counterpart, and that’s what’s referred to as a terminating decimal.
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ # 3.2: Using Derivatives to Describe Families of Functions Skills to Develop In this section, we strive to understand the ideas generated by the following important questions: • Given a family of functions that depends on one or more parameters, how does the shape of the graph of a typical function in the family depend on the value of the parameters? • How can we construct first and second derivative sign charts of functions that depend on one or more parameters while allowing those parameters to remain arbitrary constants? Mathematicians are often interested in making general observations, say by describing patterns that hold in a large number of cases. For example, think about the Pythagorean Theorem: it doesn’t tell us something about a single right triangle, but rather a fact about every right triangle, thus providing key information about every member of the right triangle family. In the next part of our studies, we would like to use calculus to help us make general observations about families of functions that depend on one or more parameters. People who use applied mathematics, such as engineers and economists, often encounter the same types of functions in various settings where only small changes to certain constants occur. These constants are called parameters. We are already familiar with certain families of functions. For example, $f (t) = a \sin(b(t − c)) + d$ is a stretched and shifted version of the sine function with amplitude $$a$$, period $$2π b$$, phase shift $$c$$, and vertical shift $$d$$. We understand from experience with trigonometric functions that a affects the size of the oscillation, b the rapidity of oscillation, and c where the oscillation starts, as shown in Figure 3.13, while d affects the vertical positioning of the graph. In addition, there are several basic situations that we already understand completely. For instance, every function of the form y = mx + b is a line with slope m and y-intercept (0, b). Note that the form y = mx + b allows us to consider every possible line by using two parameters (except for vertical lines which are of the form x = a). Further, we understand that the value of m affects the line’s steepness and whether the line rises or falls from left to right, while the value of b situates the line vertically on the coordinate axes. For other less familiar families of functions, we would like to use calculus to understand and classify where key behavior occurs: where members of the family are increasing or decreasing, concave up or concave down, where relative extremes occur, and more, all Figure 3.13: The graph of f (t) = a sin(b(t − c)) + d based on parameters a, b, c, and d. in terms of the parameters involved. To get started, we revisit a common collection of functions to see how calculus confirms things we already know. Preview Activity $$\PageIndex{1}$$ Let a, h, and k be arbitrary real numbers with a , 0, and let f be the function given by the rule f (x) = a(x − h) 2 + k. 1. What familiar type of function is f? What information do you know about f just by looking at its form? (Think about the roles of a, h, and k.) 2. Next we use some calculus to develop familiar ideas from a different perspective. To start, treat a, h, and k as constants and compute f 0 (x). 3. Find all critical numbers of f . (These will depend on at least one of a, h, and k.) 4. Assume that a < 0. Construct a first derivative sign chart for f . 5. Based on the information you’ve found above, classify the critical values of f as maxima or minima. ./ Describing families of functions in terms of parameters Given a family of functions that depends on one or more parameters, our goal is to describe the key characteristics of the overall behavior of each member of the familiy in terms of those parameters. By finding the first and second derivatives and constructing first and second derivative sign charts (each of which may depend on one or more of the 176 parameters), we can often make broad conclusions about how each member of the family will appear. The fundamental steps for this analysis are essentially identical to the work we did in Section 3.1, as we demonstrate through the following example. Example $$\PageIndex{1}$$ Consider the two-parameter family of functions given by g(x) = axe−bx , where a and b are positive real numbers. Fully describe the behavior of a typical member of the family in terms of a and b, including the location of all critical numbers, where g is increasing, decreasing, concave up, and concave down, and the long term behavior of g. Solution. We begin by computing g 0 (x). By the product rule, g 0 (x) = ax d dx f e −bx g + e −bx d dx [ax], and thus by applying the chain rule and constant multiple rule, we find that g 0 (x) = axe−bx(−b) + e −bx(a). To find the critical numbers of g, we solve the equation g 0 (x) = 0. Here, it is especially helpful to factor g 0 (x). We thus observe that setting the derivative equal to zero implies 0 = ae−bx(−bx + 1). Since we are given that a , 0 and we know that e −bx , 0 for all values of x, the only way the preceding equation can hold is when −bx + 1 = 0. Solving for x, we find that x = 1 b , and this is therefore the only critical number of g. Now, recall that we have shown g 0 (x) = ae−bx(1−bx) and that the only critical number of g is x = 1 b . This enables us to construct the first derivative sign chart for g that is shown in Figure 3.14. Figure 3.14: The first derivative sign chart for g(x) = axe−bx . Note particularly that in g 0 (x) = ae−bx(1 − bx), the term ae−bx is always positive, so the sign depends on the linear term (1 − bx), which is zero when x = 1 b . Note that this line has negative slope (−b), so (1 − bx) is positive for x < 1 b and negative for x > 1 b . Hence we can not only conclude that g is always increasing for x < 1 b and decreasing for x > 1 b , but also that g has a global maximum at ( 1 b , g( 1 b )) and no local minimum. We turn next to analyzing the concavity of g. With g 0 (x) = −abxe−bx + ae−bx, we differentiate to find that g 00(x) = −abxe−bx(−b) + e −bx(−ab) + ae−bx(−b). Combining like terms and factoring, we now have g 00(x) = ab2 xe−bx − 2abe−bx = abe−bx(bx − 2). Similar to our work with the first derivative, we observe that abe−bx is always positive, Figure 3.15: The second derivative sign chart for g(x) = axe−bx . and thus the sign of g 00 depends on the sign of (bx − 2), which is zero when x = 2 b . Since (bx − 2) represents a line with positive slope (b), the value of (bx − 2) is negative for x < 2 b and positive for x > 2 b , and thus the sign chart for g 00 is given by the one shown in Figure 3.15. Thus, g is concave down for all x < 2 b and concave up for all x > 2 b . Finally, we analyze the long term behavior of g by considering two limits. First, we note that limx→∞ g(x) = limx→∞ axe−bx = limx→∞ ax e bx . Since this limit has indeterminate form ∞ ∞ , we can apply L’Hopital’s Rule and thus find that limx→∞ g(x) = 0. In the other direction, limx→−∞ g(x) = limx→−∞ axe−bx = −∞, since ax → −∞ and e −bx → ∞ as x → −∞. Hence, as we move left on its graph, g decreases without bound, while as we move to the right, g(x) → 0. 178 All of the above information now allows us to produce the graph of a typical member of this family of functions without using a graphing utility (and without choosing particular values for a and b), as shown in Figure 3.16. Figure 3.16: The graph of g(x) = axe−bx . We note that the value of b controls the horizontal location of the global maximum and the inflection point, as neither depends on a. The value of a affects the vertical stretch of the graph. For example, the global maximum occurs at the point ( 1 b , g( 1 b )) = ( 1 b , a b e −1 ), so the larger the value of a, the greater the value of the global maximum. The kind of work we’ve completed in Example 3.3 can often be replicated for other families of functions that depend on parameters. Normally we are most interested in determining all critical numbers, a first derivative sign chart, a second derivative sign chart, and some analysis of the limit of the function as x → ∞. Throughout, we strive to work with the parameters as arbitrary constants. If stuck, it is always possible to experiment with some particular values of the parameters present to reduce the algebraic complexity of our work. The following sequence of activities offers several key examples where we see that the values of different parameters substantially affect the behavior of individual functions within a given family. Activity $$\PageIndex{2}$$ Consider the family of functions defined by p(x) = x 3 − ax, where a , 0 is an arbitrary constant. 1. Find p 0 (x) and determine the critical numbers of p. How many critical numbers does p have? 2. Construct a first derivative sign chart for p. What can you say about the overall 179 behavior of p if the constant a is positive? Why? What if the constant a is negative? In each case, describe the relative extremes of p. 3. Find p 00(x) and construct a second derivative sign chart for p. What does this tell you about the concavity of p? What role does a play in determining the concavity of p? 4. Without using a graphing utility, sketch and label typical graphs of p(x) for the cases where a > 0 and a < 0. Label all inflection points and local extrema. 5. Finally, use a graphing utility to test your observations above by entering and plotting the function p(x) = x 3 − ax for at least four different values of a. Write several sentences to describe your overall conclusions about how the behavior of p depends on a. C Activity $$\PageIndex{3}$$ Consider the two-parameter family of functions of the form h(x) = a(1 − e −bx), where a and b are positive real numbers. 1. Find the first derivative and the critical numbers of h. Use these to construct a first derivative sign chart and determine for which values of x the function h is increasing and decreasing. 2. Find the second derivative and build a second derivative sign chart. For which values of x is a function in this family concave up? concave down? 3. What is the value of limx→∞ a(1 − e −bx)? limx→−∞ a(1 − e −bx)? 4. How does changing the value of b affect the shape of the curve? 5. Without using a graphing utility, sketch the graph of a typical member of this family. Write several sentences to describe the overall behavior of a typical function h and how this behavior depends on a and b. C Activity $$\PageIndex{4}$$ Let L(t) = A 1 + ce−kt , where A, c, and k are all positive real numbers. 1. Observe that we can equivalently write L(t) = A(1 + ce−kt) −1 . Find L 0 (t) and explain why L has no critical numbers. Is L always increasing or always decreasing? Why? 2. Given the fact that L 00(t) = Ack2 e −kt ce−kt − 1 (1 + ce−kt) 3 , 180 find all values of t such that L 00(t) = 0 and hence construct a second derivative sign chart. For which values of t is a function in this family concave up? concave down? 3. What is the value of limt→∞ A 1 + ce−kt ? limt→−∞ A 1 + ce−kt ? 4. Find the value of L(x) at the inflection point found in (b). 5. Without using a graphing utility, sketch the graph of a typical member of this family. Write several sentences to describe the overall behavior of a typical function L and how this behavior depends on A, c, and kcritical number. (f) Explain why it is reasonable to think that the function L(t) models the growth of a population over time in a setting where the largest possible population the surrounding environment can support is A. ### Summary In this section, we encountered the following important ideas: • Given a family of functions that depends on one or more parameters, by investigating how critical numbers and locations where the second derivative is zero depend on the values of these parameters, we can often accurately describe the shape of the function in terms of the parameters. • In particular, just as we can created first and second derivative sign charts for a single function, we often can do so for entire families of functions where critical numbers and possible inflection points depend on arbitrary constants. These sign charts then reveal where members of the family are increasing or decreasing, concave up or concave down, and help us to identify relative extremes and inflection points. critical number
# Practical Geometry Exercise 18.3 ### Question: 1 Construct a quadrilateral ABCD in which AB = 3.8 cm, BC = 3.4 cm, CD = 4.5 cm, AD = 5 cm and ∠B = 80°. ### Solution: Steps of construction: Step I: Draw AB = 3.8 cm. Step II: Construct ABC=80°. Step III: With B as the center and radius 3.4 cm, cut off BC = 3.4 cm. Step IV: With C as the center and radius 4.5 cm, draw an arc. Step V: With A as the center and radius 5.3 cm, draw an arc to intersect the arc drawn in Step IV at D. Step VI: Join AD, BC and CD to obtain the required quadrilateral. ### Question: 2 Construct a quadrilateral ABCD, given that AB = 8 cm, BC = 8 cm, CD =10 cm, AD =10 cm and ∠A=45°. ### Solution: Steps of construction: Step I: Draw AB = 8 cm. Step II: Construct BAD=45°. Step III: With A as the centre and radius 10 cm, cut off AD =10 cm. Step IV: With D as the centre and radius 10 cm, draw an arc. Step V: With B as the centre and radius 8 cm, draw an arc to intersect the arc drawn in Step IV at C. Step VI: Join BC and CD to obtain the required quadrilateral. ### Question: 3 Construct a quadrilateral ABCD in which AB = 7.7 cm, BC = 6.8 cm, CD = 5.1 cm, AD = 3.6 cm and ∠C=120°. ### Solution: Steps of construction: Step I: Draw DC = 5.1 cm. Step II: Construct DCB=120°. Step III: With C as the center and radius 6.8 cm, cut off BC = 6.8 cm. Step IV: With B as the center and radius 7.7 cm, draw an arc. Step V: With D as the center and radius 3.6 cm, draw an arc to intersect the arc drawn in Step IV at A. Step VI: Join AB and AD to obtain the required quadrilateral. ### Question: 4 Construct a quadrilateral ABCD in which AB = BC = 3 cm, AD = CD = 5 cm and ∠B=120°. ### Solution: Steps of construction: Step I: Draw AB = 3 cm. Step II: Construct ABC=120°. Step III: With B as the center and radius 3 cm, cut off BC = 3 cm. Step IV: With C as the center and radius 5 cm, draw an arc. Step V: With A as the center and radius 5 cm, draw an arc to intersect the arc drawn in Step IV at D. Step VI: Join AD and CD to obtain the required quadrilateral. ### Question: 5 Construct a quadrilateral ABCD in which AB = 2.8 cm, BC = 3.1 cm, CD = 2.6 cm and DA =3.3 cm and ∠A=60°. ### Solution: Steps of construction: Step I: Draw AB = 2.8 cm. Step II: Construct BAD=60°. Step III: With A as the center and radius 3.3 cm, cut off AD = 3.3 cm. Step IV: With D as the center and radius 2.6 cm, draw an arc. Step V: With B as the center and radius 3.1 cm, draw an arc to intersect the arc drawn in Step IV at C. Step VI: Join BC and CD to obtain the required quadrilateral. ### Question: 6 Construct a quadrilateral ABCD in which AB = BC = 6 cm, AD = DC = 4.5 cm and ∠B=120°. ### Solution: Steps of construction: Step I: Draw AB = 6 cm. Step II: Construct ABC=120° Step III: With B as the centre and radius 6 cm, cut off BC = 6 cm. Now, we can see that AC is about 10.3 cm which is greater than AD + CD = 4.5 + 4.5 = 9 cm. We know that sum of the lengths of two sides of the triangle is always greater than the third side but here, the sum of AD and CD is less than AC. So, construction of the given quadrilateral is not possible. ### Course Features • Video Lectures • Revision Notes • Previous Year Papers • Mind Map • Study Planner • NCERT Solutions • Discussion Forum • Test paper with Video Solution r
# Leonora is factoring a trinomial. the factors of the trinomial are shown on the model. which trinomial did she factor [Solved] x 4x 12 Step-by-step rationalization: We are able to see within the vertical fringe of the mannequin that one issue is (x 2). Within the horizontal fringe of the mannequin, the opposite issue is (x-6). Multiplying the 2 elements, the trinomial is (x 2)(x-6) = x(x)-6(x) 2(x) 2(-6) = x-6x 2x-12 Combining like phrases, weve x-4x-12 x 4x 12 Step-by-step rationalization: We are able to see within the vertical fringe of the mannequin that one issue is (x 2). Within the horizontal fringe of the mannequin, the opposite issue is (x-6). Multiplying the 2 elements, the trinomial is (x 2)(x-6) = x(x)-6(x) 2(x) 2(-6) = x-6x 2x-12 Combining like phrases, weve x-4x-12 x 4x 12 Step-by-step rationalization: We are able to see within the vertical fringe of the mannequin that one issue is (x 2). Within the horizontal fringe of the mannequin, the opposite issue is (x-6). Multiplying the 2 elements, the trinomial is (x 2)(x-6) = x(x)-6(x) 2(x) 2(-6) = x-6x 2x-12 Combining like phrases, weve x-4x-12 Then the reply is letter D You multiply the x to the x. Then you definitely multiply the x to those, and the x to the destructive ones. You mix the -1x to 1x to get -4x, as a result of -6 2 is -4. Lastly you multiply the destructive ones to those, and mix like phrases to get -12. Because the quadratic type is, you plug in a, b, and c with 1. -4, and -12, so it appears to be like like this: x 4x 12 Step-by-step rationalization: We are able to see within the vertical fringe of the mannequin that one issue is (x 2). Within the horizontal fringe of the mannequin, the opposite issue is (x-6). Multiplying the 2 elements, the trinomial is (x 2)(x-6) = x(x)-6(x) 2(x) 2(-6) = x-6x 2x-12 Combining like phrases, weve x-4x-12 Get Answer for [Review] HP 83A \| CF283AD \| 2 Toner Cartridges \| Black \| Works with HP LaserJet Pro M201dw, M125nw, M127fn, M225 series [Solved]
## Weekly Overview Weekly Topics The focus of this week’s instruction is to deepen students’ understanding of: • Dividing decimals, • Mental strategies for multi-digit whole number division, • Partial quotients, and • Making like unit pictorially. Materials Needed Standards Covered 5.NBT.A.1 Recognize that in a multi-digit number, a digit in one place represents 10 times as much as it represents in the place to its right and 1/10 of what it represents in the place to its left. 5.NBT.B.6 Find whole-number quotients and remainders of whole numbers with up to four-digit dividends and two-digit divisors, using strategies based on place value, the properties of operations, and/or the relationship between multiplication and division. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models. 5.NBT.B.7 Add, subtract, multiply, and divide decimals to hundredths, using concrete models or drawings and strategies based on place value, properties of operations, and/or the relationship between operations; assess the reasonableness of answers using estimation strategies. (Limit division problems so that either the dividend or the divisor is a whole number.) 5.NF.A.1 Add and subtract fractions with unlike denominators (including mixed numbers) by replacing given fractions with equivalent fractions in such a way as to produce an equivalent sum or difference of fractions with like denominators. For example, 2/3 + 5/4 = 8/12 + 15/12 = 23/12. (In general a/b + c/d = (ad + bc)/bd.) 5.NF.A.2 Solve contextual problems involving addition and subtraction of fractions referring to the same whole, including cases of unlike denominators. Use benchmark fractions and number sense of fractions to estimate mentally and assess the reasonableness of answers. For example, recognize an incorrect result 2/5 + 1/2 = 3/7, by observing that 3/7 < 1/2. Representations • Tape Diagrams: Tape diagrams are also called “bar models” and consist of a simple bar drawing that students make and adjust to fit a word or computation problem. They then use the drawing to discuss and solve the problem. • Place Value Chart: A place value chart is a diagram that helps us to find and compare the place value of the digits in numbers. • Standards Algorithm: standard algorithm or method is a specific method of computation which is conventionally taught for solving mathematical problems • Rectangular Fraction Model: a rectangle that represents the whole amount, and divide it into equal parts. Each part is a unit fraction. • >, <, = (greater than, less than, equal to) • Digit (any of the numbers 0 to 9; e.g., what is the value of the digit in the tens place?) • Hundredths (as related to place value) • Place value (the numerical value that a digit has by virtue of its position in a number) • Tenths (as related to place value) • Unit form (e.g., 3.21 = 3 ones 2 tenths 1 hundredth) • Word form (e.g., one hundred thirty-five) • Decimal (a fraction whose denominator is a power of ten and whose numerator is expressed by figures placed to the right of a decimal point) • Divisor (the number by which another number is divided) • Equation (a statement that the values of two mathematical expressions are equal) • Equivalence (a state of being equal or equivalent) • Estimate (approximation of the value of a quantity or number) • Multiple (a number that can be divided by another number without a remainder like 15, 20, or any multiple of 5) • Product (the result of multiplying numbers together) • Quotient (the answer of dividing one quantity by another) • Remainder (the number left over when one integer is divided by another) • Renaming (decomposing or composing a number or units within a number) • Area models (e.g., an array) • Number bond • Place value disks • Benchmark fraction (e.g., 12 is a benchmark fraction when comparing 13 and 35) • Like denominators (e.g., 18 and 58) • Unlike denominators (e.g., 18 and 17) ## Materials List The following materials list will be used for the entire four weeks: Materials List. ## Weekly Assessment The links in this document are for users who cannot access the links in the lesson plans.
# GRE Math: Percentiles and Quartiles Fact: An 8 year old boy who is 4’5″ (53 inches) tall is in the 86th percentile for height for his age. What on earth does that mean? Well, the percentile of an individual tells you what percent of the population has a value of a variable is below that individual’s value of the variable. For example, to say that a 4’5″ 8 year-old boy is in the 86th percentile for height for his age, we are saying: gather together all 8 year-old boys on Earth, and measure their heights; if you sort out all the 8 year-old boys who have a height less than 4’5″, they will comprise approximately 86% of the population. That boy is taller than 86% of other boys his age – that means he’s in the 86% percentile. Percentiles is a relatively unlikely topic to see on the GRE, but if it does show up, here are a few handy facts to have up your sleeve. ## Biggest and Smallest A few details to clarify. The individual with the lowest value of the variable, with the minimum value, is not bigger than anyone, so the lowest percentile, the percentile of the rock-bottom minimum, is the 0th percentile. If my score is in the 0th percentile, then I am not higher than anyone. What’s trickier is the maximum score. If my score is the highest score, I am higher than everybody else, but that’s not the 100th percentile, because in order to be higher than 100% of the population, higher than everyone, I would have to have a score higher than my own score: a paradox! In fact, for this very reason, there’s no such thing as a 100th percentile. The person with the highest score is higher than everybody else, but not higher than herself, so she’s in the 99th percentile. If we are sticking with whole numbers, the 99th percentile is the highest possible percentile. If we go to decimals, we can get higher with the 99.9th percentile (1 out of a 1000), the 99.99th percentile (1 out of 10000), etc. ## Median and Quartiles The median is the middle of a list: the median divides a list into an “upper half” and a “lower half.” This means, the median is higher than the lower half of the population, higher than 50%, so the median is the 50th percentile. Now, we have to be careful here. On a list with only three members — e.g. {2, 4, 7} — the median is the middle number, here 4, but that number is higher than only one number out of three — so 4 is the 33rd percentile of that list. In a technical sense, the median is not always the 50th percentile. In some sense, though, that’s a specious objection. When there are only 3 members on a list, nobody in their right mind talks about percentiles. When the total number is less than a few hundred, there’s seldom talk of a percentile. Percentiles, by their very nature, are a way to make sense of tens of thousands, even millions of individuals. How many 8 year-old boys are there on Earth? Who knows, but it’s certainly a very very large number. That’s where percentiles are used in practice. When the number of folks in the group is that large, then for all intents and purposes,the median is the 50th percentile. If you are familiar with the idea of quartiles, then the first quartile is the 25th percentile and the third quartile is the 75th percentile, again, when the group sizes are truly huge. ## Practice Questions 1) Sasha took a nationwide standardized test that is graded on a scale from 20 to 60. Sasha got one of the best scores recorded on that this test. Column A Column B Sasha’s score the percentile of Sasha’s score (A) The quantity in Column A is greater. (B) The quantity in Column B is greater. (C) The two quantities are equal. (D) The relationship cannot be determined from the information given. 2) Alice took nationwide standardize test that is graded on a scale from 0 to 100. Alice scored the highest score recorded on this test. Column A Column B Alice’s score the percentile of Alice’s score (A) The quantity in Column A is greater. (B) The quantity in Column B is greater. (C) The two quantities are equal. (D) The relationship cannot be determined from the information given. 3) A large distribution of score is normally distributed Column A score that’s one standard deviation above the mean Column B score that has the 80th percentile (A) The quantity in Column A is greater. (B) The quantity in Column B is greater. (C) The two quantities are equal. (D) The relationship cannot be determined from the information given. ## Practice Questions Answers and Explanations (1) B; (2) D; (3) A; 1) We know that Sasha is near the top of the scoring distribution, so that would mean a score with a percentile close to the 99th percentile. Because of the scoring scale, the score is not going to be above 60, so the percentile is clearly bigger. Answer = B. 2) Alice got the highest score, so by definition, that’s the 99th percentile. What we don’t know is: how hard was this test? What score was the highest score? If it was a particularly challenging test, it could be that the highest score anyone achieved was only, say, a 73. In that case, the percentile would be greater. If, on the other hand, it was possible to get a perfect score, and Alice did in fact do that, then her score of a 100 would be greater than the percentile. We don’t have enough information to decide. Answer = D. 3) Here, it might be helpful to brush up on Normal Distribution. On a normal distribution, it’s always true that 68% of the populations lies within one standard deviation of the mean. That means, half of that, 34%, lie between the mean and one standard deviation above the mean. The score that is one standard deviation is higher than the 34% between the mean and one standard deviation above the mean, as well as than the 50% below the mean. That means, a score that lies one standard deviation above the mean is the 50 + 34 = 84th percentile. Thus, it’s higher than a score in the 80th percentile. Answer = A. ### 41 Responses to GRE Math: Percentiles and Quartiles 1. Amisha September 4, 2019 at 7:51 am # In que 3 we need score in option A not percentile?? • Magoosh Test Prep Expert December 6, 2019 at 7:09 pm # Hi Amisha! A score that is one standard deviation above the mean is better than 84% of the scores. That means the score is 84th percentile, which is bigger than 80th percentile. This problem depends on the 34-13.5-2 rule. You can read all about it on this post: https://magoosh.com/statistics/understanding-normal-distribution/ 2. Jitesh March 9, 2016 at 11:17 am # Thank you Mike, you are the best 🙂 Make’s everything easy to understand. 3. Tanya February 3, 2016 at 7:54 pm # 4. Aman Tukrel September 16, 2015 at 5:23 pm # Hello Mike. I have a question. If there are 500 people taking a test of 100 marks, and the test is really easy. What if all 500 of them score 100/100? I know it’s an extreme, hypothetical case, but in what percentile would, say X(1 out of these 500 people) be? • Bg June 29, 2017 at 4:43 am # Hi Aman, Percentiles, by definition, is a “relative” measure of scores, height or any other parameter. If I say I have scored, say 99th percentile, that means I have scored above 99% of the students who gave that test. Similarly, if I have scored, say 0th percentile, then this means I haven’t scored better than any “single” student who gave that test i.e. I am the least scorer in that particular test. Now, back to your question, the value of X will be 0th percentile (despite getting a perfect score on the test) because I haven’t outperformed even a “single” student who gave that test. • Magoosh Test Prep Expert June 29, 2017 at 11:31 am # That’s correct, Bg. Every student would be at 0th percentile. Which is weird and wouldn’t appear on the test. But it’s also a great thought exercise to better understand what percentiles are. 🙂 Thanks, Aman, for posting this question, and thanks to Bg for providing such a thoughtful response. 5. Akshay August 23, 2015 at 5:53 am # Hi Mike, Regarding question 1, Sasha got one of best scores means she was not the highest.What if it turns out that only 1 person got highest marks and rest got equal marks i.e Highest marks – 60 Sasha marks – 59 “nationwide” score of remaining candidates was also 59 then Sasha’s percentile will be 0? and the answer would be “D” Please correct me if I am wrong • Rasitha July 15, 2016 at 8:20 am # The problem is, one would not say Sasha got one of the highest marks if her percentile was at least was not above 90% Think of this, if Sasha actually got the lowest marks out of all the results(59 in your example), would you say Sasha was one of the students who got the highest marks? 6. Ed June 25, 2015 at 5:18 am # Hi Mike If we follow Question 1 on the same grounds as Question 2, then option D seems very likely for the first question as well. In Question 1, it is possible that almost all the students in test got one of the highest scores: that would result in a low percentile score even for the toppers, since the number of students who got lower than the highest score is very less. Hence, I chose option D for Question 1. Isn’t that right? 7. Sharath May 30, 2015 at 2:25 pm # Hi Mike, Is there something like the 100th percentile? If it does exist, does it mean that the person who secured that percentile is the only one to have got the maximum score or can there be more than one to have gotten the max. score? • Mike McGarry June 2, 2015 at 5:35 pm # Sharath, I’m happy to answer you question. 🙂 The answer is NO! You are more likely to see a unicorn shoveling snow in Tahiti than to hear anyone who knows anything about mathematics talk about the idea of a 100th percentile. The 100th percentile involves a profound logical fallacy: if an individual had a 100% percentile, then that individual would have score higher than everyone’s score, but that individual is part of everyone, so this individual would have to have a score higher than his own score. A score higher than itself: that is impossible! There is no number on the number line that is higher than itself! For that very reason, the chimerical idea of a 100th percentile is wholly and utterly impossible. Does this make sense? Mike 🙂 8. Prague September 21, 2014 at 10:40 pm # Hi Mike, Great post. Got two quick questions for you though. 1. Say I have no. of data(n) = 36 and was asked to find 95th percentile. So, 0.9536 = 34.2. Which value would I pick? 35th, or 34th, or average of the two? 2. Let’s say I have no. of data(n) = 10 and was asked to find 90th percentile. So, 0.9010 = 9. Which value would I pick? 9th, or average of the 9th and 10th? • Mike September 22, 2014 at 10:13 am # Dear Prague, I’m happy to respond. 🙂 My friend, with all due respect, understand that there is something a bit nonsensical about the scenarios you propose. The reason percentiles exist is to make sense of truly gigantic pools of data. For example, hundreds of thousands of people take the GRE each year, and percentiles are a very handy way to say where an individual candidate places with respect to this extremely large pool. In practice, nobody is going to use percentiles for a group of a size under 100. If someone is second-highest in a set of 36, it’s much easier just to say that than to compute a percentile. Technically, in a group of 36, the 95th percentile simply would not exist. It is a completely meaningless idea. The person with the highest “score” (whatever we are measuring) is higher than 35 other people, that is, higher than 35/36 = 97.222% of the group. The next-highest person is higher than 34 other people, that is higher than 34/36 = 94.444% of the group. Technically, those would be the percentile ranks of those two positions; there are no people, no scores, between those two, so there are no percentiles between those two. Unlike a mean or a median or a quartile, a percentile is always attached to a real individual score. That’s the technical answer, but again, understand that in the big picture, there is something absolutely ludicrous about discussing percentiles in such a small sample. It’s like asking for the shoe-size of an emu or the heart-rate of a lawn-mower. It’s not a question that makes any sense when you really think about it. Why on earth would we ever say, “Sue was in the 97.222th percentile in her class” when we could just say, “In a class of 36, Sue got the highest score“? The point of percentiles, the point of all statistical measures, is to add clarity and insight to otherwise hard to understand data; the point is not to create gratuitous confusion for the sake of inserting an irrelevant concept into another otherwise simple scenario. Does all this make sense? Mike 🙂 • Amit November 9, 2014 at 6:56 pm # Nicely Explained…………….Mike.! • Mike November 10, 2014 at 9:45 am # Amit, You are quite welcome! 🙂 I’m very glad you found this helpful! Best of luck to you! Mike 🙂 9. mori May 28, 2014 at 12:06 pm # There are two explanation for almost the same question in ETS and Manhattan. Is there anyone who can help me? • Mike May 28, 2014 at 1:35 pm # Dear Mori, This blog is not a place to ask for questions about outside material. Here’s what I’ll recommend. You can post your question as a new thread on GMAT Club, in the Magoosh forum: http://gmatclub.com/forum/magoosh-324/ Even though that’s a GMAT forum, we will answer your GRE math questions. Does this make sense? Mike 🙂 10. akanksha October 8, 2013 at 10:21 pm # Hi Mike Can you please explain this question: A large distribution of score is normally distributed Column A: score that’s one standard deviation above the mean Column B: score that has the 80th percentile • Mike October 9, 2013 at 10:15 am # Dear Akanksha https://magoosh.com/gre/2012/normal-distribution-on-the-gre/ That will explain why the answer is (A). Mike 🙂 • Sadia June 30, 2015 at 9:29 am # Hi, for question number 3– I just also read up on Normal Distribution, but I still do not understand why the answer is A and not B. I understand that a score that is one SD above the mean is in the 34th percentile. But I do no understand where the 50 comes from. And also if a score is 34% below the mean would that mean that it is 50-34= 16? Thank you! • swapnil August 6, 2015 at 2:27 am # in the option A-the score that’s one standard deviation above 34% above the mean that gives us 50+34 that is 84th percentile that greater than 80th.and in a normal distribution of 100 values the median is the middle value that is 50th value . 11. Abhishek September 4, 2013 at 5:56 am # Hi Mike, Just a question, if 10 people appear for a exam the highest percentile can be 90 ((100)-100/10) and if 1000 people give an exam highest percentile would be 99.9 ((100)-100/1000). Please let me know if my understanding is right or wrong • Mike September 4, 2013 at 9:53 am # Dear Abhishek, Yes, that’s quite right. Mike 🙂 • BHargav September 8, 2013 at 6:36 am # hi Mike,In dat case…if the list consists as u explained in the above (2,4,7).,the 50th percentile will be the meadian 4 na,but how it could be 33rd percentile…explain me.?? • Mike September 8, 2013 at 12:06 pm # BHargav, The percentile is a measure that is used to discuss the position of an individual within a population — a population consisting of thousands or hundreds of thousand or more. In that context, the only context in which percentile is truly meaningful, the median is always the 50th percentile. It’s absurd to talk about the percentile of a list with only three members. The GRE will not ask about percentile in a context in which it is absurd. Does this make sense? Mike 🙂 12. VAMSEE September 2, 2013 at 3:10 am # Hi Mike, The definition for percentile given by you is “the percentile of an individual tells you what percent of the population has a value of a variable is below that individual’s value of the variable. ” But what about the values that are equal to the value being considered? For example consider the below values: 1,2,3,4,5,6,6,6,7,8 What is the percentile of value 6 among the values given above? Is it 50 or 70 • Mike September 2, 2013 at 11:20 am # Dear Vamsee, That’s a great question. The percentile of 6 is the 50th percentile, because each 6 is above just 50% of the population. Having said that, keep in mind that there’s something a bit absurd about talking about percentiles when there are only 10 on the list. The use of primary percentiles is to help keep track of where a particular number falls within a population of hundreds or thousands or more. If you have 10 numbers, you can just look at the whole list — you don’t need fancy metrics to tell you what’s where. Does this make sense? Mike 🙂 13. Arun August 19, 2013 at 9:48 am # Hey Mike, Fist of all, excellent article!! I have a query regarding the first question. What if all of the test takers got the “same” highest score? Then Sasha would be in the zero percentile zone as nobody is technically under her, right? I know I am being stupid here and the case is only theoretical. But should we have to take such cases into account? Please forgive my ignorance if I am asking a really stupid question! • Mike August 19, 2013 at 10:22 am # Arun, In that question, the test is “nationwide”, which implies that thousands (if not hundreds of thousands) of people take it. This is a subtle thing about populations and variation among population. If I give a test to class of ten students, yes, it’s possible that everyone will get the same score: that would be a very poorly written test. In the real world, it cost millions of dollars to sponsor a nationwide test. In practice, no one is going to spend millions of dollar to give a test, and then not even check the quality, so that the test is so poor that everyone gets the same score. If ETS did that even once, they would be out of business! The “everyone gets the same score” scenario is a mathematical abstraction that ignores the hard tangible economic realities of the situation. Does this make sense? Mike 🙂 • P R Siddharthan September 18, 2014 at 5:39 am # I am sorry to say this, but it does not. Even I hosted a word wide test online (on edX opencourseware) only 2 took the test. So the highest score got a percentile of 50 only . How can general knowledge on how a nation wide test is conducted influence my answer. Should I expect such ambiguous questions in GRE? • Mike September 18, 2014 at 2:21 pm # Dear PR, My friend, we are getting into some subtleties of English here. Yes, anything online is available to anyone in the world with internet, so, say, the reply I am writing to you now — technically, anyone in the world could see it, but it would grossly misleading to say I have written a “worldwide” response. The words “worldwide” or “nationwide” don’t merely connote the possibility that a large number of people could have access to it: by contrast, these words connote an actual use or recognition or fame or popularity that is, in fact, wide spread among a large number of people. They connote that a substantial fraction of the relevant population actually does use or see something. It is factually correct to say that, for example, Coca Cola has worldwide popularity or recognition, or that Coke is a worldwide company. People really do recognize that logo everywhere. Magoosh has customers all over the world, but it would be grossly misleading to say that we have “worldwide” popularity — the latter would suggest that almost wherever you went, vast majority of people would recognize the brand. When Gandhi marched to the sea or when Nelson Mandala was released from prison, those legitimately were events of worldwide significant: they actually meant a great deal, and were deeply meaningful to, a large number of people on every part of the planet. If I say or do something and post it on social media, theoretically, anyone in the world could look at it, but even if a couple dozen friends of mine in different corners of the world say that it is meaningful, it would egregiously deceptive for me to say that that my activity was an event of worldwide significance. To call something “worldwide,” it must be true for a very large number of people in many different parts of the world. Much in the same way, the word “nationwide” does not merely mean that anyone in the nation could take the test, that maybe 12 or 15 people dispersed across the country have taken it. That would be a deceptively misleading use of the what the word “nationwide” means. The word “nationwide” always connotes popularity and widespread usage throughout the nation. There’s no specific number associated with the word, but if only a few dozen are using the item, then “nationwide” would a patently incorrect word, and using it could very well be construed as a deliberate way to deceive. Hilary Clinton legitimately has nationwide recognition in America: almost everyone in America, and many people in other parts of the world, know exactly who she is. People may like her or not, but everyone recognized that name. My blogs are on the web, and theoretically, everyone in America could read my blogs and find out about me, but it would be patently false for me to claim “nationwide” recognition for myself. Again, if I were to make the unfounded claim that Magoosh had “worldwide recognition,” that easily could be construed as a deliberate attempt to mislead or misrepresent on my part. People would be completely justified in saying that I was lying if I were to say that. Even there there are people all over the world that recognize and use Magoosh, the sheer numbers and sheer percentage of the relevant populations are not nearly enough to justify the wide level of inclusion that the word “worldwide” implies. The GRE never intentionally misleads students, so if they used the word “nationwide” or “worldwide,” they would only use them in their legitimate senses, the sense of sizable chunk of the population making use of something. The way they are used here models the use they could have on the GRE. Does all this make sense? Mike 🙂 • Manasi January 5, 2016 at 10:22 am # Your answers make good sense. In fact, I read your response inspite of already having a clear notion of what ‘nationwide’ meant in the question, just for the pleasure of reading it! Your explanations are very stimulating and cool 🙂 And, thank you for your main post on percentiles and quartiles. It cleared all my confusions on the topic! 14. VAMSEEDHAR REDDY August 7, 2013 at 4:05 am # 2) Alice took nationwide standardize test that is graded on a scale from 0 to 100. Alice the highest score recorded on that this test. In the above question, the mentioned looks similar to gre, isn’t it. Gre is graded on a scale 260 to 340. In the gre the topper gets a score of perfect 340 which is 99th percentile. Doesnt’ it mean that grading is done on the basis of percentiles. But according to your answer, if the exam is tough,the score may be low. please clarify the process of gradation you mentioned. • Mike August 7, 2013 at 9:34 am # Dear Vamseedhar, You are correct, highest percentile is 99%, so for most tests the highest grade is 99th percentile. This question is asking us to compare the numerical grade, to the numerical percentile. For example, for the GRE, the number for the highest score (340) would always be higher than the number for the highest percentile, because 340 > 99. If a test when from, say, 1 to 12, then in all likelihood, the number for the highest score (12) would be less than the number for the highest percentile, 12 < 99. For a test that goes from 0 to 100, Maybe Alice got a score of 100, which was 99th percentile, so (score) > (percentile), or maybe Alice's score was only 78 but it was still 99th percentile, so then (score) < (percentile). We are comparing one number to another number. Does this make sense? Mike 🙂 • VAMSEE August 8, 2013 at 11:45 pm # Dear Mike, Thank you for quick reply. But your answer does not clarify my doubt. What I said was that in an exam like gre a 99 percentiler gets a perfect score. In the question given since the grade is scaled from 0 to 100 a 99 percentiler should get 100 if it is similar to gre. May be the question does not clearly mention if it was absolute grading or relative grading(like gre). In case of relative grading, in the question given, the score of the student will be greater than percentile,I think. I hope you clarify this. • Mike August 9, 2013 at 10:34 am # Dear VAMSEE, First of all, I’m not sure what grading scheme you have in mind when you say “absolute grading”, but anything that involves percentiles, by definition, is comparing you to the rest of the population and hence is relative. Percentiles are always always always about comparing you to the rest of the population. They never have any meaning apart from that. To know the percentile of any single score, on any test, you have to have information about how the entire rest of the population did. Question #2 says: “Alice scored the highest score recorded on this test.” We don’t know the numerical value of that — maybe 100, or maybe something much lower. We do know, because this score is the highest score, it’s the 99th percentile. Does this make sense? Mike 🙂 15. Soumya May 22, 2012 at 8:14 am # Hi Mike. My original thoughts for the first question was similar to the answer given here, that Sasha’s score is probably in the the 90th percentile or above. Then I realized that the number of test takers was not included (and that thinking she was in the 90th percentile was an arbitrary guess). So, how do we really know what percentile Sasha is in? What if there were two test takers? That would mean that Sasha could either be in the 0th or 50th percentile. Right? This would mean that the answer should be D. Perhaps I read the question wrongly and made a mistake in my analysis. What do you think? This is a similar thought process used in your explanation for Question 2 (but the opposite situation). • Mike May 23, 2012 at 9:52 am # Dear Soumya — In both questions, the word “nationwide” was designed to connote that a large number of people took the test — maybe 1000, maybe 1000000, but some large number. It simply wouldn’t make sense to call the test “nationwide” if only two people took it —do you see what I mean? Furthermore, the fact that Sasha got “one of the highest score” means there was more than one highest score. That makes it impossible for there to have been only two test takers, and again suggests a large test-taking pool. Therefore, we know some reasonably large number of people took the test, and if Sasha got one of the highest scores, it’s well above the 90th percentile. Does this make sense? Mike 🙂 • Soumya May 23, 2012 at 11:10 am # Hi Mike, Yes, that definitely makes sense. This is the kind of little mistake I keep making! How frustrating. I got a 790 on the SAT Math back in the day and now I cannot break 155-160 on practice GREs (which has been verified by my Magoosh math score estimate), all because of this silly lack of attention. On that note — how would you compare SAT and GRE math? Is this dissonance in my scores expected or abnormal? I should note that in that time, I had completed both my Bachelor’s and Master’s in Engineering in a very rigorous field at a highly ranked university. Perhaps I need to go back to practicing mental math and just focus on attention to detail. Any enlightenment in this matter would be appreciated. Anyways, thanks for the reply! Soumya • Mike May 23, 2012 at 2:25 pm # Dear Souyma, First of all, about attention and lack thereof, take a look at these two earlier blogs: 1) https://magoosh.com/gmat/2012/overcome-gmat-exam-anxiety-breathe/ 2) https://magoosh.com/gmat/2012/beating-gmat-stress/ GRE math is, on average, slightly harder than SAT; the content is almost identical, but the GRE has a lower concentration of easy questions and a higher concentration of hard questions. For someone with degrees in Engineering, the difficulty level of the GRE math simply is not an issue. Instead, it’s all about attention to detail, catching distinctions, not overlooking key phrases, etc. In short,it’s not math you need to learn as much as mindfulness, which is discussed in those two posts. Does that make sense? Mike 🙂 Magoosh blog comment policy: To create the best experience for our readers, we will only approve comments that are relevant to the article, general enough to be helpful to other students, concise, and well-written! 😄 Due to the high volume of comments across all of our blogs, we cannot promise that all comments will receive responses from our instructors. We highly encourage students to help each other out and respond to other students' comments if you can! If you are a Premium Magoosh student and would like more personalized service from our instructors, you can use the Help tab on the Magoosh dashboard. Thanks!
