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Enable contrast version # TutorMe Blog ## What Makes Two Shapes Similar Figures? Inactive Jana Russick March 01, 2021 Two figures are considered to be "similar figures" if they have the same shape, congruent corresponding angles (meaning the angles in the same places of each shape are the same) and equal scale factors. Equal scale factors mean that the lengths of their corresponding sides have a matching ratio. Knowing how to identify similar figures makes it easier to prove geometric theorems and postulates. There's a difference between similar and congruent figures. Two shapes are congruent when they are the same exact size and have the same angle measurements. Similar figures, on the other hand, do not have to be the same size. Below is an example of similar shapes: Although they are different sizes, triangle ABC and triangle DEF are considered similar triangles because they have proportional shapes and angles. Triangle ABC is simply an enlargement of DEF. ## Finding the Vector Magnitude of a Line Segment Inactive Jana Russick February 28, 2021 Vector magnitude is the distance between the initial point and terminal point of a directed line segment. Here is a picture of vector AB: The length of the vector, in this case, is expressed as absolute value AB (\|AB\|). ## How To Find the Vertex of a Parabola Inactive Andrew Lee February 25, 2021 To find the vertex of a parabola, you first need to know how to graph quadratic equations. When graphing these, remember that every quadratic function can be put into a standard form (more on this later). This allows you to find the leading coefficient and solve for the x-intercepts. The x-intercept and y-intercept are points on the graph where the parabola intersects the x-axis or y-axis. Putting the quadratic function into standard form will also let you find the axis of symmetry, the line that runs through the vertex and divides the parabola in half. You can then find the x-coordinate and y-coordinate of the vertex, which is the highest or lowest point on a parabola. ## What Is the Pythagorean Theorem and When Is It Used? Inactive Jana Russick February 24, 2021 What is the Pythagorean theorem? It’s a trigonometry equation used to find the length of one side of a right triangle. Though similar concepts had been discovered by the Babylonians, Greek Mathematician Pythagoras was the first person to come up with a geometric proof about how the sum of the squares of the lengths can determine the side lengths of a right triangle. Pythagoras determined that when three squares are arranged so that they form a right angle triangle, the largest of the three squares must have the same area as the other two squares combined. In the picture below, you can see how the sum of the squares creates the right triangle ABC. This realization about the area of the squares led to the Pythagoras theorem: Squares are different from other parallelograms and trapezoids because all their sides are equal lengths. So since squares are made up of four equal sides, you can see that each individual square makes up a side of the right triangle. The length of the largest square, which we'll call length c, is the length of the hypotenuse. (The hypotenuse is the longest side of a right triangle.) The smaller squares make up the other two sides of the right triangle. ## Here’s How Vector Subtraction Works Inactive Jana Russick February 24, 2021 A vector is a term used to define any line segments with a specified starting and ending point. All vectors are drawn at an angle and have a specified direction. Learning to subtract vectors is helpful when you need to see how much one vector must travel to get to the other. Vector subtraction is the process of subtracting the coordinates of one vector from the coordinates of a second vector. See the example below. The coordinates of vector a are marked as (3,3) and the coordinates of vector b as (1, 2). When subtracting vectors, you must take the first vector quantities and subtract the second quantity. Let's subtract vector b from vector a: Your resultant vector coordinates for this particular example are (2, 1).
# How to Solve Linear Differential Equation A linear equation or polynomial, with one or more terms, consisting of the derivatives of the dependent variable with respect to one or more independent variables is known as a linear differential equation. A general first-order differential equation is given by the expression: dy/dx + Py = Q where y is a function and dy/dx is a derivative. The solution of the linear differential equation produces the value of variable y. Examples: • dy/dx + 2y = sin x • dy/dx + y = ex ## Linear Differential Equations Definition A linear differential equation is defined by the linear polynomial equation, which consists of derivatives of several variables. It is also stated as Linear Partial Differential Equation when the function is dependent on variables and derivatives are partial. A differential equation having the above form is known as the first-order linear differential equation where P and Q are either constants or functions of the independent variable (in this case x) only. Also, the differential equation of the form, dy/dx + Py = Q, is a first-order linear differential equation where P and Q are either constants or functions of y (independent variable) only. To find linear differential equations solution, we have to derive the general form or representation of the solution. ### Non-Linear Differential Equation When an equation is not linear in unknown function and its derivatives, then it is said to be a nonlinear differential equation. It gives diverse solutions which can be seen for chaos. ## Solving Linear Differential Equations For finding the solution of such linear differential equations, we determine a function of the independent variable let us say M(x), which is known as the Integrating factor (I.F). Multiplying both sides of equation (1) with the integrating factor M(x) we get; M(x)dy/dx + M(x)Py = QM(x) …..(2) Now we chose M(x) in such a way that the L.H.S of equation (2) becomes the derivative of y.M(x) i.e. d(yM(x))/dx = (M(x))dy/dx + y (d(M(x)))dx … (Using d(uv)/dx = v(du/dx) + u(dv/dx) M(x) /(dy/dx) + M(x)Py = M (x) dy/dx + y d(M(x))/dx M(x)Py = y dM(x)/dx 1/M'(x) = P.dx Integrating both sides with respect to x, we get; log M (x) = $$\begin{array}{l} \int Pdx (As \int \frac {f'(x)}{f(x)} ) = log f(x) \end{array}$$ ⇒ M(x) = $$\begin{array}{l} e^{\int Pdx} \end{array}$$ I.F Now, using this value of the integrating factor, we can find out the solution of our first order linear differential equation. Multiplying both the sides of equation (1) by the I.F. we get $$\begin{array}{l} e^{\int Pdx}\frac{dy}{dx} + yPe^{\int Pdx} = Qe^{\int Pdx} \end{array}$$ This could be easily rewritten as: $$\begin{array}{l} \frac {d(y.e^{\int Pdx})}{dx} = Qe^{\int Pdx} (Using \frac{d(uv)}{dx} = v \frac{du}{dx} + u\frac{dv}{dx} ) \end{array}$$ Now integrating both the sides with respect to x, we get: $$\begin{array}{l} \int d(y.e^{\int Pdx }) = \int Qe^{\int Pdx}dx + c \end{array}$$ $$\begin{array}{l} y = \frac {1}{e^{\int Pdx}} (\int Qe^{\int Pdx}dx + c )\end{array}$$ where C is some arbitrary constant. ### How to Solve First Order Linear Differential Equation Learn to solve the first-order differential equation with the help of steps given below. 