# 2002 AIME II Problems/Problem 12 ## Problem A basketball player has a constant probability of $.4$ of making any given shot, independent of previous shots. Let $a_n$ be the ratio of shots made to shots attempted after $n$ shots. The probability that $a_{10} = .4$ and $a_n\le.4$ for all $n$ such that $1\le n\le9$ is given to be $p^aq^br/\left(s^c\right)$ where $p$, $q$, $r$, and $s$ are primes, and $a$, $b$, and $c$ are positive integers. Find $\left(p+q+r+s\right)\left(a+b+c\right)$. ## Solutions ### Solution 1 We graph the $10$ shots on a grid. Suppose that a made shot is represented by a step of $(0,1)$, and a missed shot is represented by $(1,0)$. Then the basketball player's shots can be represented by the number of paths from $(0,0)$ to $(6,4)$ that always stay below the line $y=\frac{2x}{3}$. We can find the number of such paths using a Pascal's Triangle type method below, computing the number of paths to each point that only move right and up. $[asy] size(150); for (int i=0;i<7;++i) {draw((i,0)--(i,4));} for (int i=0;i<5;++i) {draw((0,i)--(6,i));} draw((0,0)--(6,4),dashed); label("1",(0,0),SE,fontsize(8)); label("1",(1,0),SE,fontsize(8)); label("1",(2,0),SE,fontsize(8)); label("1",(2,1),SE,fontsize(8)); label("1",(3,0),SE,fontsize(8)); label("2",(3,1),SE,fontsize(8)); label("2",(3,2),SE,fontsize(8)); label("1",(4,0),SE,fontsize(8)); label("3",(4,1),SE,fontsize(8)); label("5",(4,2),SE,fontsize(8)); label("1",(5,0),SE,fontsize(8)); label("4",(5,1),SE,fontsize(8)); label("9",(5,2),SE,fontsize(8)); label("9",(5,3),SE,fontsize(8)); label("1",(6,0),SE,fontsize(8)); label("5",(6,1),SE,fontsize(8)); label("14",(6,2),SE,fontsize(8)); label("23",(6,3),SE,fontsize(8)); label("23",(6,4),SE,fontsize(8)); [/asy]$ Therefore, there are $23$ ways to shoot $4$ makes and $6$ misses under the given conditions. The probability of each possible sequence occurring is $(.4)^4(.6)^6$. Hence the desired probability is $$\frac{23\cdot 2^4\cdot 3^6}{5^{10}},$$ and the answer is $(23+2+3+5)(4+6+10)=\boxed{660}$. ### Solution 2 The first restriction is that $a_{10} = .4$, meaning that the player gets exactly 4 out of 10 baskets. The second restriction is $a_n\le.4$. This means that the player may never have a shooting average over 40%. Thus, the first and second shots must fail, since $\frac{1}{1}$ and $\frac{1}{2}$ are both over $.4$, but the player may make the third basket, since $\frac{1}{3} \le .4$ In other words, the earliest the first basket may be made is attempt 3. Using similar reasoning, the earliest the second basket may be made is attempt 5, the earliest the third basket may be made is attempt 8, and the earliest the fourth basket may be made is attempt 10. Using X to represent a basket and O to represent a failure, this 'earliest' solution may be represented as: OOXOXOOXOX To simplify counting, note that the first, second, and tenth shots are predetermined. The first two shots must fail, and the last shot must succeed. Thus, only slots 3-9 need to be counted, and can be abbreviated as follows: XOXOOXO The problem may be separated into five cases, since the first shot may be made on attempt 3, 4, 5, 6, or 7. The easiest way to count the problem is to remember that each X may slide to the right, but NOT to the left. First shot made on attempt 3: XOXOOXO XOXOOOX XOOXOXO XOOXOOX XOOOXXO XOOOXOX XOOOOXX Total - 7 First shot made on attempt 4: Note that all that needs to be done is change each line in the prior case from starting with "XO....." to "OX.....". Total - 7 First shot made on attempt 5: OOXXOXO OOXXOOX OOXOXXO OOXOXOX OOXOOXX Total - 5 First shot made on attempt 6: OOOXXXO OOOXXOX OOOXOXX Total - 3 First shot made on attempt 7: OOOOXXX Total - 1 The total number of ways the player may satisfy the requirements is $7+7+5+3+1=23$. The chance of hitting any individual combination (say, for example, OOOOOOXXXX) is $\left(\frac{3}{5}\right)^6\left(\frac{2}{5}\right)^4$ Thus, the chance of hitting any of these 23 combinations is $23\left(\frac{3}{5}\right)^6\left(\frac{2}{5}\right)^4 = \frac{23\cdot3^6\cdot2^4}{5^{10}}$ Thus, the final answer is $(23+3+2+5)(6+4+10)=\boxed{660}$ ### Solution 3 Note $a_{10}=.4$. Therefore the player made 4 shots out of 10. He must make the 10th shot, because if he doesn't, then $a_9=\frac{4}{9}>.4$. Since $a_n\leq .4$ for all $n$ less than 11, we know that $a_1=a_2=0$. Now we must look at the 3rd through 9th shot. Now let's take a look at those un-determined shots. Let's put them into groups: the 3rd, 4th, and 5th shots in group A, and the 6th, 7th, 8th, and 9th shots in group B. The total number of shots made in groups A and B must be 3, since the player makes the 10th shot. We cannot have all three shots made in group A, since $a_5\leq .4$. Therefore we can have two shots made, one shot made, or no shots made in group A. Case 1: Group A contains no made shots. The number of ways this can happen in group A is 1. Now we must arrange the shots in group B accordingly. There are four ways to arrange them total, and all of them work. There are $\textbf{4}$ possibilities here. Case 2: Group A contains one made shot. The number of ways this could happen in group A is 3. Now we must arrange the shots in group B accordingly. There are six ways to arrange them total, but the arrangement "hit hit miss miss" fails, because that would mean $a_7=\frac{3}{7}>.4$. All the rest work. Therefore there are $3\cdot5=\textbf{15}$ possibilities here. Case 3: Group A contains two made shots. The number of ways this could happen in group A is 2 (hit hit miss doesn't work but the rest do). Now we must arrange the shots in group B accordingly. Note hit miss miss miss and miss hit miss miss fail. Therefore there are only 2 ways to do this, and there are $2\cdot 2=\textbf{4}$ total possibilities for this case. Taking all these cases into account, we find that there are $4+15+4=23$ ways to have $a_{10} = .4$ and $a_n\leq .4$. Each of these has a probability of $.4^4\cdot.6^6=\frac{2^4\cdot 3^6}{5^{10}}$. Therefore the probability that we have $a_{10} = .4$ and $a_n\leq .4$ is $\frac{23\cdot2^4\cdot3^6}{5^{10}}$. Now we are asked to find the product of the sum of the primes and the sum of the exponents, which is $(23+2+3+5)(4+6+10)=33\cdot20=\boxed{660}$.
# SIMILAR FIGURES WORKSHEET Similar figures worksheet : Worksheet given in this section is much useful to the students who would like to practice problems on similar figures. Two figures are similar if one can be obtained from the other by a sequence of translations, reflections, rotations, and dilations. Similar figures have the same shape but they may have different sizes. If two figures are similar, there must be a sequence of translations, reflections, rotations, and/or dilations that can transform one to the other. ## Similar figures worksheet - Problems 1. Identify a sequence of transformations that can transform figure A into figure B. Say whether the figures are congruent. And also, say whether they are similar. 2. Identify a sequence of transformations that can transform figure X into figure Y. Include a reflection. Say whether the figures are congruent. And also, say whether they are similar. 3. In problem 2, identify a sequence of transformations that can transform figure X into figure Y. Include a rotation. ## Similar figures worksheet - Solution Problem 1 : Identify a sequence of transformations that can transform figure A into figure B. Say whether the figures are congruent. And also, say whether they are similar. Solution : Step 1 : Both figures are squares whose orientations are the same, so no reflection or rotation is needed. Step 2 : Figure B has sides twice as long as figure A, so a dilation with a scale factor of 2 is needed. Step 3 : Figure B is moved to the right and above figure A, so a translation is needed. Step 4 : A sequence of transformations that will accomplish this is a dilation by a scale factor of 2 centered at the origin followed by the translation (x, y) ----> (x + 4, y + 6) The figures are not congruent, but they are similar. Problem 2 : Identify a sequence of transformations that can transform figure X into figure Y. Include a reflection. Say whether the figures are congruent. And also, say whether they are similar. Solution : Step 1 : The orientation of figure Y is reversed from that of figure X, so a reflection over the y-axis is needed. Step 2 : Figure Y has sides that are half as long as figure X, so a dilation with a scale factor of 1/2 is needed. Step 3 : Figure B is moved to the right and above figure A, so a translation is needed. Step 4 : Figure Y is moved above figure X, so a translation is needed. Step 5 : A sequence of transformations that can accomplish this is a dilation by a scale factor of 1/2 centered at the origin, followed by the reflection (x, y) ---> (-x , y) followed by the translation (x, y) ---> (x , y + 5) The figures are not congruent, but they are similar. Problem 3 : In example 2, identify a sequence of transformations that can transform figure X into figure Y. Include a rotation. Solution : Step 1 : The orientation of figure Y is reversed from that of figure X, so a rotation of 180º is needed. Step 2 : Figure Y has sides that are half as long as figure X, so a dilation with a scale factor of 1/2 is needed. Step 3 : Figure Y is moved above figure X, so a translation is needed. Step 4 : Figure Y is moved above figure X, so a translation is needed. Step 5 : A sequence of transformations that can accomplish, this is a rotation of 180º about the origin, followed by a dilation by a scale factor of 1/2 centered at the origin, followed by the translation (x, y) ---> (x , y + 5) The figures are not congruent, but they are similar. After having gone through the stuff given above, we hope that the students would have understood "Similar figures worksheet". Apart from the stuff given on "Similar figures worksheet", if you need any other stuff in math, please use our google custom search here. WORD PROBLEMS HCF and LCM word problems Word problems on simple equations Word problems on linear equations Algebra word problems Word problems on trains Area and perimeter word problems Word problems on direct variation and inverse variation Word problems on unit price Word problems on unit rate Word problems on comparing rates Converting customary units word problems Converting metric units word problems Word problems on simple interest Word problems on compound interest Word problems on types of angles Complementary and supplementary angles word problems Double facts word problems Trigonometry word problems Percentage word problems Profit and loss word problems Markup and markdown word problems Decimal word problems Word problems on fractions Word problems on mixed fractrions One step equation word problems Linear inequalities word problems Ratio and proportion word problems Time and work word problems Word problems on sets and venn diagrams Word problems on ages Pythagorean theorem word problems Percent of a number word problems Word problems on constant speed Word problems on average speed Word problems on sum of the angles of a triangle is 180 degree OTHER TOPICS Profit and loss shortcuts Percentage shortcuts Times table shortcuts Time, speed and distance shortcuts Ratio and proportion shortcuts Domain and range of rational functions Domain and range of rational functions with holes Graphing rational functions Graphing rational functions with holes Converting repeating decimals in to fractions Decimal representation of rational numbers Finding square root using long division L.C.M method to solve time and work problems Translating the word problems in to algebraic expressions Remainder when 2 power 256 is divided by 17 Remainder when 17 power 23 is divided by 16 Sum of all three digit numbers divisible by 6 Sum of all three digit numbers divisible by 7 Sum of all three digit numbers divisible by 8 Sum of all three digit numbers formed using 1, 3, 4 Sum of all three four digit numbers formed with non zero digits Sum of all three four digit numbers formed using 0, 1, 2, 3 Sum of all three four digit numbers formed using 1, 2, 5, 6
Question 98 # Avinash invested ₹6,400 in three business at 3%, 5% and 7% per year with simple interest. At the end of the year, he received the same amount from three business. The money invested at 3% is ( up to two decimals): Solution => Avinash invested ₹6,400 in three business at 3%, 5% and 7% per year with simple interest. Simple interest formula is $$SI\ =\ \frac{PRT}{100}$$ Where P is principal, R is rate of interest and T is time duration in years Here, Let us assume the amount invested in three businesses is P1, P2 and P3. Therefore, the interest after 1 year because of simple interest will be 3P1/100 , 5P2/100 and 7P3/100 These are equal, Hence => 3P1 = 5P2 = 7P3 It is also known that total investment done in three business is equal to 6400 P1 + P2 + P3 = 6400 Substituting values of P2 and P3 in terms of P1 P1+3P1/5 + 3P1/7 = 6400 We need to calculate the value of P1 (35+21+15)P1/35 = 6400 => 71P1/35 =6400 => P1 is approximately 3200 (Option B is correct answer)
## Multiplication of Three Digit Numbers (3 x 3 Digits) Hi Kids, Here, let us see how to multiply 2 three digit numbers. For example, let us consider the numbers 228 and 121. Write the numbers one below the other, as given in the below image. As we did in the previous multiplication problems, we need to first multiply 1 of 121 with 228, in this math. Let us do that one at a time. Multiply the “ones” first. Multiplying 8 with 1, we get 8. Write 8 in the answer “ones” column. Next, multiplying 1 with 2, we get 2. Write 2 in the “tens” column. Now multiplying 1 with 2, we get 2. Write 2 in the “hundreds” column. As done in the previous multiplication problems, let us multiply 2 of three digit number 121 with 228. Write 0 below 8 in the answer column. Multiplying 2 and 8, we get 16. Write 6 in the answer column and carry over 1 to the “tens” column. Multiply 2 with 2 of 228. We get 4. Add the carry over 1 with 4. We get 5. Write 5 in the answer column. Again multiplying 2 and 2, we get 4. Write it in the answer column. We get 4560. Now we have 1 remaining in 121 to be multiplied with 228. Since 1 is the third digit of the three digit number 121, we need to write 0 in both “ones” and “tens” place of the answer column. Multiplying 1 and 8, we get 8 in this multiplication problem. Write it below 5 in the answer column. Multiplying 1 and 2, we get 2. Write 2 in the answer column. Again multiplying 1 and 2, we get 2. Write 2 in the answer column. Now add the digits in order, from “ones” column. Thus we get the answer 27588 by multiplying 228 and 121. Follow the above steps for similar problems to understand the concept better.
# Advanced Algebra Notes ## Presentation on theme: "Advanced Algebra Notes"— Presentation transcript: Section 3.1 Solving Linear Systems by Graphing Solving Linear Systems by Graphing System of two linear equations A in two variables x and y, also called linear system, consists of two equations that can be written in the following form. When we talk about finding a we are looking for the ordered pair (x,y) that will satisfy each equation. It must work with both! Ax + By = C Dx + Ey = F solution Example 1 Graph the linear system and estimate the solution. Then check the solution algebraically. 4x+y=8 2x-3y=18 𝑦=−4𝑥+8 −3𝑦=−2𝑥+18 𝑦= −2𝑥+18 −3 (3,−4) 𝑦= 2 3 𝑥−6 Example 2 INFINITE SOLUTIONS Graph the linear system and estimate the solution. 4x-3y=8 8x-6y=16 −3𝑦=−4𝑥+8 𝑦= 4 3 𝑥− 8 3 INFINITE SOLUTIONS −6𝑦=−8𝑥+16 −3𝑦=−4𝑥+8 𝑦= 4 3 𝑥− 8 3 Example 3 𝑦=−2𝑥+4 PARALLEL 𝑦=−2𝑥+1 NO SOLUTION Graph the linear system and estimate the solution 2x+y=4 2x+y=1 𝑦=−2𝑥+4 PARALLEL 𝑦=−2𝑥+1 NO SOLUTION When lines have the same slope they are parallel. They will not cross! They will not have a solution! Example 4 Graph the linear system and estimate the solution 5x-2y=-10 2x-4y=12 −2𝑦=−5𝑥−10 𝑦= 5 2 𝑥+ 10 2 𝑦= 5 2 𝑥+5 −4𝑦=−2𝑥+12 𝑦= 1 2 𝑥−3 (−4,−5) Closing: How do you solve a system of linear equations graphically? Our solution is the point where our lines intersect. It is a coordinate point (𝑥,𝑦) If our lines do not cross, they are parallel meaning they have no solution. If our lines are the same line, they are Intersect at every point meaning they have infinite solutions. Our solution has to work for both equations!!!
# Section 12.4 Question 1 ## How do you find the optimal dimensions of a product? The size and shape of a product influences its functionality as well as the cost to construct the product. If the dimensions of a product are designed to minimize the cost of materials used to construct the product, then the objective function models the cost of the material in terms of the dimensions of the product. In Example 1, we find an objective function for the cost of materials to build a rectangular enclose. By finding the relative minimum of this objective function, we are able to find the dimensions of the enclosure that costs the least amount. ### Example 1 Minimize Cost of Materials A farmer is fencing a rectangular area for two equally sized pens. These pens share a divider that will be constructed from chicken wire costing \$0.60 per foot. The rest of the pen will be built from fencing costing \$1.10 per foot. If the pens should enclose a total of 400 square feet, what overall dimensions should the pens have to minimize fencing costs? Solution To find the minimum fencing costs, we must formulate a function that describes the cost of fencing as a function of some variable. In this problem, we have two variables we could use. The width or length of the enclosed region can be the variable. In this example, we’ll choose the variable to be the width of the region. Informally, the cost of fencing is Cost = Cost of Width Components + Cost of Length Components Each of these terms describes the cost of individual components of the fencing around the enclosed area. The first term matches the parts of the fencing along the top and bottom of the figure costing \$1.10 per foot. The second term matches the two parts of fencing along the sides costing \$1.10 per foot as well as the divider costing \$0.60 per foot. A table is useful to help us recognize the function describing the cost of the fencing. In this table, each column will represent the quantities that will vary, the two terms in the cost of fencing as well as the total cost of fencing. The table below contains five columns for these quantities and five blank rows in which we’ll place some values. Start by entering several combinations for length and width that yield 400 square feet. For example, a pen that is 5 feet by 80 feet encloses 400 square feet. Another possibility is a pen that is 10 feet by 40 feet. In general, if the width is w then the length is . Now let’s calculate the costs involved in fencing a pen that is 5 feet by 80 feet. The width components will cost and the length components will cost to give a total cost of The cost involved in fencing a pen that is 10 feet by 40 feet is calculated in a similar manner. The width components will cost and the length components will cost to give a cost of This expression is almost identical to the earlier expression except amounts of fencing are different. We can continue to calculate other sized pens to fill out a table of values. The last row in the table, the width is labeled as w and the corresponding length must be to ensure the area is 400. The other entries in the last row use the width and length to match the pattern in the rows above. Based on this pattern, we can define the cost function C(w) as a function of the width w, We can simplify the function to We’ll need to take the derivative of this function to find the relative minimum. It is easiest to take the derivative with the power rule for derivatives if we rewrite the second term with a negative exponent. With this modification, the cost function is The derivative is To find the critical points, we need to find where the derivative is equal to zero or undefined. Set the derivative equal to zero and solve for w: To solve this quadratic equation, we could use the quadratic formula. However, it is easier to isolate w2: The negative critical value is not a reasonable dimension for the width of the pen so we can ignore it. If the derivative is rewritten as , we observe that it is undefined at w = 0. Like negative values, a width of zero is not a reasonable dimension for a pen. Ignoring this value, we have only one critical value, w ≈ 22.56, in the domain of this problem. This critical value may correspond to a relative minimum or a relative maximum. We can use the first derivative test to determine which type of extrema this critical value matches. If we test the first derivative on either side of the critical value, we can establish where C(w) is increasing and decreasing. This first derivative test indicates that the function is decreasing and then increasing so w ≈ 22.56 corresponds to a relative minimum. The expression for the length is . Using the width, w ≈ 22.56 feet, the length is calculated as approximately 17.73 feet. These dimensions yield a minimum total cost of For some businesses, the goal of the design process is not a product that costs as little as possible. Instead, they maximize or minimize some characteristic of the product with respect to the dimensions of the product. For instance, a newspaper publisher might maximize the area of the printed page with respect to the margins and dimensions of the page. By doing this, they maximize the area used to print news as well as the area used for advertisements. In the next example, we find the dimensions a piece of carry-on baggage that maximizes the volume inside the bag and meets airline requirements for the dimensions. ### Example 2 Maximize Volume Most airlines charge to check baggage on flights. To avoid these charges, passengers pack as much as possible into their carry-on bags. However, airlines also limit the size of these bags. American Airlines limits the linear dimensions (defined as the sum of the length, width and height) to 45 inches. A manufacturer wishes to produce a carry-on bag whose linear dimensions are 45 inches. The shape of the bag is a rectangular solid, like the one pictured, below whose ends are squares. What are the dimensions of the bag if its volume is to be as large as possible? Solution We want to maximize the volume of the carry-on bag. To help us understand the relationships between the dimensions, let’s look at a particular carry-on bag. Suppose the dimensions of the square ends are 5 inches by 5 inches. Since the sum of the length, width, and height must be 45 inches, we know the length must be 45 – 5 – 5 = 35 inches. The volume of this carry-on bag is the product of the length, width, and height so this bag is Let’s look at a second possibility. Suppose the dimensions of the square end are 10 inches by 10 inches. The length must be inches. The volume of this carry-on bag is Notice that these dimensions result in a larger volume. We can continue this process to see if the volume continues to increase as the square end gets larger. To keep track of the information, let’s enter this information into a table. As the dimensions of the square end increases, the volume rises and then falls. It appears that the optimal dimensions are around 15 inches by 15 inches by 15 inches. To find a more exact answer, we need to come up with an expression for the volume. We do this by identifying the variable as the width w and look for pattern in each column of the table. If the width corresponds to w, so must the height since the ends are square. If we subtract these dimensions from 45, we get the length . The volume is the product of these dimensions. Using this pattern, define the volume V asThis simplifies to It is much easier to take the derivative of this function once it has been simplified. Using the basic rules for derivatives, the derivative is calculated as This derivative is set equal to zero to find the critical values for the function. Since the derivative is a polynomial, there are no values of w for which the derivative is undefined. Additionally, all critical values must be positive since the width must be a positive number. Set the derivative equal to zero and solve for w: Only the critical value at w = 15 is a reasonable dimension of the carry-on. Now let’s use the first derivative test to determine if the critical value is a relative minimum or a relative maximum: The volume function increases initially and then decreases after w = 15. This matches the behavior we saw in the last column of the table. It also tells us that the critical value corresponds to a relative maximum. The optimal solution occurs when the square end is 15 inches by 15 inches and the length is 45 – 15 – 15 or 15 inches. The volume at these dimensions is (15 inches)(15 inches)(15 inches) or 3375 in3.
# How do you find the domain of f(x) = sqrt ( x- (3x^2))? Mar 18, 2018 The domain of $f \left(x\right)$ is $x \in \left[0 , \frac{1}{3}\right]$ #### Explanation: What's under the $\sqrt{}$ sign is $\ge 0$ Here, $f \left(x\right) = \sqrt{x - 3 {x}^{2}}$ Therefore, $x - 3 {x}^{2} \ge 0$ $x \left(1 - 3 x\right) \ge 0$ Let $g \left(x\right) = x \left(1 - 3 x\right)$ Build the sign chart $\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a a}$$0$$\textcolor{w h i t e}{a a a a a a a}$$\frac{1}{3}$$\textcolor{w h i t e}{a a a a a a}$$+ \infty$ $\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a a a a a}$$-$$\textcolor{w h i t e}{a a a}$$0$$\textcolor{w h i t e}{a a a}$$+$$\textcolor{w h i t e}{a a a a a a}$$+$ $\textcolor{w h i t e}{a a a a}$$1 - 3 x$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a}$color(white)(aaaa)+$\textcolor{w h i t e}{a a}$$0$$\textcolor{w h i t e}{a a a a}$$-$ $\textcolor{w h i t e}{a a a a}$$g \left(x\right)$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a}$$0$$\textcolor{w h i t e}{a a a}$$+$$\textcolor{w h i t e}{a a}$$0$$\textcolor{w h i t e}{a a a a}$$-$ Therefore, $g \left(x\right) \le 0$, $\implies$, $x \in \left[0 , \frac{1}{3}\right]$ graph{sqrt(x-3x^2) [-0.589, 1.096, -0.307, 0.536]}
# How do you prove the statement lim as x approaches 1 for [ x + 1 ] / [ 2x + 1 ] = 2/3 using the epsilon and delta definition? Mar 24, 2016 To prove that ${\lim}_{x \to c} f \left(x\right) = L$ for some function $f$, we must show that for any $\epsilon > 0$ there exists a $\delta > 0$ such that if $\| x - c \| < \delta$ then $\| f \left(x\right) - L \| < \epsilon \|$ Proof: Let $\epsilon > 0$ be arbitrary, and let $\delta = \min \left(1 , 3 \epsilon\right)$. Then, if $\| x - 1 \| < \delta$, note that $\| x - 1 \| < 1$ and thus $\| 2 x + 1 \| > 1$ (see below for details). With that, we have, for $\| x - 1 \| < \delta$: $\| \frac{x + 1}{2 x + 1} - \frac{2}{3} \| = \| \frac{3 x + 3}{3 \left(2 x + 1\right)} - \frac{4 x + 2}{3 \left(2 x + 1\right)} \|$ $= \| \frac{- x + 1}{3 \left(2 x + 1\right)} \|$ $= \| \frac{x - 1}{3 \left(2 x + 1\right)} \|$ $= \| x - 1 \frac{\|}{3 \| 2 x + 1 \|}$ $< \| x - 1 \frac{\|}{3}$ $< \frac{3 \epsilon}{3}$ $= \epsilon$ Therefore, as we have shown that a $\delta$ exists for any $\epsilon$ such that $\| x - 1 \| < \delta \|$ implies $\| \frac{x + 1}{2 x + 1} - \frac{2}{3} \| < \epsilon$, it is the case that ${\lim}_{x \to 1} \frac{x + 1}{2 x + 1} = \frac{2}{3} \text{ }$ To see where we got $\| 2 x + 1 \| < 1$, starting from $\| x - 1 \| < 1$ we have: $\| x - 1 \| < 1$ $\implies - 1 < x - 1 < 1$ $\implies - 2 < 2 x - 2 < 2$ $\implies 1 < 2 x + 1 < 5$ $\implies 1 < \| 2 x + 1 \| < 5$ $\therefore \| 2 x + 1 \| > 1$
# Which ratio is equivalent to 7 3 Therefore, 6:14 and 9:21 are equivalent ratios of 3:7. ## How do you calculate equivalent ratios? • Enter a Ratio into the equivalent ratio calculator, for example, you could enter 7:25 • Select the number of equivalent ratios that you would like to see in the table of results • The equivalent ratio calculator will calculate as you type and produce a lis of equivalent ratios in a table below the calculator More items… ## How to calculate the ratio between two numbers? • We want to work out \$20 shared in the ratio of 1:3. • Step 1 is to work out the total number of parts in the ratio. • 1 + 3 = 4, so the ratio 1:3 contains 4 parts in total. • Step 2 is to divide the amount by the total number of parts in the ratio. • \$20 ÷ 4 = \$5. • Each of the four parts of the ratio is worth \$5. More items… ## What is 1 13/8 simplified? It can be written as 1.625 in decimal form (rounded to 6 decimal places). Therefore, 13/8 simplified to lowest terms is 13/8. Equivalent fractions: 26 / 16 39 / 24 65 / 40 91 / 56 ## Which ratio is equivalent to 13 15? Two fractions are equivalent when they are both equal when written in lowest terms. The fraction 2630 is equal to 1315 when reduced to lowest terms. To find equivalent fractions, you just need to multiply the numerator and denominator of that reduced fraction ( 1315) by the same natural number, ie, multiply by 2, 3, 4, 5, 6 … ## What is the ratio 7/8 equivalent to? 78 is 7 divided by 8, which equals 0.875. So an equivalent fraction is another fraction that also equals 0.875. To find this fraction, just take any number and multiply both the numerator and denominator by this number. ## What is the equivalent ratio of 2 ratio 7? ∴ the two equivalent ratios are 4:14 and 6:21. ## How do you find the equivalent ratio? Thus, to find a ratio equivalent to another we have to multiply the two quantities, by the same number. Another way to find equivalent ratios is to convert the given ratio into fraction form and then multiply the numerator and denominator by the same number to get equivalent fractions. ## What ratio is 4/7 equivalent to? 1 Answer. An equivalent ratio of 4 : 7 is 12 : 21. ## What does 7/8 equal as a fraction? 14/16Decimal and Fraction Conversion ChartFractionEquivalent Fractions3/86/1624/645/810/1640/647/814/1656/641/92/188/7223 more rows ## What does the ratio 7 2 mean? A ratio of 7 to 2 simply means that for every 7 of something, there are 2 of something else, with a total of 9. ## What is the ratio of 7 to 1? Ratio of 7 to 1 (7:1) A ratio of 7 to 1 can be written as 7 to 1, 7:1, or 7/1. Furthermore, 7 and 1 can be the quantity or measurement of anything, such as students, fruit, weights, heights, speed and so on. A ratio of 7 to 1 simply means that for every 7 of something, there are 1 of something else, with a total of 8. ## Why is the equivalent ratio? 0:368:14How to Find Equivalent Ratios – YouTubeYouTubeStart of suggested clipEnd of suggested clipSo let’s take a look at number 1 here where we have 8 to 12 now equivalent ratios are just likeMoreSo let’s take a look at number 1 here where we have 8 to 12 now equivalent ratios are just like equivalent fractions. We use multiplication or division. Whatever you do to one number of the ratio. You ## What is the ratio 8 to 2 equivalent to? Since the simplest form of the fraction 8/2 is 4/1, the simplest form of the ratio 8:2 is also 4:1. ## What is the 3rd equivalent fraction to 4 7? Decimal and Fraction Conversion ChartFractionEquivalent Fractions3/76/1427/634/78/1436/635/710/1445/636/712/1454/6323 more rows ## What is 7/4 as a fraction? 1 347/4 = 74 = 1 34 = 1.75 Spelled result in words is seven quarters (or one and three quarters). ## What does 7 4 ratio mean? A ratio of 7 to 4 simply means that for every 7 of something, there are 4 of something else, with a total of 11. In the box below, you can enter a number in either box and we will calculate the other number to keep the ratio of 7 to 4. ## What is the ratio of 2 to 4? 4 = 1:2Note that the ratio 2 to 4 is said to be equivalent to the ratio 1 to 2, that is 2:4 = 1:2. ## Which ratio is equivalent to the ratio 2 5? So, 2:5 is equivalent to 6:15. ## What is the ratio 8 to 2 equivalent to? Since the simplest form of the fraction 8/2 is 4/1, the simplest form of the ratio 8:2 is also 4:1. ## What are equivalent ratios to 2 3? Answer: 4/6, 6/9, 8/12, 10/15 … are equivalent to 2/3. All those fractions obtained by multiplying both the numerator and denominator of 2/3 by the same number are equivalent to 2/3. All equivalent fractions get reduced to the same fraction in their simplest form. ## How to find equivalent ratios? As we previously mentioned, Equivalent Ratios are two ratios that express the same relationship between numbers. The Equivalent Ratio Calculator provides a table of equivalent ratios that have the same relationship between each other and directly with the ratio you enter into the calculator. We will look at how to calculate equivalent ratios shortly, first lets look at how to use the free online equivalent ratio calculator: 1 Enter a Ratio into the equivalent ratio calculator, for example, you could enter 7:25 2 Select the number of equivalent ratios that you would like to see in the table of results 3 The equivalent ratio calculator will calculate as you type and produce a lis of equivalent ratios in a table below the calculator 4 [Optional] Print or email the Table of Equivalent Ratios for later use ## What is a ratio? A ratio is a direct comparison of one number against another. A ratio calculator looks to define the relationship that compares between those two numbers ## What is the importance of ensuring the right ratio of students to teachers? Education: ensuring the right ratio of students to teachers is key for effective learning. Class sizes in terms of the ratio of pupils to a teacher is a common ratio concern. ## Where are ratios used? Ratios are used everywhere, from cooking with your favourite recipes to building housing, here are some common applications of ratios in everyday life: ## Is there a formula for equivalent ratios? As equivalent ratios have the same value there is technically no equivalent ratio formula but the following equivalent ratio formula will help you with the manual math calculations. ## What is a Ratio? A contrast, which exists between two particular numbers, is defined as ratio. Our ratio calculator is developed to compute this contrast and find out the relationship between these numbers. ## How to keep numbers in direct relation? In order to keep numbers in direct relation you should first divide or multiply, which depends on your task, them in the ratio. Therefore, a ratio of 8/6 is an equivalent ratio of 4/3: in that particular ratio calculation, you should just multiply 4, as well as 3, by 2. ## How to convert a ratio to fractions? To convert a part-to-part ratio to fractions: Add the ratio terms to get the whole. Use this as the denominator. Convert the ratio into fractions. Each ratio term becomes a numerator in a fraction. Therefore, in the part-to-part ratio 1 : 2, 1 is 1/3 of the whole and 2 is 2/3 of the whole. ## When to simplify ratios? Simplify ratios or create an equivalent ratio when one side of the ratio is empty. ## What is part to part ratio? A part-to-part ratio states the proportion of the parts in relation to each other. The sum of the parts makes up the whole. The ratio 1 : 2 is read as “1 to 2.” This means of the whole of 3, there is a part worth 1 and another part worth 2. ## How many characters can a ratio calculator accept? This ratio calculator will accept integers, decimals and scientific e notation with a limit of 15 characters. ## What is Ratio? A ratio is a quantitative relationship between two numbers that describe how many times one value can contain another. Applications of ratios are fairly ubiquitous, and the concept of ratios is quite intuitive. This could likely be demonstrated by giving a child half as many cookies as his sister. While the child may not be able to voice the injustice using ratios, the raucous protestations that would most likely ensue should make it immediately obvious that he is well aware he has received 1:2 as many cookies as his sister, conceptually, if not mathematically. ## What is aspect ratio? The aspect ratio is the ratio of a geometric shape’s sizes in different dimensions. In the case of a rectangle, the aspect ratio is that of its width to its height. Although aspect ratios are widely used in applications such as tire sizing, paper sizing, and standard photographic print sizes, some of the most frequent uses of aspect ratios involve computer screen dimensions, mobile phone screens, and video sizes. As such, below is a list of typical computer screen/video resolutions and aspect ratios.
We can visualize addition using sets: or the number line: Starting at location 2 we advance 3 steps to end in location 5 We can only simplify the addition of unit-less items, or of items that have the same units. For example, we can simplify the addition of unit-less items: $$5 + 3 = 8$$ and the addition of items with the same units $$5\mbox{ cookies } + 3\mbox{ cookies } = 8\mbox{ cookies }$$ but we cannot simplify the addition of unit-less items to items with units: $$5 + 3\mbox{ cookies } = 5 + 3\mbox{ cookies }$$ or the addition of items with different units: $$5\mbox{ cookies } + 3\mbox{ marbles } = 5\mbox{ cookies } + 3\mbox{ marbles }$$ Although the following two forms describe the same addition process, sometimes it is easier to solve a problem using one form than the other. $$a + b = x$$ is the same as \begin{align} a&\\ \underline{+\,\,b}&\\ x& \end{align} ### The Associative law of addition It says that when we add three numbers, it does not matter which two numbers we add first: P1. $a + (b + c) = (a + b) + c$ Since regrouping does not affect the result of addition, we usually omit the parentheses but, in reality, the parentheses are implicitly there; it just happens that it does not matter where they are. For example, it does not matter if we mean $$a + b + c = (a + b) + c$$ or $$a + b + c = a + (b + c)$$ because in both cases the result is the same. From here we get that \begin{align} a + b + c + d & = ((a + b) + c) + d \\ & = (a + (b + c)) + d \\ & = (a + b) + (c + d) \\ & = a + ((b + c) + d) \\ & = a + (b + (c + d)) \\ \end{align} When adding numbers with multiple digits, it is often useful to use a decimal decomposition, e.g., $$1234 = 1000 + 200 + 30 + 4$$ We can use this decomposition to simplify additions, e.g., \begin{align} 17 + 34 &= (10 + 7) + (30 + 4) \\ &= (10 + 30) + (7 + 4) \\ &= 40 + 11 \\ &= 51 \\ \end{align} The U.S. standard addition algorithm adds the digits from right to left, “carrying” sums greater than a digit to the next place value: \begin{align} &1\,\,\,\,\,\,\,\,\,1\\ &785,927\\ +\,\,&\underline{262,055}\\ 1,&047,982\\ \end{align} The ‘carry’ is more significant than the number that we are writing under the line; to avoid forgetting to write the ‘carry’, get in the habit of saying, say, 7 + 5 = 12; then write the 12, in that order: first the 1, as a carry, and then the 2, under the line. Another way to add the numbers is adding by place values; this is longer but avoids errors of forgotten carries. \begin{align} &\,\,\,\,\,\,\,785,927\\ &\underline{+\,\,262,055}\\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,12\hspace{.2in}⇐ 7 + 5\\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,70\hspace{.2in}⇐ 20 + 50\\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,900\hspace{.2in}⇐ 900 + 0\\ &\,\,\,\,\,\,\,\,\,\,\,\,\,7,\hspace{.48in}⇐ (5 + 2) \times 1,000\\ &\,\,\,\,\,\,\,140,\hspace{.48in}⇐ (80 + 60) \times 1,000\\ &\underline{+\,\,900,\hspace{.45in}}⇐ (700 + 200) \times 1,000\\ &\,1,047,982\\ \end{align} ### Writing sums of sequences When writing a sum of a sequence we usually use dots, or the symbol for a summation: \begin{align} \sum_{i = 1}^{10}i &= 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10\\ &= 1 + 2 + \dots + 10 \end{align} The same applies for infinite sums: $$\sum_{i = 1}^{\infty}i = 1 + 2 + 3 + \dots$$ An easy way to add sequences of numbers is using Gaussian addition. Say that we want to add the numbers from 1 to 5: $$1 + 2 + 3 + 4 + 5 = V$$ We can obtain the same result if the write the sequence backwards: $$5 + 4 + 3 + 2 + 1 = V$$ Now, we pair up the terms of each sum and add the equations to obtain an equation equal to two times our original desired result V: \begin{align} (1 + 2 + 3 + 4 + 5 = V)&\\ \underline{+ (5 + 4 + 3 + 2 + 1 = V)}&\\ 6 + 6 + 6 + 6 + 6 = 2V& \end{align} Because of the symmetry of the columns, the vertical pairs add up to the same number; we simplify the equation to $$6 \times 5 = 2V$$ and, finally, solve for V: $$V = \frac{6 \times 5}{2} = 15$$ In general, the sum of all the numbers from 1 to n is We can use Gaussian addition to add many other sequences. For example, let’s add the odd numbers from 20 to 50: \begin{align} (21 + 23 + \cdots + 47 + 49 = V)&\\ \underline{+ (49 + 47 + \cdots + 23 + 21 = V)}&\\ 70 + 70 + \cdots + 70 + 70 = 2V& \end{align} Since there are 15 odd numbers from 20 to 50, we have that $$\frac{70 \times 15}{2} = 525$$ Yeah? show me. ### Sets A set is a collection of things; we can make sets of anything: the set of characters on a page, or the set of children in a room, or the set of animals that fly; each thing in a set is called a member or an element of the set. Sets are useful to group things that share some characteristics, so we can study and understand them better; for example, we can place certain animals into the set of birds, or the set of mammals, or the set of fish. We do the same thing with numbers: to study and understand them better, we divide them in sets, like Natural numbers (a.k.a. ℕ): we use them to count, like 1, 2, 3,…, etc. There is no universal agreement about whether the natural numbers should include 0 so we have the following more specific sets: • Whole numbers (a.k.a. non-negative integers) = {0, 1, 2, 3, …} • Counting numbers (a.k.a. positive integers) = {1, 2, 3 …} Integer numbers (a.k.a. ℤ, from the german word “Zahlen” that means “to count”): This set contains numbers with no fractions, i.e., the zero, the counting numbers and their negatives: $$ℤ = {… -3, -2, -1, 0, 1, 2, 3, …}$$ Rational numbers (a.k.a. ℚ, from the word “Quotient”): These numbers are ‘rational’ because they come from ‘ratios’. This set contains all the numbers that are ratios of two integer numbers, and all their corresponding decimals, whether their representation is finite, or periodic infinite. Some examples are: • 0 = 0/1 • 5 = 5/1 • -1/2 = -0.5 (finite decimal) • 1/3 = 0.33333.. (infinite periodic decimal) • 1/7 = 0.14285714285714285… (infinite periodic decimal) Irrational numbers (a.k.a. ℝ\ℚ, meaning ‘real minus rationals’): This set contains all the numbers that are not the ratio of two integer numbers; all their decimal representations are infinite and non-periodic. Some examples are: • $\sqrt{2} = 1.4142135623730951…$ • $\sqrt{3} = 1.7320508075688772…$ • $\pi = 3.141592653589793…$ • $e= 2.718281828459045…$ • $(1 + \sqrt{5})/2 = 1.618033988749895…$ a.k.a. the golden ratio Real numbers (a.k.a. ℝ): This set contains all the rationals and irrationals, i.e., all the numbers in the number line. In summary we have the following relationship: ### Problems 5 + 3 8 + 9 3 + 8 7 + 5 12 + 34 74 + 15 19 + 83 49 + 72 79 + 145 127 + 63 734 + 125 251 + 84 4. Add using both the U.S. standard and decimal decomposition algorithms \begin{align} &1,234,567\\ +\,&\underline{2,468,013} \end{align} 5. Find the following sum using both term-by-term addition and Gaussian addition. $$1 + 2 + \dots + 10 = ?$$ 6. Use Gaussian addition to add the odd numbers from 10 to 50 $$11 + 13 + \dots + 49 = ?$$ 7. Find $f + g$ for f = 2x + xy + 4 g = 4x – xy + y + 5 “take all the time you want” problems: For the following problems use… P1. $a + (b + c) = (a + b) + c$Associativity to show that … 8. $((p + q) + r) + s = p + (q + (r + s))$ 9. $(p + (q + r)) + s = (p + q) + (r + s)$ ### Challenge – mental math We want to find $a+b = c$. The result $c$ doesn’t change if we add $0$ to $a+b$, and $0$ is simply the addition of a number and its opposite, i.e., $d + (-d) = 0$. Hence: \begin{align} c &= a + b\\ &= (a + b) + 0\\ &= (a + b) + (d + (-d))\\ &= (a + d) + (b + (-d))\\ \end{align} We can simplify additions by choosing $d$ wisely, in a way that simplifies one of the operands. For example, to add $494 + 337$, we can add $d=6$ to $494$ and $d=(-6)$ to $337$; this turns 494 into 500, and simplifies the sum: \begin{align} 494 + 337&= (494 + 6) + (337+(-6))\\ &= (494 + 6) + (337-6)\\ &= 500 + 331\\ &= 831 \end{align} Likewise, \begin{align} 503 + 215&= (503 +(-3)) + (215+3)\\ &= (503 – 3) + (215+3)\\ &= 500 + 218\\ &= 718 \end{align} We are moving a portion of one of the operands to the other. We add the numbers in each column; if the sum of the digits in the column to the right is greater than ten, then add one more. For example: \begin{align} &\,\,\,\,\,\,\,785,007\\ &\underline{+\,\,262,015}\\ &\,\,\,10\,\,\,\,\,\,\,\,\,\hspace{.48in}⇐ (7+2) + 1 \mbox{ (add 1 because }8+6\ge 10\mbox{)}\\ &\,\,\,\,\,\,\,\,\,4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\hspace{.2in}⇐ 8 + 6 = 14 \mbox{ (discard the 1)}\\ &\,\,\,\,\,\,\,\,\,\,\,\,\,7\,\,\,\,\,\,\,\,\,\,\,\hspace{.2in}⇐ 5 + 2 = 7 \mbox{ (don’t add 1 because }0 + 0\lt 10\mbox{)}\\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\hspace{.2in}⇐ 0 + 0 = 0 \mbox{ (don’t add 1 because }0 + 1\lt 10\mbox{)}\\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\,\,\,\hspace{.2in}⇐ (0 + 1)+1 = 2 \mbox{ (add 1 because }7+5\ge 10\mbox{)}\\ &\underline{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2}\hspace{.15in}⇐ 7 + 5 = 12 \mbox{ (discard the 1)}\\ &\,1,047,022\\ \end{align} The result of the sum of each column has to fit in the number of places of the result. All the results are constrained to one digit, except for the left-most result which can use two. That is why we discard the 1’s of every column sum except for the left-most column, if any. Try to add from left to right the numbers that we added in the previous section: \begin{align} &\,\,\,\,\,\,\,494\\ &\underline{+\,\,337} \end{align} and \begin{align} &\,\,\,\,\,\,\,503\\ &\underline{+\,\,215} \end{align} If it went well, try to add \begin{align} &\,\,\,\,\,\,\,934,532,520\\ &\underline{+\,\,740,629,157} \end{align}
Equilibrium in 2D Video Lessons Concept # Problem: A bag of cement weighing Fg = 325 N hangs in equilibrium from three wires as shown in the figure. Two of the wires make angles θ1 = 60.0° and θ2 = 40.0° with the horizontal. Assuming the system is in equilibrium, find the tensions T1, T2, and T3 in the wires. ###### FREE Expert Solution We're asked for the tension in each of the wires supporting the bag of cement. This is a 2D Equilibrium type of problem with multiple forces in different angles. As usual for equilibrium problems, we'll follow these steps: 1. Draw a free body diagram for each point of interest 2. Set up our equilibrium equations 3. Solve for the target. In two dimensions, we use equilibrium equations in component form: $\overline{){\mathbf{\sum }}{{\mathbit{F}}}_{{\mathbit{x}}}{\mathbf{=}}{\mathbf{0}}\phantom{\rule{0ex}{0ex}}{\mathbf{\sum }}{{\mathbit{F}}}_{{\mathbit{y}}}{\mathbf{=}}{\mathbf{0}}}$ Anytime we're working with forces that aren't at 90° angles to each other, we'll also need some of the equations to convert magnitude-angle notation to components. Magnitude: $\overline{)\mathbf{\|}\stackrel{\mathbf{⇀}}{\mathbit{F}}\mathbf{\|}{\mathbf{=}}\sqrt{{{\mathbit{F}}_{\mathbit{x}}}^{\mathbf{2}}\mathbf{+}{{\mathbit{F}}_{\mathbit{y}}}^{\mathbf{2}}}}$ Angle: Components of a force: 84% (242 ratings) ###### Problem Details A bag of cement weighing Fg = 325 N hangs in equilibrium from three wires as shown in the figure. Two of the wires make angles θ1 = 60.0° and θ2 = 40.0° with the horizontal. Assuming the system is in equilibrium, find the tensions T1, T2, and T3 in the wires.
# 2005 USAMO Problems/Problem 5 ## Problem 5 Let $n$ be an integer greater than 1. Suppose $2n$ points are given in the plane, no three of which are collinear. Suppose $n$ of the given $2n$ points are colored blue and the other $n$ colored red. A line in the plane is called a balancing line if it passes through one blue and one red point and, for each side of the line, the number of blue points on that side is equal to the number of red points on the same side. Prove that there exist at least two balancing lines. ## Solutions Consider the convex hull of the the $2n$ points, or the points that would form the largest convex polygon. If the points in the convex hull contain both red and blue points, then there must be at least 2 edges of the graph of the convex hull such that the edge connects a blue and a red point. Drawing a line through those points would give a balancing line, as we have n-1 blue points and n-1 red points on one side, and 0 points on the other. Therefore it suffices to show that there exist at least 2 balancing lines when the convex hull is colored all the same color. Pick a random point on the convex hull, and without loss of generality we can say it is blue (if there are no red we can change all the colors, and end up with an equivalent setup). Consider a line going through it and not any other points. As we rotate the line clockwise, we encounter the red points in some order. Let the ith point encountered be $R_i$. Let $b_i$ and $r_i$ be the number of points encountered before $R_i$. Then $r_i=i-1$. Define a sequence $f(i)=b_i-r_i$. Then $f(1)=b_i-(1-1)=b_i\geq0$ $f(n)=b_n-(n-1)=b_n-n+1\leq0$, because we can only encounter up to n-1 blue points. Thus, $f(i)$ goes from negative to positive as $i$ goes from 1 to $n$. We can also see that $f(i)$ can only increase by one for each change in i, so we know $f(i)$ must be 0 for some value of $i$, and so there is a balancing line for every point on the convex hull. Since a polygon must have at least 3 vertices, there must be at least $3>2$ balancing lines for the set of points when the convex hull is all the same color, and the statement is true as we desired.
# 2015 AIME I Problems/Problem 4 (diff) ← Older revision \| Latest revision (diff) \| Newer revision → (diff) ## Problem Point $B$ lies on line segment $\overline{AC}$ with $AB=16$ and $BC=4$. Points $D$ and $E$ lie on the same side of line $AC$ forming equilateral triangles $\triangle ABD$ and $\triangle BCE$. Let $M$ be the midpoint of $\overline{AE}$, and $N$ be the midpoint of $\overline{CD}$. The area of $\triangle BMN$ is $x$. Find $x^2$. ## Diagram $[asy] pair A = (0, 0), B = (16, 0), C = (20, 0), D = (8, 8sqrt(3)), EE = (18, 2sqrt(3)), M = (9, sqrt(3)), NN = (14, 4*sqrt(3)); draw(A--B--D--cycle); draw(B--C--EE--cycle); draw(A--EE); draw(C--D); draw(B--M--NN--cycle); dot(A); dot(B); dot(C); dot(D); dot(EE); dot(M); dot(NN); label("A", A, SW); label("B", B, S); label("C", C, SE); label("D", D, N); label("E", EE, N); label("M", M, NW); label("N", NN, NE); [/asy]$ Diagram by RedFireTruck (talk) 18:52, 15 February 2021 (EST) ## Solution Let point $A$ be at $(0,0)$. Then, $B$ is at $(16,0)$, and $C$ is at $(20,0)$. Due to symmetry, it is allowed to assume $D$ and $E$ are in quadrant 1. By equilateral triangle calculations, Point $D$ is at $(8,8\sqrt{3})$, and Point $E$ is at $(18,2\sqrt{3})$. By Midpoint Formula, $M$ is at $(9,\sqrt{3})$, and $N$ is at $(14,4\sqrt{3})$. The distance formula shows that $BM=BN=MN=2\sqrt{13}$. Therefore, by equilateral triangle area formula, $x=13\sqrt{3}$, so $x^2$ is $\boxed{507}$. ## Solution 2 Use the same coordinates as above for all points. Then use the Shoelace Formula/Method on triangle $BMN$ to solve for its area. ## Solution 3 Note that $AB=DB=16$ and $BE=BC=4$. Also, $\angle ABE = \angle DBC = 120^{\circ}$. Thus, $\triangle ABE \cong \triangle DBC$ by SAS. From this, it is clear that a $60^{\circ}$ rotation about $B$ will map $\triangle ABE$ to $\triangle DBC$. This rotation also maps $M$ to $N$. Thus, $BM=BN$ and $\angle MBN=60^{\circ}$. Thus, $\triangle BMN$ is equilateral. Using the Law of Cosines on $\triangle ABE$, $$AE^2 = 16^2 + 4^2 - 2\cdot 16\cdot 4\cdot\left(-\frac{1}{2}\right)$$ $$AE = 4\sqrt{21}$$ Thus, $AM=ME=2\sqrt{21}$. Using Stewart's Theorem on $\triangle ABE$, $$AM\cdot ME\cdot AE + AE\cdot BM^2 = BE^2\cdot AM + BA^2\cdot ME$$ $$BM = 2\sqrt{13}$$ Calculating the area of $\triangle BMN$, $$[BMN] = \frac{\sqrt{3}}{4} BM^2$$ $$[BMN] = 13\sqrt{3}$$ Thus, $x=13\sqrt{3}$, so $x^2 = 507$. Our final answer is $\boxed{507}$. Admittedly, this is much more tedious than the coordinate solutions. I also noticed that there are two more ways of showing that $\triangle BMN$ is equilateral: One way is to show that $\triangle ADB$, $\triangle BMN$, and $\triangle ECB$ are related by a spiral similarity centered at $B$. The other way is to use the Mean Geometry Theorem. Note that $\triangle BCE$ and $\triangle BDA$ are similar and have the same orientation. Note that $B$ is the weighted average of $B$ and $B$, $M$ is the weighted average of $E$ and $A$, and $N$ is the weighted average of $C$ and $D$. The weights are the same for all three averages. (The weights are actually just $\frac{1}{2}$ and $\frac{1}{2}$, so these are also unweighted averages.) Thus, by the Mean Geometry Theorem, $\triangle BMN$ is similar to both $\triangle BAD$ and $\triangle BEC$, which means that $\triangle BMN$ is equilateral.
# What Is 8/29 as a Decimal + Solution With Free Steps The fraction 8/29 as a decimal is equal to 0.275. A decimal number can be either repetitive or non-repetitive. Repeating numbers in decimals means that they will be having the same number forever, hence non-terminating. Non-repetitive numbers have non-repeating decimals after each place. Here, we are more interested in the division types that result in a Decimal value, as this can be expressed as a Fraction. We see fractions as a way of showing two numbers having the operation of Division between them that result in a value that lies between two Integers. Now, we introduce the method used to solve said fraction to decimal conversion, called Long Division, which we will discuss in detail moving forward. So, let’s go through the Solution of fraction 8/29. ## Solution First, we convert the fraction components, i.e., the numerator and the denominator, and transform them into the division constituents, i.e., the Dividend and the Divisor, respectively. This can be done as follows: Dividend = 8 Divisor = 29 Now, we introduce the most important quantity in our division process: the Quotient. The value represents the Solution to our division and can be expressed as having the following relationship with the Division constituents: Quotient = Dividend $\div$ Divisor = 8 $\div$ 29 This is when we go through the Long Division solution to our problem. Given is the Long Division process in Figure 1: Figure 1 ## 8/29 Long Division Method We start solving a problem using the Long Division Method by first taking apart the division’s components and comparing them. As we have 8 and 29, we can see how 8 is Smaller than 29, and to solve this division, we require that 8 be Bigger than 29. This is done by multiplying the dividend by 10 and checking whether it is bigger than the divisor or not. If so, we calculate the Multiple of the divisor closest to the dividend and subtract it from the Dividend. This produces the Remainder, which we then use as the dividend later. Now, we begin solving for our dividend 8, which after getting multiplied by 10 becomes 80. We take this 80 and divide it by 29; this can be done as follows: 80 $\div$ 29 $\approx$ 2 Where: 29 x 2 = 58 This will lead to the generation of a Remainder equal to 80 – 58 = 22. Now this means we have to repeat the process by Converting the 22 into 220 and solving for that: 220 $\div$ 29 $\approx$ 7 Where: 29 x 7 = 203 This, therefore, produces another Remainder which is equal to 220 – 203 = 17. Now we must solve this problem to Third Decimal Place for accuracy, so we repeat the process with dividend 170. 170 $\div$ 29 $\approx$ 5 Where: 29 x 5 = 145 Finally, we have a Quotient generated after combining the three pieces of it as 0.275, with a Remainder equal to 25. Images/mathematical drawings are created with GeoGebra.
# How do you solve the equation abs(2(1/3+1/2x))=1? Nov 26, 2017 $x = \frac{1}{3} \text{ or } x = - \frac{5}{3}$ #### Explanation: $\text{distribute the factor}$ rArr\|(2(1/3+1/2x)\|=\|2/3+x\| $\text{the value inside the absolute value function can be}$ $\textcolor{b l u e}{\text{positive or negative}}$ $\textcolor{b l u e}{\text{first solution}}$ $\frac{2}{3} + x = 1 \Rightarrow x = 1 - \frac{2}{3} = \frac{1}{3}$ $\textcolor{b l u e}{\text{second solution}}$ $- \left(\frac{2}{3} + x\right) = 1$ $\Rightarrow - \frac{2}{3} - x = 1$ $\Rightarrow - x = 1 + \frac{2}{3} = \frac{5}{3} \Rightarrow x = - \frac{5}{3}$ $\textcolor{b l u e}{\text{As a check}}$ x=1/3" $\Rightarrow \| \frac{2}{3} + \frac{1}{3} \| = \| 1 \| = 1$ $x = - \frac{5}{3}$ $\Rightarrow \| \frac{2}{3} - \frac{5}{3} \| = \| - 1 \| = 1$ Nov 26, 2017 Use the piecewise definition of the absolute value function to separate the equation into two equations and then solve each equation. #### Explanation: The piecewise definition of the absolute value function is: \|f(x)\| = {(f(x); f(x) >= 0),(-f(x); f(x) < 0):} In this case $f \left(x\right) = 2 \left(\frac{1}{3} + \frac{1}{2} x\right)$ Substitute into the definition: \|2(1/3+1/2x)\| = {(2(1/3+1/2x); 2(1/3+1/2x) >= 0),(-2(1/3+1/2x); 2(1/3+1/2x) < 0):} Simplify the domain restrictions: \|2(1/3+1/2x)\| = {(2(1/3+1/2x); x >= -2/3),(-2(1/3+1/2x); x < -2/3):} Separate the given equation into two equations with its respective domain restriction: 2(1/3+1/2x) = 1; x >=-2/3 and -2(1/3+1/2x) = 1; x < -2/3 Multiply the second equation by -1: 2(1/3+1/2x) = 1; x >=-2/3 and 2(1/3+1/2x) = -1; x < -2/3 Distribute the two in both equations: 2/3+x = 1; x >=-2/3 and 2/3+x = -1; x < -2/3 Subtract $\frac{2}{3}$ from both sides of both equations: x = 1/3; x >=-2/3 and x = -5/3; x < -2/3 The domain restrictions can be dropped, because neither equation violates them: $x = \frac{1}{3}$ and $x = - \frac{5}{3}$
Span, Subspaces, and Reduction by Justin Skycak on The span of a set of vectors consists of all vectors that can be made by adding multiples of vectors in the set. We can often reduce a set of vectors to a simpler set with the same span. This post is part of the book Justin Math: Linear Algebra. Suggested citation: Skycak, J. (2019). Span, Subspaces, and Reduction. In Justin Math: Linear Algebra. https://justinmath.com/span-subspaces-and-reduction/ The span of a set of vectors consists of all vectors that can be made by adding multiples of vectors in the set. For example, the span of the set $\left\lbrace \left< 1,0 \right>, \left< 0,1 \right> \right\rbrace$ is just the entirety of the 2-dimensional plane: any vector $\left< x,y \right>$ in this plane can be made by adding $x \left< 1,0 \right> + y \left< 0,1 \right>$. For instance, $\left< 5,-2 \right>$ can be written as $5\left< 1,0 \right> - 2 \left< 0,1 \right>$. Similarly, the span of the set $\left\lbrace \left< 1,1 \right>, \left< -1,1 \right> \right\rbrace$ is also the entirety of the 2-dimensional plane: any vector $\left< x,y \right>$ in this plane can be made by adding $\left( \frac{x+y}{2} \right) \left< 1,1 \right> + \left( \frac{x-y}{2} \right) \left< 1,-1 \right>$. This is a little less obvious, but it’s true: for example, $\left< 3,-7 \right>$ can be written as $-2 \left< 1,1 \right> + 5 \left< 1,-1 \right>$. The span of the set $\left\lbrace \left< 1,1 \right>, \left< 2,2 \right> \right\rbrace$, however, is just a 1-dimensional line within the 2-dimensional plane: it contains only vectors of the form $k\left< 1,1 \right>$, where $k$ is a constant. To see why this is, observe what happens when we try to add multiples of the vectors: \begin{align} &a \left< 1,1 \right> + b \left< 2,2 \right> \\ &= a \left< 1,1 \right> + 2b \left< 1,1 \right> \\ &= (a+2b) \left< 1,1 \right> \\ &= k \left< 1,1 \right> \end{align} Subspaces of Two-Dimensional Space The span of $\left\lbrace \left< 1,1 \right>, \left< 2,2 \right> \right\rbrace$ is a 1-dimensional line within the 2-dimensional plane. So, we say that the span forms a 1-dimensional subspace of the 2-dimensional plane. In 2 dimensions, it turns out that any set of two vectors that are multiples of one another will span a line, and any set of two vectors that are NOT multiples of one another will span the entire space. For example, the set $\left\lbrace \left< 2,-3 \right>, \left< -4,6 \right> \right\rbrace$ spans a line because $\left< -4,6 \right> = -2 \left< 2,-3 \right>$. On the other hand, the set $\left\lbrace \left< 7,4 \right>, \left< 1,-2 \right> \right\rbrace$ spans the entire space because the vectors cannot be written as multiples of each other. But just because two vectors are multiples, doesn’t mean they can’t be included in a set that spans the space. For example, the set $\left\lbrace \left< 1,1 \right>, \left< 2,2 \right> \right\rbrace$ spans only a line, but if we add include a third vector $\left< 1,2 \right>$ that is not a multiple of the original two vectors, then the set $\left\lbrace \left< 1,1 \right>, \left< 2,2\right>, \left< 1,2 \right> \right\rbrace$ spans the entire plane. Subspaces of N-Dimensional Space Now, let’s generalize these ideas to N dimensions. It might be tempting to think that in general, a set of vectors will span the entire space provided there are some three vectors that aren’t multiples of one another. But this isn’t always true. For example, consider the set $\left\lbrace \left< 1,0,0 \right>, \left< 0,1,0 \right>, \left< 1,1,0 \right> \right\rbrace$. None of these vectors are multiples of each other, but there is no way to combine the vectors to reach a point whose third component is not zero. The issue here is that the third vector is the sum of the first two vectors. As a result, the third vector is redundant – we can already reach any point using the first two vectors, that we can reach using the third vector. The set, then, has the same span as the set of just the first two vectors. It covers just a plane, a 2-dimensional subspace of 3-dimensional space. The vectors span the plane $x_3=0$. No matter how we combine vectors, the third component will always be $0$ – so we cannot reach any points above or below the plane, by adding multiples of vectors in the set. Independence In general, the dimension of the span of a set of vectors is equal to the number of independent vectors that remain after we remove the dependent vectors. A vector is said to be dependent if it can be written as a sum of multiples of other vectors in the set. The labeling of vectors as independent or dependent depends on the order in which the vectors are considered, but regardless of order, removing all dependent vectors will leave the same number of independent vectors, even if the independent vectors themselves are different for different orders. For example, in the set $\left\lbrace \left< 1,0,0 \right>, \left< 0,1,0 \right>, \left< 1,1,0 \right> \right\rbrace$ we can start by looking at the first vector, $\left< 1,0,0 \right>$. This vector is dependent since it can be produced by subtracting the other two vectors: $\left< 1,0,0 \right> = \left< 1,1,0 \right> - \left< 0,1,0 \right>$. Removing this vector from the set yields the reduced set $\left\lbrace \left< 0,1,0 \right>, \left< 1,1,0 \right> \right\rbrace$, which contains two independent vectors and thus cannot be reduced any further. Since the fully reduced set has two independent vectors, it spans a 2-dimensional plane, and since the original set has the same span as the reduced set, the original set also spans the same 2-dimensional plane. Alternatively, beginning with the original set $\left\lbrace \left< 1,0,0 \right>, \left< 0,1,0 \right>, \left< 1,1,0 \right> \right\rbrace$ we can start by looking at the second vector, $\left< 0,1,0 \right>$. This vector is dependent since it can be produced by subtracting the other two vectors: $\left< 0,1,0 \right> = \left< 1,1,0 \right> - \left< 1,0,0 \right>$. Removing this vector from the set yields the reduced set $\left\lbrace \left< 1,0,0 \right>, \left< 1,1,0 \right> \right\rbrace$, which contains two independent vectors and thus cannot be reduced any further. Again, since the fully reduced set has two independent vectors, it spans a 2-dimensional plane, and since the original set has the same span as the reduced set, the original set also spans the same 2-dimensional plane. The last alternative, beginning with the original set $\left\lbrace \left< 1,0,0 \right>, \left< 0,1,0 \right>, \left< 1,1,0 \right> \right\rbrace$, is to start by looking at the third vector, $\left< 1,1,0 \right>$. This vector is dependent since it can be produced by adding the other two vectors: $\left< 1,1,0 \right> = \left< 0,1,0 \right> + \left< 1,1,0 \right>$. Removing this vector from the set yields the reduced set $\left\lbrace \left< 1,0,0 \right>, \left< 0,1,0 \right> \right\rbrace$, which contains two independent vectors and thus cannot be reduced any further. Again, since the fully reduced set has two independent vectors, it spans a 2-dimensional plane, and since the original set has the same span as the reduced set, the original set also spans the same 2-dimensional plane. Maximum Number of Independent Vectors Since the number of independent vectors in a set tells us the dimension of the span of that set, we can make the general conclusion that a set of N-dimensional vectors can have at most N independent vectors. The N-dimensional vectors reside in N-dimensional space, so the largest space they can possibly span is their full N-dimensional space. Consequently, it’s not possible for the set of vectors to contain more than N independent vectors – otherwise, they would need to span a space of more than N dimensions. For example, consider the following set of vectors: \begin{align} \left\lbrace \left< 1,1 \right>, \left< 1,-2 \right>, \left< -3,1 \right>, \left< 2,3 \right> \right\rbrace \end{align} Looking at the first two vectors $\left< 1,1 \right>$ and $\left< 1,-2 \right>$, we see that these two vectors are independent since they are not multiples of each other. As a result, the span of the vectors must have a dimension of at least 2. But the vectors reside in 2-dimensional space, so their span is limited to at most 2 dimensions. Thus, we can conclude that the set of vectors spans exactly 2 dimensions, and that the third and fourth vectors in the set must be dependent, without even needing to check whether they can be written as sums of multiples of the first two vectors. Now, consider the following set of vectors: \begin{align} \left\lbrace \begin{matrix} \left< 1,-1,1,2 \right> \\ \left< 2,0,4,2 \right> \\ \left< -1,3,3,2 \right> \\ \left< 3,-4,1,4 \right> \end{matrix} \right\rbrace \end{align} We can tell the first two vectors $\left< 1,-1,1,2 \right>$ and $\left< 2,0,4,2 \right>$ are independent since they aren’t multiples of each other, but it’s harder to see whether the remaining two vectors $\left< -1,3,3,2 \right>$ and $\left< 3,-4,1,4 \right>$ are independent because we also have to make sure they can’t be written as sums of multiples of other vectors in the set. Reduction To make it easier for us to tell whether these vectors are independent, we can reduce the set of vectors to a simpler set with the same span, by adding multiples of vectors from each other. To begin the process of reduction, we can add multiples of the first vector to the other vectors so that we eliminate the first component from each of the other vectors. • • The second vector has a first component of $2$, so we can eliminate it by adding $-2$ times the first vector. • • The third vector has a first component of $-1$, so we can eliminate it by adding $1$ times the first vector. • • The fourth vector has a first component of $3$, so we can eliminate it by adding $-3$ times the first vector. \begin{align} \left\lbrace \begin{matrix} \left< 1,-1,1,2 \right> \\ \left< 2,0,4,2 \right> - 2\left< 1,-1,1,2 \right> \\ \left< -1,3,3,2 \right>+1\left< 1,-1,1,2 \right> \\ \left< 3,-4,1,4 \right>-3\left< 1,-1,1,2 \right> \end{matrix} \right\rbrace \end{align} The resulting set of vectors is shown below. \begin{align} \left\lbrace \begin{matrix} \left< 1,-1,1,2 \right> \\ \left< 0,2,2,-2\right> \\ \left< 0,2,4,4 \right> \\ \left< 0,-1,-2,-2 \right> \end{matrix} \right\rbrace \end{align} Since we only added multiples of vectors, we haven’t changed the span at all. But now all of the first components are zero EXCEPT for the first component in the first vector, so we can see that the first vector cannot be written as a sum of other vectors in the set. All the other vectors have zero in their first component, so every time we add multiples of them, the result will still have zero in the first component. To check whether the second vector is independent, we can add multiples of the second vector to the remaining vectors to eliminate their second components. But to make this easier, we can start by rescaling (i.e. multiplying) the second vector to have a second component of $1$. Its second component is currently $2$, so to convert $2$ to $1$, we need to multiply by $\frac{1}{2}$. \begin{align} \left\lbrace \begin{matrix} \left< 1,-1,1,2 \right> \\ \frac{1}{2} \left< 0,2,2,-2\right> \\ \left< 0,2,4,4 \right> \\ \left< 0,-1,-2,-2 \right> \end{matrix} \right\rbrace \end{align} \begin{align} \left\lbrace \begin{matrix} \left< 1,-1,1,2 \right> \\ \left< 0,1,1,-1 \right> \\ \left< 0,2,4,4 \right> \\ \left< 0,-1,-2,-2 \right> \end{matrix} \right\rbrace \end{align} Now, we can add multiples of the second vector to the third and fourth vectors so that we eliminate their second components. • • The third vector has a second component of $2$, so we can eliminate it by adding $-2$ times the second vector. • • The fourth vector has a second component of $-1$, so we can eliminate it by adding $1$ times the second vector. \begin{align} \left\lbrace \begin{matrix} \left< 1,-1,1,2 \right> \\ \left< 0,1,1,-1 \right> \\ \left< 0,2,4,4 \right> - 2\left< 0,1,1,-1 \right> \\ \left< 0,-1,-2,-2 \right> + 1\left< 0,1,1,-1 \right> \end{matrix} \right\rbrace \end{align} \begin{align} \left\lbrace \begin{matrix} \left< 1,-1,1,2 \right> \\ \left< 0,1,1,-1 \right> \\ \left< 0,0,2,6 \right> \\ \left< 0,0,-1,-3 \right> \end{matrix} \right\rbrace \end{align} Clearly, the second vector cannot be written as a sum of multiples including the first vector, since including the first vector would cause the first component to become nonzero. And the second vector cannot be written as a sum of multiples of the third and fourth vectors, either, because no combination of them can produce a nonzero second component. So the second vector must be independent. To determine whether the third and fourth vectors are independent, we can repeat the usual process once more. First, we’ll rescale the third vector by $\frac{1}{2}$ so that its first component is $1$. \begin{align} \left\lbrace \begin{matrix} \left< 1,-1,1,2 \right> \\ \left< 0,1,1,-1 \right> \\ \frac{1}{2}\left< 0,0,2,6 \right> \\ \left< 0,0,-1,-3 \right> \end{matrix} \right\rbrace \end{align} The result is shown below. \begin{align} \left\lbrace \begin{matrix} \left< 1,-1,1,2 \right> \\ \left< 0,1,1,-1 \right> \\ \left< 0,0,1,3 \right> \\ \left< 0,0,-1,-3 \right> \end{matrix} \right\rbrace \end{align} The fourth vector has a third component of $-1$, so we can eliminate it by adding $1$ times the third vector. \begin{align} \left\lbrace \begin{matrix} \left< 1,-1,1,2 \right> \\ \left< 0,1,1,-1 \right> \\ \left< 0,0,1,3 \right> \\ \left< 0,0,-1,-3 \right>+1\left< 0,0,1,3 \right> \end{matrix} \right\rbrace \end{align} Our final result is shown below. \begin{align} \left\lbrace \begin{matrix} \left< 1,-1,1,2 \right> \\ \left< 0,1,1,-1 \right> \\ \left< 0,0,1,3 \right> \\ \left< 0,0, 0,0 \right>\end{matrix} \right\rbrace \end{align} We see that the first three vectors are independent, whereas the fourth vector is dependent since it is a multiple of every vector in the set (you can multiply any other vector by $0$ to obtain the fourth vector). As a result, our set spans a 3-dimensional subspace of 4-dimensional space. Exercises Tell the dimension of the span of the set of vectors. (You can view the solution by clicking on the problem.) \begin{align} 1) \hspace{.5cm} \left\lbrace \left< 1,1 \right>, \left< 2,0 \right> \right\rbrace \end{align} Solution: \begin{align} 2 \end{align} \begin{align} 2) \hspace{.5cm} \left\lbrace \left< 1,1 \right>, \left< 2,2 \right> \right\rbrace \end{align} Solution: \begin{align} 1 \end{align} \begin{align} 3) \hspace{.5cm} \left\lbrace \left< 1,1 \right>, \left< 2,0 \right>, \left< 3,2 \right> \right\rbrace \end{align} Solution: \begin{align} 2 \end{align} \begin{align} 4) \hspace{.5cm} \left\lbrace \left< 1,2,3 \right>, \left< 3,2,1 \right> \right\rbrace \end{align} Solution: \begin{align} 2 \end{align} \begin{align} 5) \hspace{.5cm} \left\lbrace \begin{matrix} \left< 1,2,3 \right> \\ \left< 3,2,1 \right> \\ \left< 1,0,0 \right> \end{matrix} \right\rbrace \end{align} Solution: \begin{align} 3 \end{align} \begin{align} 6) \hspace{.5cm} \left\lbrace \begin{matrix} \left< 1,2,3 \right> \\ \left< 3,2,1 \right> \\ \left< 1,1,1 \right> \end{matrix} \right\rbrace \end{align} Solution: \begin{align} 2 \end{align} \begin{align} 7) \hspace{.5cm} \left\lbrace \begin{matrix} \left< 1,1,1 \right> \\ \left< -1,-1,-1 \right> \\ \left< 2,2,2 \right> \end{matrix} \right\rbrace \end{align} Solution: \begin{align} 1 \end{align} \begin{align} 8) \hspace{.5cm} \left\lbrace \begin{matrix} \left< 1,0,-1 \right> \\ \left< 1,0,1 \right> \\ \left< 0,1,1 \right> \end{matrix} \right\rbrace \end{align} Solution: \begin{align} 3 \end{align} \begin{align} 9) \hspace{.5cm} \left\lbrace \begin{matrix} \left< 1,1,0 \right> \\ \left< 0,1,1 \right> \\ \left< 1,0,1 \right> \\ \left< 0,1,0 \right> \end{matrix} \right\rbrace \end{align} Solution: \begin{align} 3 \end{align} \begin{align} 10) \hspace{.5cm} \left\lbrace \begin{matrix} \left< 1,2,1,0 \right> \\ \left< 4,3,3,1 \right> \\ \left< 3,4,3,3 \right> \\ \left< 4,0,0,0 \right> \end{matrix} \right\rbrace \end{align} Solution: \begin{align} 4 \end{align} \begin{align} 11) \hspace{.5cm} \left\lbrace \begin{matrix} \left< 1,2,3,4,5 \right> \\ \left< 1,2,1,2,1 \right> \\ \left< -1,2,-1,2,-1 \right> \\ \left< 1,0,1,0,1 \right> \end{matrix} \right\rbrace \end{align} Solution: \begin{align} 3 \end{align} \begin{align} 12) \hspace{.5cm} \left\lbrace \begin{matrix} \left< 0,1,2,3,4 \right> \\ \left< 1,2,3,4,5 \right> \\ \left< 2,3,4,5,6 \right> \\ \left< 3,4,5,6,7 \right> \\ \left< 4,5,6,7,8 \right> \\ \left< 5,6,7,8,9 \right> \end{matrix} \right\rbrace \end{align} Solution: \begin{align} 2 \end{align} \begin{align} 13) \hspace{.5cm} \left\lbrace \begin{matrix} \left< 1,-1,2,-2,3 \right> \\ \left< 1,1,0,0,0 \right> \\ \left< 0,1,1,0,0 \right> \\ \left< 0,0,1,1,0 \right> \\ \left< 0,0,0,1,1 \right> \\ \left< 0,1,0,1,0 \right> \\ \left< 1,0,1,0,1 \right> \end{matrix} \right\rbrace \end{align} Solution: \begin{align} 5 \end{align} \begin{align} 14) \hspace{.5cm} \left\lbrace \begin{matrix} \left< 1,3,3,1,0 \right> \\ \left< 1,4,6,4,1 \right> \\ \left< 0,1,3,3,1 \right> \\ \left< 1,2,1,2,1 \right> \end{matrix} \right\rbrace \end{align} Solution: \begin{align} 3 \end{align} This post is part of the book Justin Math: Linear Algebra. Suggested citation: Skycak, J. (2019). Span, Subspaces, and Reduction. In Justin Math: Linear Algebra. https://justinmath.com/span-subspaces-and-reduction/ Tags:
R Functions Related to Binomial Distributions To find the binomial probability of seeing exactly $x$ successes in $n$ trials, where the probability of success on any one trial is $p$. (i.e., ${}_n C_k p^x q^{n-x}$ where $q = 1-p$), we can use the following R function: dbinom Usage dbinom(x, size = n, prob = p) Example To find the probability that one flips exactly 4 heads in 8 tosses of a fair coin, one could use the following: > dbinom(4, size = 8, prob = 1/2) [1] 0.2734375 If one needs to find the probability that the number of successes (with probability $p$) seen in $n$ trials is $x$ or less, one can use the following cumulative probability distribution function: pbinom Usage pbinom(x, size = n, prob = p) Examples Suppose there are $12$ multiple choice questions in an English class quiz. Each question has five possible answers, and only one of them is correct. One way to find the probability of having four or less correct answers if a student attempts to answer every question at random, would be to do the following: > dbinom(0, size=12, prob=0.2) + + dbinom(1, size=12, prob=1/5) + + dbinom(2, size=12, prob=1/5) + + dbinom(3, size=12, prob=1/5) + + dbinom(4, size=12, prob=1/5) [1] 0.9274445 However, it will be far quicker to use the cumulative probability function for the binomial distribution: > pbinom(4, size=12, prob=1/5) [1] 0.9274445 If we instead wanted the probability of the student getting somewhere between $4$ and $8$ (inclusive) questions correct, we can use a difference of two cumulative probabilities, as the below illustrates: > pbinom(8, size=12, prob=1/5) - pbinom(3, size=12, prob=1/5) [1] 0.2053689 R also has a function that lets you simulate the outcome of a random variable $X$ that follows a binomial distribution: rbinom Usage rbinom(x, size = n, prob = p) Note: unlike the previous functions, here $x$ represents the number of realizations of your random variable you wish to produce. Examples The two examples below independently simulate $12$ trials where the probability of success is $1/5$ and return the number of successes seen. Note, there is a random element to this function, so it can (and does) return different values when you run it at different times. > rbinom(1,size=12,prob=1/5) [1] 2 > rbinom(1,size=12,prob=1/5) [1] 4 If you want to run this experiment several times, you just alter the first parameter to the function. Below, we run the $12$ trials $6$ times, returning how many successes are seen each time. > rbinom(6,size=12,prob=1/5) [1] 3 4 5 2 2 2
dark # Lesson 7 Homework 5.5 Answer Key Algebra can be a tricky subject, but it’s essential to succeed in higher-level math and sciences. Understanding algebra concepts is crucial, and practicing the problems will help you gain mastery. As a student, it’s understandable to experience difficulties with homework, but don’t worry. In this blog, we’ll share the answer key for Lesson 7 Homework 5.5, which will help you with your algebra practice. ### Blog Body: Algebra often involves dealing with radicals, which can appear complex at first glance. The first problem in Lesson 7 Homework 5.5 focuses on simplifying radical expressions. To tackle this, ensure you understand the rules for combining and simplifying radicals. The answer key serves as your guide, demonstrating how to subtract and simplify radicals, paying close attention to the index and like terms. By following these steps, you can simplify radical expressions efficiently. #### 2. Multiplying and Dividing Radical Expressions: The second problem delves into the multiplication and division of radical expressions. This task requires a firm grasp of the order of operations (PEMDAS) and the rules for handling radicals. The answer key illustrates the step-by-step process of simplifying expressions by multiplying and dividing like terms while maintaining the integrity of the radicals. Learning from this key will empower you to confidently navigate these algebraic challenges. Moving forward, the third problem involves adding and subtracting fractions with radicals. This skill is crucial for mastering algebraic manipulations. The answer key breaks down the process, emphasizing the importance of a common denominator and providing clear instructions on simplifying numerators and denominators before combining. Following these guidelines will enhance your ability to handle fractions containing radicals effectively. #### 4. Operations with Rational Expressions: Rational expressions can appear daunting due to their combination of variables, fractions, and integers. The fourth problem requires factoring and canceling common factors to simplify expressions. The answer key showcases the necessary steps to factor, cancel, and simplify, all while avoiding common errors. Utilize this key to strengthen your skills in operating with rational expressions. #### 5. Simplifying Complex Fractions: The final problem introduces complex fractions, a concept that might appear intricate at first glance. The answer key guides you through the process of simplifying complex fractions step by step. By tackling the numerator and denominator separately and applying fraction simplification techniques, you can conquer this challenge with confidence. ### Conclusion: Algebra can be challenging at first, but with practice, you’ll gain mastery. The answer key for Lesson 7 Homework 5.5 will help you simplify radical expressions, multiply, divide, add and subtract fractions, operate with rational expressions, and simplify complex fractions. Remember that persistence and practice are essential in mastering algebra concepts. Keep on practicing, and soon, you’ll ace your algebra exam! ## Lesson 7 Homework Practice Equivalent Expressions Answer Key ##### Related Posts Introduction Enter the exciting world of online gambling with VBlink Casino APK Download, a virtual casino platform that…
Courses Courses for Kids Free study material Offline Centres More Store # A weight $\omega \;$ is suspended from the midpoint of a rope, whose ends are at the same level. In order to make the rope perfectly horizontal, the force applied to each of its ends must be(A) Less than $\omega \;$(B) Equal to $\omega \;$(C) Equal to $2\omega \;$(D) Infinitely large Last updated date: 03rd Aug 2024 Total views: 66.9k Views today: 1.66k Verified 66.9k+ views Hint: We draw the free body diagram of the following system to solve this question. We find the sum of forces acting upwards and downwards and equate them since there is horizontal force. Both the horizontal components of the two sides of the string get cancelled. So, the weight acting downwards is equal to the sum of upward forces. The angle between the string and the x-axis should be zero for the body to be horizontal. Substituting this condition in the final equation we get the answer. Complete step by step solution: The free body diagram of the system is The weight is at the centre of the string so the tension is distributed equally on both sides of the string. The tension force on both the sides of the string is resolved into $T = Tcos\theta + T\sin \theta$ As shown in the diagram The horizontal components are equal and opposite in direction so they get cancelled. $Tcos\theta - Tcos\theta = 0$ Since the body is at equilibrium the sum of upwards forces is equal to downwards force (weight of the body) Forces acting upwards are the vertical components of the tension force $T\sin \theta + T\sin \theta = 2T\sin \theta$ Weight of the body is $\omega \$ Upward force is equal to downwards force $\omega \ = 2T\sin \theta$ For the rope to be perfectly horizontal $\theta$ should be $0^\circ$ Substituting $\theta = 0^\circ$ $\omega \ = 2T\sin \theta$ $\omega \ = 2T\sin 0^\circ$ $T = \dfrac{{\omega \ }}{{2\sin 0^\circ }} = \dfrac{{\omega \ }}{0}$ Anything by zero is infinite Hence the force applied on each end should be infinitely large. Option (D) infinitely large is the correct answer. Note: The angle $\theta$ is also equal to the angle between the tension force and the Horizontal which is the reason we can resolve the tension force with angle $\theta$ Students might go wrong by taking tensions in the two sides of the string with different values. The tension remains the same on both the sides because the weight is being suspended from the center.