1. Rearrange the terms of the given equation in the form dy/dx + Py = Q where P and Q are constants or functions of the independent variable x only. 1. To obtain the integrating factor, integrate P (obtained in step 1) with respect to x and put this integral as a power to e. $$\begin{array}{l} e^{\int Pdx} \end{array}$$ = I.F 1. Multiply both the sides of the linear first-order differential equation with the I.F. $$\begin{array}{l} e^{\int Pdx} \frac{dy}{dx} + yPe^{\int Pdx} = Qe^{\int Pdx} \end{array}$$ 1. The L.H.S of the equation is always a derivative of y × M (x) i.e. L.H.S = d(y × I.F)/dx d(y × I.F)dx = Q × I.F 1. In the last step, we simply integrate both the sides with respect to x and get a constant term C to get the solution. ∴ y × I.F = $$\begin{array}{l} \int Q × I.F dx + C \end{array}$$ , where C is some arbitrary constant Similarly, we can also solve the other form of linear first-order differential equation dx/dy +Px = Q using the same steps. In this form P and Q are the functions of y. The integrating factor (I.F) comes out to be and using this we find out the solution which will be (x) × (I.F) = $$\begin{array}{l} \int Q × I.F dy + c \end{array}$$ Now, to get a better insight into the linear differential equation, let us try solving some questions. where C is some arbitrary constant. ## Solved Examples Example 1: Solve the LDE = dy/dx = [1/(1+x3)] – [3x2/(1 + x2)]y Solution: The above mentioned equation can be rewritten as dy/dx + [3x2/(1 + x2)] y = 1/(1+x3) Comparing it with dy/dx + Py = O, we get P = 3x2/1+x3 Q= 1/1 + x3 Let’s figure out the integrating factor(I.F.) which is $$\begin{array}{l} e^{\int Pdx} \end{array}$$ ⇒I.F = $$\begin{array}{l} e^{\int \frac {3x^2}{1 + x^3}} dx = e^{ln (1 + x^3)} \end{array}$$ ⇒I.F. = 1 + x3 Now, we can also rewrite the L.H.S as: d(y × I.F)/dx, d(y × (1 + x3)) dx = [1/(1 +x3)] × (1 + x3) Integrating both the sides w. r. t. x, we get, ⇒ y × ( 1 + x3) = x ⇒ y = x/(1 + x3) ⇒ y = [x/(1 + x) + C Example 2: Solve the following differential equation: dy/dx + (sec x)y = 7 Solution: Comparing the given equation with dy/dx + Py = Q We see, P = sec x, Q = 7 Now lets find out the integrating factor using the formula $$\begin{array}{l} e^{\int Pdx} \end{array}$$ = I.F $$\begin{array}{l} e^{\int secdx} \end{array}$$ = I.F. ⇒I.F. = $$\begin{array}{l} e^{ln \|sec x + tan x \|} = sec x + tan x \end{array}$$ Now we can also rewrite the L.H.S as d(y × I.F)/dx}, i.e . d(y × (sec x + tan x )) d(y × (sec x + tan x ))/dx = 7(sec x + tan x) Integrating both the sides w. r. t. x, we get, $$\begin{array}{l} \int d ( y × (sec x + tan x )) = \int 7(sec x + tan x) dx \end{array}$$ $$\begin{array}{l} \Rightarrow y × (sec x + tan x) = 7 (ln\|sec x + tan x\| + log \|sec x\| ) \end{array}$$ ⇒ y = $$\begin{array}{l} \frac {7(ln\|sec x + tan x\| + log\|sec x\| }{(sec x + tan x)} + c \end{array}$$ Example 3: A curve is passing through the origin and the slope of the tangent at a point R(x,y) where -1<x<1 is given as (x4 + 2xy + 1)/(1 – x2). What will be the equation of the curve? Solution: We know that the slope of the tangent at (x,y) is, tanƟ= dy/dx = (x4 + 2xy + 1)/1 – x2 Reframing the equation in the form dy/dx + Py = Q , we get dy/dx = 2xy/(1 – x2) + (x4 + 1)/(1 – x2) dy/dx – 2xy/(1 – x2) = (x4 + 1)/(1 – x2) Comparing we get P = -2x/(1 – x2) Q = (x4 + 1)/(1 – x2) Now, let’s find out the integrating factor using the formula. $$\begin{array}{l} e^{\int Pdx} \end{array}$$ = I.F $$\begin{array}{l} e^{\int \frac{-2x}{1-x^2}}dx = e^{ln (1 – x^2)} = 1 – x^2 \end{array}$$ I.F Now we can also rewrite the L.H.S as $$\begin{array}{l} \frac {d(y × I.F)}{dx}, \end{array}$$ i.e $$\begin{array}{l} \frac{d(y × (1 – x^2))}{dx} = \frac{x^4 + 1}{1 – x^2} × 1 – x^2 \end{array}$$ Integrating both the sides w. r. t. x, we get, $$\begin{array}{l} \int d(y × (1 – x^2)) = \int \frac{x^4 + 1}{1 – x^2} × (1 – x^2 )dx \end{array}$$ $$\begin{array}{l} \Rightarrow y × (1 – x^2) = \int x^4 + 1 dx \end{array}$$ ……(1) x (1 – x2) = x5/5 + x + C ⇒ y = x5/5 + x/(1 – x2) + C It is the required equation of the curve. Also as the curve passes through origin; substitute the values as x = 0, y = 0 in the above equation. Thus, C = 0. Hence, equation of the curve is: ⇒ y = x5/5 + x/(1 – x2) ## Frequently Asked Questions – FAQs ### What is a linear differential equation? A linear equation or polynomial, with one or more terms, consisting of the derivatives of the dependent variable with respect to one or more independent variables is known as a linear differential equation. ### What is the example of a linear differential equation? Examples of linear differential equations are: xdy/dx+2y = x2 dx/dy – x/y = 2y dy/dx + ycot x = 2x2 ### How to solve the first order differential equation? First write the equation in the form of dy/dx+Py = Q, where P and Q are constants of x only Find integrating factor, IF = e∫Pdx Now write the solution in the form of y (I.F) = ∫Q × I.F C ### What is the difference between linear and nonlinear equations? A linear equation will always exist for all values of x and y but nonlinear equations may or may not have solutions for all values of x and y. ### What is the difference between linear and nonlinear differential equations? A linear differential equation is defined by a linear equation in unknown variables and their derivatives. A nonlinear differential equation is not linear in unknown variables and their derivatives.
# Direction Cosines & Direction Ratios of a Line View Notes ## Concepts of Direction Ratios of a Line in 3D Cartesian Coordinate System Vector analysis is a part of the advanced subjects of mathematics and physics. It is used in different concepts to calculate the angle of the components of a vector quantity plotted on a 3D Cartesian system. The direction cosines & direction ratios of a line are a bunch of concepts developed to analyze the vector components and their directions based on the three axes of the Cartesian system. It is a very crucial part of analytical geometry that finds exceptional uses in advanced physics subjects studied in the engineering subjects. To understand the concept of direction ratios of a line, continue reading this section. ### What is a Position Vector? We all know that a vector quantity has both magnitude and direction. In fact, the direction property of a vector quantity differentiates it from a scalar quantity. If we consider point A in a 3D system and O as the origin then OA is the magnitude of this quantity and the direction from O to A is the other property. Upon segregating the components of the vector quantity based on the three axes X, Y, and Z as l, m, and n respectively, we can write down the direction cosines as: X = l\|$\vec{r}$\| Y = m\|$\vec{r}$\| Z = n\|$\vec{r}$\| The expressions l\|$\vec{r}$\|, m\|$\vec{r}$\|, and n\|$\vec{r}$\| are considered to the direction ratios and are expressed by a, b, and c respectively. A direction cosine can be calculated for a vector by dividing the coordinate of that vector by the length of the same vector. In fact, adding the square of all the direction cosines of a vector will be equal to 1. Once you learn how to determine the direction cosines, you can easily understand the meaning of the terms used in the formula used to find the direction cosines of the line joining the points. ### Things to Remember Related to Direction Cosines When the line segment OP is extended in the 3D Cartesian coordinate system, we will have to consider the supplement of the direction angles mentioned in the diagram to calculate the direction cosines and direction ratios. When the same line is reversed, the vector will also signify the opposite direction. The calculation will then be adjusted based on the changes in the direction of the vector quantity. In a similar context, when a vector line does not pass through the origin O, another line segment parallel to the vector line is drawn that passes through O. This simple adjustment is done for easy calculation of the direction ratios of a line. This line segment should be parallel and of the same length as the vector quantity plotted on the 3D Cartesian coordinate system. This ensures that the angles made by the vector will be similar to that of the constructed line passing through the origin. Hence, it will ease our calculations as we can use the same formula. ### How to Calculate the Direction Cosines & Direction Ratios of a Line? If you follow the elaboration of the concept, you can understand the following things. If r is the magnitude of vector OP then, X = rcos Î± Y = rcos Î² Z = rcos Î³ From this equation, we can conclude: r = âˆš {(x â€“ 0)2 + (y â€“ 0)2 + (z â€“ 0)2} This equation becomes: r = âˆš (x2 + y2 + z2) When the value of x, y, and z are introduced as per the previous concept then, the equation becomes: X = rcos Î± = l\|$\vec{r}$\| Y = rcos Î² = m\|$\vec{r}$\| Z = rcos Î³ n\|$\vec{r}$\| On further replacing using the unit vector concept and representing the value of â€˜râ€™ with those components; we can conclude that the direction cosines of the respective angles of the vector OP are nothing but the coefficients of the same unit vector. This happens when the rectangular components of the unit vector are considered. These are the direction cosines and direction ratios of a vector plotted on a 3D Cartesian coordinate system. In fact, when we square and add the direction cosines, we find the result as 1. This answer can be concluded when we replace the values of the components on every axis by the direction cosines and then square it. As per the Pythagorean representation of a 3D Cartesian coordinate system, the answer will come as 1. Here the trigonometric identities will also be used. To understand these concepts properly, you need to draw and correlate every explanation given in this segment niftily. This step will help you grab the concepts well. You will be able to find the direction cosines of the line joining the points manually and will find the values of the components without misreading them.