#### Need Help? Get in touch with us # Use Strategies to Add and Subtract Decimal Sep 27, 2022 ## Key Concepts • Addition and subtraction of decimals by lining up the decimals • Adding decimals using properties of numbers and also Number Line • Subtracting decimals using number line and also using partial difference method ### Introduction In this chapter, we will learn about adding and subtracting decimals just like adding and subtracting whole numbers and using commutative and associative properties of numbers, partial addition, and subtraction, and also using number lines. ## Addition of Decimals Using strategies Decimals can be added like the way we add whole numbers. There are different ways of adding decimals. 1. Lining them using place values 2. Using properties of numbers (Commutative property and Associative property) 3. Using number line 4. Using partial sums ### Let us learn one by one #### Lining them using place value In this method, we follow the following steps: Write down the numbers one under the other, with the decimal points lined up Add zeros to the right of the number so that the number to be added are of same digits For example, Step 1: Line up the decimals Step 2: Add zeros to the right side of the number if needed ### Using properties of numbers The two properties of numbers are: Commutative property of addition means that you can switch the order of any of the numbers in an addition, the answer remains the same. For example: Sum of 4.2 + 3.5 = 7.7 By changing the order of the addends, 3.5 + 4.2 = 7.7 That is, 4.2 + 3.5 = 3.5 + 4.2 = 7.7 • Associative property of addition means that you can change the groupings of numbers being added and it does not change the result. For example, (2.3 + 4.6) + 7.4 = 2.3 + (4.6 + 7.4) Let us check LHS = (2.3 + 4.6) + 7.4 = 14.3 RHS = 2.3 + (4.6 + 7.4) = 14.3 ⸫ (2.3 + 4.6) + 7.4 = 2.3 + (4.6 + 7.4) = 14.3 #### Using number line To add decimals using number line, labelling the number line with decimals is very important. We know how to label a number line using whole numbers. The number line with decimals will look like this Here, we started labelling the number line with zero and increase by 0.1. We can label the number line increased by 0.25 or 0.5 or 0.75 etc. We can also start the number line with the numbers given in the question. For example: We can start labelling the number line starting with 3.6 and increase by 0.1 Then start adding 0.8 to 3.6, we get 4.4 ⸫ 3.6 + 0.8 = 4.4 #### Using Partial sums break the numbers up in the individual places and add. For example: ⸫ 4.65 + 2.76 = 7.41 ### Subtraction of Decimals Using strategies Decimals can be subtracted just like the subtraction of whole numbers. The different ways of subtracting decimals are: 1. Lining them using place values 2. Using number line 3. Using partial differences ### Let us learn one by one #### Lining them using place values To subtract decimals by lining them using place values, we follow the following steps: Step 1: Line up the decimal points in a column. When needed add a zero to the left of the number to match the number of digits. Step 2: Start on the right, and subtract each column in turn. Remember, we are subtracting digits in the same place value position. Step 3: If the digit you are subtracting is bigger than the digit you are subtracting from, you have to borrow a group of ten from the column to the left. For example: Subtract 4.65 – 2. 49 ### Using number line To subtract decimal numbers using a number line. Start on the far-right side on the number line and label it backwards by tenths. This is nothing but counting back. For example: Subtract 4.8 – 0.9 Draw the number line labelling backward starting from 4.8 Then count backwards by tenths, 9 times ⸫ 4.8 – 0.9 = 3.9 #### Using Partial differences Using partial differences helps you to subtract numbers that are difficult to subtract in one step in your head. The steps to be followed while subtracting numbers using partial differences are as follows: • When subtracting using partial differences, we write the numbers one below the other and start subtracting from left to right • Then we start subtracting the whole number part by place values, then subtract the tenths digit and hundredths digit vice versa. • If the number to be subtracted is greater than the number to be subtracted from, then swap the numbers in head, subtract the smaller one from the bigger one and put the negative sign in the answer. • When writing these results in the answer column subtract the negative numbers from the number with positive sign. For example: • Subtract 43.85 – 21.63 Now, subtracting by place values 40 – 20 = 20 3 – 1 = 0.8 – 0.6 = 0.2 0.05 – 0.03 = 0.02 22.22 • Subtract 54.85 – 31.56 Now, subtracting by place values 50 – 30 = 20 4 – 1 = 3 0.8 – 0.5 = 0.3 0.05 – 0.06 = – 0.01 (⸪0.06 > 0.05) Since there is a negative number, subtract that from the positive number • 0.3 – 0.01 è 0.30 – 0.01 = 0.29 ⸫ 54.85 – 31.56 = 20 + 3 + 0.29 = 23.29 ## Exercise 1. Add 6. 5+ 3.3 using number line 2. Casey runs 9.5 miles, 13.2 miles then 11.5 miles the first week and 11.5 miles, 13.2 miles and 9.5 miles the next week. Which property is represented by Casey’s two weeks of running? 3. Name the property illustrated below. 3.2+(a+5.6)=(3.2+a)+5.6 4. Add 15.67 + 11.74 using partial sums 5. Subtract 9.8 – 7.6 using number line 6. Ms. Gracie is an electrician and has a length of wire that is 54.7m long . She has another length of wire that is 16.7 m long . How much longer is one wire than the other. Use any method to solve this problem 7. Subtract 25.32 – 13.26 using partial differences. ### What have we learned • Adding Decimals using Number line, Lining up the decimal point, Partial sums and using properties of numbers. • Subtracting decimals using Number line, lining up the decimals and partial differences #### Addition and Multiplication Using Counters & Bar-Diagrams Introduction: We can find the solution to the word problem by solving it. Here, in this topic, we can use 3 methods to find the solution. 1. Add using counters 2. Use factors to get the product 3. Write equations to find the unknown. Addition Equation: 8+8+8 =? Multiplication equation: 3×8=? Example 1: Andrew has […] #### Dilation: Definitions, Characteristics, and Similarities Understanding Dilation A dilation is a transformation that produces an image that is of the same shape and different sizes. Dilation that creates a larger image is called enlargement. Describing Dilation Dilation of Scale Factor 2 The following figure undergoes a dilation with a scale factor of 2 giving an image A’ (2, 4), B’ […] #### How to Write and Interpret Numerical Expressions? Write numerical expressions What is the Meaning of Numerical Expression? A numerical expression is a combination of numbers and integers using basic operations such as addition, subtraction, multiplication, or division. The word PEMDAS stands for: P → Parentheses E → Exponents M → Multiplication D → Division A → Addition S → Subtraction Some examples […]
## How Do You Calculate Present Value in an Annuity? The example in the text book on present value asks how much would you need to deposit today to end up with \$2,000,000? The example on the homework is easier. ## What is the Difference Between Future Value and Present Value? In many investment problems, you are given an amount of money and asked what will it accumulate to in a certain amount of time at some interest rate. Essentially, these problems are asking you to find the future value of the amount of money. Depending on how that money accumulates, you might use one of several different formulas. If the problem specifies simple interest, you would use where an amount PV accumulates to a future value FV in t years at a rate of r. If the problem specifies compound interest, you would use where i is the interest rate per period and n is the number of periods. It is also possible for a problem to specify continuous compounding in which case you would use where r is the continuous rate and t is the number of years. For a future value problem, the quantities on the right side of each of these equations will be specified so that you can calculate the future value FV. In a present value problem, you will be given the amount in the future FV and asked to find the amount you would start with to get to that amount. Let’s look at two possible problems. Problem What would need to be deposited today to reach a value 0f 100,000 dollars in 10 years at a rate of 3% per year compounded annually? Solution The first thing we need to realize is which formula we need to use. Since it says the interest is compounded annually, we need In the compound interest formula we are given that the future value is 100,000 dollars, the interest rate per period is i = 0.03, and the money will accumulate over 10 periods. Putting these values into the compound interest formula yields Solving this equation for PV yields This means that the present value of 100,000 dollars is approximately 74,409.39 dollars when compounded annually for 10 years at an annual rate of 3%. Now let’s change the problem up and see how this affects the solution process. Problem What is the present value of 100,000 dollars compounded continuously at a rate of 3% per year? Solution On the surface, this is almost the same problem as the one above except that it specifies continuous interest. This means we need to start from instead of the compound interest formula. Putting in the values in the problem yields Isolating the present value PV gives In each of these problems, we need to solve for the present value PV but we start from a different formula that depends on the how interest is earned. ## How Do You Solve for the Number of Years in the Compound Interest Formula? In another MathFAQ, I looked at how you can find the rate in the compound interest formula. Now let’s look at an example where we solve for the number of years n. This problem is different because what we are looking for appears in a power. Problem Suppose \$5000 is deposited in an account that earns 2% compound interest that is done annually. In how many years will there be \$6000 in the account. Solution This problem requires the use of the compound interest formula, This formula applies when interest is earned on an annual basis and the interest is earned once a year. Let’s look at the quantities in the problem statement: • \$5000 is deposited in an account > P = 5000 • that earns 2% compound interest that is done annually > r = 0.02 • Will there be \$6000 in the account > A = 6000 Putting these values into the formula above gives us Unlike other problems where we solve for P or r, here we need to solve for the power in the right hand side, n. Solving for a value in the power requires the property of logarithms, log(yx) = x logy. It allows us to move the n in the power and change it to a multiplier. But before we can apply this property, we isolate the factor containing the n: Now take the logarithm of both sides of the equation: This gives us or n ≈ 9.21 years. In WolframAlpha, we could evaluate the logs as follows. ## How Do You Solve For The Rate In The Compound Interest Formula? Problems that ask you to solve for the rate r in the compound interest formula require the use of roots or creative use of exponents. Let’s look at an example. Problem Suppose 5000 dollars is deposited in an account that earns compound interest that is done annually. If there is 7000 dollars in the account after 2 years, what is the annual interest rate? Solution The easiest way to approach this problem is to use the compound interest formula, This formula applies when interest is earned on an annual basis and the interest is earned once a year. Let’s look at the quantities in the problem statement: • 5000 dollars is deposited in an account > P = 5000 • If there is 7000 dollars in the account after 2 years > A = 7000 and n = 2 Putting these values into the formula above gives us We need to find the annual interest rate r. Since the r is hidden in the parentheses, we start by isolating the parentheses. To get at the r, we need to remove the square on the parentheses. Using a calculator to do the square root, we get r ≈ 0.183 or 18.3%. Although most calculators have a square root key, when removing powers it is often useful to raise both sides to a power. For instance, we could remove the square by raising both sides to the ½ power. When you raise a power to another power, you multiply the exponents 2 ∙ ½ = 1. The right side simply becomes 1 + r. Now we can solve for r: Using the power key on your calculator gives the same answer as before. Make sure the 1/2 power is entirely in the power. You can make sure this happens using parentheses: (7000/5000)^(1/2)-1. Now what if the interest is earned over six years instead of two years? Instead of a square on the parentheses we now have a sixth power. To solve for r in this equation, we follow similar steps. The root can be computed on a graphing calculator using the MATH button or put into WolframAlpha: Either method gives r ≈ 0.577 or 5.77%. Notice that the annual interest is lower when it is earned over a longer period of time. If we use a 1/6 power to solve for r, we would carry out the steps below: Using a 1/6 power on your calculator gives the same answer as above. ## How Do You Find Compound Interest Future Value In Google Sheets? Spreadsheets have several built in functions for working with compound interest and annuities. To use these functions, we’ll start with a standard sheet. This worksheet contains the variables used throughout Chapter 8. These variables correspond to these letter used in the text. • Number of periods is n • Annual interest rate is r • Payment is R • Present value is P • Future value is A • Periods per year is m Values given in a problem will be entered in column B. Values calculated by the spreadsheet will be entered in column C. We will also assume that amounts paid out are negative and amounts received are positive.
# Rhombus – Definition with Examples A rhombus is an intriguing geometric form that appears in art, architecture, and the natural world, among other places. This article will teach us how to compute the area and perimeter of a rhombus as well as examine its characteristics. Additionally, we’ll provide you some practice questions to help you solidify your understanding of this adaptable structure. Now let’s explore the fascinating world of rhombuses and learn about their special qualities with Brighterly! ## What is a Rhombus? One kind of quadrilateral, or anything with four sides, is a rhombus. The equal length of each of a rhombus’s four sides is what distinguishes it. It is sometimes referred to as an equal quadrilateral. The square is another equilateral quadrilateral that you might be familiar with. Not all rhombuses are squares, even if all squares are rhombuses. Other characteristics of a square, such as all angles being equal, are not always true of a rhombus. ## What Does a Rhombus Look Like? A rhombus resembles a diamond or an oblique square. Imagine taking a square and pushing one set of opposite corners towards each other, causing the square to slant. As a result, the rhombus will have two sets of parallel sides. Visit Brighterly’s Worksheets to learn more about rhombuses and other quadrilaterals. ## Properties of a Rhombus There are several important properties of a rhombus that distinguish it from other quadrilaterals: 1. All sides are equal in length: This is the defining characteristic of a rhombus. 2. Opposite sides are parallel: This means that if you were to extend the sides indefinitely, they would never intersect. 3. Opposite angles are equal: The angles formed by two adjacent sides are called the interior angles of the rhombus. In a rhombus, the angles across from each other will have the same measure. 4. Diagonals bisect each other at right angles: The diagonals are the lines connecting opposite vertices (corners) of the rhombus. In a rhombus, the diagonals will intersect at a 90-degree angle and divide each other into two equal parts. ### Properties Of The Rectangle Rhombus And Square Worksheet Answers PDF Properties Of The Rectangle Rhombus And Square Worksheet Answers ### Properties Of Rectangles Rhombuses And Squares Worksheet Answers PDF Properties Of Rectangles Rhombuses And Squares Worksheet Answers The rhombus plays a significant role in the curriculum of math for children, and Brighterly has some of the most interactive and enjoyable worksheets to help kids learn about this essential concept. ## Area of a Rhombus To calculate the area of a rhombus, we need to know the lengths of its diagonals. The formula for finding the area is: Area = (diagonal1 × diagonal2) / 2 To calculate the area, simply multiply the lengths of the two diagonals and divide the result by 2. You can practice calculating the area of a rhombus using Brighterly’s Rhombus Area Calculator. ## Perimeter of Rhombus Calculating the perimeter of a rhombus is quite simple. Since all the sides are equal in length, you can multiply the length of one side by 4: Perimeter = side × 4 ## Practice Questions on Rhombus 1. Identify a rhombus in a group of quadrilaterals. 2. Calculate the area of a rhombus with diagonals of 12 cm and 16 cm. 3. Find the perimeter of a rhombus with a side length of 5 cm. Properties Of Rectangles Rhombuses And Squares Worksheet Properties Of Rhombuses Rectangles And Squares Worksheet ## Conclusion A rhombus is a fascinating and versatile shape that can be found in many aspects of our daily lives. By understanding its properties and learning how to calculate its area and perimeter, you’ll have a deeper appreciation for this unique quadrilateral. We hope this article has helped you gain a better understanding of rhombuses, their properties, and how to work with them in various mathematical problems. Remember to practice your new skills and reinforce your understanding through exercises and real-world applications. ### Is a square a rhombus? Yes, a square is a special type of rhombus. All squares are rhombuses because they have all four sides equal in length. However, not all rhombuses are squares, as squares have additional properties such as all angles being equal (90 degrees). ### How is a rhombus different from a parallelogram? Both rhombuses and parallelograms have opposite sides parallel, but a rhombus has all four sides equal in length, while a parallelogram only requires that opposite sides are equal in length. ### Can a rhombus have right angles? Yes, a rhombus can have right angles. In fact, if a rhombus has all its angles equal to 90 degrees, it is a square. ### How do you find the side length of a rhombus if you know the diagonals? If you know the diagonals of a rhombus, you can use the Pythagorean theorem to find the side length. Since the diagonals of a rhombus bisect each other at right angles, they form four congruent right triangles. Using half the length of each diagonal as the legs of the right triangle, you can find the side length (the hypotenuse) using the formula: Side length = √((diagonal1/2)^2 + (diagonal2/2)^2) Information Sources
Courses Courses for Kids Free study material Offline Centres More Store # A pole is painted red and white. The red portion is $1.8m$ and the white portion is three times as long as the red portion. How long is the pole?A). $5.4m$B). $7.2m$C). $3.6m$D). None of these Last updated date: 14th Jul 2024 Total views: 346.5k Views today: 5.46k Verified 346.5k+ views Hint: Type of questions are basically based on linear equations in one variable. For the type of problems according to concept we have to assume the unknown value to some variable, then relate it with the given value according to the condition given in the question, through where we will get the relation in unknown and known value and by solving we will get the unknown value. Complete step-by-step solution: As in our case we have a red and white portion, as the red portion length is $1.8m$and white portion is three times that of red. So we can say that if we multiply three with the length of the red portion we will get the value equal to that of white portion, from here we will get the relation between unknown and known value and by further solving we will get the white portion length. So moving ahead with the question we have; Length of red portion given $=1.8m$ Let the length of white portion be $=x$(which is variable) So according to the question we have white portion three times that of red, so in order to equate both if we multiply the red portion with three then we will have the red portion equal to that of red. So we can write it as; white portion$=$$3\times$red portion \begin{align} & white\text{ }portion=3\times red\text{ }portion \\ & x=3\times 1.8 \\ &\Rightarrow x=5.4 \\ \end{align} Hence white portion is $5.4m$. Now we find the length of pole which will be = Red portion+white painted portion=$1.8+5.4=7.2$ Hence the pole is 7.2m long, so the option B is the correct option. Note: For the type of question we just had to find any relation between the values, on further solving we will get the answer. Moreover, as we had multiplied by three in the red portion to equate, we can also divide white portion by three to equate it to the red portion; that will also be right.
# Tag Archives: Daniel Luevanos ## Solving Equations In Two Variables Directions: Using the digits 1 to 9 at most one time each, fill in the boxes so x = y. Source: Arnav Gulati and Daniel Luevanos ## Create a System of Two Equations Directions: Using the digits 1 to 30, at most one time each, fill in the boxes to create a system of two linear equations where (3, 2) is the solution to the system. Source: Daniel Luevanos ## Negative Exponents – Closest to Zero Directions: Using the digits 0 to 9, at most one time each, fill in the boxes to make a result that is as close to zero as possible. Source: Daniel Luevanos ## Negative Exponents Directions: Using the digits 0 to 9, at most one time each, place a digit in each box to make a true statement. Source: Daniel Luevanos ## Linear Inequalities in Two Variables Directions: Create 5 ordered pairs using the whole digits 0 – 9 exactly one time each. Then, create a linear inequality such that: 1. Two of the ordered pairs are solutions to the linear inequality. 2. Two of the ordered pairs are not solutions to the linear inequality. 3. One of the ordered pairs is on the boundary line but … ## Multi-Step Equations – Smallest (or Largest) Solution Directions: Using the digits 1 to 9, at most one time each, create an equation where x has the smallest (or greatest) possible value. Source: Daniel Luevanos ## Multi-Step Equations – Positive (or Negative) Solution Directions: Using the digits 1 to 9, at most one time each, create an equation where x has a positive (or negative) value. Source: Daniel Luevanos ## Converting Fractions to Repeating Decimals Directions: Using the digits 0 to 9, at most one time each, fill in the boxes so that the fraction equals the repeating decimal. Source: Daniel Luevanos
# Exploring the Marvels of Algebra: Formulas, Calculations, and Real-Life Applications Algebra is a powerful mathematical tool that enables us to solve a wide range of problems and understand the relationships between variables. In this article, we embark on a journey through the realm of algebra, exploring its formulas, calculations, and real-life applications that bring its concepts to life. At the heart of algebra lie fundamental formulas that help us solve equations, simplify expressions, and unravel the mysteries of unknown quantities. Let’s delve into some essential algebraic formulas and their practical significance: The quadratic formula is a powerful tool for finding the solutions of quadratic equations. For a quadratic equation in the form ax^2 + bx + c = 0, the quadratic formula states that x = (-b ± √(b^2 – 4ac))/(2a). This formula allows us to calculate the roots of quadratic equations, which find applications in various fields like physics, engineering, and finance. ##### Binomial Expansion The binomial expansion formula enables us to expand expressions of the form (a + b)^n, where a and b are constants and n is a positive integer. The formula states that (a + b)^n = C(n, 0)a^n b^0 + C(n, 1)a^(n-1) b^1 + … + C(n, n-1)a^1 b^(n-1) + C(n, n)a^0 b^n. This formula is particularly useful in probability theory, statistics, and calculus. ##### Distance Formula The distance formula helps us calculate the distance between two points in a coordinate plane. For two points (x1, y1) and (x2, y2), the distance formula is given by d = √((x2 – x1)^2 + (y2 – y1)^2). This formula finds applications in geometry, navigation, and physics, allowing us to measure distances accurately. ##### Simplifying Expressions: Algebra provides various techniques for simplifying expressions, such as combining like terms, using the distributive property, and factoring. These techniques help us simplify complex expressions into more manageable forms, making calculations and problem-solving more efficient. ##### Pythagorean Theorem: The Pythagorean theorem, a^2 + b^2 = c^2, forms the basis of right triangle geometry. It helps us find the length of a side in a right triangle when the lengths of the other two sides are known. This theorem has widespread applications in fields such as architecture, engineering, and navigation. ##### Laws of Exponents: The laws of exponents, such as multiplying exponents with the same base (a^m * a^n = a^(m+n)) and raising a power to a power (a^m^n = a^(mn)), are indispensable in simplifying and manipulating exponential expressions. These laws play a crucial role in areas like finance, population growth, and scientific calculations. ### Real-Life Applications: 1. Financial Planning: Algebra is essential in financial planning, helping individuals and businesses make informed decisions regarding investments, loans, and budgeting. By using algebraic concepts like compound interest, present value, and future value, we can calculate returns, plan for retirement, and analyze investment opportunities. 2. Engineering and Physics: In engineering and physics, algebra plays a pivotal role in solving complex equations that govern principles like Newton’s laws of motion, electrical circuits, and fluid dynamics. By applying algebraic techniques, engineers and physicists can design structures, optimize systems, and predict physical phenomena. 3. Data Analysis and Statistics: Algebraic concepts are foundational to data analysis and statistics. Formulas for calculating mean, median, and standard deviation, as well as linear regression equations, allow us to analyze and interpret data, make predictions, and draw meaningful conclusions. ### Math Calculation Example Let’s consider a practical example involving financial planning. Suppose you want to determine the future value of an investment. Using the formula FV = PV(1 + r)^n, where FV is the future value, PV is the present value, r is the interest rate, and n is the number of periods, you can calculate the future value of your investment accurately. ##### Example 1 For instance, if you invest \$10,000 at an annual interest rate of 5% for 5 years, the future value can be calculated as FV = \$10,000(1 + 0.05)^5 = \$12,762.82. ##### Example 2 Math Calculation Example: Let’s apply the quadratic formula to solve a quadratic equation. Consider the equation 2x^2 + 5x – 3 = 0. Using the quadratic formula, we have: x = (-5 ± √(5^2 – 42(-3)))/(22) = (-5 ± √(25 + 24))/4 = (-5 ± √49)/4 = (-5 ± 7)/4 So, the solutions to the quadratic equation are x = (-5 + 7)/4 = 1/2 and x = (-5 – 7)/4 = -3. ##### Example 3 Let’s consider an example of calculating compound interest using the formula A = P(1 + r/n)^(nt), where A is the final amount, P is the principal, r is the interest rate, n is the number of times interest is compounded per year, and t is the number of years. Suppose you invest \$10,000 at an annual interest rate of 5% compounded annually for 3 years. Applying the formula, we have A = \$10,000(1 + 0.05/1)^(13) = \$11,576.25. ##### Question: 5x + 2 = 2x + 17 Answer: 5x + 2 – 2x = 2x + 17 – 2x 3x + 2 = 17 3x + 2 – 2 = 17 – 2 3x = 15 x = 5 Solve the quadratic equation 2x^2 – 5x + 2 = 0. Solution: Using the quadratic formula, we can find the roots: x = (-(-5) ± √((-5)^2 – 4(2)(2)))/(2(2)) = (5 ± √(25 – 16))/(4) = (5 ± √(9))/(4) = (5 ± 3)/(4) Therefore, the solutions are x = 2/4 = 0.5 and x = 8/4 = 2. ##### Pythagorean Theorem Find the length of the hypotenuse in a right triangle with sides measuring 3 and 4 units. Solution: Applying the Pythagorean theorem, we have: c^2 = 3^2 + 4^2 = 9 + 16 = 25 Taking the square root of both sides, we get: c = √(25) = 5 Therefore, the length of the hypotenuse is 5 units. ##### Slope-Intercept Form Find the equation of a line with a slope of 2 and a y-intercept of -3. Solution: Using the slope-intercept form, we have: y = 2x – 3 Therefore, the equation of the line is y = 2x – 3. Formula: ax^2 + bx + c = 0 Example: Solve the equation 2x^2 + 5x – 3 = 0. Solution: Using the quadratic formula x = (-b ± √(b^2 – 4ac)) / 2a, we can calculate the values of x: x = (-5 ± √(5^2 – 42(-3))) / (22) Simplifying the equation gives us x = (-5 ± √(25 + 24)) / 4 x = (-5 ± √49) / 4 So the solutions are x = (-5 + 7) / 4 = 2/4 = 1/2 and x = (-5 – 7) / 4 = -12/4 = -3. ##### Algebraic Formula: Pythagorean Theorem Formula: a^2 + b^2 = c^2 Example: Find the length of the hypotenuse in a right triangle with side lengths 3 and 4 units. Solution: Using the Pythagorean theorem, we have 3^2 + 4^2 = c^2. Simplifying the equation gives us 9 + 16 = c^2. Hence, c^2 = 25, and taking the square root, we find c = 5. ##### Algebraic Formula: Linear Equation Formula: y = mx + b Example: Given a line with slope m = 2 and y-intercept b = 3, find the value of y when x = 4. Solution: Substituting the values into the equation y = 2x + 3, we have y = 2(4) + 3 = 8 + 3 = 11. Algebra, with its wealth of formulas and calculations, empowers us to solve complex problems, make predictions, and understand the world around us. By mastering algebraic concepts and applying them to real-life situations, we unlock a powerful tool that opens doors to countless opportunities. So let us embark on this algebraic journey, where equations become solvable, expressions simplify, and mathematical principles shape our understanding of the universe. Photo credit: papazachariasa via Pixabay Math Formulas and Examples for Everyday Use
What is Parallel Traces in Math? What is Parallel Lines in Math? It really is a very difficult query and there are different techniques you possibly can use to answer it. By far the most general procedure would be the equation of https://www.domyhomeworkfor.me/ the line, with and while not directions. You can then simplify this equation by finding a straight line. But this is not the sole technique and there is an extra distinct tactic you could use. This 2nd methodology employs triangle identities. To clear up an issue of parallel lines, to start with determine if the line is parallel to another straight line. Then find out precisely what is the fourth energy belonging to the angle concerning the parallel lines. The fourth strength on the angle is generally identified by dividing the angle through the length with the parallel strains. The fourth potential on the angle could very well be uncovered as follows. https://engineering.purdue.edu/EPICS Multiply the size of your traces along with the angle, then divide the quotient by two. It may well be published as follows: angle = -2 x length squared In what the heck is perpendicular traces in math, you require to look for the fourth power. If the angle is called two-thirds of the circle, the square root of two is a single. It is similar to applying the Pythagorean theorem. The triangle need to be sq. in shape and should possess a duration of 3 times the length belonging to the hypotenuse. In the event the hypotenuse is referred to as the radius, then the sq. in the length on the triangle is just one. It’s always easier to determine the size. It is actually 3 times the radius. The triangle can be described as 3 sided figure with a few sides and it’s a certain angle within the center within the a few sides. The triangle defines two angles, with both sides experiencing two sides. It’s two frequent angles. These angles are named proper angles. There are other ways to solve parallel lines trouble. A person is termed the most suitable Triangle Option. In the Parallel Traces in Math, the method really is a issue of three. Quite simply, you can get 3 sides. The edges should always have equivalent lengths additionally, the hypotenuse could be equivalent towards length on the facet using a best suited angle. The final methodology is referred to as the Steeper Suitable Triangle Way. When you concentrate on an issue of perpendicular traces, the third way is the most appropriate and is to simplify the equation of the straight line with and not having instructions. Solving the Equation of a Parallel Traces in Math is fairly very easy. There are some dissimilarities while using the way made use of for your other approaches. The most crucial variance may be the proven fact that you’ll have to find the fourth strength. From the procedure employed for triangle and quadratic features, this phase may be dealt with immediately. On the solution employed for parallel strains, you will have to discover the fourth electric power. The last methodology is recognized as the Steeper Suitable Triangle Process. This site uses Akismet to reduce spam. Learn how your comment data is processed.
# FINDING THE nth TERM IN A SEQUENCE ## Formula's for the nth term ALGEBRA Here I intend to look at what is known as the nth term sequence. In doing so we will look at some examples and find the nth term which is known in mathematical terms by a "u" followed by a subscript(small) "n". So what exactly is the connection( the pattern) between the "n" and the "un". If we take a closer look at the un we can see that it goes up in two's (2's). (Position) U 1 2 3 4 5 ( Term) Un 3 5 7 9 11 n x 2 2 4 6 8 10 3 + 2 = 5 5 + 2 = 7 7 + 2 = 9 9 + 2 = 11 Therefore we can write down that un = 2n, in other words we are doubling the number. Looking again we may need to adjust that 2n slightly; 2n means 2 x n therefore, 2 x 1 = 2 2 x 2 = 4 2 x 3 = 6 2 x 4 = 8 2 x 5 = 10 These are the numbers we have got after multiplyingn x 2, you will have noticed that they are 1 less in each case than the numbers in the unbracket, therefore we need to add on 1 in the formula. Our formula now looks like this un = 2n + 1 (where the un is the nth term) 2 x n = ? + 1 = un 2 x 1 = 2 + 1 = 3 2 x 2 = 4 + 1 = 5 2 x 3 = 6 + 1 = 7 2 x 4 = 8 + 1 = 9 2 x 5 = 10 + 1 = 11 Answer:- un = 2n + 1 This is the type of question for the nth term, could be asked in an exam paper; Find the nth term for the sequence; 2, 6, 10, 14, 18 ... (etc)? To begin with we need to change this sequence into a table! The "n" is the position and the “un” is the term. The position simply represents the number of digits in a sequence, in the above there are "5" [ 2(which is 1), 6(which is 2), 10(which is 3), 14(which is 4), 18( finally this is number 5 )] (Position) U 1 2 3 4 5 ( Term) Un 2 6 10 14 18 n x 4 4 8 12 16 20 So what exactly is the connection (the pattern ) between the "n" and the "un", in this question? If we take a closer look at the "un" we can see that it is going up in four's (4's). 2 + 4 = 6 6 + 4 = 10 10 + 4 = 14 14 + 4 = 18 Therefore we can write down that un = 4n (which means 4 x n) 4 x 1 = 4 4 x 2 = 8 4 x 3 = 12 4 x 4 = 16 4 x 5 = 20 These are the numbers we have got after multiplying n x 4, you will have noticed that there are 2 more in each case than the numbers in the un term, therefore we need to subtract 2 in the formula. Our formula now looks like this; un = 4n - 2 (where the unis the nth term) 4 x n = ? - 2 = un 4 x 1 = 4 – 2 = 2 4 x 2 = 8 – 2 = 6 4 x 3 = 12– 2 = 10 4 x 4 = 16– 2 = 14 4 x 5 = 20– 2 = 18 Answer:- un = 4n - 2 In this example we need to find the nth term for this sequence; Find the nth term for the sequence; 3, 6, 11, 18, 27 ... (etc)? To begin with we need to change this sequence into a table! This example is quite similar to the one above!!!! LOOK CAREFULLY:- The "n" is the position and the “un“ is the term Again there are five numbers and they are holding the first five POSITIONS The position simply represents the number of digits in a sequence, in the above there are "5" [ 3(which is 1), 6(which is 2), 11(which is 3), 18(which is 4), 27( finally this is number 5 )] (Position) U 1 2 3 4 5 ( Term) Un 3 6 11 18 27 n2=(n x n) 1 4 9 16 25 So what exactly is the connection (the pattern ) between the "n" and the "un", in this question? Is there a set pattern? Are they going up in 2's or 4's? --- NO they are not! 3 + 3 = 6 6 + 5 = 11 11 + 7 = 18 18 + 9 = 27 In a case like this we would square (n2) the numbers and see what that brings us too. Therefore we can write down that un = n2 (which means n x n) 1 x 1 = 1 2 x 2 = 4 3 x 3 = 9 4 x 4 = 16 5 x 5 = 25 These are the numbers we have got after multiplying n x n, you will have noticed that they are 2 less in each case than the numbers in the un term, therefore we need to add 2 in the formula. Our formula now looks like this; un = n2 + 2 ( where the un is the nth term) n x n = ? + 2 = un 1 x 1 = 1 + 2 = 3 2 x 2 = 4 + 2 = 6 3 x 3 = 9 + 2 = 11 4 x 4 = 16 + 2 = 18 5 x 5 = 25 + 2 = 27 Answer:- un = n2 + 2 Another favourite question in exams is as follows; The nth term of a sequence is represented by this formula:- un = 3n + 2 (i) Find the first 4 terms (ii) To find the 49th term ( i ) Find the first 4 terms Our formula is:- 3n + 2 = un Therefore our 1st term is u1 = 3 x 1 = 3 + 2 = 5 2nd term u2 = 3 x 2 = 6 + 2 = 8 3rd term u3 = 3 x 3 = 9 + 2 = 11 4th term u4 = 3 x 4 = 12 + 2 = 14 Our first four terms are; u1 , u2 , u3 , and u4 They end up as 5, 8, 11, and 14 \ First 4 terms are: 5, 8, 11, 14. (Answer:-) ( ii ) We have been asked to find the 49th term. 49th term: This will be known as u49 As before in part ( i ) the formula is; un = 3n + 2 Therefore we will replace the “n” with the 49 u= 3n + 2 u49 = (3 x 49) = 147 + 2 = 149 4 0 0 2 ## Popular 25 2 • ### The Number 9 \| The Secret Knowledge of The Ancients Number Nine Code 911 111 0 of 8192 characters used • Bubble 7 years ago what would be the nth term for 3,4,6,9...? • empeen 7 years ago Really good one,very neatly explained with relevent examples Same way you can postAP.,g.P. and H.P. also. I like it. • megan 7 years ago what's the first five sequences of this nth term, 3n + 4? • tish. 7 years ago 7, 10, 13, 16, 19 :) • Oli 7 years ago What about the nth term for 1,2,4,8,15,26,42... It goes up in 1,2,4,7,11,16 and that goes up in 1,2,3,4,5 please post back quickly, all I can work out is how to find what it goes up in innitially the formulae for that is as follows: Xn= n(n-1)/2+1 can't find how to make the actual original sequence. • amber 7 years ago what is nth term for squared numbers • dana 7 years ago whaat is the nth term of 4 6 8 10 • ,.jnmkl.jh 7 years ago What is the nth term of this sequnce 5 10 15 20 25 • ronak 7 years ago what will be the nth term for 1, 12, 24, 36, 48 • eivlys 7 years ago n2/n+1 • Raz 7 years ago find a formula for 95, 90, 85,80 interms of n • dog 7 years ago gd website what will be the nth term in this sequence 2,3,6,7,10,11...... • Jk 7 years ago Help!!! What is the nth term in the sequence 107,68,29,-10,-49???pease help me!!! • bbSTAR 7 years ago tn=a(r^(n-1)) where: tn = general term a = first term r = common ratio n = position of the term • shwetha 7 years ago if Tn=5-3n then find Tn+1 and Tn+2 • Author t.elia 7 years ago from Northern Ireland whats the first five sequences of this nth term, 3n + 4? 3n + 4 3 x 1 = 3 + 4 = 7 3 x 2 = 6 + 4 = 10 3 x 3 = 9 + 4 = 13 3 x 4 = 12 + 4 = 16 3 x 5 = 15 + 4 = 19 • Author t.elia 7 years ago from Northern Ireland Thankyou Empeen just trying to help. • Author t.elia 7 years ago from Northern Ireland Shweta, if Tn=5-3n then find Tn+1 and Tn+2 Tn=5-3n then T(n + 1) = 5 - 3(n + 1) = 5 - 3n - 3 = -3n + 2 So Tn + 1 = -3n + 2 (Ans) And Tn +2 = 5 - 3(n + 2) = 5 - 3n - 6 = - 3n - 1 So Tn + 2 = - 3n - 1 (Ans) • Author t.elia 7 years ago from Northern Ireland JK, What is the nth term in the sequence 107,68,29,-10,-49? Tn = a + (n - 1) d a = 107 d = 68 - 107 = -39 Tn = 107 + (n - 1) x -39 ...= 107 - 39n + 39 ...= 146 - 39n (Ans) • Author t.elia 7 years ago from Northern Ireland Ronak, what will be the nth term for 1, 12, 24, 36, 48 a = 1 d = 12 Tn = 1 + (n - 1) x 12 ....= 1 + 12n - 12 ....= 12n - 11 ......12n - 11 is the nth term • Author t.elia 7 years ago from Northern Ireland Eivlys find the 9th term n2/n+1 Tn = n2 / n + 1 ....= 9squared / 9 + 1 ....= 81 / 10 ....= 8.1 (Ans) • Author t.elia 7 years ago from Northern Ireland Raz, find a formula for 95, 90, 85,80 in terms of n Tn = a + (n - 1) x d ..a = 95 ..d = 90 - 95 = -5 Tn = 95 + (n - 1) x -5 ....= 95 - 5n + 5 ....= 100 - 5n (Ans) • Author t.elia 7 years ago from Northern Ireland ,.jnmkl.jh (Position) U 1 2 3 4 5 ( Term) Un 5 10 15 20 25 ..n x 5 Therefore we can write down that un = 5n U1 = 5 = 5 x 1 U2 = 10 = 5 x 2 U3 = 15 = 5 x 3 U4 = 20 = 5 x 4 U5 = 25 = 5 x 5 un = n x 5 = 5n (Ans) un = 5n • Author t.elia 7 years ago from Northern Ireland dana, (Position) U 1 2 3 4 ( Term) Un 4 6 8 10 n x 2 4 + 2 = 6 6 + 2 = 8 8 + 2 = 10 Therefore we can write down that un = 2n Looking again we may need to adjust that 2n slightly; 2 x 1 = 2 2 x 2 = 4 2 x 3 = 6 2 x 4 = 8 These are the numbers we have got after multiplying n x 2, you will have noticed that they are 2 less in each case than the numbers in the un bracket, therefore we need to add on 2 in the formula. Our formula now looks like this un = 2n + 2 (where the un is the nth term) 2 x n = ? + 1 = un 2 x 1 = 2 + 2 = 4 2 x 2 = 4 + 2 = 6 2 x 3 = 6 + 2 = 8 2 x 4 = 8 + 2 = 10 Answer:- un = 2n + 2 • Author t.elia 7 years ago from Northern Ireland Bubble, U 1 2 3 4 Un 3 4 6 9 U1 = 3 U2 = 3 + (2-1) = 4 U3 = 4 + (3-1) = 6 U4 = 6 + (4-1) = 9 So Un = U(n-1) + (n-1) Hope this helps • Author t.elia 7 years ago from Northern Ireland Bubble, U 1 2 3 4 Un 3 4 6 9 U1 = 3 U2 = 3 + (2-1) = 4 U3 = 4 + (3-1) = 6 U4 = 6 + (4-1) = 9 So Un = U(n-1) + (n-1) Hope this helps • Author t.elia 7 years ago from Northern Ireland For further help follow the link in the hub "Let me solve your maths problems". Assistance can be found for Bodmas HCF LCM Prime and Composite numbers Square Root Cube Root Place Values Expanded Notation Rounding off Numbers Writing numbers as a product of their Prime Numbers • Author t.elia 7 years ago from Northern Ireland This is just an ordinary sequence(not an arithematic one) The values go upwards in differences of 1's and 3's. 2 3 6 7 10 11 --- --- --- --- --- (Difference is) 1 3 1 3 1 so at some point in a sequence like this values would be ommitted and you would have to find them. • Ema 7 years ago • Author t.elia 7 years ago from Northern Ireland Ema, This is not an arithematical sequence nor is it a geometric one, it is a general sequence that is alternating. In this question "d" is not unique; t2-t1 = -2 -1 = -3 t3-t2 = 3 -(-2) = 3 + 2 = 5 So the answer to this sequence is; nth term is -n (if "n" is even) ............+n (if "n" is odd) Hope this helps, thanks Telia • Nico 7 years ago i'm trying to work out the nth term for the following: 5 8 11 14 i know there's three difference between each number, but i'm getting confused as how to work this out... thanks :) • Author t.elia 7 years ago from Northern Ireland Nico, (Position) U..1....2....3....4 ( Term).......5....8....11...14 you are correct the common difference(d) is "3" Therefore; Go back to your table and put in your values in the 3n formula and see what else is required to solve it. d x u 3 x 1 = 3 + ? = 5 3 x 2 = 6 + ? = 8 By now you'll see that you need to add on 2 So your formula is going to be 3n + 2 • Author t.elia 7 years ago from Northern Ireland un = 3n + 2 (where the un is the nth term) • Author t.elia 7 years ago from Northern Ireland Oli. (i) 1, 2, 4, 8 Firstly we will put these values into a table n 1 2 3 4 un 1 2 4 8 ? ? ? (now to find the differences) 1 2 4 (The number is being doubled each time, this suggests that the rule involves 2 to the power of something.) un = ? un = 2n This time the formula involves subtraction of the power Therefore; un = 2n - 1 21 - 1 = n0 = 20 which equals 1 22 - 1 = n1 = 21 which equals 2 23 - 1 = n2 = 22 which equals 4 24 - 1 = n3 = 23 which equals 8 if you check back to the above table the formula is correct un = 2n - 1 • Author t.elia 7 years ago from Northern Ireland That answer is 2 (to the power of n - 1) Telia • Pien 7 years ago can you find the rule for the nth term for the following sequence on numbers? 5,27,79,173 ive looked at it for a bit and i can only come up with this rule, which only works for the first 3 numbers: 6n^4/2n - n^n-1 +5 • Author t.elia 7 years ago from Northern Ireland Pien, The rule you are looking for is 2n^3 + 3n^2 - (n - 1) Hope this helps • waqas 7 years ago let 1,3,2,5......find 6th term • Author t.elia 7 years ago from Northern Ireland The 6th term in this sequence is 7. 5th is 3 • Krishnasingh 6 years ago Some comments are not satisfactory. Also in many cases only nth term rule is given. how can we find the process? • Author t.elia 6 years ago from Northern Ireland Krishnasingh If you read through the above hub observing and understanding the tables you'll be able to follow the process to find the nth term. Say your common difference is 4 then you are working with 4n plus or minus something to find the rest of your formula to solve the equation. It takes a little bit of patience and practise. If there is no visible difference then try squaring the numbers in the "u" bracket and find out what else you have to add or subtract to find the formula. Please read over the worked examples again and you'll see not only is the nth term rule there but so is the process. Good Luck • bubba #9 6 years ago What is the ninth term in the following sequence? 6, 10, 14, 18, 22,... • Author t.elia 6 years ago from Northern Ireland Bubba, First start by finding the nth term. Your difference is 4 so that tells you the formula is going to start as 4n + or - something. So do out your table as shown in the worked examples above ...1...2...3...4...5 ...6..10..14..18..22 4 x 1 = 4 + ? = 6 (? = 2) 4 x 2 = 8 + 2 = 10 4 x 3 = 12 + 2 = 14 So the formula is 4n + 2 Now use the formula to find the 9th term U(9) = 4 x 9 = 36 36 + 2 = 38 Therefore 9th term = 38 Hope this helps. • rix 6 years ago what is the nth sequence of 1,4,12,32,80 • Author t.elia 6 years ago from Northern Ireland Rix Im not finding a common difference here have you wrote this question down properly? Please check • Georgia 6 years ago What is the rule of this sequence? 23, 19, 13, 8 • ema 6 years ago what would be the nth term for the sequence 0,6,14,24,36,50 ?????? • Author t.elia 6 years ago from Northern Ireland Georgia, Your question doesn't follow any rule have you wrote it down properly? • Author t.elia 6 years ago from Northern Ireland Ema, tn = 6, 8, 10, 12, 14 a = 6 d = 2 tn = a + (n-1) 2 tn = a + 2n - 2 • Author t.elia 6 years ago from Northern Ireland The nth term your looking for EMA is tn = 2n + 4 • Elias Jacobo 6 years ago what is the next term of 1,6,4,9,7 • Author t.elia 6 years ago from Northern Ireland Elias, 1..6..4..9..7..12..10 • Jordan Newland 6 years ago what is the first five terms for the nth term of 16-3n • Author t.elia 6 years ago from Northern Ireland Jordan, 16 - (3 x 1) = 13 16 - (3 x 2) = 10 16 - (3 x 3) = 7 16 - (3 x 4) = 4 16 - (3 x 5) = 1 So first five terms are as follows 13..10..7..4..1 • ari h 6 years ago two terms of an arithmetic sequence are a8=76 and a28=296 what's the rule for the nth term? • Author t.elia 6 years ago from Northern Ireland Hi ari h, an = a + (n - 1) d now put in n = 8, an = 76 76 = a + (8 - 1) d 76 = a + 7d (equation 1) an = a + (n - 1) d now put n = 28, an = 296 296 + a + (28 - 1) d 296 = a + 27d ( equation 2) Take the equations away from each other 296 = a + 27d 76 = a + 7d -------------- 220 = 20d Then find common denominator and cancel down. 20d = 220 ----------- 20 20 (CANCEL DOWN) The two 20's cancel each other leaving "d" on its own on the left hand side of the equation. On the right hand side of the equation 20 divides into 220 = 11 Therefore d = 11 a + 7d = 76 d = 11 a + 7 (11) = 76 a + 77 = 76 a = -1 Therefore a = -1, d = 11 Last Step in this Equation; an = a + (n - 1) d an = -1 + (n - 1) 11 an = -1 + 11n - 11 an = 11n - 12 So the nth term for a8 = 76 and a28 = 296 is; an = 11n - 12 Hope this helps. • Humairaa Patel 6 years ago This site is cool but it would be cooler if you added this place which says want to find out the nth term of something type it in here and you will get your answer • Author t.elia 6 years ago from Northern Ireland Thanks Humaira Patel, I will always answer irrespective of the fact that sometimes the question might not be asked in the correct hub. I am always glad when a student shows enough intution to ask a question if they are stuck or unsure of something. • Tom 6 years ago Help please how is this worked out first term 4 term to term rule x3 +1 • Author t.elia 6 years ago from Northern Ireland Tom, un = 3x + 1 .....3 x 1 = 3 + 1 = 4 .....3 x 2 = 6 + 1 = 7 .....3 x 3 = 9 + 1 = 10 etc Hope this helps • Varsha 6 years ago 5A(n+1) + 1 = 5A(n) + n Given A(5)=55. Find A(55) I am using the brackets to denote that the numbers in the brackets are at the base. i.e A(n+1) denotes n+1 th term. • Author t.elia 6 years ago from Northern Ireland Hi Varsha, Using the brackets to denote that the numbers in the brackets are at the base. 5A(n + 1) + 1 = 5A(n) + n Given A(5) = 55 .......A(55) = ? 5A(n + 1) + 1 = 5A(n) + n Take the 1 over the = sign to the other side and it becomes -1 (change side ,change sign) 5A(n + 1) = 5A(n) + n - 1 Now divide each side by 5 to get A(n + 1) on its own; Therefore A(n + 1) = 5A(n)/5 + (n - 1 /5) The 5's cancel each other out and your left with the formula; A(n + 1) = A(n) + (n - 1 / 5) So use this formula to solve A(55) We know A(5) = 55 So we will find A(6)....up til A(55) This will have to be done manually as the difference is 0.2 and it is compoundly added on to the next total. OK so far! So lets take the Formula we found and start plugging in the values to solve this equation. A(n + 1) = A(n) + (n - 1 / 5) A(5) = 55 A(5 + 1) = A(5) + (5 - 1 / 5) .........= A(5) + (4 / 5) .........= 55 + 0.8 = 55.8 _________________________________________________________ A(6 + 1) = A(6) + (6 - 1 / 5) .........= A(6) + (5 / 5) .........= 55.8 + 1 = 56.8 _________________________________________________________ A(7 + 1) = A(7) + (7 - 1 / 5) .........= A(7) + (6 / 5) .........= 56.8 + 1.2 = 58 and so on until A(54 + 1) = A(54) + (54 - 1 / 5) .........= A(54) + (53 / 5) .........= 329 + 10.6 = 339.6 At the start of this equation I was able to find the (n - 1 / 5) to be 10.6 but had to do the whole equation manually to find the 339.6 value. Therefore A(55) = 339.6 Hope this helps • Varsha 6 years ago Thank you for your quick response. Sorry to bother you again ...but require one clarification... Did you calculate every term to arrive to A(55)or you calculated it directly by some formula. i.e how you got A(54) as 329? • Author t.elia 6 years ago from Northern Ireland Hi Varsha, Sorry but there is no Formula for this aspect of the equation - each term has to be calculated individually to arrive at A(55). Is this a homework question? • Varsha 6 years ago I am giving some entrance exam. This was one of the question in the mock test which has a time constraint. Anyways will have to skip such questions as they are too time consuming or will have to search some better method for solving this. • Author t.elia 6 years ago from Northern Ireland Hi Varsha, If you practice this question you should be ok, but please don't skip it as this is one of the question which would separate the "men from the boys" so to speak. The first part of this question is relevantly straight forward- do that then come back to finish it, this will show them you know what you are doing. It seperates the "A" students from the "C" students. Good Luck • Varsha 6 years ago Hi Was trying for long to find some different method to solve this. Check this out.... Using {} these brackets to denote the actual brackets. 5A(n + 1) + 1 = 5A(n) + n A(n+1) = A(n) + 1/5 {n-1} A(6)= A(5)+ 1/5 {5-1} ....= 55 + 1/5 {4} Similarly A(7) = A(6)+ 1/5{6-1} .... = 55 + 1/5 {5} + 1/5 {4} .... = 55 + 1/5 {4 + 5} So A(55) = 55 + 1/5 {4+5+6+7......+52+53} ... = 55 + 1/5 {53x54/2 - {3+2+1} ... = 55 + 1/5 {1431 -6} ... = 55 + 285 ... = 340 which is close to 339.6 What is your opinion? Is this a correct method? • Author t.elia 6 years ago from Northern Ireland Hi Varsha, Sorry for the delay was out. OK here goes , I made a slight mistake of 0.4 at the A(40) part of this equation for that I apologise. The answer should have been 202 at that point and not 201.6 as I had noted in my notepad. As a result after plugging in your idea I have got this to work as per your answer, A(n + 1) = A(n) + [n - 1 / 5] + d ...........A(55) + [4+5+6+7+8+9....+53 - 1 / 5] + d ...........A(55) = [1425 - 1 / 5] + 0.2 ..................284.8 + 0.2 = 285 ..................285 + 55 = 340 0.2 being the difference each time Absolutely brilliant Varsha Good Luck • Author t.elia 6 years ago from Northern Ireland Although we come up with a formula Varsha you still need to work out the first 3 or 4 terms manually to see if there is any correlation within the problem. My idea was to find the first term of the series and then using the Arithmatic progression formula to find the value of A(55) - for that the difference between two terms has to be constant - so we don’t have a choice here but to do it manually:( If you work out a few more of them you will find you are always adding 55 at the end in the equation plus the difference of 0.2 Good Luck with the entrance exam. • Varsha 6 years ago Ya dats true.. Will have to do that much of calculations...No other option :( Anyways, Thanks a lot. Sorry but will keep on bothering you till my exams. • Author t.elia 6 years ago from Northern Ireland Hi Varsha, Glad to be of a help to you and keep in contact - if you need anything else clarified before your exam don't hesitate to contact me. Good Luck in your exam studies. • Sheena 6 years ago what's is the nth term for 38, 33, 28, 23, 18 and 640, 320, 160, 80 I',m having a hard time with this thank you • varsha 6 years ago Hi, "Need i more help from you. I tried searching sin theta table for multiples of 5 on internet which is in fractions. Like sin theta 75 degree, 80 degree etc in fractions. But the search showed sin theta table in decimals. Do you have one with fractions? My purpose is to learn these fractions so as to answer area of triangle questions by the 1/2 ab sin theta formula, where a & b are the sides and theta is the angle between them. Also are there any other trigonometric formulas for finding areas of geometric figures?" • ssss!! 6 years ago i really need to know this today and right now b4 i go off.. what's the first 4 terms of the number sequence using the nth term rule which is: N3+3 ??? the first number 3 is actually supposed to be over the n and small... • Author t.elia 6 years ago from Northern Ireland SSSS, Do you mean N cubed + 3 If so then, 1 x 1 x 1 = 1 + 3 = 4 2 x 2 x 2 = 8 + 3 = 11 3 x 3 x 3 = 27 + 3 = 30 4 x 4 x 4 = 64 + 3 = 67 Please clarify if this is correct • Author t.elia 6 years ago from Northern Ireland Sheena, tn = a + (n - 1) d a = 38 d = -5 tn = 38 + (n - 1) -5 tn = 38 -5n + 5 tn = 43 - 5n (answer) In the next one the sequence is 640,...320,...160,...80 1st term = 640 2nd term = 320 = 640 / 2 = 1st term / 2 3rd term = 160 = 320 / 2 = 2nd term / 2 4th term = 80 = 160 / 2 = 3rd term / 2 so we can say nth term = n - 1 / 2 (n - 1)th term is previous to nth term • Author t.elia 6 years ago from Northern Ireland Varsha, When theta is 0...sin theta is 0 theta is 30 degrees...sin theta is 1/2 theta is 45 degrees...sin theta is 1/squareroot 2 theta is 60 degrees...sin theta is squareroot 3 / 2 theta is 90 degrees...sin theta is 1 theta is 120 degrees...sin theta is sq root 3 / 2 theta is 135 degrees...sin theta is sq root 2 / 2 theta is 150 degrees...sin theta is 1 / 2 theta is 180 degrees...sin theta is 0 theta is 210 degrees...sin theta is -1 / sq root 2 theta is 225 degrees...sin theta is -sq root 2 / 2 theta is 240 degrees...sin theta -sq root 3 / 2 theta is 270 degrees...sin theta is -1 theta is 300 degrees...sin theta is - sq root 3 / 2 theta is 315 degrees...sin theta is - sq root 2 / 2 theta is 330 degrees...sin theta is -1 / 2 theta is 360 degrees...sin theta is 0 Hope this helps Varsha • varsha 6 years ago Thanks a lot.. • mike 6 years ago does this sequence diverge or converge, and if it converges, what is the limit? (2n+1)/(1-3sqrt(n))?? thanks • ????????????? 6 years ago wow ! • sebastian 6 years ago what is the nth term of the sequence 6,14,24,36? is it a quadratic? • michael 6 years ago what is the general term of 1,2,7,11,16,22? thank you • michael 6 years ago Is there any technique to get the general terms? plsss help me. huuuh. thank you. • andy 6 years ago Here's a weird one we can't figure out: 0, 2, 4, 8, 24, 220 What's the next term and rule? I need to figure the first 12 terms. Pls answer to ahofer@nstsystems.com if possible. • michaelneLa 6 years ago help me plsss.. our final exam is coming up.. can you help for the general term of this no. 3,3,6,6,9,9,12,12 thanks in advance • Author t.elia 6 years ago from Northern Ireland Hi Mike. Sorry about the delay in answering your question but Ive been very busy. Now to get caught up! This sequence diverges. (2n + 1) / (1 - 3 sqrt n) rationalise the denominator (2n + 1) ( 1 + 3 sqrt n) / (1 - 3 sqrt n)(1 + 3 sqrt n) = (2n + 1) (1 + 3 sqrt n) / 1 - 9n = 2n + 2n.3 sqrt n + 1 + 3 sqrt n / 1 - 9n = 2n + 6(n)to the power 3/2 + 1 + 3 sqrt n Divide each term by the highest power of n (e.g) (n to the power of 3/2) = 2n / n to the power 3/2 + 6n to the power 3/2 / n to the power 3/2 + 1 / n to the power 3/2 + 3 sqrt n / n to the power of 3/2 all over 1 / n to the power 3/2 - 9n / n to the power of 3/2 = 2 / n to the power of 1/2 + 6 + 1 / n to the power of 3/2 + 3 / n squared, all over 1 / n to the power of 3/2 - 9 / n to the power 1/2 put the limit n arrow infinity (use sign for infinity which looks like an eight sideways) = 2 / infinity + 6 + 1 / infinity + 3 / infinity all over 1 / infinity - 9 / infinity = 0 + 6 + 0 + 0 / 0 - 0 = 6 / 0 = infinity Therefore the limit of this sequence is infinity and clearly this is a diverging sequence. Hope this helps and Good Luck. • Author t.elia 6 years ago from Northern Ireland Hi Sebastian, what is the nth term of the sequence 6,14,24,36? is it a quadratic? NO 6 = 6 x 1 = (6 + 1) x (1 + 1) 14 = 7 x 2 = (6 + 2) x ( 1 + 2) 24 = 8 x 3 = (6 + 3) x ( 1 + 3) 36 = 9 x 4 so nth term {6 + (n - 1)} x {1 + (n - 1)} ...........= (6 + n - 1) (1 + n - 1) ...........= (n + 5)(n) ...........= n squared + 5n Hope this helps and good luck • Author t.elia 6 years ago from Northern Ireland Hi Michael, what is the general term of 1,2,7,11,16,22? The pattern is as follows; 1 + 0 = 1 1 + 1 = 2 2 + 2 = 4 4 + 3 = 7 7 + 4 = 11 11 + 5 = 16 16 + 6 = 22 22 + 7 = 29 So next term is 29 ( 8th term = 29) 9th term = 8th term + 8 ..........29 + 8 = 37 Therefore the general term is; Term (n + 1) = Term (n) + n T (n + 1) = T (n) + n Hope this helps, Good Luck There is no magic formula for working out sequences except work out their pattern, tedious as it is. • Author t.elia 6 years ago from Northern Ireland Hi MichaelneLa, can you help for the general term of this no. 3,3,6,6,9,9,12,12 Series questions are set out like this double 3's, double 6's etc so general term is 3n + 3, 3n + 3 3 x 1 = 3 + 3 = 6 3 x 2 = 6 + 3 = 9 3 x 3 = 9 + 3 = 12 Term you are looking is 3n + 3 Hope this helps and Good Luck • Author t.elia 6 years ago from Northern Ireland Hi Caroll, What's the nth term of 1, 1/2, 3, 1/4, 5, 1/6, 7, 1/8, 9, 1/10, 11 From the above sequence the 9th term is 9 The sequence is found out by the following method; Every odd number increases by 2 First term = 1 Third term = 3 etc Whereas 2 is being added to the demoninator each time to get the next even term. nth term is; = 1 + (n - 1) 2 = 1 + 2n - 2 = 2n - 1 Therefore 2n - 1 is the nth term for odd terms. For the even terms 1/2, 1/4, 1/6, the nth term is 1/(2n) that is written as a fraction 1 over 2n Hope that helps and Good Luck • bob lee 6 years ago nive post. really helped. • turkish 6 years ago what is the nth term of 6, 8, 10, 12, 14, 16... ?? • Jai 6 years ago t.elia: Your solution (4 posts above mine) for "What's the nth term of 1, 1/2, 3, 1/4, 5, 1/6, 7, 1/8, 9, 1/10, 11" is incorrect. It should be n for odd numbers and 1/n for even. • Author t.elia 6 years ago from Northern Ireland Dear Jai, Im sorry you have made a BIG mistake in your calculations. Please re-check them, not for my sake but for your own educational purposes.The answer you posted above is WRONG! My Calculations are CORRECT The answer for the workings of the nth term of this equation are as I stated; Therefore 2n - 1 is the nth term for odd terms. For the even terms 1/2, 1/4, 1/6, the nth term is 1/(2n) that is written as a fraction 1 over 2n Many Thanks for the post Telia • Author t.elia 6 years ago from Northern Ireland Hi Turkish, The nth term you are looking for to solve this equation is; 2n + 4 Regards Telia • Iltbbas 6 years ago What the nth term for 15,21,27,33,39 and 19,13,7,1,-5 • hi! I need help! 6 years ago rule of 14,23,32,41,50? and ule of 1,6,11,16,21 please1 thanks! • Author t.elia 6 years ago from Northern Ireland Hi lltbbas, Tn = a + (n-1) d a = 15 d = 21 - 15 = 6 Tn = 15 + (n-1) x 6 Tn = 15 + 6n - 6 Tn = 9 + 6n (ANS) Now for the second one; Tn = a + (n-1) d a = 19 d = 13 - 19 = -6 Tn = 19 + (n-1) x -6 Tn = 19 - 6n + 6 Tn = 25 - 6n (ANS) Good Luck • Author t.elia 6 years ago from Northern Ireland Hi , Tn = a + (n-1) d a = 14 d = 23 - 14 = 9 Tn = 14 + (n-1) x 9 Tn + 14 + 9n - 9 Tn = 5 + 9n (ANS) The next one is; Tn = a + (n-1) d a = 1 d = 6 - 1 = 5 Tn = 1 + (n-1) x 5 Tn = 1 + 5n - 5 Tn = -4 + 5n (ANS) Good Luck and hope this helps. • Abhie 6 years ago What is the 10th term of the Tn = n (n + 5)? • lui 6 years ago 20,4,16,37 • ahmed 6 years ago • Author t.elia 6 years ago from Northern Ireland Dear Hafsa, I never judge a book by its cover. Telia • iamhappy 6 years ago whta i the the nth term formula for 3,6,9,12,15,18 ......Please!! • Author t.elia 6 years ago from Northern Ireland Dear iamhappy, Your common difference is 3, therefore this tells you to write down 3n Now if you work it out using the layout of the tables at the start of this hub you will find the following; 3 x 1 = 3 3 x 2 = 6 3 x 3 = 9 3 x 4 = 12 Good luck • Author t.elia 6 years ago from Northern Ireland Hi chipcat, 3n + 4 what is the 8th term of this question? 3n = ( 3 x n)replace the "n" with the "8" Therefore; 3 x n 3 x 8 = 24 + 4 = 28 So 8th term of this question is 28 • Daisy 6 years ago What would the nth term of the sequence 1,13,37,73,... be? Can someone please explain with steps. Thank you! • Author t.elia 6 years ago from Northern Ireland Hi Daisy, You need to follow this method to complete this math; 1, 13, 37,73..... The general form of this sequence is; an^2 + bn + c (^2 means squared)STEP 1 of this equation Constant = 13 - 1 = 12 a = constant/2 (constant divided by 2) = 12 / 2 = 6 a = 6 Plug a = 6 into STEP 1 an^2 + bn + c (^2 means squared)STEP 1 6n^2 + bn + c (STEP 2 of this equation) for n = 1 Next 6(1) + b(1) + c = 1 6 + b + c = 1 b + c = -5 (STEP 3 of this equation) Now n = 2 in STEP 2 6(2)^2 + b(2) + c = 13 2 x 2 = 4 x 6 = 24 24 + 2b + c = 13 2b + c = -11 ( STEP 4) STEP 3 = b + c = -5 STEP 4 = 2b + c = -11 (-) (-) ------------------------ = -b/-1 = 6/-1 Cancel down b= -6 Plug 6 = -6 (in @ STEP 3) b + c = -5 -6 + c = -5 c = -5 + 6 c = 1 Plug a = 6, b = -6, & c = 1 (into STEP 1 an^2 + bn + c 6n^2 - 6n + 1 ( Answer) Good Luck...Hope this helps. • Daisy 6 years ago Hey! Thank you so much :) That helped a lot, its so clearly explained! Thanks again! • hellp 6 years ago find the a8 for the sequence a1=-2, an=3a(n-1) +6 • Author t.elia 6 years ago from Northern Ireland Hi hellp... Given:- a(1) = -2 an = 3a(1) (n - 1) + 6 Solution:- an = 3a (n - 1) + 6 Plug n = 8 and a(1) = -2 a(8) = 3(-2) (8 - 1) + 6 a(8) = -6 (7) +6 a(8) = -6 x 7 + 6 a(8) = -42 + 6 a(8) = -36 Hope this helps & Good Luck • Ash 5 years ago find the nth term for 1/5, 2/6, 3/7, 4/8, 5/9... ? I can see the pattern but I'm not sure how to express it Help? :) • brookie 5 years ago position: 1 2 3 4 5 6 value of : 5 14 23 32 ?? term • killah 5 years ago • 5 years ago help me to solve this: (28% - 8%) / 1.4 = ? • R.Purnalingam 5 years ago What is the nth term of the sequence 0,1,1,2,2,3,3,4,4,5,5......? • Nikee 5 years ago The nth term of a sequence is 100-2n What is the 1st term, 15th term and 30th term? • megan 5 years ago hi what is the sequence and the result in nth term for these numbers please, 23, 16, 9, 2 Thanks • megan 5 years ago hi again i posted a minute ago also can you tell me what the rule is as well as the sequence and the result in nth term thanks • amar 5 years ago what's the nth term of 4,8,16 • Bitsy 5 years ago I am looking at this pattern: 7,0,0,0...I need to find the next 2 numbers in the pattern and I can't even see the rule for the pattern. Can anyone help? • Arpita 5 years ago how do i get the first five numbers in the sequence when 34-5n is the nth term rule? • trung 5 years ago what is the nth term of pattern 20,44,76,116.... i know it is adding multiples of 8 each time • Rohit 5 years ago What is the formula for 1,2,3,4,5,6 11,9,7,5,3,1 • exam 5 years ago I have an exam tommorow. Can you plz tell me the formula for: 20, 17, 14, 11, 8, • Author t.elia 5 years ago from Northern Ireland EXAM, The formula you need to use to solve this equation is as follows; Tn = a + (n - 1) d Therefore a = 20 d = 17 Take d away from a 17 - 20 = -3 Now plug the results into your formula; Tn = a + (n - 1) d ...= 20 +(n - 1) -3 (multiply out the brackets) ...= 20 -3n + 3 ...= 23 - 3n Good Luck in your Exams tomorrow. • Author t.elia 5 years ago from Northern Ireland Arpita, How to find the first 5 numbers in the sequence 34 - 5n -5 x n -5 x 1 = -5 .......= -5 + 34 = 29 -5 x 2 = -10 .......= -10 + 34 = 24 -5 x 3 = -15 .......= -15 + 34 = 19 -5 x 4 = -20 .......= -20 + 34 = 14 -5 x 5 = -25 .......= -25 + 34 = 9 So therefore the first 5 numbers in the sequence are ; 29 , 24 , 19 , 14 , 9 Good Luck hope you understand this. • Author t.elia 5 years ago from Northern Ireland MEGAN The formula you need to use is; Tn = a + (n - 1) d a = 23 d = 16 16 - 23 = -7 Now plug the results into your formula; Tn = 23 + (n - 1) d Tn = 23 + (n - 1) -7 (multiply out the brackets) ...= 23 -7n + 7 ...= 30 - 7n (nth term rule) Therefore; -7 x 1 = -7 .......= -7 + 30 = 23 -7 x 2 = -14 .......= -14 + 30 = 16 -7 x 3 = -21 .......= -21 + 30 = 9 -7 x 4 = -28 .......= -28 + 30 = 2 -7 x 5 = -35 .......= -35 + 30 = -5 etc Good Luck • Author t.elia 5 years ago from Northern Ireland BROOKIE Use the following formula; Tn = a + (n - 1) d a = 5 d = 14 14 - 5 = 9 Therefore plug the results into the formula as follows; Tn = a + (n - 1) d ...= 5 + (n - 1) 9 (multiply out the brackets) ...= 5 + 9n - 9 9 x (1) = 9 - 4 = 5 9 x (2) = 18 - 4 = 14 9 x (3) = 27 - 4 = 23 9 x (4) = 36 - 4 = 32 9 x (5) = 45 - 4 = 41 etc Good Luck • Author t.elia 5 years ago from Northern Ireland NIKEE The nth term of a sequence is 100-2n What is the 1st term, 15th term and 30th term? 100-2n First Term 2 x (1) = 2 100 - 2 = 98 Fifteenth Term 2 x (15) = 30 100 - 30 = 70 Thirtith Term 2 x (30) = 60 100 - 60 = 40 1st Term = 98 15th Term = 70 30th Term = 40 Good Luck • Author t.elia 5 years ago from Northern Ireland ROHIT You need to follow this formula for the numbers 11,9,7,5,3,1 Tn = a + (n - 1) d a = 11 d = 9 9 - 11 = -2 Tn = a + (n - 1) d Tn = 11 + (n - 1) -2 (multiply out brackets) Tn = 11 -2n = 2 ...= 13 - 2n (ANSWER)( nth term rule) Good Luck • Bob 5 years ago what are the next 5 nth terms for 4n+4?!?!?!?!!?!? thanks. • Author t.elia 5 years ago from Northern Ireland BOB The next 5 terms for the nth term rule 4n + n are; 4 x (n) + 4 Term 1 4 x (1) = 4 + 4 = 8 Term 2 4 x (2) = 8 + 4 = 12 Term 3 4 x (3) = 12 + 4 = 16 Term 4 4 x (4) = 16 + 4 = 20 Term 5 4 x (5) = 20 + 4 = 24 Hope this helps what would be the next 3 terms for o.25n+5 and the 10th term • Author t.elia 5 years ago from Northern Ireland I take it this is 0.25n + 5 Workings of First Term; 0.25n + 5 = 0.25 x (n) + 5 0.25 x (1) = 0.25 + 5 = 5.25 ( Ans) Workings of Second Term; 0.25 x (2) = 0.50 + 5 = 5.5 ( Ans) Workings of Third Term; 0.25 x (3) = 0.75 + 5 = 5.75 ( Ans) Workings of Tenth Term; 0.25 x (10) = 2.5 = 5 = 7.5 ( Ans) Good Luck • sophie 5 years ago • Author t.elia 5 years ago from Northern Ireland SOPHIE For the sequence 95,90,85,80... The formula you need to use is as follows; Tn = a + (n - 1) d a = 95 d = 90 90 - 95 = -5 Now plug this into your formula; Tn = a + (n - 1) d Tn = 95 + (n - 1) -5 (multiply out the brackets) Tn = 95 -5n + 5 Tn = 100 - 5n ( Answer) 100 - 5n ( This means 5 multiplied by n) 100 - 5 x 1 100 - 5 = 95 Next 100 - 5n 100 - 5 x 2 100 - 10 = 90 etc Good Luck • James 5 years ago Hi please can you tell me the nth term for 36 42 48 54 60 -8 -3 2 7 12 1.2 1.4 1.6 1.8 2 • sam 5 years ago how do you do like 5/12 = n/6 • Bill 5 years ago Hi please can you tell me the nth term for 36 42 48 54 60 -8 -3 2 7 12 1.2 1.4 1.6 1.8 2 • dell336 5 years ago Please can you tell me the sequence pattern for 3 15 35 63 • Emma 5 years ago Can you please tell me the sequence pattern for 2, 8, 14, 20 AND 8, 5, 2, -1 please?? Also can you give the answer for tomorrow (26 September)? • Author t.elia 5 years ago from Northern Ireland SAM 5/12 = n/6 You will need to cross multiply here That will give you; 12n = 30 ..n = 30/12 (Cancel Down by dividing by 6) ..n = 5/2 Good Luck • Author t.elia 5 years ago from Northern Ireland P ( 28% - 8%)/ 1.4 = (20%)/ 1.4 = 20 over 100/ 1.4 = 0.2/ 1.4 = 1/7 (Ans) • Author t.elia 5 years ago from Northern Ireland EMMA the sequence pattern for 2, 8, 14, 20 Use the following formula; Tn = a + (n - 1 ) d a = 2 d = 8 8 - 2 = 6 Tn = a + (n - 1) d ...= 2 + (n - 1 )6 ...= 2 + 6n - 6 ...= 6n - 4 (Ans) 6n - 4 6(1) - 4 = 6 - 4 = 2 6(2) - 4 = 12 - 4 = 8 6(3) - 4 = 18 - 4 = 14 6(4) - 4 = 24 - 4 = 20 Good Luck • Author t.elia 5 years ago from Northern Ireland EMMA the sequence pattern for 8, 5, 2, -1 Use the following formula; Tn = a + (n - 1 ) d a = 8 d = 5 5 - 8 = -3 Tn = a + (n - 1 ) d Tn = 8 + (n -1) -3 ...= 8 - 3n + 3 ...= -3n + 11 (Ans) -3n + 11 -3(1) + 11 = -3 + 11 = 8 -3(2) + 11 = -6 + 11 = 5 -3(3) + 11 = -9 + 11 = 2 -3(4) + 11 = -12 + 11 = -1 Good Luck • Author t.elia 5 years ago from Northern Ireland BILL the nth term for 36,42,48,54,60 Use the following formula; Tn = a + (n - 1 ) d a = 36 d = 42 42 - 36 = 6 Tn = a + (n - 1 ) d ...= 36 + (n - 1) 6 ...= 36 + 6n - 6 ...= 30 + 6n (Ans) 30 + 6n 6(1) + 30 = 6 + 30 = 36 6(2) + 30 = 12 + 30 = 42 6(3) + 30 = 18 + 30 + 48 6(4) + 30 = 24 + 30 = 54 6(5) + 30 = 30 + 30 = 60 Good Luck • Author t.elia 5 years ago from Northern Ireland BILL the nth term for -8 -3 2 7 12 Use the following formula; Tn = a + (n - 1 ) d a = -8 d = -3 -3 - (-8) = 5 Tn = a + (n - 1 ) d ...= -8 + (n - 1) 5 ...= -8 + 5n -5 ...= -13 + 5n -13 + 5n 5(1) - 13 = 5 - 13 = -8 5(2) - 13 = 10 - 13 = -3 5(3) - 13 = 15 - 13 = 2 5(4) - 13 = 20 - 13 = 7 5(5) - 13 = 25 - 13 = 12 Good Luck • Author t.elia 5 years ago from Northern Ireland BILL the nth term for 1.2,1.4,1.6,1.8,2 Use the following formula; Tn = a + (n - 1 ) d a = 1.2 d = 1.4 1.4 - 1.2 = 0.2 Tn = a + (n - 1 ) d ...= 1.2 + (n -1) 0.2 ...= 1.2 + 0.2n - 0.2 ...= 1 + 0.2n (Ans) 1 + 0.2n 0.2(1) + 1 = 0.2 + 1 = 1.2 0.2(2) + 1 = 0.4 + 1 = 1.4 0.2(3) + 1 = 0.6 + 1 = 1.6 0.2(4) + 1 = 0.8 + 1 = 1.8 0.2(5) + 1 = 1.0 + 1 = 2.0 = 2 Good Luck • Author t.elia 5 years ago from Northern Ireland DELL336 the sequence pattern for 3 15 35 63 Sum of n terms = n/2 (2a + (n - 1) d ) Tn = 3 + 12 + 20 + 28... (add following term to get next number in sequence) Tn = 3 + (n - 1/ 2) (2(12) + (n - 2) 8) Tn = 3 + (n - 1/ 2) 24 + 8n -16 Tn = 3 + (n - 1/ 2) + 8n + 8 Tn = 3 + 8(n - 1 )(n + 1)/ 2 Tn = 3 + 4(n^2 - 1) Tn = 3 + 4n^2 -4 Tn = 4n^2 - 1 (Ans) 4n^2 - 1 4(1^2) - 1 = 4 - 1 + 3 4(2^2) - 1 = 16 - 1 = 15 4(3^2) - 1 = 36 - 1 = 35... Good Luck • Michelle 5 years ago Please could u help me find the nth term for: Q 1, 1/2, 1/8, 1/4, ..... this is the only part of the sequence i have sorry . • George 5 years ago nth term formula for 8, 19, 34, 53, 76, 103 • need help 5 years ago what is the formula of the sequence 10,8,5,1 • Oliver 5 years ago What is are the next three terms for this sequence: 1,2,2,4,8 Im really confused! • Author t.elia 5 years ago from Northern Ireland MICHELLE, The nth term for 1, 1/2, 1/4, 1/8... 1/2^0 = 1 1/2^1 = 1/2 1/2^2 = 1/4 1/2^3 = 1/8 1/2^4 = 1/16 Therefore;- tn = 1/2^n-1 Good Luck • Author t.elia 5 years ago from Northern Ireland GEORGE, nth term formula for 8, 19, 34, 53, 76, 103 The difference is 4 at the second level, letting you know you are dealing with a squared number in the answer. tn = 2^n2 + 5n + 1 (Answer) Good Luck • Author t.elia 5 years ago from Northern Ireland NEED HELP, what is the formula of the sequence 10,8,5,1 Yn = X - n yn = final term n = number of terms X = initial term Explanation with Example n = 2, X = 10 yn = x - n ....10 - 2 = 8 Yn = X - n (Answer) Good Luck • Author t.elia 5 years ago from Northern Ireland OLIVER, What is are the next three terms for this sequence: 1,2,2,4,8 1 x 2 = 2 2 x 2 = 4 2 x 4 = 8 1st term x 2nd term = Next term 4 x 8 = 32 8 x 32 = 8192 32 x 8192 = 262,2144 Good Luck • saf 5 years ago ok i know this is random but if you start with 2 and double each time then what is the number if the nth term is n+3 then 105-5n 2n-0.5v thanx • renee 5 years ago what are the first 5 nth term of 3n+6 • Bex 5 years ago What's the nth term for: -1 +1 -1 +1 -1. Doing my GCSEs and stuck on homework. • Russ 5 years ago Can somebody help me with the sequence of 1 8 9 12 please...are there 2 sequences? What follows after 12? Much appreciated! • Me 5 years ago WHAT IS THE FORMULA FOR tHIS SEQUENCE PLEAAAASE! 