### Home > CCA2 > Chapter 5 > Lesson 5.2.5 > Problem5-124 5-124. Some of the following algebraic fractions have common denominators and some do not. Add or subtract the expressions and, if possible, simplify. 1. $\frac { 3 } { ( x - 4 ) ( x + 1 ) } + \frac { 6 } { x + 1 }$ $\frac{3}{(x-4)(x+1)}+\frac{6}{x+1}\cdot \frac{x-4}{x-4}$ $\frac{3+6(x-4)}{(x-4)(x+1)}$ $\frac{6x-21}{(x-4)(x+1)}$ 1. $\frac { 5 } { 2 ( x - 5 ) } + \frac { 3 x } { x - 5 }$ See part (a). 1. $\frac { x } { x ^ { 2 } - x - 2 } - \frac { 2 } { x ^ { 2 } - x - 2 }$ Try factoring after subtracting. 1. $\frac { x + 2 } { x ^ { 2 } - 9 } - \frac { 1 } { x + 3 }$ Factor to find the common denominator, then use the strategy from part (a) to simplify. $\frac{5}{x^2-9}$ If your answer was $-\frac{1}{x^2-9},$ check the step where you subtracted. $−(x−3)=x+3$
What Do The Stars Say About Hugh Laurie? (11/09/2019) How will Hugh Laurie get by on 11/09/2019 and the days ahead? Let’s use astrology to perform a simple analysis. Note this is just for fun – don’t get too worked up about the result. I will first find the destiny number for Hugh Laurie, and then something similar to the life path number, which we will calculate for today (11/09/2019). By comparing the difference of these two numbers, we may have an indication of how smoothly their day will go, at least according to some astrology experts. PATH NUMBER FOR 11/09/2019: We will analyze the month (11), the day (09) and the year (2019), turn each of these 3 numbers into 1 number, and add them together. Here’s how it works. First, for the month, we take the current month of 11 and add the digits together: 1 + 1 = 2 (super simple). Then do the day: from 09 we do 0 + 9 = 9. Now finally, the year of 2019: 2 + 0 + 1 + 9 = 12. Now we have our three numbers, which we can add together: 2 + 9 + 12 = 23. This still isn’t a single-digit number, so we will add its digits together again: 2 + 3 = 5. Now we have a single-digit number: 5 is the path number for 11/09/2019. DESTINY NUMBER FOR Hugh Laurie: The destiny number will consider the sum of all the letters in a name. Each letter is assigned a number per the below chart: So for Hugh Laurie we have the letters H (8), u (3), g (7), h (8), L (3), a (1), u (3), r (9), i (9) and e (5). Adding all of that up (yes, this can get tedious) gives 56. This still isn’t a single-digit number, so we will add its digits together again: 5 + 6 = 11. This still isn’t a single-digit number, so we will add its digits together again: 1 + 1 = 2. Now we have a single-digit number: 2 is the destiny number for Hugh Laurie. CONCLUSION: The difference between the path number for today (5) and destiny number for Hugh Laurie (2) is 3. That is less than the average difference between path numbers and destiny numbers (2.667), indicating that THIS IS A GOOD RESULT. But don’t get too happy yet! As mentioned earlier, this is not at all guaranteed. If you want a reading that people really swear by, check out your cosmic energy profile here. Check it out now – what it returns may blow your mind. Abigale Lormen Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene. Latest posts by Abigale Lormen (see all) Abigale Lormen Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
## Friday, March 13, 2009 ### Math problems of the week: 6th grade Connected Math vs. Singapore Math 1. From "Bit and Pieces II" unit of the 6th grade Connected Mathematics curriculum (Bits and Pieces, II. p. 48): Problem 4.4 Work with your group to develop at least one algorithm for adding fractions and at least one algorithm for subtracting fractions. You might want to look back over the first three problems in this investigation and discuss how each person in your group thought about them. Look for ideas that you think will help you develop algorithms for adding and subtracting fractions that always work, even with mixed numbers Test your algorithms on a few problems, such as these: 5/8 + 7/8 3/5+5/3 3 3/4 + 7 2/9 3/4 - 1/8 5 4/6 - 2 1/3 5/6 - 1/4 If necessary, make adjustments to your algorithms until you think they will work all the time. Write up a final version of each algorithm. Make sure they are neat and precise so others can follow them. Problem 4.4 Follow-Up 1. Exchange your addition algorithm with that of another group. Test the other group's plan. Write a paragraph explaining how your algorithm and the other group's algorithm are alike and how they are different. 2. Exchange your subtraction algorithm with that of another group (a different group from the group you exchanged with in part 1). Test the other group's plan. Write a paragraph explaining how your algorithm and the other group's algorithm are alike and how they are different. 2. From the "Fractions" unit of the 6th grade Singapore Math curriculum (Primary Mathematics 6B, p. 14): Exercise 7 1. A shopkeeper had 150 lb of rice. He sold 2/5 of it and packed the remainder equally into 5 bags. Find the weight of the rice in each bag. 2. Peter had 400 stamps. 5/8 of them are U.S. stamps and the rest are Canadian stamps. He gave 1/5 of the U.S. stamps to his friend. How many stamps did he have left? 3. Kyle gave 2/7 of his money to his wife and spent 3/5 of the remainder. If he had \$300 left, how much money did he have at first? 4. 2/3 of the beads in a box are red, 1/4 are yellow and the rest are blue. There are 42 more red beads than blue beads. How many beads are there altogether? 3. Extra Credit Work with a group and write a paragraph about (1) which problem set involves more higher level thinking, and (2) which one is more irritating. Then exchange your answer to the first question with one group and and your answer to the second question with another group, and write two paragraphs explaining how your answers are alike and how they are different. #### 1 comment: bky said... When the curriculum has students invent algorithms for basic mathematical operations, to me the message is this: none of this really matters -- that's why we're letting 10-year olds make the decisions. What if you teach them how to add fractions and build it up by: (1) giving a good foundation of what a/b means (b partitions of [0,1], count up a of them), (2) using that foundation to show, very common sensibly at this point, how to add fractions with like denominators, then (3) give a good foundation for understanding equivalent representation of fractions (e.g. why 2/3 = 4/6), and then (4) very naturally lead to the general algorithm for adding fractions with different denominators ...? (that sentence started out as a question) Then the message is that adding fractions is so important (and the students are so important) that we want the kids to really be able to do it and understand how it works.
# Electronic calculators may not be used. This text is not the text of solutions of exam papers! Here we will discuss the solutions of the exampapers. Save this PDF as: Size: px Start display at page: Download "Electronic calculators may not be used. This text is not the text of solutions of exam papers! Here we will discuss the solutions of the exampapers." ## Transcription 1 EXAM FEEDBACK INTRODUCTION TO GEOMETRY (Math 20222). Spring 2014 ANSWER THREE OF THE FOUR QUESTIONS If four questions are answered credit will be given for the best three questions. Each question is worth 20 marks. Electronic calculators may not be used This text is not the text of solutions of exam papers! Here we will discuss the solutions of the exampapers. 1. Throughout this question {e, f, g} is an orthonormal basis in E 3. (a) Explain what is meant by saying that two bases in E 3 have the same orientation. Consider the ordered triple {e + f, e f, g}. Show that this triple is a basis. Show that the bases {e, f, g} and {e + f, e f, g} have opposite orientations. Let an ordered triple {a, b, c} be a basis in E 3. Explain why the bases {a, b, c} and {e, f, g} have the same orientation or the bases {a, b, c} and {e + f, e f, g} have the same orientation. (b) Consider the ordered triple of vectors { 3f + 4g e, 5 }, a, [7 marks] where a is such a vector that this triple is an orthonormal basis which has the orientation opposite to the orientation of the basis {e, f, g}. Find vector a (express it as a linear combination of the vectors e, f, g). (c) Formulate the Euler Theorem about rotations. [4 marks] 2 Let P be a linear orthogonal operator on E 3 such that it preserves orientation and Find P (e). P (f) = g, P (g) = e. We know that, due to the Euler Theorem, P is a rotation operator. Find the axis and angle of this rotation. Discussion of first question a) Students have no special problems when solving this question. Some students have proved the fact that the ordered triple {e+f, e f, g} is a basis checking straightforwardly that these vectors are linear independent. Some students just deduced this fact from the non-degeneracy of transition matrix. Almost all students answered right, that the bases have opposite orientation since the transition matrix has negative determinant. b) Many students solved this problem in a following way: if a = xe + yf + zg then using condition of orthonormality they calculated coefficients x, y and z and calculated the vector a up to a sign: a = ± ( ) 3g 4f 5. Then the final answer follows from the fact that new basis has opposite orientation. This is right solution. Much more nice solution is based on the fact that orhonormality implies that vector a must be equal up to a sign to vector product of vectors e and 3f+4g. The orientation arguments imply that 5 ( ) 3f + 4g 4f 3g a = e =. 5 5 Few students solved this question in this way. c) P (e) has to have unit length and it is orthogonal to vectors P (f) = g and P (g) = e since P is orthogonal operator. Hence P (e) = ±f. Orientation arguments imply that P (e) = f. The same can be deduced in matrix language: The matrix of operator P in the x 0 1 orthonormal basis e, f g is y 0 0 Analysing orthogonality condition of matrix we z 1 0 come to x = z = 0 and y = ±1. Then we see that y = 1 since determinant is equal to 1 (preservation of orinetation) The standard application of Euler Theorem implies that axis of rotation is directed along the vector e + f + g (this is eigenvector) and angle φ = 2π. 3 Problems: As usual standard mistake was confusing the notion of orthonormal operator and operator with determinant 1 (or in the language of matrices confusion between orthogonal matrices and matrices with determinant 1.) Many (too many students!!!???) could not caluclate angle φ such that cos φ = 1 2 (without help of calculator) 3 There is really very beautiful way to calculate angle φ which follows from the fact that P 3 is identity matrix (why?), i.e. 3φ = 2π. Helas, nobody on exam noticed this. 2. (a) Give the definition of a differential 1-form on E n. Let f be a function on E 2 given by f(r, ϕ) = r 2 sin 2ϕ, where r, ϕ are polar coordinates in E 2. Calculate the value of the 1-form ω = df on the vector field A = ϕ. Express the 1-form ω in Cartesian coordinates x, y (x = r cos ϕ, y = r sin ϕ). Give an example of a non-zero vector field B such that the 1-form ω vanishes on the vector field B at all points of E 2. (b) Consider in E 2 the differential 1-form ω = xdy ydx and the upper-half of the circle C defined by equations x 2 + y 2 = 4x and y 0. Give examples of two different parameterisations of the curve C such that these parameterisations have opposite orientations. Calculate the integral C ω. How does the answer depend on a parameterisation? [7 marks] (c) Show that the integral of the differential form ω = xdy+ydx over the circle x 2 +y 2 = 1 is equal to zero. Will the answer change if instead of this circle we consider another closed curve in E 2. Justify your answer. Discussion of second question [4 marks] a) Not a difficult question. Just again I would like to focus attention on the fact that it is much easier to calculate ω = df in Cartesian coordinates in the following way: first to calculate f = r 2 sin 2ϕ = 2r 2 sin ϕ cos ϕ = 2xy in Cartesian coorinates. Then we come immediately to the answer ω = df = d(2xy) = 2xdy + 2ydx. Some students have chosen more difficult way: they first calculated ω in polar coordinates then have tried to express this 1-form in Cartesian coordinates. Many of them failed in these calculations which 4 need much more time and experience. b) Students did good this question. Just a remark: many students wrote two parameterisations with different orientations but did not justify the answer: you need to show that velocity changes the sign (i.e. reparameterisation has negative derivative) or explicitly show that during motion the starting point becomes final and vice versa. When writing parameterisation do not forget to fix the interval of parameter changing: c) Students who have guessed that the form ω = xdy + ydx is exact (ω = d(xy)) have no any problem to answer this question: exactness implies that integral over arbitrary closed curve vanishes. Unfortunately many students did not notice the exactness of the form. Much more worse was that some students have claimed the wrong statement that for an arbitrary form integral over closed curve is equal to zero. General problem: still many students confuse notation for differential d and vector field : The expression a x + b y is for vector field, the expression adx + bdy it is 1-form not a vector field!!! 