1 8 9 12 • Author t.elia 5 years ago from Northern Ireland RUSS & ME, Here is help with the sequence for your formula; 1 , 8 , 9 , 12 1^2, 2^3, 3^2 , (232) = 12 So the first number is squared = 1 ...the second number is cubed = 8 ...the third number is squared = 9 ...the fourth is the powers multiplied (232) = 12 The fifth number will be 5^2 = 25 The sixth number will be 6^3 = 216 The seventh number is 7^2 = 49 The eighth will be the powers multiplied (232) = 12 and so on ...... Hope this helps Good Luck • dom 5 years ago 3,8,15,24,35 what is the rule for that? im stuck ! • David 5 years ago Excuse me, would you help me with 50,48,46,44 and 3,8,15,24,35 and 53,56,61,68 please, thanks. • Author t.elia 5 years ago from Northern Ireland DOM & DAVID, you have both asked for the same problem here; 3, 8, 15, 24, 35 First difference is 5, 7, 9, 11 Second difference is 2 The Sum of the n terms is written as Sn and we know that the nth term = Tn = Sn - (Sn-1) Tn = a + (n - 1) d Tn = n/2 [ 2a + (n - 1) d ] Tn = n/2 [ 23 + (n - 1) 2] Tn = n/2 x 6 + n/2 x (n - 1)2 (cancel down) Tn = 3n + n^2 - n Tn = n^2 + 2n (Answer) Good Luck • Author t.elia 5 years ago from Northern Ireland DAVID 50, 48, 46, 44 Tn = a + (n - 1) d a = 50 d = 48 48 - 50 = -2 Tn = a + (n - 1) d Tn = 50 + (n - 1) -2 Tn = 50 - 2n + 2 Tn = 52 - 2n (Answer) Good Luck • Author t.elia 5 years ago from Northern Ireland DAVID 53, 56, 61, 68 first difference is 3, 5, 7 second difference is 2 Tn = a + (n - 1) d The Sum of the n terms is written as Sn and we know that the nth term = Tn = Sn - (Sn-1) Tn = 53 + 3 + 5 + 7 Tn = 53 + n/2[ 23 + (n - 1) 2] Tn = 53 + n/2 x 6 + 2(n - 1) (Cancel down) Tn = 53 + 3n + n^2 - n Tn = n^2 + 2n + 53 (Answer) Good Luck • Mahmoud Shahoud 5 years ago Hi , You're smart , but all comments don't help me . can you help me . I have this sequence . 1 1 1 1 1 1 1 1 1 2 I want to calculate the 11st element and 12nd element I want Math Equations . I have a product gave me the answer but I want the way . 11st term = 11 12nd term = 56 • liv 5 years ago 26-4n? • Author t.elia 5 years ago from Northern Ireland LIV 26-4n 26 - 4 x N 26 - 4 x (1) = 22 26 - 4 x (2) = 18 26 - 4 x (3) = 14 etc Hope this helps Good Luck • bubble 543 5 years ago what would the answer be for the nth term for a sequence is n-2/ n squared write the first three terms of the sequence • oscar 5 years ago • Author t.elia 5 years ago from Northern Ireland Oscar, 3 x 1 = 3 + 4 = 7 3 x 2 = 6 + 4 = 10 3 x 3 = 9 + 4 = 13 etc Good Luck & hope this helps • shreyansh 5 years ago The value of (2/3)(3/4)(4/5).....(1-1/n)= • Jamie 5 years ago Hi, can you help me. If i have these 4 numbers, 12 15 18 21, what's the sequente to get the nth term? • Author t.elia 5 years ago from Northern Ireland Hi Jamie • Paul 5 years ago My homework sheet says that i have to write a rule in terms of n, i don't know what that means, the website for my homework is http://docs.google.com/viewer?a=v&q=cache:ap2t.../maths/docs/building_number_patterns.doc+building+investigative+skills&hl=en&gl=uk&pid=bl&srcid=ADGEESgdc2qKUJbnfZK9K6L22BqBGd27GIaVEhDj6m_QE5xBedMtgqj5QD7hfd1vx0armiQYpFn4mPs83vidpckpAmTdbMgP_2PR7u30s46xICSkUbnCcSq1TaTrXxr-ywh0FKdy2oaH&sig=AHIEtbScDxgftte_31i6WwekjrGnqCDRgA, i also don't get question 4 • Author t.elia 5 years ago from Northern Ireland Hi Paul, this link you provided isn't working. • student 5 years ago hello plaese help me find Recurrence Relations t(n)=t(3n/5)+t(2n/5)+n • Author t.elia 5 years ago from Northern Ireland Ogleoo, 48 - 4n 48 - (4 x 1)= 48 - 4 = 44 48 - (4 x 2)= 48 - 8 = 40 48 - (4 x 3)= 48 - 12 = 36 etc I hope you find this easy to follow Good Luck • ogleoo 5 years ago thank you so much • ogleoo 5 years ago i keep forgetting that n means term number!!! • MAHI 5 years ago WHAT IS AP? • Author t.elia 5 years ago from Northern Ireland MAHI, Arithematic Progression:- Now add a number d, (for "difference"). We get a1 + d and the first 2 terms in our sequence are: a1, a1 + d For the next term, let's add d to that last term and we have a1 + 2d. Our sequence is now: a1, a1 + d, a1 + 2d We continue this process and the resulting set of numbers is called an arithmetic progression (AP) or arithmetic sequence Good Luck Telia Sorry posted the comment and wasn't signed in. • Author t.elia 5 years ago from Northern Ireland Simba, What are you working on multiplication?? Please clarify. Telia • Marc 5 years ago Find the next term of the progression 24,12,8,..... • miley 5 years ago how do you find the equation and 50th term of 1,4,9,16,25,36,48 • Kieran 5 years ago What is the nth term for this sequence (or similar ones): 2, 7, 15, 26. It goes up like this: +5, +8, +9. • Carl 5 years ago Hi would you be able to help me with the general term (tn) I guess, I doesn't know the word in english... well here are the numbers, 2,7,15,26,40 thanks :) • eivlys 5 years ago Find the next term of the progression 2 4 6 14 23 • rohit choudahry 5 years ago 0,2,4,6,8,12,12,20,16,? can u explain the trick behind this? • grace 5 years ago what would be the nth term of 40,38,36,34,32... • joyce 5 years ago what would be the nth term of 30,28,26,24,22... and what would be the nth term of 40,37,34,31,28... • joyce 5 years ago what would be the nth term of 30,28,26,24,22.. • arihant 5 years ago what is the nth term of 1+(1+2)+(1+2+3).....to n terms • Simon 5 years ago I'm stuck on this question so could you help me out? All terms in a geometric sequence are positive integers, and the first three ter4ms are n, n+3, and 2n+6. What is the fourth term of this sequence? • Filip 5 years ago -1 , -2 , -1 , 2 , 7 , 14 what's the nth term please? • Vin 5 years ago What's The Next Three Terms of the Arithmetic Sequence? 6,12,18,24,__,___,___? • akanksha raj 5 years ago • shubham 5 years ago iam confuse between nth and n terms of A.P • Ronak 5 years ago what will be the next term to the series 2,3,4,9,16... and how it will be???..... • Geralyn. 5 years ago the radiator of the truck contains 12 gallons of water. We draw off 3 gallons and replace it with alcohol, then we draw off 3 gallons of the mixture then replaced it by alcohol. until 8 drawings and replacements have been dine. How much alcohol is in the final mixture? • t.elia 5 years ago how bout sixth term of 3/4,3,12... • Anne David 5 years ago what is the formula of sixth term of 3/4,3,12... • Anne David 5 years ago and the tenth term of 0.4,0.04,0.004,,, plz give me a formula for this • sita 5 years ago how to find the 40 term of a sequance with term to term rule add 5and the 1st term of 6 • 5 years ago 5 x 4 x 3 ... -5 x -2 x - 4 Write out the correct form of the sequence • 5 years ago What is the Nth Term to: -1, 7, 18, 29, 42, 57 and 2, 9, 18, 29, 42, 57? • anonymous 5 years ago given the general form an^2+bn+c find a pls help with the above • Charlott3_Fay3 5 years ago What is 2-3G and 7-1/2n??? I'm really confused and it's annoying me please answer tonight xxxx • Jim 5 years ago The nth term of a sequence is the l.c.m of integers from 1 to n. Which term of the sequence is the first one that's divisible by 100? • alicia 5 years ago what is the 75th term 8, 14, 20, 26, 32 • waleedrehman 5 years ago plz find the indicated term of the following,1,3/2,5/4,7/8,...a7. • john 5 years ago terms: 1,2,3,4,5,...n values: 1,2,4,8,16,... • brittany 5 years ago Can Anyone Find What's The 6th Term For 5 , 7 , 11 .... • paul 5 years ago in an AP the fifth term is 8 and the nineth term is 14. fin the first term. • DIANA 5 years ago WHAT IS THE FORMULER FOR THE SEQUENCE; 22,18,14,10? • dmndmz 5 years ago Greetings ;) What is the formula for the sequence 16, 29, 48, 52, 54 and 5 ? • Fazla 5 years ago Maa''m i can't understand general term.. I don't knw why.. If u can help me in dis.. it will be a great help.. -2,4,-8,16.. can u help me 2 find da general term of dis set?? • williamovely@yahoo.com 5 years ago a series is s uch that the sum of the first nth therms is 2n^2 minus 3n. 1.how to Prove that the series is arithmatic 2.find the value of the 18th term • CHRISTY 5 years ago nth term of sequence ?7, ?21, 3?7..... • SURUTHI 5 years ago HI, pleasssssssse solve this problem ,i am not getting the solution anywhere in the net. "the sum of the first n terms of an ap are in the ratio (3n +8):(7n + 15), find the ratio of the 3 rd terms" ans 23:50. • SURUTHI 5 years ago HI, pleasssssssse solve this problem ,i am not getting the SOLUTION anywhere in the net. "the sum of the first n terms of an ap are in the ratio (3n +8):(7n + 15), find the ratio of the 3 rd terms" {ans=23:50}. • Jimmy 5 years ago nth term of sequence the sequence is 4, 12, 24 This is what i gett 4 12 24 40 \ / \ / \ / 8 12 16 \ / \ / 4 4 But I dont know how to make a formula from this help • Kalyanakrishnan 5 years ago give solution to this sum sum of the series 2+4(33)+6(55)+.........+22(21*21) • suruthi 5 years ago if the sum of 4 consecutive terms in a gp is 60 and if the t1,18,t4 are in ap find the terms ans(4,8,16,18) • Subhasish Raha 5 years ago what is the missing number in the following series 9,19,40,___,146? • Lisa 5 years ago trying to find out the correct sequence for if n+1/3 • Helmi 5 years ago how to solve this U 1 2 3 4 5 Un 1 4 10 22 46 • fawzy 5 years ago plz define in detail what do we do with sequences not having a common difference E.G 0..-1..-8..-27..-64. And also do you happen to know any formula like Tn= A+d(n-1)+ 1/2[(n-1)(n-2)c] which claims to work on everything. If Yes than please explain the formula. • yaya 5 years ago hi am year 7 what is the 5 terms and the 10th term of 2n+2 • ragu 5 years ago In the sequence, 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 when we add any three numbers we have to get the answer 30 • Pavan Kumar 5 years ago 24 24 28 46 94 what is the next sequence • chris 5 years ago An arithmetic sequence has first term a=6 and common difference d=12. How many terms of this sequence must be added to get 5040? • Adam Saleh 5 years ago The nth term of a sequence is n^2-6n-4, find the sum of 3rd and 4th terms • Kyle 5 years ago Which statement describes how to find the nth term in the sequence? 12, 29, 46, 63, 80, ... A.add 5 times the term number and 12 B.subtract 17 times the term number from 12 C.subtract 5 from 17 times the term number D.add 5 and 17 times the term number • luiza 5 years ago look at the equation n+3=12. Use iot to work out the value of n-3 • met 5 years ago what are the next two numbers 4, 5, 14, 15, 40, 41, _, _, ? • timmi2000 5 years ago wondering if anyone could help me with these questions, and help me work it out as i do not seem to get it. 1) find the 8th term of each of these sequences a) 2,5,8,11 b) 3,7,11,15 c) 2,10,50,250 d) 2,6,18,54 2) find the sum of the first 10 terms of the following series a)3,8,13,18 b)1,3,9,27 3) the first term of an arithmetic series is 1. the common difference is 6 find the 10th term of the series, the sum of the first N terms is 7400 find N • Howard 5 years ago what is the pattern of these numbers? a)2,3,6,85,68 thank you • Kiki 5 years ago Given the progression 7, 12, 17 and 22. What is the expression for the nth term of the progression? • Cherry 4 years ago • ray 4 years ago what's 5 ways to find the 77th term? • tom 4 years ago what's the nth term for 10, 40, 90, 160, 250, 360, 490... • kola 4 years ago 1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20=210 • Naseem Choudhury 4 years ago what is the term of 25-4n • becky 4 years ago how would I solve for the nth term: 1+2=(2x3)/2 1+2+3=(3X4)/2 1+2+3+4=(4x5)/2 • Rues 3 years ago 24,54,103,.... • Telamon 3 years ago what is the first 5 terms and the 100th term for the sequence whose rule is 0.5n - 0.5 • Avicii 3 years ago Does anybody on this goddamn blog answer questions!!!?? • pooja 3 years ago This website is boaring • wow 3 years ago • thalene 3 years ago the sum of the 17 term of an arithmetic sequence is 2414 and the last term is 256. find the first term • amberb 2 years ago HELP ASAP the first 5 terms and the 100th term for the sequence 0.5n-0.5 • john 2 years ago find the nth term of 2n+5. find the frist term,second term and common difference
# How to you find the general solution of sqrt(1-4x^2)y'=x? Nov 8, 2016 $y = - \frac{1}{4} \sqrt{1 - 4 {x}^{2}} + C$ #### Explanation: Changing the notation from Lagrange's notation to Leibniz's notation we have: $\sqrt{1 - 4 {x}^{2}} \frac{\mathrm{dy}}{\mathrm{dx}} = x$ which is a First Order separable DE which we can rearrange as follows: $\mathrm{dy} = \frac{x}{\sqrt{1 - 4 {x}^{2}}} \mathrm{dx} \implies \int \mathrm{dy} = \int \frac{x}{\sqrt{1 - 4 {x}^{2}}} \mathrm{dx}$ To integrate the RHS we need to use a substitution Let $u = 1 - 4 {x}^{2} \implies \frac{\mathrm{du}}{\mathrm{dx}} = - 8 x \implies - \frac{1}{8} \mathrm{du} = x \mathrm{dx}$ Se we can now substitute and integrate to get our DE solution: $y = \int \frac{- \frac{1}{8}}{\sqrt{u}} \mathrm{du}$ $\therefore y = - \frac{1}{8} \int {u}^{- \frac{1}{2}} \mathrm{du}$ $\therefore y = - \frac{1}{8} {u}^{\frac{1}{2}} / \left(\frac{1}{2}\right) + C$ $\therefore y = - \frac{1}{4} {u}^{\frac{1}{2}} + C$ $\therefore y = - \frac{1}{4} \sqrt{1 - 4 {x}^{2}} + C$
Suggested languages for you: Americas Europe Q41. Expert-verified Found in: Page 352 ### Precalculus Mathematics for Calculus Book edition 7th Edition Author(s) James Stewart, Lothar Redlin, Saleem Watson Pages 948 pages ISBN 9781337067508 # Logarithmic Equations Use the definition of the logarithmic function to find x.(a) ${\mathrm{log}}_{2}\left(\frac{1}{2}\right)=x$ (b) ${\mathrm{log}}_{10}x=-3$ 1. The required value of x is $-1$. 2. The required value of x is 0.001. See the step by step solution ## Part a. Step 1. Given. The given equation is ${\mathrm{log}}_{2}\left(\frac{1}{2}\right)=x$. ## Part a. Step 2. To determine. We have to find the value of x using the definition of the logarithmic function. ## Part a. Step 3. Calculation. We’ll use the definition of the logarithmic function ${\mathrm{log}}_{a}c=b\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{a}^{b}=c$. Comparing ${\mathrm{log}}_{2}\left(\frac{1}{2}\right)=x$ with ${\mathrm{log}}_{a}c=b\text{\hspace{0.17em}}$ we get $a=2,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}b=x,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}c=\frac{1}{2}$. So, the equivalent exponential form is: ${a}^{b}=c$ or, ${2}^{x}=\frac{1}{2}$ or, ${2}^{x}={2}^{-1}$ or, $x=-1$ [Equated the exponents, since the bases are same] Hence, the required value of x is $-1$. ## Part b. Step 1. Given. The given equation is ${\mathrm{log}}_{10}x=-3$. ## Part b. Step 2. To determine. We have to find the value of x using the definition of the logarithmic function. ## Part b. Step 3. Calculation. We’ll use the definition of the logarithmic function ${\mathrm{log}}_{a}c=b\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{a}^{b}=c$. Comparing ${\mathrm{log}}_{10}x=-3$ with ${\mathrm{log}}_{a}c=b$ we get $a=10,\text{\hspace{0.17em}\hspace{0.17em}}b=-3,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}c=x$. So, the equivalent exponential form is: ${a}^{b}=c$ or, ${10}^{-3}=x$ or, $\frac{1}{1000}=x$ or, $0.001=x$ Hence, the required value of x is 0.001.
$$\require{cancel}$$ # 05. Analyzing a More Complex Motion ### Analyzing a More Complex Motion Let’s re-visit our scenario, although this time the light turns green while the car is slowing down: The driver of an automobile traveling at 15 m/s, noticing a red-light 30 m ahead, applies the brakes of her car. When she is 10 m from the light, and traveling at 8.0 m/s, the light turns green. She instantly steps on the gas and is back at her original speed as she passes under the light. Our first step in analyzing this motion should be to draw a motion diagram. I’ve noted on the motion diagram the important events that take place during the motion. Notice that between the instant she hits the brakes and the instant she steps on the gas the acceleration is negative, while between the instant she steps on the gas and the instant she passes the light the acceleration is positive. Thus, in tabulating the motion information and applying the kinematic relations we will have to be careful not to confuse kinematic variables between these two intervals. Below is a tabulation of motion information using the coordinate system established in the motion diagram. Event 1: She hits the brakes. t1 = 0 s r1 = 0 m v1 = +15 m/s a12 = Event 2: She steps on the gas. t2 = r2 = +20 m v2 = +8.0 m/s a23 = Event 3: She passes the light t3 = r3 = +30 m v3 = +15 m/s First, notice that during the time interval between “hitting the brakes” and “stepping on the gas” there are two kinematic variables that are unknown. Recall that by using your two kinematic relations, you should be able to determine these values. Second, notice that during the second time interval again two variables are unknown. Once again, the two kinematic relations will allow you to determine these values. Thus, before I actually begin to do the algebra, I know the unknown variables can be determined! First let’s examine the motion between hitting the brakes and stepping on the gas: Now, using these results, examine the kinematics between stepping on the gas and passing the light. Note that the initial values of the kinematic variables are denoted by ‘2’ and the final values by ‘3’, since we are examining the interval between event 2 and event 3. We now have a complete kinematic description of the motion.
# What Is The Formula For Mass Using Density And Volume Why Study Math? – Solving Linear Systems by Linear Combinations You are searching about What Is The Formula For Mass Using Density And Volume, today we will share with you article about What Is The Formula For Mass Using Density And Volume was compiled and edited by our team from many sources on the internet. Hope this article on the topic What Is The Formula For Mass Using Density And Volume is useful to you. Muc lục nội dung ## Why Study Math? – Solving Linear Systems by Linear Combinations Now that we have seen how to solve a system of linear equations using the substitution method, let’s move on to a more convenient method called linear combinations. With this method (also known as addition-subtraction), we eliminate one variable by adding an appropriate multiple of one equation. We can then eliminate one variable and solve for the other. Once this is done, we use the second equation to solve for the second variable. This method can be made algorithmic and therefore the steps to be followed to solve the system using linear combinations are listed here: Step 1: arrange equations with like terms into columns. Step 2: Multiply one or both equations to get coefficients that are opposites of one variable. Step 3: Add the equations from the previous step. Combining like terms eliminates one variable and allows you to solve for another. Step 4: Substitute the value obtained in the previous step into one equation and solve for the other variable. Step 5: Check the solution in each of the original equations. To illustrate the algorithm, we solve the following system: 4x + 3y = 16 and 2x – 3y = 8. We first arrange these two equations into columns so that similar variables line up. So we have 4x + 3y = 16 2x – 3y = 8 Since we see that the coefficients of the y terms are opposites, there is no need to multiply the equations to get this form. So we add the two so the y terms disappear. So we have 6x = 24. Solving for x, we have x = 4. Substituting this value into the first equation, suppose we get 4(4) + 3y = 16 or 16 + 3y = 16 or 3y = 0 or y = 0. Checking , substituting these values into each of the original equations yields a true statement. Therefore, the solution is x = 4 and y = 0, or the point (4, 0) as the point of intersection of these two lines on the coordinate plane. Let’s see how we can use the method of linear combinations to model a historical problem. According to legend, the famous Greek mathematician Archimedes used the relationship between an object’s weight and volume to determine that fraud had occurred in the manufacture of the golden crown. This was done on the principle of volume displacement. You see, if the crown is pure gold, it should displace the same volume as an equal amount of gold. The following problem also uses the concept of density. By definition, the density of an object is equal to its mass divided by its volume. The density of gold is 19 grams per cubic centimeter. The density of silver is 10.5 grams per cubic centimeter. We will use these facts in the next problem. The problem: Suppose a gold crown suspected of containing silver weighs 714 grams and has a volume of 46 cubic centimeters. What percentage of the crown was silver? To solve this, we observe that the volume of gold plus the volume of silver must equal the total volume given by the value 46. Since we know the density of both gold and silver, and we know that density times volume equals mass, we have that gold density times gold volume plus silver density times silver volume equals total weight. Let G = volume of gold and S = volume of silver. Now we can translate the problem into mathematics and a linear system. We have G + S = 46 and 19G + 10.5S = 714. Putting these equations in the columns, we get G + S = 46 19G + 10.5S = 714. Now multiply the first equation by -19 to get the opposite coefficients for G. So it is -19G + -19S = -874 19G + 10.5S = 714. Adding the two equations, we get – 8.5S = -160. Dividing both sides by -8.5 gives us S = 18.8, which we can round to 19. Thus, the volume of silver is 19 cubic centimeters, and the percentage of silver in the krone is 19/46, or 41% with full percent accuracy. . Keep this method in mind the next time someone tries to pawn you pure gold, when the reality is something quite different. Beware of fools gold! ## Question about What Is The Formula For Mass Using Density And Volume If you have any questions about What Is The Formula For Mass Using Density And Volume, please let us know, all your questions or suggestions will help us improve in the following articles! The article What Is The Formula For Mass Using Density And Volume was compiled by me and my team from many sources. If you find the article What Is The Formula For Mass Using Density And Volume helpful to you, please support the team Like or Share! Rate: 4-5 stars Ratings: 9808 Views: 3219449 5 ## Search keywords What Is The Formula For Mass Using Density And Volume What Is The Formula For Mass Using Density And Volume way What Is The Formula For Mass Using Density And Volume tutorial What Is The Formula For Mass Using Density And Volume What Is The Formula For Mass Using Density And Volume free #Study #Math #Solving #Linear #Systems #Linear #Combinations Source: https://ezinearticles.com/?Why-Study-Math?—Solving-Linear-Systems-by-Linear-Combinations&id=1152015 Bài viết đã được tạo 2736 ## How Many Ounces Of Formula For A 10 Month Old How to Calculate Your Light Savings From Replacing Incandescent Bulbs Bắt đầu nhập từ khoá bên trên và nhấp enter để tìm kiếm. Nhấn ESC để huỷ. Trở lên trên
# CBSE 9th Maths ## Polynomials Polynomial: A polynomial in one variable x is an algebraic expression of the form p(x) = a0xn + a1xn-1 + a2xn-2 + … an Types of Polynomial: (i) Constant polynomial: A polynomial containing one term only, consisting of a constant is called a constant polynomial. Generally, each real number is a constant polynomial. (ii) Zero polynomial: A polynomial consisting of one term, namely zero only, is called a zero polynomial. (iii) Monomial: Polynomials having only one term are called monomials (iv) Binomial: Polynomials having only two terms are called binomials (v) Trinomial: Polynomials having only three terms are called trinomials The degree of a Polynomial: Highest power of the variable in a polynomial is the degree of the polynomial. (a) In one variable: In case of a polynomial in one variable, the highest power of the variable is called the degree of the polynomial. (b) In two or more variables: In case of a polynomial in more than one variable, the sum of the powers of the variables in each term is taken up and the highest sum so obtained is called the degree of the polynomial. Linear Polynomial: A polynomial of degree one is called a linear polynomial. e.g., x + 5 is a linear polynomial in x, y and z. Quadratic Polynomial: A polynomial of degree two is called a quadratic polynomial. e.g., 2x2 + x+ 1 Cubic Polynomial: A polynomial of degree three is called a cubic polynomial. e.g., 2y3 + 3 Value of a Polynomial: Value of a polynomial p(x) at x = a is p(a). e.g., If p(x) = x2 + 2x + 6 then, at x = 2, p(2) = 22 + 2 × 2 + 6 = 14 Zeroes of a Polynomial: Zeroes of a polynomial p(x) is a number a such that p(a) = 0. • Zero may be a zero of a polynomial. • Every linear polynomial has one and only one zero. • Zero of a polynomial is also called the root of the polynomial. • A non-zero constant polynomial has no zero. • Every real number is a zero of the zero polynomial. • A polynomial can have more than one zero. The maximum number of zeroes of a polynomial is equal to its degree. Remainder Theorem: Let p (x) be any polynomial of degree n greater than or equal to one (n ≥ 1) and let a be any real number. If p (x) is divided by the linear polynomial x – a, then the remainder is p (a). Dividend = (Divisor × Quotient) + Remainder Factor Theorem: Let q(x) be a polynomial of degree n ≥ 1 and a be any real number, then (i) (x – a) is a factor of q (x), if q(a) = 0 and (ii) q(a) = 0, if x – a is a factor of q (x). Algebraic Identities: An algebraic identity is an algebraic equation that is true for all values of the variable occurring in it. Some algebraic identities are given below • (x + y)2 = x2 + 2xy + y2 • (x – y)2 = x2 – 2xy + y2 • x2 – y2 = (x + y)(x – y) • (x + a)(x + b)= x2 + (a + b)x + ab • (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx • (x + y)3 = x3 + y3 + 3xy(x + y) • (x – y)3 = x3 – y3 – 3xy(x – y) • x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx) If x + y + z = 0, then x3 + y3 + z3 = 3xyz
# AP Statistics Curriculum 2007 Normal Prob (Difference between revisions) Revision as of 04:51, 26 October 2009 (view source)IvoDinov (Talk \| contribs) (added a link to the Problems set)← Older edit Revision as of 16:26, 4 February 2011 (view source)IvoDinov (Talk \| contribs) m (→General Normal Distribution: typo)Newer edit → Line 8: Line 8: * The area: $\mu -2\sigma < x < \mu+2\sigma = 0.9772 - 0.0228 = 0.9544$ * The area: $\mu -2\sigma < x < \mu+2\sigma = 0.9772 - 0.0228 = 0.9544$ * The area: $\mu -3\sigma < x < \mu +3\sigma= 0.9987 - 0.0013 = 0.9974$ * The area: $\mu -3\sigma < x < \mu +3\sigma= 0.9987 - 0.0013 = 0.9974$ - * Note that the [http://en.wikipedia.org/wiki/Inflection_point inflection points] ($f ''(x)=0$)of the (general) Normal density function are $\pm \sigma$. + * Note that the [http://en.wikipedia.org/wiki/Inflection_point inflection points] ($f ''(x)=0$) of the (general) Normal density function are $\pm \sigma$. [[Image:SOCR_EBook_Dinov_RV_Normal_013108_Fig6.jpg\|500px]] [[Image:SOCR_EBook_Dinov_RV_Normal_013108_Fig6.jpg\|500px]] ## General Advance-Placement (AP) Statistics Curriculum - Non-Standard Normal Distribution and Experiments: Finding Probabilities Due to the Central Limit Theorem, the Normal Distribution is perhaps the most important model for studying various quantitative phenomena. Many numerical measurements (e.g., weight, time, etc.) can be well approximated by the normal distribution. While the mechanisms underlying natural processes may often be unknown, the use of the normal model can be theoretically justified by assuming that many small, independent effects are additively contributing to each observation. ### General Normal Distribution The (general) Normal Distribution, N(μ,σ2), is a continuous distribution that has similar exact areas, bound in terms of its mean, like the Standard Normal Distribution and the x-axis on the symmetric intervals around the origin: • The area: μ − σ < x < μ + σ = 0.8413 − 0.1587 = 0.6826 • The area: μ − 2σ < x < μ + 2σ = 0.9772 − 0.0228 = 0.9544 • The area: μ − 3σ < x < μ + 3σ = 0.9987 − 0.0013 = 0.9974 • Note that the inflection points (f''(x) = 0) of the (general) Normal density function are $\pm \sigma$. • General Normal density function $f(x)= {e^{{-(x-\mu)^2} \over 2\sigma^2} \over \sqrt{2 \pi\sigma^2}}.$ • General Normal cumulative distribution function $\Phi(y)= \int_{-\infty}^{y}{{e^{{-(x-\mu)^2} \over 2\sigma^2} \over \sqrt{2 \pi\sigma^2}} dx}.$ • The relation between the Standard and the General Normal Distribution is provided by these simple linear transformations (Suppose X denotes General and Z denotes Standard Normal Random Variables): $Z = {X-\mu \over \sigma}$ converts general normal scores to standard (Z) values. $X = \mu +Z\times\sigma$ converts standard scores to general normal values. ### Examples #### Systolic Arterial Pressure Example Suppose that the average systolic blood pressure (SBP) for a Los Angeles freeway commuter follows a Normal distribution with mean 130 mmHg and standard deviation 20 mmHg. Denote X to be the random variable representing the SBP measure for a randomly chosen commuter. Then $X\sim N(\mu=130, \sigma^2 =20^2)$. • Find the percentage of LA freeway commuters that have a SBP less than 100. That is compute the following probability: p=P(X<100)=? (p=0.066776) • If normal SBP is defined by the range [110 ; 140], and we take a random sample of 1,000 commuters and measure their SBP, how many would be expected to have normal SBP? (Number = 1,000P(110<X<140)= 1,000*0.532807=532.807). • What is the 90th percentile for the SBP? That is what is xo, so that P(X < xo) = 0.9? • What is the range of SBP values that contain the central 80% of the SBPs for all commuters? That is what are xo,x1, so that P(x0 < X < x1) = 0.8 and ${x_o+x_1\over2}=\mu=130$ (i.e., they are symmetric around the mean)? (xo = 104,x1 = 156) ### Assessing Normality How can we tell if data collected from a process or experiment we observe is normally distributed? There are several methods for checking normality: • Why do we care if the data is normally distributed? Having evidence that the data we are analyzing is normally distributed allows us to use the (General) Normal distribution as a model to calculate the probabilities of various events and assess significant observations. • Example: Suppose we are given the heights for 11 women. • First we need to show that there is no evidence suggesting that the Normal and Data distributions are significantly distinct. • Then, we want to use the normal distribution to make inference on women heights. If the height of a randomly chosen woman is measured, how likely is that she'll be taller than 60 inches? 70 inches? Between 55 and 65 inches? Height (in.) 61 62.5 63 64 64.5 65 66.5 67 68 68.5 70.5
# Do Now 1.) 2). 5/15/2014 10-3 C Surface Area of Cylinders. ## Presentation on theme: "Do Now 1.) 2). 5/15/2014 10-3 C Surface Area of Cylinders."— Presentation transcript: Do Now 1.) 2) 5/15/2014 10-3 C Surface Area of Cylinders Parts of a cylinder A cylinder has 2 main parts. A rectangle and A circle – well, 2 circles really. Put together they make a cylinder. The Soup Can Think of the Cylinder as a soup can. You have the top and bottom lid (circles) and you have the label (a rectangle – wrapped around the can). The lids and the label are related. The circumference of the lid is the same as the length of the label. Area of the Circles Formula for Area of Circle A=  r 2 = 3.14 x 3 2 = 3.14 x 9 = 28.26 But there are 2 of them so 28.26 x 2 = 56.52 units squared The Rectangle This has 2 steps. To find the area we need base and height. Height is given (6) but the base is not as easy. Notice that the base is the same as the distance around the circle (or the Circumference). Key Concept Example 2 Find the surface area of the cylinder. Use 3.14 for . Round to the nearest tenth. Since the diameter of the cylinder is 12 meters, use 6 meters for the radius. S.A. = 2πrh + 2πr 2 S.A. ≈ 2(3.14)(6)(8) + 2(3.14)(6) 2 S.A. ≈ 301.44 + 226.08 S.A. ≈ 527.52 The surface area is about 527.5 square meters. Example 3 A candle is 8 centimeters tall and has a diameter of 4 centimeters. It has a paper label around its side only. Find the area of the candle that is covered by the label. Round to the nearest tenth. Since only the side of the cylinder is covered by the label, the formula is adjusted so that only the curved surface is found. S.A. = 2πrhArea of the curved surface of a cylinder Answer: The area covered by the label is about 100.5 square centimeters. S.A. ≈ 2(3.14)(2)(8) Replace r with 2 and h with 8. S.A. ≈ 100.48 Simplify. Example 4 SA = (  d x h) + 2 (  r 2 ) = (3.14 x 22 x 14) + 2 (3.14 x 11 2 ) = (367.12) + 2 (3.14 x 121) = (967.12) + 2 (379.94) = (967.12) + (759.88) = 1727 cm 2
Survey * Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project Document related concepts Vincent's theorem wikipedia , lookup Location arithmetic wikipedia , lookup Elementary mathematics wikipedia , lookup Transcript ```Warm Up 11/9/2015  Who makes it, has no need of it. Who buys it, has no use for it. Who uses it can neither see nor feel it. What is it? A Coffin!!! Squares & Square Roots Perfect Squares Square Number  Also called a “perfect square”  A number that is the square of a whole number  Can be represented by arranging objects in a square. Square Numbers Square Numbers 1x1=1 2x2=4 3x3=9  4 x 4 = 16 Square Numbers 1x1=1 2x2=4 3x3=9  4 x 4 = 16 Activity: Calculate the perfect squares up to 152… Square Numbers 1x1=1  9 x 9 = 81 2x2=4  10 x 10 = 100 3x3=9  11 x 11 = 121  4 x 4 = 16  12 x 12 = 144  5 x 5 = 25  13 x 13 = 169  6 x 6 = 36  14 x 14 = 196  7 x 7 = 49  15 x 15 = 225  8 x 8 = 64 Activity: Identify the following numbers as perfect squares or not. i. ii. iii. iv. v. vi. 16 15 21 36 64 71 Activity: Identify the following numbers as perfect squares or not. 16 = 4 x 4 ii. 15 iii. 21 iv. 36 = 6 x 6 v. 64 = 8 x 8 vi. 71 i. Squares & Square Roots Square Roots Finding a root of a number is the inverse operation of raising a number to a power. index n a The index defines the root to be taken. Square Root  A number which, when multiplied by itself, results in another number.  Ex: 5 is the square root of 25. 5 = 25 Every positive number has two square roots, one positive and one negative   When you calculate the square root of a number on a calculator, only the positive square root appears. This is the principal square root.  Principal Square RootThe non-negative square root of a number. Activity: Find the principal square roots of the following numbers  Warm Up 11/10/15  What gets wetter and wetter the more it dries? A towel!!! a b ab ONLY when a≥0 and b≥0 For Example: 9  16  9 16  144  12 9  16  3  4  12 Equal Simplify the following expressions  Simplifying Square Roots Write the following as a radical (square root) in simplest form: 36 is the biggest perfect square that divides 72. Simplify. 72  36  2  36 2  6 2 Rewrite the square root as a product of roots. 27  9  3  9 3  3 3 Ignore the 5 multiplication until the end. 5 32  5 16  2  5 16 2  5  4  2  20 2 Warm Up 11/12/15 A) 16 4 B) 8 C) 7 E )4 63  12 2 2 D) 75 7 5 3 F ) 128  8 2 Warm Up Part 2 11/12/15  A man is pushing his car along, and when he reaches a hotel he shouts “I’m bankrupt!” Why? He’s playing Monopoly!!! a  b a b ONLY when a≥0 and b>0 For Example: 64 16 64 16  64 16   4 2 8 4 2 Equal Simplify the expressions  The Square Root of a Fraction Write the following as a radical (square root) in simplest form: Take the square root of the numerator and the denominator 3 3 3   2 4 4 Simplify. Simplify the expressions  Warm Up 11/13/15 Define the following (you can look these up on your computer if you don’t know them!): Rational Number any real # that can be expressed as the quotient of 2 integers Irrational Numberany real # that cannot be expressed as a ratio of integers. What can run but never walks, has a mouth but never talks, has a bed but never sleeps? A river!!!! Rationalizing a Denominator The denominator of a fraction cannot contain a radical. To rationalize the denominator (rewriting a fraction so the bottom is a rational number) multiply by the Simplify the following expressions: 5 2 5 2 5 2    2 2 2 2 2   6 3 6 3 3 2 3 2 3 6 3 3       2 35 5 15 53 3 5 3 5 3 6   Why do we rationalize the denominator?  The main reason we do this is to have a standard form in which certain kinds of answers can be written. That makes it easier for us as teachers to check answers, and for the students to check their own
# What are the angles? Classification, types and names Angles are fundamental geometric figures in various branches of mathematics, particularly geometry. Angles indicate the difference in inclination between two lines or two planes and is measured in degrees or radians. In geometry, we refer to this concept to refer to the set that forms the vertex and the two edges that form it. In geometry there are many figures that are defined according to the shape and dimension of their angles. For example, regular polygons have all angles equal. Among other regular polygons we can highlight the square (with all its edges at ninety degrees), the pentagon, or the equilateral triangle. ## What are the angles? Angles are a measure that indicates the inclination of a straight line with respect to another. In three-dimensional space, they are also used to determine the difference in inclination between two planes. When we talk about geometric figures, an angle is the part of the figure made up of two sides that intersect at a point called the vertex. For example, a square is made up of four angles that measure 90 degrees on the inside and 270 degrees on the outside. This measurement can be expressed in degrees or radians. A circle is divided into 360º or 2π radians forming a complete angle. ## angle classification These geometric elements can be classified in different ways: • The angles according to their position between them can be classified as: adjacent, consecutive, opposite angles by the vertex, interior and exterior of a geometric figure. • According to the arithmetic relations between them. In this sense, they can be congruent, complementary, supplementary or conjugated. • Determined by two parallels and a transversal: these can be parallel lines cut by a secant, corresponding and alternate. • The angles according to their measure are differentiated by the position of their ends and can be: right, acute, obtuse and flat. ## What are the types of angles? • The right angle is the one that has a measure of 90°, and is formed when the two lines intersect perpendicularly. • An acute angle is one that has a measure less than 90°. • An obtuse angle is one that has a measure greater than 90° but less than 180°. • The straight or flat angle is the one that has a measure equal to 180 °. Depending on the relationship they have with other angles they receive the following names: • Complementary angles: the sum of their degrees is equal to 90º. • Consecutive: are angles that share a vertex and a side. • Adjacent angles: they also share a vertex and a side but also their two add up to 180º. • Convex angle: is between 0 and 180 degrees. • Concave angle: It is more than 180 degrees and therefore not convex. • Interior: The interior angle is the angle between two consecutive edges of a closed geometric figure measured from the interior of the figure. • Exterior: is formed by two consecutive sides of a figure measured from the outside of the figure. ## Formula to convert radians to degrees Angles can be measured in degrees or radians. To convert from one unit to another, we use the formula: • degrees = radians * 180 / π. • radians = degrees * π / 180. A complete circle consists of 360 degrees and 2π radians. This means that 1 radian is equal to 360 / 2π = 57.29578 degrees. Therefore, we can say that: • 1 degree = π / 180 ≈ 0.0174 radians. • 1 radian ≈ 57.29578 degrees. Author: Publication Date: February 8, 2023 Last Revision: February 8, 2023
Area And Perimeter Of Square Perimeter and area are important concepts when it comes to understanding squares. Let’s break them down in a simple way. Perimeter is the distance around a shape. Imagine you’re walking around the edge of a garden. The perimeter tells you how much distance you would cover. The word “perimeter” comes from Greek words that mean “measure” and “around.” Area, on the other hand, refers to the space inside a shape. It’s like the floor space inside a garden. The word “area” comes from a Latin word that means “open space.” This teaching guide aims at simplifying the concept of the perimeter and area of a square shape and helping educators further by providing teaching resources like animated stories, and impressive posters along with hands-on activities. ## Square : Now, let’s focus specifically on squares. A square is a special shape with four equal sides and four equal angles. It’s like a closed box with all sides being the same length. ### Perimeter of a square : To find the perimeter of a square, you simply add up the lengths of all four sides. Since all sides of a square are equal, you can calculate the perimeter by multiplying the length of one side by four. If ‘a’ denotes the side of the square, then the length of each side of a square is ‘a’ units Perimeter of the square = PQ+ QR + RS + SP = (a + a + a + a) units = 4a units The perimeter of the square = 4a units ### Area of a square : When it comes to the area of a square, it tells you how much space is inside the square. It’s like the total floor space it covers. • By counting the unit squares covered by the square : To calculate the area of a two-dimensional shape, we can break it down into smaller squares. These squares are called unit squares, and each one takes up one square unit of space. By counting the total number of unit squares in the shape, we can determine its area. The area of a shape is measured using square units. In the case of a square, we can find its area by using unit squares. Here, the area of the square PQRS = 9 square units. • By using the formula : To find the area, you multiply the length of one side by itself. In other words, you square the length of a side. In mathematical terms, we can represent the area of a square as “A = side × side” or “A = side².” Here, “A” represents the area, and “side” represents the length of one side. Area = side × side Here, in the figure side = 3 units A = 3 x 3 square units A = 9 square units So, whether you want to calculate the distance around a square (perimeter) or find out how much space it covers (area), understanding these concepts can help you work with squares more effectively. Posters Teaching area and perimeter of a square with kid-friendly, clear, and easy-to-understand posters from Uncle Math School by Fun2Do Labs Stories Ignite kids’ curiosity with engaging stories for role play and skits, making the learning of this concept an exciting and effective experience. Teaching area and perimeter of a square through stories from Uncle Math School by Fun2Do Labs : Text of Stories Activities Learning the area and perimeter of a square can be made enjoyable by incorporating interactive games and activities. ### Draw And Determine! This is a simple activity to help children practise areas and perimeters in an easy way. The activity can be carried out in the following steps : • Instruct kids to grab 2 square-shaped objects, like a table coaster and a handkerchief. • Let them measure the sides of both square-shaped objects using a measuring tape or ruler and record them. • Next, determine the area and perimeter of these objects by using the formula : 1. Area = s x s 2. Perimeter = 4 x side Worksheets Help your kids practise the area and perimeter of a square with interesting and engaging fun worksheets and solutions from Uncle Math by Fun2Do Labs. Worksheet 330 : Perimeter Of Square Solution 330 : Perimeter Of Square Worksheet 333 : Area And Perimeter Solution 333 : Area And Perimeter Worksheet 334 : Area And Perimeter Solution 334 : Area And Perimeter