3. (a) Describe what is meant by a natural parameter on a curve in E n. Find a natural parameter for the following interval of the straight line: { x = t C :, 0 < t <. y = 3t + 2 Explain why if a curve is given in natural parameterisation, then velocity and acceleration vectors are orthogonal to each other at any given point of the curve. (b) Give the definition of the curvature of a curve in E n. [6 marks] For a curve in E 2, write down the formula for the curvature in terms of an arbitrary parameterisation. Calculate the curvature k(t) of the parabola r(t): x = t, y = t 2. Calculate the curvature of the parabola y = x 2 + px + q at the vertex of this parabola. (c) 5 On the sphere x 2 +y 2 +z 2 = 1 in E 3 find at least two curves C 1 and C 2 such that curvature of the curve C 1 is equal to 1 at all points of this curve and curvature of the curve C 2 is equal to 1000 at all points of this curve. Discussion of third question [5 marks] a) No specific problems with this question. b) Here I would like to discuss how to find curvature of parabola y = x 2 + px + q at vertex. Students (almost all) calculted in the previous subsection the curvature of parabola x = t, y = t 2 2 ; it is equal to k(t) =. For parabola y = x 2 the vertex is at (1+4t 2 ) 3/2 the point x = y = 0 and curvature at this point is equal to 2. Now notice that translations y y + a, x x + b transform parabola y = x 2 into the parabola y = x 2 + px + q and they transform vertex to the vertex. iit is easy to see that translations do not change the curvature (why?). Hence the curvature of the parabola y = x 2 + px + q at the vertex is the same that the curvature of parabola y = x 2 at the vertex. It is equal to 2. This is the shortest and nicest solution of the question. Unfortunately nobody came to this solution. Many students calculated curvature of an arbitrary point of parabola y = x 2 + px + q, noticed that vertex is at the point x = p and after some calculations some students (but 2 not many) came to the right answer. c) This question was not considered as an easy question. First of all the solution: The circle of raidus r has the curvature k = 1. So we have to find two circles C r 1 and C 2 on the sphere x 2 + y 2 + z 2 = 1 with radii r 1 and r 2 such that r 1 = 1 and r 2 = 1. For curve 1000 C 1 we can take any great circle on the sphere { e.g. intersection of the plane z = 0 with the sphere x 2 + y 2 + z 2 x 2 + y 2 = 1 = 1, i.e. the curve, or intersection of the plane z = 0 { z 2 + y 2 = 1 x = 0 with the sphere, i.e. the curve (In fact you can take an intersection x = 0 of the sphere with any plane ax + by + cz = 0 (abc 0) passing via origin.) For curve C 2 you can take intersection of sphere with an arbitrary plane which is on the distance h = 1 r2 2 = from the origin. E.g. you can take the plane z = { , x 2 + y 2 = 10 6 i.e. C 2 : z = Many students tried to attack this question. Some students noticed that they have to find a circle of the appropriate radius. Many of these students found the curve C 1 and only few students found the curve C 2 on the sphere. Remark Two years ago on 2012 on exam it was suggested the following problem: to find the curve C on the cylindre x 2 + y 2 = 1 such that the curvature of this curve is equal to k = 1. One can consider as a curve C a helix. Unfortunately too many students instead 100 of solving the exam problem and find a circle on the sphere (the simplest possible curve 6 with non-zero curvature) tried desperately to use the solution of former exam problem (helix on the cylinder) and tried to find in vain helix on the sphere. 4. (a) Explain what is meant by the shape operator for a surface r = r(u, v) in E 3 by defining its action on an arbitrary tangent vector to the surface. Explain why the value of the shape operator is also a tangent vector to the surface. [6 marks] (b) Consider a sphere of radius R in E 3 : x = R sin θ cos ϕ r(θ, ϕ): y = R sin θ sin ϕ z = R cos θ. Show that n(θ, ϕ) = r(θ,ϕ) R is a normal unit vector to the sphere at the point r(θ, ϕ). Let p be an arbitrary point on this sphere and X be an arbitrary tangent vector at the point p. Calculate the action of shape operator on the vector X. Calculate the Gaussian and mean curvatures of the sphere. (c) On the sphere in part (b) consider three points A, B, C with Cartesian coordinates A (0, 0, R), B (R, 0, 0) and C (R cos ϕ, R sin ϕ, 0), where ϕ is an angle such that 0 < ϕ < π. Consider the isosceles triangle ABC formed by arcs of great circles which 2 connect these points. Calculate the integral of the Gaussian curvature of the sphere over the interior of this triangle. Discussion of fourth question [5 marks] a) Students have not specific problems to answer this question. b) Many students who solved this problem checked by straightforward calcualtions the conditions that n is a unit vector and it is orthogonal to the surface, i.e. (n, n) = 1 and (n, r θ ) = (n, r ϕ ) = 0. Sure this is right solution. On the other hand there is is another 7 nicer way to deal with this question (its idea was discussed during tutorials) which based on the relation (r, r) = R 2. for points of sphere. Differentiating this relation along θ and ϕ we come to the conditions (r, r θ ) = (r, r ϕ ) = 0, This implies the answer. We see that n θ = r θ θ R and n θ = rϕ, i.e. shape operator S is proportional to identity ϕ R operator, SX = 1 X (up to a sign). Almost all students who were solving this problem R calculated the matrix of shape operator but many of them did not write the action of operator on arbitrary vector. c) We see that Area of triangle ABC Area of hemisphere = ϕ 2πR S 2 ABC = R 2 ϕ K = R 2 1 ϕ S ABC R = ϕ. 2 This is all. I was really surprised that almost nobody solved this problem. Again as in the previous problem many students instead solving the problem were trying to revivre a different problem from the quesiont 4 of exam paper in 2012 year which sound similar. ### 1.7 Cylindrical and Spherical Coordinates 56 CHAPTER 1. VECTORS AND THE GEOMETRY OF SPACE 1.7 Cylindrical and Spherical Coordinates 1.7.1 Review: Polar Coordinates The polar coordinate system is a two-dimensional coordinate system in which the ### 88 CHAPTER 2. VECTOR FUNCTIONS. . First, we need to compute T (s). a By definition, r (s) T (s) = 1 a sin s a. sin s a, cos s a 88 CHAPTER. 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# Mathematics Common Core Cluster. Mathematics Common Core Standard. Domain Save this PDF as: Size: px Start display at page: ## Transcription 1 Mathematics Common Core Domain Mathematics Common Core Cluster Mathematics Common Core Standard Number System Know that there are numbers that are not rational, and approximate them by rational numbers. Number System Know that there are numbers that are not rational, and approximate them by rational numbers. NS.8.1 Know that numbers that are not rational are called irrational. Understand informally that every number has a decimal expansion; for rational numbers show that the decimal expansion repeats eventually, and convert a decimal expansion which repeats eventually into a rational number. NS.8.2 Use rational approximations of irrational numbers to compare the size of irrational numbers, locate them approximately on a number line diagram, and estimate the value of expressions (e.g., π^2). For example, by truncating the decimal expansion of 2 (square root of 2), show that 2 is between 1 and 2, then between 1.4 and 1.5, and explain how to continue on to get better approximations. Expressions & Equations Work with radicals and integer exponents. EE.8.1 Know and apply the properties of integer exponents to generate equivalent numerical expressions. For example, 3^2 3^( 5) = 3^( 3) = 1/(3^3) = 1/27. Expressions & Equations Work with radicals and integer exponents. EE.8.2 Use square root and cube root symbols to represent solutions to equations of the form x^2 = p and x^3 = p, where p is a positive rational number. Evaluate square roots of small perfect squares and cube roots of small perfect cubes. Know that 2 is irrational. Page 1 of 35 2 Expressions & Equations Work with radicals and integer exponents. EE.8.3 Use numbers expressed in the form of a single digit times an integer power of 10 to estimate very large or very small quantities, and to express how many times as much one is than the other. For example, estimate the population of the United States as 3 10^8 and the population of the world as 7 10^9, and determine that the world population is more than 20 times larger. Expressions & Equations Work with radicals and integer exponents. EE.8.4. Perform operations with numbers expressed in scientific notation, including problems where both decimal and scientific notation are used. Use scientific notation and choose units of appropriate size for measurements of very large or very small quantities (e.g., use millimeters per year for seafloor spreading). Interpret scientific notation that has been generated by technology. Expressions & Equations Understand the connections between proportional relationships, lines, and linear equations. EE.8.5 Graph proportional relationships, interpreting the unit rate as the slope of the graph. Compare two different proportional relationships represented in different ways. For example, compare a distance-time graph to a distance-time equation to determine which of two moving objects has greater speed. Expressions & Equations Understand the connections between proportional relationships, lines, and linear equations. EE.8.6 Use similar triangles to explain why the slope m is the same between any two distinct points on a non-vertical line in the coordinate plane; derive the equation y =mx for a line through the origin and the equation y = mx + b for a line intercepting the vertical axis at b. Page 2 of 35 3 Expressions & Equations Analyze and solve linear equations and pairs of simultaneous linear equations. EE.8.7 Solve linear equations in one variable. Expressions & Equations Give examples of linear equations in one variable with one solution, infinitely many solutions, or no solutions. EE.8.7.a Show which of these possibilities is the case by successively transforming the given equation into simpler forms, until an equivalent equation of the form x = a, a = a, or a = b results (where a and b are different numbers). Expressions & Equations Give examples of linear equations in one variable with one solution, infinitely many solutions, or no solutions. EE.8.7.b Solve linear equations with rational number coefficients, including equations whose solutions require expanding expressions using the distributive property and collecting like terms. Expressions & Equations Analyze and solve linear equations and pairs of simultaneous linear equations. EE.8.8 Analyze and solve pairs of simultaneous linear equations. Expressions & Equations Analyze and solve linear equations and pairs of simultaneous linear equations. EE.8.8.a Understand that solutions to a system of two linear equations in two variables correspond to points of intersection of their graphs, because points of intersection satisfy both equations simultaneously. Expressions & Equations Analyze and solve linear equations and pairs of simultaneous linear equations. EE.8.8.b Solve systems of two linear equations in two variables algebraically, and estimate solutions by graphing the equations. Solve simple cases by inspection. For example, 3x + 2y = 5 and 3x + 2y = 6 have no solution because 3x + 2y cannot simultaneously be 5 and 6. Functions Define, evaluate, and compare functions. F.8.1 Understand that a function is a rule that assigns to each input exactly one output. The graph of a function is the set of ordered pairs consisting of an input and the corresponding output. (Function notation is not required in Grade ) Page 3 of 35 4 Functions Define, evaluate, and compare functions. F.8.2 Compare properties of two functions each represented in a different way (algebraically, graphically, numerically in tables, or by verbal descriptions). For example, given a linear function represented by a table of values and a linear function represented by an algebraic expression, determine which function has the greater rate of change. Functions Define, evaluate, and compare functions. F.8.3 Interpret the equation y = mx + b as defining a linear function, whose graph is a straight line; give examples of functions that are not linear. For example, the function A = s^2 giving the area of a square as a function of its side length is not linear because its graph contains the points (1,1), (2,4) and (3,9), which are not on a straight line. Functions Use functions to model relationships between quantities. F.8.4 Construct a function to model a linear relationship between two quantities. Determine the rate of change and initial value of the function from a description of a relationship or from two (x, y) values, including reading these from a table or from a graph. Interpret the rate of change and initial value of a linear function in terms of the situation it models, and in terms of its graph or a table of values. Functions Use functions to model relationships between quantities. F.8.5 Describe qualitatively the functional relationship between two quantities by analyzing a graph (e.g., where the function is increasing or decreasing, linear or nonlinear). Sketch a graph that exhibits the qualitative features of a function that has been described verbally. Geometry Understand congruence and G.8.1 Verify experimentally the properties of rotations, similarity using physical models, reflections, and translations: transparencies, or geometry -- a. Lines are taken to lines, and line segments to line software. segments of the same length. -- b. Angles are taken to angles of the same measure. -- c. Parallel lines are taken to parallel lines. Page 4 of 35 5 Geometry Understand congruence and G.8.2 Understand congruence and similarity using physical similarity using physical models, models, transparencies, or geometry software. Understand transparencies, or geometry that a two-dimensional figure is congruent to another if the software. second can be obtained from the first by a sequence of rotations, reflections, and translations; given two congruent figures, describe a sequence that exhibits the congruence between them. Geometry Understand congruence and G.8.3 Describe the effect of dilations, translations, rotations similarity using physical models, and reflections on two-dimensional figures using transparencies, or geometry coordinates. software. Geometry Understand congruence and G.8.4 Understand that a two-dimensional figure is similar to similarity using physical models, another if the second can be obtained from the first by a transparencies, or geometry sequence of rotations, reflections, translations, and dilations; software. given two similar two-dimensional figures, describe a sequence that exhibits the similarity between them. Page 5 of 35 6 Geometry Understand congruence and G.8.5 Use informal arguments to establish facts about the similarity using physical models, angle sum and exterior angle of triangles, about the angles transparencies, or geometry created when parallel lines are cut by a transversal, and the software. angle-angle criterion for similarity of triangles. For example, arrange three copies of the same triangle so that the three angles appear to form a line, and give an argument in terms of transversals why this is so. Geometry Understand and apply