# How to do Matrix Multiplication If we think about how to do matrix multiplication, it’s straightforward when we just want to multiply a matrix by a single number. 2 × \begin{bmatrix} 4 & 1 & 2 \\ 3 & 5 & 1 \end{bmatrix} = \begin{bmatrix} 2\times4 & 2\times1 & 2\times2 \\ 2\times3 & 2\times5 & 2\times1 \end{bmatrix} = \begin{bmatrix} 8 & 2 & 4 \\ 6 & 10 & 2 \end{bmatrix} We just multiply each element of the matrix by the number. Things are a little bit more complex when we want to multiply matrices together however. ## Multiplying Matrices Together We can look at the case of multiplying a 2 x 2 matrix by another 2 x 2 matrix. \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} × \begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{bmatrix} = \begin{bmatrix} a_{11}b_{11}+a_{12}b_{21} & a_{11}b_{12}+a_{12}b_{22} \\ a_{21}b_{11}+a_{22}b_{21} & a_{21}b_{12}+a_{22}b_{22} \end{bmatrix} Now to see an example using numbers, with . representing multiplication, this looks like: \begin{bmatrix} 1 & 3 \\ {\text{-}}4 & 1 \end{bmatrix} × \begin{bmatrix} 5 & 2 \\ 1 & 3 \end{bmatrix} = \begin{bmatrix} 1.5+3.1 & 1.2+3.3 \\ {\text{-}}4.5+1.1 & {\text{-}}4.2+1.3 \end{bmatrix} = \begin{bmatrix} 8 & 11 \\ {\text{-}}19 & {\text{-}}5 \end{bmatrix} The result of multiplying two 2 x 2 matrices together was another 2 x 2 matrix. But generally the size of the new matrix resulting from the multiplication, depends on how many rows and columns are in the matrices being multiplied. Along with the fact that matrices can only be multiplied together if one matrix has the same number of columns as the other does rows. ## How to do Matrix Multiplication, Steps A matrix can be denoted by: ( ROWS × COLUMNS ). So if we have 2 matrices ( m × n ) and ( p × q ), and if we want to multiply them together, ( m × n ) × ( p × q ). The first step is to check that n and p are equal. Which is checking that the number of columns in the first matrix equals the amount of rows in the second. Then if this is satisfied, the matrices can be multiplied, and the resulting matrix from the multiplication will be of size ( m × q ). This will probably look a bit clearer with an example involving numbers. Example (1.1) \begin{bmatrix} 2 & 4 & 3 \\ 1 & 2 & 1 \end{bmatrix} × \begin{bmatrix} 2 & 4 \\ 3 & 1 \\ 5 & 4 \end{bmatrix} Solution This is a ( 2 × 3 ) matrix multiplying a ( 3 × 2 ) matrix. So the first matrix has the same number of columns as the second matrix has rows, which is what we need. We can also see that the resulting matrix from multiplication should be a ( 2 × 2 ) matrix. \begin{bmatrix} 2 & 4 & 3 \\ 1 & 2 & 1 \end{bmatrix} × \begin{bmatrix} 2 & 4 \\ 3 & 1 \\ 5 & 4 \end{bmatrix} = \begin{bmatrix} 2.2+4.3+3.5 & 2.4+4.1+3.4 \\ 1.2+2.3+1.5 & 1.4+2.1+1.4 \end{bmatrix} = \begin{bmatrix} 4+12+15 & 8+4+12 \\ 2+6+5 & 4+2+4 \end{bmatrix} = \begin{bmatrix} 31 & 24 \\ 13 & 10 \end{bmatrix} ### NOTE: Matrix Order of Multiplication It’s important to pay attention to order when learning how to do matrix multiplication. As if you have two matrices A and B, generally A×BB×A. We can look at the same matrices that we multiplied above, but this time reverse the order. \begin{bmatrix} 2 & 4 \\ 3 & 1 \\ 5 & 4 \end{bmatrix} × \begin{bmatrix} 2 & 4 & 3 \\ 1 & 2 & 1 \end{bmatrix} This order now gives us a ( 3 × 2 ) × ( 2 × 3 ) matrix multiplication. Which will produce a ( 3 × 3 ) matrix. \begin{bmatrix} 2 & 4 \\ 3 & 1 \\ 5 & 4 \end{bmatrix} × \begin{bmatrix} 2 & 4 & 3 \\ 1 & 2 & 1 \end{bmatrix} = \begin{bmatrix} 2.2+4.1 & 2.4+4.2 & 2.3+4.1 \\ 3.2+1.1 & 3.4+1.2 & 3.3+1.1 \\ 5.2+4.1 & 5.4+4.2 & 5.3+4.1 \end{bmatrix} = \begin{bmatrix} 4+4 & 8+8 & 6+4 \\ 6+1 & 12+2 & 9+1 \\ 10+4 & 20+8 & 15+4 \end{bmatrix} = \begin{bmatrix} 8 & 16 & 10 \\ 7 & 14 & 10 \\ 14 & 28 & 19 \end{bmatrix} ( When practicing how to do matrix multiplication it’s handy to be able to quickly check your answers. A handy matrix multiplication calculator is available to use at the matrix reshish website. ) ## Identity Matrix Multiplication A specific case to consider when learning how to do matrix multiplication, is that multiplication of a matrix with the identity matrix, regardless of order, results in the same matrix. \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} × \begin{bmatrix} 4 & 3 \\ 1 & 9 \end{bmatrix} = \begin{bmatrix} 4+0 & 3+0 \\ 0+1 & 0+9 \end{bmatrix} = \begin{bmatrix} 4 & 3 \\ 1 & 9 \end{bmatrix} \begin{bmatrix} 4 & 3 \\ 1 & 9 \end{bmatrix} × \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 4+0 & 0+3 \\ 1+0 & 0+9 \end{bmatrix} = \begin{bmatrix} 4 & 3 \\ 1 & 9 \end{bmatrix} This example was just with a simple ( 2 × 2 ) matrix, but the result will be the same with larger matrices also. ## Properties of Matrix Multiplication To round off this how to do matrix multiplication page we’ll display a short list of properties of matrix multiplication as a summary. 1) A × ( BC ) = AB × C ( associative property ) 2) cA × B = A × cB ( where c is a constant ) 3) AI = A , IA = A ( where I is the identity matrix ) 1. Home 2. › 3. Algebra 2 4. › Matrix Multiplying
We have seen that the distributive law can be used to rewrite a product involving brackets as an expression without brackets. For instance, the product $$3(a+2)$$ can be rewritten as $$3a+6$$; this is called the expanded form of the expression and the process is called expansion. The process of writing an algebraic expression as a product of two or more algebraic factors is called factorisation. Factorisation is the reverse process to expansion. In the example above, $$3(a+2)$$ is the factorised form of the algebraic expression. ### Factorise using common factors If each term in the algebraic expression to be factorised contains a common factor, then this common factor is a factor of the entire expression. To find the other factor, we divide each term by the common factor. The common factor is placed outside brackets. For this reason, the process is sometimes called 'taking the common factor outside the brackets'. #### Example 1 Factorise $$4a+12$$. #### Solution 4 is a common factor of $$4a$$ and 12 thus $$4a+12=4(a+3)$$ #### Example 2 Factorise $$3x+9$$. #### Solution 3 is a common factor of $$3x$$ and 9 thus $$3x+9=3(x+3)$$ #### Example 3 Factorise $$x^2-4x$$. #### Solution $$x$$ is a common factor of $$x^2$$ and $$4x$$ thus $$x^2–4x=x(x-4)$$ #### Example 4 Factorise $$2xy+4x^2-6xz$$. #### Solution $$2x$$ is a common factor of $$2xy, 4x^2\ \text{and}\ 6xz$$ thus $$2xy+4x^2-6xz =2x(y+2x-3z)$$
Do you recognize that once you main point the two consecutive numbers 10 and 11, you get 110? In this chapter, we will calculate the components of 110, prime factors of 110, and factors the 110 in pairs along with solved examples for a much better understanding. You are watching: What is the prime factorization of 110 Factors of 110: 1, 2, 5, 10, 11, 22, 55 and 110Prime factorization of 110: 110 = 2 × 5 × 11 1 What room the components of 110? 2 How to Calculate determinants of 110? 3 Factors that 110 by element Factorization 4 Factors the 110 in Pairs 5 FAQs on determinants of 110 ## What are the factors of 110? Factors that 110 are the number which as soon as multiplied in pairs offer the product as 110. The components of 110 space those number which division it without leaving any type of remainder. So, as soon as you multiply any two entirety numbers through each other and get 110 together the answer, you deserve to say the both this numbers are the factors of 110.Observe the adhering to numbers. You obtain 110 together the product after multiply the given numbers. 1 × 110 = 110 2 × 55 = 110 5 × 22 = 110 10 × 11 = 110 11 × 10 = 110 22 × 5 = 110 55 × 2 = 110 110 × 1 = 110 So, we can say that "The determinants of 110 are all the integers that 110 can be divided into." ## How to calculate the components of 110? Let"s calculate the factors of 110, starting with the smallest whole number, i.e., 1.Divide 110 with this number. Is the remainder 0? Yes! So, us get 110/1 = 110 110 × 1 = 110 The next whole number is 2 Now, divide 110 v this number. 110/2 = 55 2 × 55 = 110 Proceeding in a comparable manner we get, 110 = 1 × 110 = 2 × 55 = 5 × 22 = 10 ×11 Finally, we gain all the components of 110 as : Hence, the factors the 110 are 1, 2, 5, 10, 11, 22, 55 and also 110. Explore factors using illustrations and interactive examples. ## Factors that 110 by prime Factorization "Prime factorization method to refer a composite number as the product of its prime factors". To acquire the element factorization the 110, we division it by its the smallest prime variable which is 2 110/2 = 55 Now, 55 is divided by its the smallest prime factor and the quotient is obtained. This procedure goes ~ above till we get the quotient together 1. The prime factorization the 110 is presented below: Hence, the prime determinants of 110 are 2, 5, and 11. ## Factors the 110 in Pairs The bag of numbers, which when multiplied provide 110, are known as the variable pairs of 110. Observe the adhering to factors of 110 in pairs. Product form of 110 Pair factors 1 × 110 = 110 (1, 110) 2 × 55 = 110 (2, 55) 5 × 22 = 110 (5, 22) 10 × 11 = 110 (10, 11) As seen in the table offered above, after 10 × 11, there are no new factors. So, that is sufficient to find components till (10,11). If us consider an adverse integers, then both the number in the pair components will it is in negative. Due to the fact that ( - ve × - ve ) = +ve, we deserve to list the element pairs that 110 as (-1, -110); (-2, -55); (-5, -22), and also (-10, -11). Important Notes: As 110 ends v the digit 0, the will have actually 5 and 10 as its factors. This hold true for all the numbers that finish with 0110 is a non-perfect square number. Thus, the will have actually even number of factors. This residential or commercial property holds true because that every non-perfect square number. See more: How Long Is 4 Kilometers In Miles Conversion Plus Calculator (Kilometers To Mi) Example 1: Peter to buy two rectangle-shaped rugs which had various dimensions, yet had the very same area of 110 inches². He wanted his friend Andrew come guess the size of the rugs. Could you help him through the 2 probable pairs other than (1,110) and also (55, 2), the would an outcome in the provided area? Solution: Andrew told that (5, 22) and also (10,11) were the two variable pairs of 110 which could be the dimensions of the two rugs. Let united state see how he defined it.Since Area that a rectangle = length × breadthFor the an initial rug, Area = 22 × 5 = 110 in² For the 2nd rug, Area =10 × 11= 110 in² Hence, the two element pairs space (22, 5) and (10, 11). Example 2: Robert"s teacher told him that -55 is one of the determinants of 110. Have the right to you aid him find the other factor? Solution: 110 = element 1 × aspect 2Hence, 110 =(-55) × element 2Factor 2 = 110/(-55) =(-2)Hence, the other factor is -2
# Algebraic Expression Algebraic expression, or variable expression, is a mathematical expression consisting of two main parts, variables and constants, joined together using mathematical operators addition, subtraction, multiplication, division, and exponentiation. Here are a few examples of algebraic expressions: • 2x + 9 • 5x + 7y + 3z – 11 • ${5x^{2}+4y-3}$ ## Parts Let us consider the algebraic expression 2x + 9 to understand its parts: Here, • x is a variable (an unknown value)Â • 9 is a constant (a fixed value) • 2x and 9 are the terms, and • 2 is a coefficient of x ### Variables A symbol whose value is unknown to us is called a variable. It is denoted by small letters and can take any value. a, b, and x are a few examples of variables. ### Constants A symbol with a fixed, definite numerical value is called a constant. 1, 2, and 50 are constants. ### Coefficients The number multiplied by the variable is called its coefficient. Let us consider the expression 5y + z. Here, the coefficient of y is 5, and the coefficient of z is 1. ### Terms A term can be a variable alone, a number, a product of two or more variables, or a product of a number and a variable. These numbers or variables are called the factors of a term. Thus, a term or a group of terms form an expression, and a term is formed by multiplying its factors. An example of a term is ${13xy^{2}}$, where 13, x, y, y are factors. 1. Like Terms: Terms containing the same variables with the same exponents are called like terms.Â 2x and 3x, 4y and -9y are some examples of like terms.Â 2. Unlike Terms: Terms containing the same variables with different exponents or containing different variables are called unlike terms. 2x and 3xy, 2x and ${5x^{2}}$ are some examples of unlike terms. In an algebraic expression, terms can be of 2 types: ## Types Based on the number of unlike terms, the algebraic expressions are classified into the following types: ### Monomial An algebraic expression consisting of only one unlike term is called a monomial expression. Examples of some monomial expressions are 3x, 4xyz, and ${2x^{2}}$. ### Binomial An algebraic expression with two unlike terms is known as a binomial expression. Examples of some binomial expressions are 2x + y, 4z + 7, and ${10x^{2}+5x^{3}}$. ### Trinomial A trinomial expression is an expression with three unlike terms. Examples of some trinomial expressions are ${7x^{2}+3x+5}$, – 6x + 3 + 3y, and 5x + 7y + 3z. Binomial and trinomial expressions are also known as multinomial expressions (an expression with more than one unlike term). ### Polynomial An expression having one or more than one unlike terms with positive integral exponents of variables is known as a polynomial expression. Some examples of polynomial expressions are 2x, 3x + 11, x + y + z – 7, etc. ### Other Types #### Numeric Expression A numeric expression containing only numbers and operations but never has any variable. Some examples of numeric expressions are 5, 50, 5 + 39, and 17 x 14. #### Equivalent Expressions Two expressions are equivalent if they are equal in value but look different. Some examples of equivalent expressions are 5x – 7y and – 7y + 5x, 3a + b and b + 3a. Now, let us understand the operations of the algebraic expressions in detail. ## Simplifying To simplify an algebraic expression, first, we must combine the like terms and then perform the basic mathematical operations (addition, subtraction, multiplication, and division) following the rule of PEMDAS. ### Combining Like Terms All the like terms are combined and solved to obtain a simplified algebraic expression. Let us consider 2x + 4y – 6x + 3y + ${3xy^{2}}$ – ${8xy^{2}}$ and reduce it to its lowest form to simplify. Grouping the like terms we get, (2x – 6x) + (4y + 3y) + (${3xy^{2}}$ – ${8xy^{2}}$) To further simplify, we need to add or subtract the like terms. To simplify the above algebraic expression further, we add or subtract all the like terms. For simplifying, we get, (2x – 6x) + (4y + 3y) + (${3xy^{2}}$ – ${8xy^{2}}$) Now, adding the expressions, we get, (2x – 6x) + 7y + (${3xy^{2}}$ – ${8xy^{2}}$) And subtracting the expressions, we get, -4x + 7y – ${5xy^{2}}$ Thus, 2x + 4y – 6x + 3y + ${3xy^{2}}$ – ${8xy^{2}}$ can be simplified as -4x + 7y – ${5xy^{2}}$. Sometimes, an expression can also have multiplication and division operators. Then, we need to multiply and divide to simplify further. ### Multiplying and Dividing Every term of the first expression is multiplied by every term of the second to multiply the algebraic expressions. Whereas for dividing the algebraic expressions, first, we factor the numerator and denominator, cancel all the possible terms, and then simplify the rest. Let us take the algebraic expressions as 2x + 10 and ${\dfrac{x-4}{2}}$ Now, multiplying the expressions, we get, (2x + 10)${\left( \dfrac{x-4}{2}\right)}$ = ${\dfrac{2x\left( x-4\right) +10\left( x-4\right) }{2}}$ = ${\dfrac{2x\times x-2x\times 4+10\times x-10\times 4}{2}}$ = ${\dfrac{2x^{2}-8x+10x-40}{2}}$ By combining the like terms, we get, ${\dfrac{2x^{2}+\left( -8x+10x\right) -40}{2}}$ Adding the like terms, we get, ${\dfrac{2x^{2}+2x-40}{2}}$ Before divisions, we must find the common factors among all the terms and represent the remaining ones. By finding the common factors, we get, ${\dfrac{2\left( x^{2}+x-20\right) }{2}}$ Now, dividing the expressions, we get, ${x^{2}+x-20}$ Thus, 2x + 10 and ${\dfrac{x-4}{2}}$ can be simplified as ${x^{2}+x-20}$. Sometimes, we use some algebraic expression formulas to easily solve the expressions or equations. ## Formulas Following is a list that shows some of the general formulas. ## Solved Examples Which algebraic expression is a polynomial (check all that apply and write the type): a) 4 – 5x + 2y b) 2x + 5 – 2(x +2) – 1 c) ${4-7x^{2}+5+2x^{2}}$ d) ${5x^{3}+2x^{2}-8x+17x^{3}-15+8x}$ Solution: a) Here, the unlike terms are 4, – 5x, 2y. Thus, 4 – 5x + 2y is a trinomial expression, and it is a polynomial. b) The given expression is 2x + 5 – 2(x +2) – 1 Now, by solving the brackets, we get, 2x + 5 – 2x – 4 – 1 By combining the like terms, we get, (2x – 2x) + (5 – 4 – 1) = 0 Thus, 2x + 5 – 2(x +2) – 1 is not a polynomial expression. c) The given expression is ${4-7x^{2}+5+2x^{2}}$ Now, by combining the like terms, we get, (4 + 5) + ${\left( -7x^{2}+2x^{2}\right)}$ = 9 – ${5x^{2}}$ Here, the unlike terms are 9, ${-5x^{2}}$. Thus, ${4-7x^{2}+5+2x^{2}}$ is a binomial expression, and it is a polynomial. d) The given expression is ${5x^{3}+2x^{2}-8x+17x^{3}-15+8x}$ Now, by combining the like terms, we get, ${\left( 5x^{3}+17x^{3}\right) +2x^{2}+\left( 8x-8x\right) -15}$ = ${22x^{3}+2x^{2}-15}$ Here, the unlike terms are ${22x^{3}}$, ${2x^{2}}$, – 15. Thus, ${5x^{3}+2x^{2}-8x+17x^{3}-15+8x}$ is a trinomial expression, and it is a polynomial. Simplify:Â a) ${\dfrac{9x^{2}y}{15x^{3}}}$ b) ${6x^{2}y+4x^{2}-3x^{2}y-2x^{2}}$ c) (x + 5)(x – 4) Solution: a) The given expression is ${\dfrac{9x^{2}y}{15x^{3}}}$ By finding all the factors, we get, ${\dfrac{3\times 3\times x\times x\times y}{3\times 5\times x\times x\times x}}$ = ${\dfrac{3y}{5x}}$ Thus, the simplified expression is ${\dfrac{3y}{5x}}$.Â b) The given expression is ${6x^{2}y+4x^{2}-3x^{2}y-2x^{2}}$ By combining all the like terms, we get, ${\left( 6x^{2}y-3x^{2}y\right) +\left( 4x^{2}-2x^{2}\right)}$ = ${3x^{2}y+2x^{2}}$ Thus, the simplified expression is ${3x^{2}y+2x^{2}}$. c) The given expression is (x + 5)(x – 4) Now, multiplying the factors, we get, x(x – 4) + 5(x – 4) = ${x^{2}-4x+5x-20}$ By combining the like terms, we get, ${x^{2}+\left( -4x+5x\right) -20}$ = ${x^{2}+x-20}$ Thus, the simplified expression is ${x^{2}+x-20}$. a) 3x + 7 and x + 2y + 9 b) ${7x^{2}+3x+5}$ and ${-3x^{2}+7x+y-9}$ Solution: a) Given, (3x + 7) + (x + 2y + 9) = 3x +7 + x + 2y + 9 By combining all the like terms, we get, (3x + x) + 2y + (7 + 9) = 4x + 2y + 16 Thus, the required sum is 4x + 2y + 16. b) Given, (${7x^{2}+3x+5}$) + (${-3x^{2}+7x+y-9}$) = ${7x^{2}+3x+5}$ + ${-3x^{2}+7x+y-9}$ By combining all the like terms, we get, (${7x^{2}}$ ${-3x^{2}}$) + (3x + 7x) + y + (5 – 9) = ${4x^{2}}$ +10x + y – 4 Thus, the required sum is ${4x^{2}}$ +10x + y – 4. Subtract: $({7x^{2}+3x+5)}$ – ${(-3x^{2}+7x+y-9)}$ Solution: Given, (${7x^{2}+3x+5}$) – (${-3x^{2}+7x+y-9}$) = ${7x^{2}+3x+5}$ + ${3x^{2}-7x-y+9}$ By combining all the like terms, we get, (${7x^{2}}$ + ${3x^{2}}$) + (3x – 7x) – y + (5 + 9) = ${10x^{2}}$ – 4x – y + 14 Thus, the required sum is ${10x^{2}}$ – 4x – y + 14.
# Polynomial Functions Example 1: A rectangle has a perimeter of 40 ft. What is the maximum area of the rectangle? Assign the width of the rectangle the variable x. The sum of the length and the width is 20. So in terms of x, the length is 20 − x . The function A (x) = the area of the rectangle depending upon its width: A (x) = x (20 − x) Remember, that area = width · length. Now to find its maximum value, find its vertex: The above is the x- coordinate of the vertex . The following is the y-coordinate of the vertex, the maximum value: The maximum area is 100 ft2 when the length and width are 10 ft each. Example 2: Let's try problem 12 on page 72. Let I (d) = the income for the fruit based on the days waited: The vertex is (5, 25000). The grower should wait 5 days. The maximum income is \$25,000. Example 3: Try problem 4 on page 73 of your text. If the perimeter is 30 in, then the sum of the length and width is 15. The entire width is 3 x, so the length is 15 − 3 x. The length of the rectangle becomes the height of the prism. To find the volume of a prism, you find the area of the base, the equilateral triangle, and multiply it by the height of the prism, 15 − 3 x. From geometry, we know each angle of an equilateral triangle measures 60°. The altitude is the height of the triangle, and it will bisect the third side. The triangle formed by drawing the altitude has angle measurements of 30°, 60°, and 90°. The length of the altitude is found by multiplying half the base, .5 x, by . Now find the area of the triangle: Area = .5 (base) (height). Area of triangular base is The volume of the prism = area of base× height: The domain of V (x) consists of the values of x that can be used in the function. Because x represents one side of the triangular base, it can't be less than or equal to zero. Also, 15 − 3 x represents the height of the prism, 15 − 3 x > 0: 0 < x < 5. If you have a graphing calculator, graph the function and trace the values of x and y within the range 0 < x < 5. What would be the maximum volume, or the largest value of y? Study Exercises Complete odd- numbered problems 1–7 in the Written Exercises section on page 71 of your text. Section 2-5: Using Technology to Approximate Roots of Polynomial Equations pp. 75–80 We have discussed that roots of polynomial equations can be found by factoring, using synthetic division, and by graphing. In this section, we will use the LOCATION PRINCIPLE to approximate roots. LOCATION PRINCIPLE If P (x) is a polynomial with real coefficients and a and b are real numbers such that P (a) and P (b) have opposite signs, then between a and b there is at least one real root r of the equation P (x) = 0. The idea behind the location principle is that zero is between positive and negative numbers. So if you place values into a function and the value of the function changes in sign, the function had a value of zero between the two values . Your text gives a program you can type into a programmable calculator to find several function values. If your calculator is not programmable, you can keep entering values into the function and keep a chart of the values. Example 1: Let's try just one problem, problem 16 on page 78 of your text. If all faces of the cube have x cm shaved off, all dimensions will be decreased by 2 x. The original volume was 3×5×4 = 60 cm3. We are looking for the number of centimeters shaved off to give half of that volume, 30 cm3. Let V (x) = the volume of a cube when x cm is shaved off each face: The domain is 0 < x < 1.5, because you cannot shave 1.5 cm or more from each side of a 3-cm length, the smallest dimension. We are looking for an x-value that will produce V (x) = 0. A table of values follows. x-value V(x)-value You can see by looking at the table that the value of V (x) changed in sign between V (.25) and V (.5). So I keep trying x-values between .25 and .4. The directions in your text state to find a value approximated to the nearest hundredth. I found that V (x) has a zero-value .39 < x < .4, and V (.39) is closer to zero than V (.4). Each face of the solid will need to be shaved .39 cm to obtain a rectangular solid half its original volume. Study Exercises Complete odd-numbered problems 1–15 in the Written Exercises section on page 78 of your text. Section 2-6: Solving Polynomial Equations by Factoring, pp. 80–85 Factoring is a very powerful tool for finding roots of polynomial equations. Some polynomial equations of a degree higher than two can be factored by grouping or by ordinary factoring methods. Example 1: Solve 4 x3 + 3 x2 − 36 x − 27 = 0. Factor by grouping into two sets of binomials : (4 x3 + 3 x2) + (−36 x − 27) = 0. Factor out the GCF of each binomial : x2 (4 x + 3) − 9 (4 x + 3) = 0. Notice a GCF of (4 x + 3). Factor it out: The solutions are , and x = 3. Example 2: Solve x3 − x2 = 20 x: The solutions are x = 0, x = 5, and x = − 4. Example 3: Solve 8 (2 x + 1)2 + 10 (2 x + 1) = 3: 8 (2 x + 1)2 + 10 (2 x + 1) − 3 = 0. This equation can be written in quadratic form if 2 x + 1 is replaced by y. Factor: Replace y with 2 x + 1: Example 4: Solve 4 x6 + 17 x3 − 15 = 0. This equation can be written in quadratic form if x3 is replaced by y: Replace y with x3: Take the cube root of both sides: Rationalize the denominator in the first solution by multiplying by : The solutions are Let's examine 18 x2 + 3 x − 10 = 0 closely. It can be factored as (6 x + 5) (3 x − 2) = 0. The solutions are and . Note the factors of 18 (6 and 3) are in the denominators of the solutions. Also the factors of 10 (5 and 2) are in the numerators of the solutions . If you were trying to find solutions to an equation you were not capable of factoring, you could make a list of possible rational roots. The possible rational roots have factors of the leading coefficient in the denominators and factors of the constant term in the numerators. The RATIONAL ROOT THEOREM states that fact. Prev Next
# 6.7 Nonterminating divisions Page 1 / 1 This module is from Fundamentals of Mathematics by Denny Burzynski and Wade Ellis, Jr. This module discusses nonterminating divisions. By the end of the module students should understand the meaning of a nonterminating division and be able to recognize a nonterminating number by its notation. ## Section overview • Nonterminating Divisions • Denoting Nonterminating Quotients ## Nonterminating divisions Let's consider two divisions: 1. $9\text{.}\text{8}÷\text{3}\text{.}5$ 2. $4÷\text{3}$ ## Terminating divisions Previously, we have considered divisions like example 1 , which is an example of a terminating division. A terminating division is a division in which the quotient terminates after several divisions (the remainder is zero ). ## Exact divisions The quotient in this problem terminates in the tenths position. Terminating divisions are also called exact divisions . ## Nonterminating division The division in example 2 is an example of a nonterminating division. A non-terminating division is a division that, regardless of how far we carry it out, always has a remainder . ## Repeating decimal We can see that the pattern in the brace is repeated endlessly. Such a decimal quotient is called a repeating decimal . ## Denoting nonterminating quotients We use three dots at the end of a number to indicate that a pattern repeats itself endlessly. $\text{4}÷\text{3}=\text{1}\text{.}\text{333}\dots$ Another way, aside from using three dots, of denoting an endlessly repeating pattern is to write a bar ( ¯ ) above the repeating sequence of digits. $\text{4}÷\text{3}=\text{1}\text{.}\overline{3}$ The bar indicates the repeated pattern of 3. Repeating patterns in a division can be discovered in two ways: 1. As the division process progresses, should the remainder ever be the same as the dividend, it can be concluded that the division is nonterminating and that the pattern in the quotient repeats. This fact is illustrated in [link] of [link] . 2. As the division process progresses, should the "product, difference" pattern ever repeat two consecutive times, it can be concluded that the division is nonterminating and that the pattern in the quotient repeats. This fact is illustrated in [link] and 4 of [link] . ## Sample set a Carry out each division until the repeating pattern can be determined. $\text{100}÷\text{27}$ When the remainder is identical to the dividend, the division is nonterminating. This implies that the pattern in the quotient repeats. $\text{100}÷\text{27}=3\text{.}\text{70370370}\dots$ The repeating block is 703. $\text{100}÷\text{27}=3\text{.}\overline{\text{703}}$ $\text{1}÷\text{9}$ We see that this “product, difference”pattern repeats. We can conclude that the division is nonterminating and that the quotient repeats. $1÷9=0\text{.}\text{111}\dots$ The repeating block is 1. $1÷9=0\text{.}\overline{1}$ Divide 2 by 11 and round to 3 decimal places. Since we wish to round the quotient to three decimal places, we'll carry out the division so that the quotient has four decimal places. The number .1818 rounded to three decimal places is .182. Thus, correct to three decimal places, $\text{2}÷\text{11}=\text{0}\text{.}\text{182}$ Divide 1 by 6. We see that this “product, difference” pattern repeats. We can conclude that the division is nonterminating and that the quotient repeats at the 6. $1÷6=0\text{.}1\overline{6}$ ## Practice set a Carry out the following divisions until the repeating pattern can be determined. $1÷3$ $0\text{.}\overline{3}$ $5÷6$ $0\text{.}8\overline{3}$ $\text{11}÷9$ $1\text{.}\overline{2}$ $\text{17}÷9$ $1\text{.}\overline{8}$ Divide 7 by 6 and round to 2 decimal places. 1.17 Divide 400 by 11 and round to 4 decimal places. 36.3636 ## Exercises For the following 20 problems, carry out each division until the repeating pattern is determined. If a repeating pattern is not apparent, round the quotient to three decimal places. $4÷9$ $0\text{.}\overline{4}$ $8÷\text{11}$ $4÷\text{25}$ 0.16 $5÷6$ $1÷7$ $0\text{.}\overline{\text{142857}}$ $3÷1\text{.}1$ $\text{20}÷1\text{.}9$ 10.526 $\text{10}÷2\text{.}7$ $1\text{.}\text{11}÷9\text{.}9$ $0\text{.}1\overline{\text{12}}$ $8\text{.}\text{08}÷3\text{.}1$ $\text{51}÷8\text{.}2$ $6\text{.}\overline{\text{21951}}$ $0\text{.}\text{213}÷0\text{.}\text{31}$ $0\text{.}\text{009}÷1\text{.}1$ $0\text{.}\text{00}\overline{\text{81}}$ $6\text{.}\text{03}÷1\text{.}9$ $0\text{.}\text{518}÷0\text{.}\text{62}$ 0.835 $1\text{.}\text{55}÷0\text{.}\text{27}$ $0\text{.}\text{333}÷0\text{.}\text{999}$ $0\text{.}\overline{3}$ $0\text{.}\text{444}÷0\text{.}\text{999}$ $0\text{.}\text{555}÷0\text{.}\text{27}$ $2\text{.}0\overline{5}$ $3\text{.}8÷0\text{.}\text{99}$ ## Calculator problems For the following 10 problems, use a calculator to perform each division. $7÷9$ $0\text{.}\overline{7}$ $8÷\text{11}$ $\text{14}÷\text{27}$ $0\text{.}\overline{\text{518}}$ $1÷\text{44}$ $2÷\text{44}$ $0\text{.}0\overline{\text{45}}$ $0\text{.}7÷0\text{.}9$ (Compare this with [link] .) $\text{80}÷\text{110}$ (Compare this with [link] .) $0\text{.}\overline{\text{72}}$ $0\text{.}\text{0707}÷0\text{.}\text{7070}$ $0\text{.}\text{1414}÷0\text{.}\text{2020}$ 0.7 $1÷0\text{.}\text{9999999}$ ## Exercise for review ( [link] ) In the number 411,105, how many ten thousands are there? 1 ( [link] ) Find the quotient, if it exists. $\text{17}÷0$ . ( [link] ) Find the least common multiple of 45, 63, and 98. 4410 ( [link] ) Subtract 8.01629 from 9.00187 and round the result to three decimal places. ( [link] ) Find the quotient. $\text{104}\text{.}\text{06}÷\text{12}\text{.}1$ . 8.6 #### Questions & Answers 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 characteristics of micro business Abigail Do somebody tell me a best nano engineering book for beginners? 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 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 what's the easiest and fastest way to the synthesize AgNP? China Cied types of nano material I start with an easy one. carbon nanotubes woven into a long filament like a string Porter many many of nanotubes Porter what is the k.e before it land Yasmin what is the function of carbon nanotubes? Cesar I'm interested in nanotube Uday what is nanomaterials and their applications of sensors. what is nano technology what is system testing? preparation of nanomaterial Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it... what is system testing what is the application of nanotechnology? Stotaw In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google Azam anybody can imagine what will be happen after 100 years from now in nano tech world Prasenjit after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments Azam name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world Prasenjit how hard could it be to apply nanotechnology against viral infections such HIV or Ebola? Damian silver nanoparticles could handle the job? Damian not now but maybe in future only AgNP maybe any other nanomaterials Azam Hello Uday I'm interested in Nanotube Uday this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15 Prasenjit how did you get the value of 2000N.What calculations are needed to arrive at it Privacy Information Security Software Version 1.1a Good Berger describes sociologists as concerned with 7hours 36 min - 4hours 50 min
Algebra 5.6 # Algebra 5.6 Télécharger la présentation ## Algebra 5.6 - - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - - ##### Presentation Transcript 1. Algebra 5.6 5.6 Slope Intercept Form 2. Learning Targets Language Goal Students will be able to write a linear equation in slope-intercept form. Math Goal Students will be able to graph a line using slope-intercept form. Essential Questions Why do we have many forms to graph a line? 3. Warm-up 4. Homework Check 5. Homework Check 6. Homework Check 7. Homework Check 8. Homework Check 9. Go over 5.1 – 5.5 Quiz 10. Steps for Graphing by using Slope and Intercept • Steps for graphing by using Slope and Intercept 1. Plot the y-intercept using the point (0, #) 2. Slope = . • The numerator moves you up or down. (Up if positive, Down if negative.) • The denominator moves you left or right. (Right if positive, Left if negative) 3. Plot a point once you have moved with the appropriate slope. 4. Draw a line connecting the two points 11. Example Type 1: Graphing by using slope and y-intercept Graph each line given the slope and y-intercept. A. Slope = ; y-int. = -2 B. Slope = -2; y-int. = 4 12. Example Type 1: Graphing by using slope and y-intercept Graph each line given the slope and y-intercept. C.Slope =2; y-int. = -3D. Slope = ; y-int. = 1 13. Example 2: Writing Linear Equations in Slope-Intercept Form • **If you know the slope of a line and the y-intercept you can write an equation that describes the line. 1. Slope is m = . 2. The y-intercept is usually in the form y-intercept = # or (0, #) That number is your b. 3. Slope intercept form is y = mx + b 14. Example 2, Part A:Writing Linear Equations in Slope-Intercept Form A. Slope = , y-int. = 6 B. Slope = -12, y-int. = 15. Example 2, Part A: Writing Linear Equations in Slope-Intercept Form C. Slope =1, y-int. = 0 D. Slope = 0, y-int. = 16. Example 2, Part B: Finding Slope Form Given Slope and a Point • Step 1: Find the y – intercept by... • Plugging in the slope and the ordered pair. Solve for b. • Step 2: Write the equation using the slope and b. • Slope Intercept Form: • y = mx + b 17. Example 2, Part B: Finding Slope Form Given Slope and a Point A. Slope = 4, (2, 5) is on the line. B. Slope = 8, (3, -1) is on the line. 18. Example 2, Part B: Finding Slope Form Given Slope and a Point C. Slope = 2, (3, 4) is on the line 19. Example 3: Using Slope-Intercept Form to Graph • Step 1: Find the slope and y-intercept from the equation. • Find m and b. Remember y = mx + b • Step 2: Plot the y-intercept • Step 3: Use slope to plot a second point • Step 4: Draw a line connecting the points 20. Example 3: Using Slope-Intercept Form to Graph A. y = 4x – 3 B. y = - x + 2 21. Example 3: Using Slope-Intercept Form to Graph C. y = 3x – 1 22. Additional Example 3: Changing Equations to Slope Intercept Form • Step 1: Write the equation in slope intercept form. • Solve the equation for y. • Move everything else to the other side of the equal to sign. • Step 2: Graph the line as previously done. 23. Additional Example 3: Changing Equations to Slope Intercept Form A. 3x + 2y = 8 B. 2y + 3x = 6 24. Additional Example 3: Changing Equations to Slope Intercept Form C. 6x + 2y = 10 25. Review 5.6 Write an equation of the line and then graph. A. Slope = 1; y intercept = 4 26. Review 5.6 Write an equation of the line and then graph. B. Slope = ; y intercept = -2 27. Review 5.6 Find m and b, then graph. C. m = __________ b = __________ 28. Review 5.6 Find m and b, then graph. D. m = __________ b = __________ 29. Review 5.6 Find Slope Intercept Form given slope and a point. E. (0, -5), m = 30. Review 5.6 Find Slope Intercept Form given slope and a point. F. (4, 6), m = 31. Review 5.6 Write an equation in slope-intercept form of the following graphs. G. H. I. 32. Lesson Quiz