the Pythagorean Theorem. G.8.6 Explain a proof of the Pythagorean Theorem and its converse. Geometry Geometry Understand and apply the Pythagorean Theorem. Understand and apply the Pythagorean Theorem. G.8.7 Apply the Pythagorean Theorem to determine unknown side lengths in right triangles in real-world and mathematical problems in two and three dimensions. G.8.8 Apply the Pythagorean Theorem to find the distance between two points in a coordinate system. Geometry Solve real-world and mathematical problems involving volume of cylinders, cones and spheres. G.8.9 Know the formulas for the volume of cones, cylinders, and spheres and use them to solve real-world and mathematical problems. Statistics & Probability Investigate patterns of association in bivariate data. SP.8.1 Construct and interpret scatter plots for bivariate measurement data to investigate patterns of association between two quantities. Describe patterns such as clustering, outliers, positive or negative association, linear association, and nonlinear association. Statistics & Probability Investigate patterns of association in bivariate data. SP.8.2 Know that straight lines are widely used to model relationships between two quantitative variables. For scatter plots that suggest a linear association, informally fit a straight line, and informally assess the model fit by judging the closeness of the data points to the line. Statistics & Probability Investigate patterns of association in bivariate data. SP.8.3 Use the equation of a linear model to solve problems in the context of bivariate measurement data, interpreting the slope and intercept. For example, in a linear model for a biology experiment, interpret a slope of 1.5 cm/hr as meaning that an additional hour of sunlight each day is associated with an additional 1.5 cm in mature plant height. Page 6 of 35 7 Statistics & Probability Investigate patterns of association in bivariate data. SP.8.4 Understand that patterns of association can also be seen in bivariate categorical data by displaying frequencies and relative frequencies in a two-way table. Construct and interpret a two-way table summarizing data on two categorical variables collected from the same subjects. Use relative frequencies calculated for rows or columns to describe possible association between the two variables. For example, collect data from students in your class on whether or not they have a curfew on school nights and whether or not they have assigned chores at home. Is there evidence that those who have a curfew also tend to have chores? Page 7 of 35 8 Algebra 1 Suggested Resources Months Taught Quarter Assessed Algeba 1 1-3, 3-8, Buckle Down, Tasks October 1st Algebra 1 1-3, 3-8, Buckle Down, Tasks October 1st Algebra 1 1-2, 8-1, 8-3, 8-4, 8-5, Buckle Down, Tasks February 2nd Algebra 1 1-2, Buckle Down, Tasks October 1st Page 8 of 35 9 Algebra 1 8-2, Course 3 w/s 12-5, Buckle Down, Tasks February 3rd Algebra 1 8-3, Course 3 w/s 12-3,w/s 12-5, Activity labs 2-8a, 12-3b, Buckle Down, Tasks, February 3rd Algebra 1 5-5, 5-6, Course 3 Activity lab 11-4a, Buckle Down, Tasks November 2nd Algebra 1 6-1, 6-2, 6-4,6-5,6-6, Buckle Down, Tasks November 2nd Page 9 of 35 10 Algebra 1 2-1,3-2,3-2,3-6,3-7,4-1, 4- January 2,4-3,4-4,7-5, Buckle Down Tasks 3rd Algebra 1 3-1, 3-2, 3-3, 3-6, 3-7, 4-1, 4-2, 4-3, 4-4, 7-5, Buckle Down, Tasks January 3rd Algebra 1 2-4, 3-2, 3-3, 3-6, 3-7, Buckle Down, Tasks January 3rd Algebra 1 7-1, 7-2, 7-3, 7-4, 7-6, Buckle Down, Tasks February 3rd Algebra 1 7-1,7-4, 7-6, Buckle Down, Tasks, February 3rd Algebra 1 7-2, 7-3, 7-4, Buckle Down, Tasks February 3rd Algebra 1 1-4, 3-1, 5-1, 5-2, Buckle Down, Tasks November 2nd Page 10 of 35 11 Algebra 1 1-4, 5-4, Buckle Down, Tasks November 2nd Algebra 1 5-3, Course 3 w/s 11-5, w/s 11-7 Buckle Down, Tasks November, January 2nd, 3rd Algebra 1 5-1, 5-2,6-3, Course 3 Activity lab 3-5b, 11-4a, Buckle Down, Tasks November 2nd Algebra 5-1, 5-2, 5-5, Course 3 w/s 11-2, w/s 11-6, Buckle Down, Tasks November, January 2nd, 3rd Course 3 w/s 3-6, w/s3-7, w/s3-8 Activity Lab 3-7a, 3-8a, Buckle Down, Tasks August 1st Page 11 of 35 12 Course 3 w/s 7-3, Buckle Down, Tasks September 1st Course 3 w/s 3-6, w/s 3-7, w/s 3-8, w/s 4-5, Buckle Down, Tasks September 1st Course 3 w/s 4-4, w/s 4-5, Buckle Down, Tasks September 1st Page 12 of 35 13 Course 3 w/s 4-4, w/s 7-2, Buckle Down, Tasks, Activity Lab 7-5a, Supplemental material September, March 1st, 4th Algebra 1 3-9, 3-9A, Buckle Down, Tasks October 2nd Algebra 1 3-9, 3-9A, Buckle Down, Tasks October 2nd Algebra #49-55, Course 3 w/s 3-4, Buckle Down, Tasks October 2nd Course 3 w/s 8-6, w/s 8-7, w/s 8-8, Buckle Down, Tasks April 4th Algebra 1 1-5,Course 3 Activity lab 9-7a, Buckle Down, Tasks December 2nd Algebra 1 6-7, Buckle Down, Tasks December 2nd Algebra 1 6-7, Buckle Down, Tasks, Supplemental material December 2nd Page 13 of 35 14 Buckle Down, Tasks, Supplemental material December 2nd Page 14 of 35 15 Page 15 of 35 16 Page 16 of 35 17 Page 17 of 35 18 Page 18 of 35 19 Page 19 of 35 20 Page 20 of 35 21 Page 21 of 35 22 Page 22 of 35 23 Page 23 of 35 24 Page 24 of 35 25 Page 25 of 35 26 Page 26 of 35 27 Page 27 of 35 28 Page 28 of 35 29 Page 29 of 35 30 Page 30 of 35 31 Page 31 of 35 32 Page 32 of 35 33 Page 33 of 35 34 Page 34 of 35 35 Page 35 of 35 ### with functions, expressions and equations which follow in units 3 and 4. 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# How to find the standard form of the equation of the specified circle given that it Is tangent to both axis and passes through ( 2 ,-1)? May 25, 2017 ${x}^{2} + {y}^{2} - 10 x + 10 y + 25 = 0$ or ${x}^{2} + {y}^{2} - 2 x + 2 y + 1 = 0$ #### Explanation: As the circle is tangent to both the axis, its centre is equidistant from both the axis and as it passes through $\left(2 , - 1\right)$, it lies in fourth quadrant and coordinates of centre are of type $\left(a , - a\right)$ and its equation is ${\left(x - a\right)}^{2} + {\left(y + a\right)}^{2} = {a}^{2}$ or ${x}^{2} + {y}^{2} - 2 a x + 2 a y + {a}^{2} = 0$ and as it passes through $\left(2 , - 1\right)$, we have ${2}^{2} + {\left(- 1\right)}^{2} - 2 a \times 2 + 2 a \times \left(- 1\right) + {a}^{2} = 0$ or ${a}^{2} - 6 a + 5 = 0$ or $\left(a - 5\right) \left(a - 1\right) = 0$ Hence $a = 5$ or $a = 1$ and equation of circle could be ${x}^{2} + {y}^{2} - 10 x + 10 y + 25 = 0$ or ${x}^{2} + {y}^{2} - 2 x + 2 y + 1 = 0$ graph{(x^2+y^2-10x+10y+25)(x^2+y^2-2x+2y+1)=0 [-1.793, 8.207, -3.94, 1.06]}
# Difference between revisions of "2011 AIME I Problems/Problem 4" ## Problem 4 In triangle $ABC$, $AB=125$, $AC=117$ and $BC=120$. The angle bisector of angle $A$ intersects $\overline{BC}$ at point $L$, and the angle bisector of angle $B$ intersects $\overline{AC}$ at point $K$. Let $M$ and $N$ be the feet of the perpendiculars from $C$ to $\overline{BK}$ and $\overline{AL}$, respectively. Find $MN$. ## Solution Extend ${MN}$ such that it intersects lines ${AC}$ and ${BC}$ at points $O$ and $Q$, respectively. Notice that ${OQ}$ is a midline; it then follows that ${OC} = 58.5$ and ${QC} = 60$. Now notice that $\triangle MQC$ and $\triangle NOC$ are both isosceles. Thus, $ON = OC = 58.5$ and $MQ = QC = 60$. Since $OQ$ is a midline, $OQ = 62.5$. We want to find $MN$, which is just $ON + MQ - OQ$. Substituting, the answer is $58.5 + 60 - 62.5 = \boxed {56}$.