# Trigonometric Functions & Their Signs (2 Ways To Remember) Trigonometric functions are often used in math (including calculus), and they also have real-life applications (such as signal processing). However, it can be tricky to remember the signs of these functions. So, how do you remember trigonometric functions and their signs? Use the memory device ASTC (all students take calculus) to label quadrants 1, 2, 3, & 4. A means all trig functions are positive and S, T, C stand for sine, tangent, & cosine. Also, cosine is the x-coordinate of a point on the unit circle, and sine is the y-coordinate of a point on the unit circle. Of course, once you know the signs of sine, cosine, and tangent in a given quadrant, you can also find the signs of their reciprocal trig functions (cosecant, secant, and cotangent). In this article, we’ll talk about ways to remember the signs of trigonometric functions, depending on what quadrant they are in (and the related angles). We’ll also talk about some special cases (when the trig functions are zero or undefined). Let’s get started. ## Trigonometric Functions & Their Signs There are two methods to help you remember the signs of trigonometric functions: • By Quadrant – use the memory device ASTC (all students take calculus) to label the 4 quadrants and find out which functions are positive or negative (more detail on this below). • By Coordinates – remember that the cosine of an angle gives us the x coordinate of the corresponding triangle, and the sine gives us the y coordinate of the corresponding triangle (more detail on this below). ### Trigonometric Functions & Their Signs By Quadrants We can use the memory device ASTC (all students take calculus) to keep track of the signs of trigonometric functions by quadrant. We assign each quadrant a letter as follows: • A = 1st Quadrant (top right): All trig functions are positive. • S = 2nd Quadrant (top left): Sine is positive (along with cosecant, the reciprocal of sine). • T = 3rd Quadrant (bottom left): Tangent is positive (along with cotangent, the reciprocal of tangent). • C = 4th Quadrant (bottom right): Cosine is positive (along with secant, the reciprocal of cosine). You can see a graph with the quadrant labels below. Notice how the memory device works: remember the phrase “all students take calculus” and take the first letter of each word: ASTC. Remember that A stands for all, S stands for sine, t stands for tangent, and c stands for cosine. We put them in the same order as the number labels for the quadrants: 1, 2, 3, 4. However, this isn’t the only way to remember and understand the signs of trig functions. ### Trigonometric Functions & Their Signs By Coordinates Another way to keep track of signs of trig functions is to use what we know about the coordinates of a point on the unit circle and the trig functions. The image below shows a point P on the unit circle, the line segment between P and O (where O is the origin), the angle between the line segment PO and the x-axis, and the coordinates of the point P. If the point P has coordinates (xP, yP) and the angle between OP and the x-axis is A, then we have: • xP = cos(A) • yP = sin(A) In other words, the cosine of the angle A gives us the x coordinate of the point P, and the sine of the angle A gives us the y coordinate of the point P (note that this only works for points on the unit circle centered at the origin). For a circle of radius R centered at the origin, the point P would have coordinates: • xP = Rcos(A) • yP = Rsin(A) So, how does this help us with the signs of trig functions? Well, we can think of a point P as being either to the left or to the right of the y-axis: • If the point P is to the right of the y-axis, then xP is positive, and so cos(A) must be positive. • If the point P is to the left of the y-axis, then xP is negative, and so cos(A) must be negative. Similarly, we can think of a point P as being either above or below of the x-axis: • If the point P is above the x-axis, then yP is positive, and so sin(A) must be positive. • If the point P is below the x-axis, then yP is negative, and so sin(A) must be negative. Once we know the signs of the sine and cosine functions, we can easily find the signs of the other trig functions: • tan(A) = sin(A) / cos(A) • sec(A) = 1 / cos(A) • csc(A) = 1 / sin(A) • cot(A) = 1 /tan(A) So, for example, if sin(A) and cos(A) have the same sign, then tan(A) must be positive. If sin(A) and cos(A) have opposite signs, then tan(A) is negative. Also: • sec(A) has the same sign as cos(A) • csc(A) has the same sign as sin(A) • cot(A) has the same sign as tan(A) Note that tangent, secant, cosecant, and cotangent can be undefined when they have zero denominators (this can happen when either sine or cosine are equal to zero – more on this later). The table below summarizes the signs of tangent, secant, cosecant, and cotangent, based on the signs of sin(A) and cos(A). ### Sine Let’s review when sine is positive, negative, zero, and undefined. #### When Is Sine Positive? Sine is positive in the 1st and 2nd quadrants, when the y-coordinate is above the x-axis. This occurs when the angle A is between 0 and π radians (0 and 180 degrees). #### When Is Sine Negative? Sine is negative in the 3rd and 4th quadrants, when the y-coordinate is below the x-axis. This occurs when the angle A is between π and 2π radians (180 and 360 degrees). #### When Is Sine 1? Sine is 1 when the angle A is π/2 radians (90 degrees). This means that the point on the unit circle lies on the y-axis (between the 1st and 2nd quadrants). Sine will also be 1 when A is any angle that is equivalent to π/2, such as: • -7π/2, -3π/2, π/2, 5π/2, 9π/2, … (4k + 1)π / 2 for any integer k Note that sine is -1 when the angle A is –π/2 radians (-90 degrees) or equivalently 3π/2 radians (270 degrees). This means that the point on the unit circle lies on the y-axis (between the 3rd and 4th quadrants). Sine will also be -1 when A is any angle that is equivalent to -π/2 radians, such as: • -5π/2, -π/2, 3π/2, 7π/2, … (4k + 3)π / 2 for any integer k #### When Is Sine 0? Sine is 0 when the angle A is 0 radians (0 degrees). This means that the point on the unit circle is on the x-axis (between the 1st and 4th quadrants). Sine is also 0 when the angle A is π radians (180 degrees). This means that the point on the unit circle lies on the x-axis (between the 2nd and 3rd quadrants). Sine will also be 0 when A is any angle that is equivalent to 0 or π, such as: • -3π, -2π, -π, 0, π, 2π, 3π … kπ for any integer k #### When Is Sine Undefined? Sine is never undefined. It can take on any value between -1 and 1, and we can use any angle as an input into the sine function. #### When Is Sine Equal To Cosine? Sine is equal to cosine when the angle A is π/4 radians (45 degrees). This means that the point on the unit circle is on the halfway point of the circumference between (0, 0) and (0, 1) in the first quadrant. Sine is also equal to cosine when the angle A is any angle that is equivalent to π/4, such as: • -15π/4, -7π/4, π/4, 9π/4, 17π/4, … (8k + 1)π / 4 for any integer k ### Cosine Let’s review when cosine is positive, negative, zero, and undefined. #### When Is Cosine Positive? Cosine is positive in the 1st and 4th quadrants, when the x-coordinate is to the right of the y-axis. This occurs when the angle A is between –π/2 and π/2 radians (-90 and 90 degrees). #### When Is Cosine Negative? Cosine is negative in the 2nd and 3rd quadrants, when the x-coordinate is to the left of the y-axis. This occurs when the angle A is between π/2 and 3π/2 radians (90 and 270 degrees). #### When Is Cosine 1? Cosine is 1 when the angle A is 0 radians (0 degrees). This means that the point on the unit circle lies on the x-axis (between the 1st and 4th quadrants). Cosine will also be 1 when A is any angle that is equivalent to 0, such as: • -4π, -2π, 0, 2π, 4π, … 2kπ for any integer k Note that cosine is -1 when the angle A is π radians (180 degrees) or equivalently -π radians (-180 degrees). This means that the point on the unit circle lies on the x-axis (between the 2nd and 3rd quadrants). Cosine will also be -1 when A is any angle that is equivalent to π radians, such as: • -3π, -π, π, 3π … (2k + 1)π for any integer k #### When Is Cosine 0? Cosine is 0 when the angle A is π/2 radians (90 degrees). This means that the point on the unit circle is on the y-axis (between the 1st and 2nd quadrants). Cosine is also 0 when the angle A is 3π/2 radians (270 degrees). This means that the point on the unit circle lies on the y-axis (between the 3rd and 4th quadrants). Cosine will also be 0 when A is any angle that is equivalent to π/2 or 3π/2, such as: • -3π/2, -π/2, π/2, 3π/2, … (2k + 1)π / 2 for any integer k #### When Is Cosine Undefined? Cosine is never undefined. It can take on any value between -1 and 1, and we can use any angle as an input into the cosine function. ### Tangent Let’s review when tangent is positive, negative, zero, and undefined. #### When Is Tangent Positive? Tangent is positive in the 1st and 3rd quadrants, when sine and cosine have the same sign (both positive or both negative). This occurs when the angle A is between 0 and π/2 radians (0 and 90 degrees) or between π and 3π/2 radians (180 and 270 degrees). #### When Is Tangent Negative? Tangent is negative in the 2nd and 4th quadrants, when sine and cosine have opposite signs (one positive, the other negative). This occurs when the angle A is between π/2 and π radians (90 and 180 degrees) or between 3π/2 and 2π radians (270 and 360 degrees). #### When Is Tangent 1? Tangent is 1 when sine and cosine have the same value (including sign). This happens when the angle A is π/4 radians (45 degrees), and sin(A) = cos(A) = √2 / 2. This means that the point on the unit circle is on the halfway point of the circumference between (0, 0) and (0, 1) in the first quadrant. Tangent will also be 1 when angle A is 5π/4 radians (225 degrees), and sin(A) = cos(A) = -√2 / 2. This means that the point on the unit circle is on the halfway point of the circumference between (-1, 0) and (0, -1) in the third quadrant. Tangent will also be 1 when A is any angle that is equivalent to π/4 or 5π/4, such as: • -7π/4, -3π/4, π/4, 5π/4, 9π/4, … (4k + 1)π / 4 for any integer k Note that tangent is -1 when sine and cosine have opposite values (one positive and one negative). This occurs when the angle A is 3π/4 radians (135 degrees). This means that the point on the unit circle is on the halfway point of the circumference between (0, 1) and (-1, 0) in the second quadrant. Tangent will also be -1 when angle A is 7π/4 radians (315 degrees). This means that the point on the unit circle is on the halfway point of the circumference between (0, -1) and (1, 0) in the fourth quadrant. Tangent will also be -1 when A is any angle that is equivalent to 3π/4 or 7π/4 radians, such as: • -9π/4, -5π/4, -π/4, 3π/4, 7π/4, … (4k + 3)π / 4 for any integer k #### When Is Tangent 0? Tangent is zero whenever sine is zero. This is due to the fact that tan(A) = sin(A) / cos(A), which means sin(A) is the numerator of tan(A). We know when sine is zero from our earlier discussion (see “When Is Sine Zero” above). Tangent is 0 when the angle A is 0 radians (0 degrees). This means that the point on the unit circle is on the x-axis (between the 1st and 4th quadrants). Tangent is also 0 when the angle A is π radians (180 degrees). This means that the point on the unit circle lies on the x-axis (between the 2nd and 3rd quadrants). Tangent will also be 0 when A is any angle that is equivalent to 0 or π, such as: • -3π, -2π, -π, 0, π, 2π, 3π … kπ for any integer k #### When Is Tangent Undefined? Tangent is undefined whenever cosine is zero. This is due to the fact that tan(A) = sin(A) / cos(A), which means cos(A) is the denominator of tan(A). We know when cosine is zero from our earlier discussion (see “When Is Cosine Zero” above). Tangent is 0 when the angle A is π/2 radians (90 degrees). This means that the point on the unit circle is on the y-axis (between the 1st and 2nd quadrants). Tangent is also 0 when the angle A is 3π/2 radians (270 degrees). This means that the point on the unit circle lies on the y-axis (between the 3rd and 4th quadrants). Tangent will also be 0 when A is any angle that is equivalent to π/2 or 3π/2, such as: • -3π/2, -π/2, π/2, 3π/2, … (2k + 1)π / 2 for any integer k ### Secant Let’s review when secant is positive, negative, zero, and undefined. #### When Is Secant Positive? Secant is positive whenever cosine is positive. This is due to the fact that sec(A) = 1 / cos(A), which means sec(A) is the reciprocal of cos(A). We know when cosine is positive from our earlier discussion (see “When Is Cosine Positive” above). Secant is positive in the 1st and 4th quadrants, when the x-coordinate is to the right of the y-axis. This occurs when the angle A is between –π/2 and π/2 radians (-90 and 90 degrees). #### When Is Secant Negative? Secant is negative whenever cosine is negative. This is due to the fact that sec(A) = 1 / cos(A), which means sec(A) is the reciprocal of cos(A). We know when cosine is negative from our earlier discussion (see “When Is Cosine Negative” above). Secant is negative in the 2nd and 3rd quadrants, when the x-coordinate is to the left of the y-axis. This occurs when the angle A is between π/2 and 3π/2 radians (90 and 270 degrees). #### When Is Secant 1? Secant is 1 whenever cosine is 1. This is due to the fact that sec(A) = 1 / cos(A), which means sec(A) is the reciprocal of cos(A). We know when cosine is 1 from our earlier discussion (see “When Is Cosine 1” above). Secant is 1 when the angle A is 0 radians (0 degrees). This means that the point on the unit circle lies on the x-axis (between the 1st and 4th quadrants). Secant will also be 1 when A is any angle that is equivalent to 0, such as: • -4π, -2π, 0, 2π, 4π, … 2kπ for any integer k Note that Secant is -1 when the angle A is π radians (180 degrees) or equivalently -π radians (-180 degrees). This means that the point on the unit circle lies on the x-axis (between the 2nd and 3rd quadrants). Secant will also be -1 when A is any angle that is equivalent to π radians, such as: • -3π, -π, π, 3π … (2k + 1)π for any integer k #### When Is Secant 0? Secant is never zero. This is due to the fact that sec(A) = 1 / cos(A), and cos(A) only takes on finite values (between -1 and 1). #### When Is Secant Undefined? Secant is undefined whenever cosine is zero. This is due to the fact that sec(A) = 1 / cos(A), and cos(A) = 0 would give us a zero denominator (undefined). We know when cosine is zero from our earlier discussion (see “When Is Cosine Zero” above). Secant is undefined when the angle A is π/2 radians (90 degrees). This means that the point on the unit circle is on the y-axis (between the 1st and 2nd quadrants). Secant is also undefined when the angle A is 3π/2 radians (270 degrees). This means that the point on the unit circle lies on the y-axis (between the 3rd and 4th quadrants). Secant will also be undefined when A is any angle that is equivalent to π/2 or 3π/2, such as: • -3π/2, -π/2, π/2, 3π/2, … (2k + 1)π / 2 for any integer k #### When Is Secant Equal To Cosecant? Secant is equal to cosecant whenever cosine is equal to sine, since: • sec(x) = csc(x) • 1 / cos(x) = 1 / sin(x) • sin(x) = cos(x) We know when sine and cosine are equal from our earlier discussion (see “When Is Sine Equal to Cosine” above). Secant is equal to cosecant when the angle A is π/4 radians (45 degrees). This means that the point on the unit circle is on the halfway point of the circumference between (0, 0) and (0, 1) in the first quadrant. Secant is also equal to cosecant when the angle A is any angle that is equivalent to π/4, such as: • -15π/4, -7π/4, π/4, 9π/4, 17π/4, … (8k + 1)π / 4 for any integer k ### Cosecant Let’s review when cosecant is positive, negative, zero, and undefined. #### When Is Cosecant Positive? Cosecant is positive whenever sine is positive. This is due to the fact that csc(A) = 1 / sin(A), which means csc(A) is the reciprocal of sin(A). We know when sine is positive from our earlier discussion (see “When Is Sine Positive” above). Cosecant is positive in the 1st and 2nd quadrants, when the y-coordinate is above the x-axis. This occurs when the angle A is between 0 and π radians (0 and 180 degrees). #### When Is Cosecant Negative? Cosecant is negative whenever sine is negative. This is due to the fact that csc(A) = 1 / sin(A), which means csc(A) is the reciprocal of sin(A). We know when sine is negative from our earlier discussion (see “When Is Sine Negative” above). Cosecant is negative in the 3rd and 4th quadrants, when the y-coordinate is below the x-axis. This occurs when the angle A is between π and 2π radians (180 and 360 degrees). #### When Is Cosecant 1? Cosecant is 1 whenever sine is 1. This is due to the fact that csc(A) = 1 / sin(A), which means csc(A) is the reciprocal of sin(A). We know when sine is 1 from our earlier discussion (see “When Is Sine 1” above). Cosecant is 1 when the angle A is π/2 radians (90 degrees). This means that the point on the unit circle lies on the y-axis (between the 1st and 2nd quadrants). Cosecant will also be 1 when A is any angle that is equivalent to π/2, such as: • -7π/2, -3π/2, π/2, 5π/2, 9π/2, … (4k + 1)π / 2 for any integer k Note that Cosecant is -1 when the angle A is –π/2 radians (-90 degrees) or equivalently 3π/2 radians (270 degrees). This means that the point on the unit circle lies on the y-axis (between the 3rd and 4th quadrants). Cosecant will also be -1 when A is any angle that is equivalent to -π/2 radians, such as: • -5π/2, -π/2, 3π/2, 7π/2, … (4k + 3)π / 2 for any integer k #### When Is Cosecant 0? Cosecant is never zero. This is due to the fact that csc(A) = 1 / sin(A), and csc(A) only takes on finite values (between -1 and 1). #### When Is Cosecant Undefined? Cosecant is undefined whenever sine is zero. This is due to the fact that csc(A) = 1 / sin(A), and sin(A) = 0 would give us a zero denominator (undefined). We know when sine is zero from our earlier discussion (see “When Is Sine Zero” above). Cosecant is 0 when the angle A is 0 radians (0 degrees). This means that the point on the unit circle is on the x-axis (between the 1st and 4th quadrants). Cosecant is also 0 when the angle A is π radians (180 degrees). This means that the point on the unit circle lies on the x-axis (between the 2nd and 3rd quadrants). Cosecant will also be 0 when A is any angle that is equivalent to 0 or π, such as: • -3π, -2π, -π, 0, π, 2π, 3π … kπ for any integer k ### Cotangent Let’s review when cotangent is positive, negative, zero, and undefined. #### When Is Cotangent Positive? Cotangent is positive whenever tangent is negative. This is due to the fact that cot(A) = 1 / tan(A), which means cot(A) is the reciprocal of tan(A). We know when tangent is positive from our earlier discussion (see “When Is Tangent Positive” above). Cotangent is positive in the 1st and 3rd quadrants, when sine and cosine have the same sign (both positive or both negative). This occurs when the angle A is between 0 and π/2 radians (0 and 90 degrees) or between π and 3π/2 radians (180 and 270 degrees). #### When Is Cotangent Negative? Cotangent is negative whenever tangent is negative. This is due to the fact that cot(A) = 1 / tan(A), which means cot(A) is the reciprocal of tan(A). We know when tangent is negative from our earlier discussion (see “When Is Tangent Negative” above). Cotangent is negative in the 2nd and 4th quadrants, when sine and cosine have opposite signs (one positive, the other negative). This occurs when the angle A is between π/2 and π radians (90 and 180 degrees) or between 3π/2 and 2π radians (270 and 360 degrees). #### When Is Cotangent 1? Cotangent is 1 whenever tangent is 1. This is due to the fact that cot(A) = 1 / tan(A), which means cot(A) is the reciprocal of tan(A). We know when tangent is 1 from our earlier discussion (see “When Is Tangent 1” above). Cotangent is 1 when sine and cosine have the same value (including sign). This happens when the angle A is π/4 radians (45 degrees), and sin(A) = cos(A) = √2 / 2. This means that the point on the unit circle is on the halfway point of the circumference between (0, 0) and (0, 1) in the first quadrant. Cotangent will also be 1 when angle A is 5π/4 radians (225 degrees), and sin(A) = cos(A) = -√2 / 2. This means that the point on the unit circle is on the halfway point of the circumference between (-1, 0) and (0, -1) in the third quadrant. Cotangent will also be 1 when A is any angle that is equivalent to π/4 or 5π/4, such as: • -7π/4, -3π/4, π/4, 5π/4, 9π/4, … (4k + 1)π / 4 for any integer k Note than cotangent will be -1 whenever tangent is -1. #### When Is Cotangent 0? Cotangent is zero whenever cosine is zero. This is due to the fact that cot(A) = cos(A) / sin(A), which means cos(A) is the numerator of tan(A). We know when cosine is zero from our earlier discussion (see “When Is Cosine Zero” above). Cotangent is 0 when the angle A is π/2 radians (90 degrees). This means that the point on the unit circle is on the y-axis (between the 1st and 2nd quadrants). Cotangent is also 0 when the angle A is 3π/2 radians (270 degrees). This means that the point on the unit circle lies on the y-axis (between the 3rd and 4th quadrants). Cotangent will also be 0 when A is any angle that is equivalent to π/2 or 3π/2, such as: • -3π/2, -π/2, π/2, 3π/2, … (2k + 1)π / 2 for any integer k #### When Is Cotangent Undefined? Cotangent is undefined whenever sine is zero. This is due to the fact that cot(A) = cos(A) / sin(A), which means sin(A) is the denominator of tan(A). We know when sine is zero from our earlier discussion (see “When Is Sine Zero” above). Cotangent is 0 when the angle A is 0 radians (0 degrees). This means that the point on the unit circle is on the x-axis (between the 1st and 4th quadrants). Cotangent is also 0 when the angle A is π radians (180 degrees). This means that the point on the unit circle lies on the x-axis (between the 2nd and 3rd quadrants). Cotangent will also be 0 when A is any angle that is equivalent to 0 or π, such as: • -3π, -2π, -π, 0, π, 2π, 3π … kπ for any integer k ## Conclusion Now you know all about the signs of trigonometric functions and how to remember them (drawing out coordinate axes with numbers and letters ASTC helps!) You can learn about the output values of sine here. You can learn about the output values of cosine here. You can learn about inverse trigonometric functions here. I hope you found this article helpful. If so, please share it with someone who can use the information. ~Jonathon
# Maths 7-11 years Some typical maths topics explained for KS2 (Key Stage 2) Maths. Explaining numbers to children It is important that pupils have a clear understanding of numbers and how to use them. Younger children start establishing a clear understanding of units then they gradually progress to Tens. Visual aids such as conkers, beads, blocks or sometimes fingers can help establish the number concept As our number system is worked in BASE 10 each number has a place value. This means that its value depends on its position. In arithmetic the positions or columns are given names. This shows for example the number 965. Th H T U (Thousands) (Hundreds) (Tens) (Units) 9 6 5 When you move from one column to the next starting with the units each number becomes 10 times larger. For the number 9 6 5 we have: U = 5 units T = 6 groups of 10 = 60 H = 9 groups of 100 = 900 In these number examples the figures underlined have the following values: 3789 3 is 3000 (three thousand) 40327 40 is 40000 (forty thousand) 65891 8 is 800 (eight hundred) An even number is a number which can be exactly divided by 2 e.g 2 4 6 8 10. All other numbers are odd numbers e.g 1 3 5 7 9 11. A prime number is a number which can only be divided exactly by itself and by one. All prime numbers are odd numbers apart from 2. 1 is not a prime number 5 7 11 13 29 and 53 are examples of prime numbers. Square numbers are numbers whose dots can be arranged in the shape of a square. e.g. 16 The names of 3D shapes will be learnt. Here are the names of the angles. ## Have fun with Numbers and Maths It is important that Maths is as much fun as possible so incidental learning with games and puzzles can reinforce all learning.
# PAIRS OF ANGLES In geometry, certain pairs of angles can have special relationships. Using our knowledge of acute, right, and obtuse angles, along with properties of parallel lines, we will begin to study the relations between pairs of angles. Complementary Angles Two angles are complementary angles if their degree measurements add up to 90°. That is, if we attach both angles and fit them side by side (by putting the vertices and one side on top of each other), they will form a right angle. We can also say that one of the angles is the complement of the other. Complement ary angles are angles whose sum is 90° Supplement ary Angles Another special pair of angles is called supplementary angles. One angle is said to be the supplement of the other if the sum of their degree measurements is 180°. In other words, if we put the angles side by side, the result would be a straight line. Supplementary angles are angles whose sum is 180° Vertical Angles Vertical angles are the angles opposite of each other at the intersection of two lines. They are called vertical angles because they share a common vertex. Vertical angles always have equal measures. ∠JKL and ∠MKN are vertical angles. Another pair of vertical angles in the picture is ∠JKM and ∠LKN. Alternate Interior Angles Alternate interior angles are formed when there exists a transversal. They are the angles on opposite sides of the transversal, but inside the two lines the transversal intersects. Alternate interior angles are congruent to each other if (and only if) the two lines intersected by the transversal are parallel. An easy way of identifying alternate interior angles is by drawing the letter "Z" (forwards and backwards) on the lines as shown below. In the figure on the left, ∠ADH and ∠GHD are alternate interior angles. Note that ∠CDH and ∠EHD are also alternate interior angles. The figure on the right has alternate interior angles that are congruent because there is a set of parallel lines. Alternate Exterior Angles Similar to alternate interior angles, alternate exterior angles are also congruent to each other if (and only if) the two lines intersected by the transversal are parallel. These angles are on opposite sides of the transversal, but outside the two lines the transversal intersects. In the figure on the left, ∠ADB and ∠GHF are alternate exterior angles. So are ∠CDB and ∠EHF. The figure on the left does not have alternate enterior angles that are congruent, but the figure on the right does. Corresponding Angles Corresponding angles are the pairs of angles on the same side of the transversal and on corresponding sides of the two other lines. These angles are equal in degree measure when the two lines intersected by the transversal are parallel. It may help to draw the letter "F" (forwards and backwards) in order to help identify corresponding angles. This method is illustrated below. Drawing the letter "F" backwards helps us see that ∠ADH and ∠EHF are corresponding angles. We have three other pairs of corresponding angles in this figure. Now that we have familiarized ourselves with pairs of angles, let's practice applying some of their properties in the following exercises. Segment Bisector Definition of Segment Bisector  A Segment Bisector is a line or a ray or a segment that divides a line segment into two equal parts. More about Segment Bisector  A segment bisector always passes through the midpoint of a line segment.  If a segment bisector crosses the segment at 900, then it is called as perpendicular bisector of the segment. Example of Segment Bisector In the given figure, the line segment DE is the bisector of the segment AC as it intersects the line segment AC at its midpoint B. Solved Example on Segment Bisector What is the length of AB, if line l is the segment bisector and AO = 6 units? Solution: Step 1: Line l divides AB into two equal parts and O is the midpoint of AB. [As the segment bisector passes through the midpoint of the segment.] Step 2: AB = 2(AO) = 2(6) = 12 [Substitute AO = 6.] Step 3: So, the length of AB is 12 units. Angle Bisectors Definition of Angle Bisector  Angle Bisector is a ray that divides an angle into two equal parts. More about Angle Bisector  Angle bisector is also called the internal angle bisector.  Angle bisectors of a triangle intersect at a point called incenter of the triangle.  A line or a line segment can also be an angle bisector at times. For example, the angle bisectors of polygons are line segments. Examples of Angle Bisector In the example shown, OA is the angle bisector. It divides 60° into two equal parts (i.e. two 30°'s). Solved Example on Angle Bisector In the given figure, is the angle bisector of ∠AOB and m∠AOC = 3y - 4 and m∠COD = 28, then find the value of y. is the angle bisector of ∠COB. If Solution: Step 1: m∠COD = 28, so m∠BOC = 56, because is an angle bisector of ∠COB. bisects ∠AOB. Step 2: m∠BOC = 56 implies that m∠AOC = 56, because Step 3: So, 3y - 4 = 56 gives y = 20°. Related Terms for Angle Bisector
# How to Calculate the Volume of a Cube Author Info Updated: March 29, 2019 A cube is a three-dimensional shape that has equal width, height, and length measurements. A cube has six square faces, all of which have sides of equal length and all of which meet at right angles.[1] Finding the volume of a cube is a snap - generally, all that's needed is to multiply the cube's length × width × height. Since a cube's sides are all equal in length, another way of thinking of a cube's volume is s3, where s is the length of one of the cube's sides. See Step 1 below for a detailed breakdown of these processes. ### Method 1 of 3: Cubing One of the Cube's Sides 1. 1 Find the length of one side of the cube. Often, in problems asking you to find the volume of a cube, you'll be given the length of one of a cube's sides. If you have this information, you have all you need to solve for the cube's volume. If you're not solving an abstract math problem but are instead attempting to find the volume of a real-life object shaped like a cube, use a ruler or measuring tape to measure the side of the cube.[2] • To better understand the process of finding the volume of a cube, let's follow along with an example problem as we go through the steps in this section. Let's say the side of the cube is 2 inches (5.08 cm) long. We'll use this information to find the volume of the cube in the next step. 2. 2 Cube the length of the side. When you've found the length of one of the cube's sides, cube this number. In other words, multiply it by itself twice. If s is the length of the side, you would multiply s × s × s (or, in simplified form, s3). This will give you the volume of your cube![3] • This process is essentially the same as finding the area of the base and then multiplying it by the cube's height (or, in other words, length × width × height), since the area of the base is found by multiplying its length and its width. Since the length, width, and height of a cube are equal, we can shorten this process by simply cubing any of these measurements. • Let's proceed with our example. Since the length of the side of our cube is 2 inches, we can find the volume by multiplying 2 x 2 x 2 (or 23) = 8. 3. 3 Label your answer with cubic units.[4] Since volume is the measure of three-dimensional space, your answer should be in cubic units by definition. Often, on math schoolwork, neglecting to label your answer with the right units can cause you to lose points on a problem, so don't forget to use the correct label! • In our example, since our original measurement was in inches, our final answer will be labelled with the units "cubic inches" (or in3). So, our answer of 8 becomes 8 in3. • If we had used a different initial unit of measurement, our final cubic units would differ. For instance, if our cube had sides with lengths of 2 meters, rather than 2 inches, we would label it with cubic meters (m3). ### Method 2 of 3: Finding Volume from Surface Area 1. 1 Find your cube's surface area. While the easiest way to find a cube's volume is to cube the length of one of its sides, it's not the only way. The length of a cube's side or the area of one of its faces can be derived from several other of the cube's properties, which means that if you start with one of these pieces of information, you can find the volume of the cube in a roundabout manner. For instance, if you know a cube's surface area, all you need to do to find its volume is to divide the surface area by 6, then take the square root of this value to find the length of the cube's sides. From here, all you'll need to do is cube the length of the side to find the volume as normal. In this section, we'll walk through this process step-by-step. • The surface area of a cube is given via the formula 6s2, where s is the length of one of the cube's sides. This formula is essentially the same as finding the 2-dimensional area of the cube's six faces and adding these values together. We'll use this formula to find the volume of the cube from its surface area.[5] • As a running example, let's say that we have a cube whose surface we know to be 50 cm2, but we don't know its side lengths. In the next few steps, we'll use this information to find the cube's volume. 2. 2 Divide the cube's surface area by 6. Since the cube has 6 faces with equal area, dividing the cube's surface area by 6 will give you the area of one of its faces. This area is equal to the lengths of two of its sides multiplied (l × w, w × h, or h × l).[6] • In our example, dividing 50/6 = 8.33 cm2. Don't forget that two-dimensional answers have square units (cm2, in2, and so on). 3. 3 Take the square root of this value. Since the area of one of the cube's faces is equal to s2 (s × s), taking the square root of this value will find you the length of one of the cube's sides. Once you have this, you have enough information to solve for the volume of the cube as you normally would.[7] • In our example, √8.33 is roughly 2.89 cm. 4. 4 Cube this value to find the cube's volume. Now that you've obtained a value for the length of the cube's side, simply cube this value (multiply it by itself twice) to find the volume of the cube as detailed in the section above. Congratulations - you've found the volume of a cube from its surface area.[8] • In our example, 2.89 × 2.89 × 2.89 = 24.14 cm3. Don't forget to label your answer with cubic units. ### Method 3 of 3: Finding Volume from Diagonals 1. 1 Divide the diagonal across one of the cube's faces by √2 to find the cube's side length. By definition, the diagonal of a perfect square is √2 × the length of one of its sides. Thus, if the only information you're given about a cube is regarding the diagonal length of one of its faces, you can find the side length for the cube by dividing this value by √2. From here, it's relatively simple to cube your answer and find the volume of the cube as described above.[9] • For instance, let's say that one of a cube's faces has a diagonal that is 7 feet long. We would find the side length of the cube by dividing 7/√2 = 4.96 feet. Now that we know the side length, we can find the volume of the cube by multiplying 4.963 = 122.36 feet3. • Note that, in general terms, d2 = 2s2 where d is the length of the diagonal of one of the cube's faces and s is the length of one of the sides of the cube. This is because, according to the Pythagorean theorem, the square of the hypotenuse of a right triangle is equal to the sums of the squares of the other two sides. Thus, because the diagonal of a cube's face and two of the sides on that face form a right triangle, d2 = s2 + s2 = 2s2. 2. 2 Square the diagonal of two opposite corners of the cube, then divide by 3 and take the square root to find the side length. If the only piece of information you're given about a cube is the length of a 3-dimensional line segment stretching diagonally from one corner of the cube to the corner opposite it, it's still possible to find the volume of the cube. Because d forms one of the sides of a right triangle that has the diagonal between the two opposite corners of the cube as a hypotenuse, we can say that D2 = 3s2, where D = the 3-dimensional diagonal between opposite corners of the cube.[10] • This is because of the Pythagorean Theorem. D, d, and s form a right triangle with D as the hypotenuse, so we can say that D2 = d2 + s2. Since we calculated above that d2 = 2s2, we can say that D2 = 2s2 + s2 = 3s2. • As an example, let's say that we know that the diagonal from one of the corners in the base of the cube to the opposite corner in the "top" of the cube is 10 m. If we want to find the volume, we would insert 10 for each "D" in the equation above as follows: • D2 = 3s2. • 102 = 3s2. • 100 = 3s2 • 33.33 = s2 • 5.77 m = s. From here, all we need to do to find the volume of the cube is to cube the side length. • 5.773 = 192.45 m3 ## Community Q&A Search • Question If the volume of a cube is 512 cubic meters, what are its side measurements? Donagan The side length is the cube root of the volume. • Question A cube-shaped structure has sides 3 feet long. If I plan to fill it with gravel to a depth of 2.5 feet, how much gravel will I need? Donagan Multiply the area of the bottom of the cube by 2.5. • Question If the length is 2 meters, the height is 1 meter and the width is 2 meters of a water tank, how many gallons of water will the cube contain? First, you need to find the volume of the tank. This can be found by multiplying the length, height, and width altogether. When you do that, you get 4 cubic meters. Doing some research, you will find that 1 cubic meter is approximately equal to about 220 imperial gallons or 264.2 US liquid gallons. If you use the imperial gallons, then the tank would hold a little less than 880 imperial gallons. But if you use the US liquid gallons, it would hold slightly less than 1,057 gallons. • Question How do you find the volume of a cube if you are given that a circle with a diameter of 14 fits in it? Good question. Assume that the circle fits perfectly into the cube. So that means the inside edges touch. You know that the circle had a diameter of 14. So if it fits perfectly into the cube, the cube would also have a side length of 14. So the volume could be easily calculated: 14 cubed, which is 2744. • Question Can you find the capacity of a tank with dimensions 8cm x 6cm x 2.5cm? Well, that isn't exactly a cube but the formula is similar. You just need to multiply the length, width, and height altogether. 8 x 6 x 2.5 is 120. And there you have it. 120cm cubed. It's cubed because the units are little cubes. • Question If the perimeter of a face of a cube is 16 mm, what is its volume? Donagan Divide the perimeter by 4. That gives you the length of one edge. Cube it. That's the volume. • Question How do I find the number of cubical boxes of side 3 cm which can be accommodated in a carton of dimensions 15 cm * 9 cm * 12 cm? Donagan 3 cm will fit into the 15 cm dimension five times; it will fit into the 9 cm dimension three times; and it will fit into the 12 cm dimension four times. 5 x 3 x 4 = 60 cubical boxes. • Question If the volume of a cube is 1728, then what is its surface area? We know that to find the volume of a cube, we cube one of its sides. So if we want to inverse or reverse that operation, we take the cube root of 1728. It happens that the cube root is 12. So we now know that the length of one side is 12. If the length of one side of the cube is 12, then the area of one face would be 144. A cube has 6 faces, so we need to multiply 144 by 6 which is 864. • Question How do I find the number of blocks in a layer of a cube? Donagan It depends on the size of the blocks. • Question How do I find the cube root? Donagan 200 characters left wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. To create this article, 81 people, some anonymous, worked to edit and improve it over time. Together, they cited 10 references. This article has also been viewed 1,471,721 times. Co-authors: 81 Updated: March 29, 2019 Views: 1,471,721 Categories: Featured Articles \| Volume Article SummaryX To calculate the volume of a cube, find the length of one of the sides of the cube. When you have this measurement, multiply it by itself 2 times to get the volume, which is called “cubing” the number. For example, if your cube has a length of 2, you would multiply 2 × 2 × 2 to get a volume of 8. Be sure to include the units cubed with your answer. To learn more, like how to find the volume if you have the cube's surface area, keep reading!
# Circular system In circular system the unit of measurement is 'radian'. As radian is a unit of measurement so there should be some constant quantity. Theorem 1: Radian is a constant angle. Proof: Let us consider a circle with center 'O' and radius 'r'. Take any point 'A' and 'P' on the circle that make an arc AP. Angle formed at the center is $\angle$ AOP = $1^{c}$. Produce AO to meet the circle at B such that $\angle$ AOB is a straight angle =$180^{0}$. Since the angles at the center of a circle is proportional to the arc subtended in it. ∴ $\frac{\angle AOP}{\angle AOB}= \frac{arc AP}{arc APB}$ ⇒ $\frac{\angle AOP}{\angle AOB}= \frac{r}{\pi r}$ ⇒ $\frac{\angle AOP}{\angle AOB}= \frac{1}{\pi }$ ⇒ $\angle AOP = \frac{1}{\pi }\angle AOB = \frac{straight angle}{\pi}$ ⇒ $1^{c}=\frac{straight angle}{\pi}$ ⇒ $1^{c}$= constant . Hence, Radian is a constant angle. Theorem 2 :The number of radians in an angle subtended by an arc of a circle at the center is equal to $\frac{arc}{radius}$. Proof: Let us consider a circle with center 'O' and radius 'r'. Take any point 'A' and 'Q' on the circle that make an arc AP. Angle formed at the center is $\angle$ AOQ = $\Theta^{c}$.Let P be any point on the arc AQ and its length is 's'. Let arc AP=r. Thus, $\angle AOP = 1^{c}$ Since the angles at the center of a circle is proportional to the arc subtended in it. $\frac{\angle AOQ}{\angle AOP}= \frac{arc AQ}{arc AP}$ ⇒ $\angle AOQ = \left ( \frac{AQ}{AP} \times 1 \right )^{c}$ ⇒ $\Theta = \frac{s}{r}$ radians. ## Example on circular system Find the length of an arc of a circle of radius 5 cm subtending central angle measuring $15^{0}$. Solution : Let 's' be the length of the arc subtending an angle $\Theta^{c}$ at the center of a circle of radius 'r'. Then, $\Theta = \frac{s}{r}$ r = 5 cm and $\Theta = 15^{0}$ First we have to convert $\Theta$ to radians. $15^{0} = \left ( 15\times \frac{\pi}{180} \right )^{c} = \left ( \frac{\pi}{12} \right )^{c}$ $\frac{\pi}{12} = \frac{s}{5}$ ∴ s =$\frac{5\pi}{12}$ cm.

Subsets and Splits

No community queries yet

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