# Exercise 6.2(Linear Inequalities ) Solve the following inequalities graphically in two-dimensional plane:(Ex 6.2 Linear Inequalities ncert math solution class 11) Question 1: x + y < 5 Solution: Given that x 0 5 y 5 0 Now, draw a dotted line x + y = 5 in the graph Now, consider x + y < 5 Select a point (0, 0). â‡’Â 0 + 0 < 5 â‡’Â 0 < 5 (this is true) âˆ´ The solution region of the given inequality is below the line x + y = 5. (That is, the origin is included in the region.) The graph is as follows: Question 2: 2x + y â‰¥ 6 Solution : Given that: 2x + y â‰¥ 6 x 0 3 y 6 0 Now, draw a solid line 2x + y = 6 in the graph Now, consider 2x + y â‰¥6 Select a point (0, 0). â‡’Â 2 Ã— (0) + 0 â‰¥Â 6 â‡’ 0 â‰¥ 6 (This is false.) âˆ´ The solution region of the given inequality is above the line 2x + y = 6. (Away from the origin.) The graph is as follows: Question 3: 3x + 4y â‰¤ 12 Solution: Given that, 3x + 4y â‰¤ 12 x 0 4 y 3 0 Now, draw a solid line 3x + 4y = 12 in the graph Now, consider 3x + 4y â‰¤ 12 Select a point (0, 0). â‡’Â 3 Ã— (0) + 4 Ã— (0)Â â‰¤ 12 â‡’ 0 â‰¤ 12 (This is true.) âˆ´ The solution region of the given inequality is below the line 3x + 4y = 12. (That is, the origin is included in the region.) The graph is as follows: Question 4: y + 8 â‰¥ 2x Solution: Given that, y + 8 â‰¥ 2x x 0 4 y -8 0 Now, draw a solid line y + 8 = 2x in the graph Now, consider y + 8 â‰¥ 2x Select a point (0, 0). â‡’Â (0) + 8Â â‰¥Â 2 Ã— (0) â‡’ 0â‰¤ 8 (This is true.) âˆ´ The solution region of the given inequality is above the line y + 8 = 2x. (That is, the origin is included in the region.) The graph is as follows: Question 5: x â€“ y â‰¤ 2 Solution: Given that, x â€“ y â‰¤ 2 x 0 2 y -2 0 Now, draw a solid line x â€“ y = 2 in the graph Now, consider x â€“ y â‰¤ 2 Select a point (0, 0). â‡’Â (0) â€“ (0) â‰¤ 2 â‡’ 0 â‰¤ 2 (This is true.) âˆ´ The solution region of the given inequality is above the line x â€“ y = 2. (That is, the origin is included in the region.) The graph is as follows: Question 6: 2x â€“ 3y > 6 Solution: Given that, 2x â€“ 3y > 6 Now, draw a dotted line 2x â€“ 3y = 6 in the graph Now, consider 2x â€“ 3y > 6 Select a point (0, 0). â‡’Â 2 Ã— (0) â€“ 3 Ã— (0) > 6 â‡’ 0 > 6 (This is false.) âˆ´ The solution region of the given inequality is below the line 2x â€“ 3y > 6. (Away from the origin.) The graph is as follows: Question 7: â€“ 3x + 2y â‰¥ â€“ 6 Solution: Given that, â€“ 3x + 2y â‰¥ â€“ 6 Now, draw a solid line â€“ 3x + 2y = â€“ 6 in the graph Now, consider â€“ 3x + 2y â‰¥ â€“ 6 Select a point (0, 0). â‡’Â â€“ 3 Ã— (0) + 2 Ã— (0) â‰¥ â€“ 6 â‡’ 0 â‰¥ â€“ 6 (This is true.) âˆ´ The solution region of the given inequality is above the line â€“ 3x + 2y â‰¥ â€“ 6. (That is, the origin is included in the region) The graph is as follows: Question 8: 3y â€“ 5x < 30 Solution: Given that, y â€“ 5x < 30 x 0 -6 y 10 0 Now, draw a dotted line 3y â€“ 5x = 30 in the graph Now, consider 3y â€“ 5x < 30 Select a point (0, 0). â‡’Â 3 Ã— (0) â€“ 5 Ã— (0) < 30 â‡’ 0 < 30 (This is true.) âˆ´ The solution region of the given inequality is below the line 3y â€“ 5x < 30. (That is, the origin is included in the region.) The graph is as follows: Question 9: y < â€“ 2 Solution: Given that, y < â€“ 2 Now, draw a dotted line y = â€“ 2 in the graph Now, consider y < â€“ 2 Select a point (0, 0). â‡’ 0 < â€“ 2 (This is false) âˆ´ The solution region of the given inequality is below the line y < â€“ 2. (That is, away from the origin.) The graph is as follows: Question 10: x > â€“ 3 Solution: Given that, x > â€“ 3 Now, draw a dotted line x = â€“ 3 in the graph Now, consider x > â€“ 3 Select a point (0, 0). â‡’Â 0 > â€“ 3 â‡’ 0 > â€“ 3 (This is true.) âˆ´ The solution region of the given inequality is right to the line x > â€“ 3. (That is, the origin is included in the region.) The graph is as follows: Ex 6.1 Linear Inequalities ncert math solution class 11 Ex 6.2 Linear Inequalities ncert math solution class 11 gmath.in
# y_n=log x_n, n =2,3,4,...and y_n-(n-1)/n y_(n-1)=1/n log n, with y_2=log sqrt2, how do you prove that x_n=(n!)^(1/n)? Jan 24, 2017 By induction #### Explanation: Note that as ${y}_{n} = \log \left({x}_{n}\right)$, we have ${x}_{n} = {e}^{{y}_{n}}$ (assuming the natural logarithm. Using a different base will not change the structure of the proof, however). Additionally, note that ${y}_{n} = \frac{1}{n} \log \left(n\right) + \frac{n - 1}{n} {y}_{n - 1}$ by the second given relation. Proof: (By induction) Base Case: For $n = 2$, we are given ${y}_{2} = \log \left(\sqrt{2}\right)$, meaning x_2 = e^log(sqrt(2)) = sqrt(2) = (2!)^(1/2) Inductive Hypothesis: Suppose that x_k = (k!)^(1/k) for some integer $k \ge 2$. Induction Step: We wish to show that x_(k+1) = [(k+1)!]^(1/(k+1)). Indeed, examining ${y}_{k + 1}$, we have ${y}_{k + 1} = \frac{1}{k + 1} \log \left(k + 1\right) + \frac{k}{k + 1} {y}_{k}$ $= \frac{1}{k + 1} \left[\log \left(k + 1\right) + k \log \left({x}_{k}\right)\right]$ =1/(k+1)[log(k+1)+klog((k!)^(1/k))] =1/(k+1)[log(k+1)+log(k!)] =1/(k+1)log(k!(k+1)) =1/(k+1)log((k+1)!) =log([(k+1)!]^(1/(k+1))) meaning x_(k+1) = e^(y_(k+1)) = [(k+1)!]^(1/(k+1)), as desired. We have supposed true for $k$ and shown true for $k + 1$, thus, by induction, x_n = (n!)^(1/n) for all integers $n \ge 2$.
# How do you solve 5(x + 4) = 10? Apr 5, 2016 $x = - 2$ #### Explanation: color(blue)(5(x+4)=10 Use distributive property color(brown)(a(b+c)=ab+ac $\rightarrow 5 x + 20 = 10$ Subtract both sides by $20$ $\rightarrow 5 x + 20 - 20 = 10 - 20$ $\rightarrow 5 x = - 10$ Divide both sides by $5$ $\rightarrow \frac{{\cancel{5}}^{1} x}{\cancel{5}} ^ 1 = - \frac{10}{5}$ color(green)(rArrx=-2 Check $5 \left(- 2 + 4\right) = 10$ $5 \left(2\right) = 10$ $10 = 10$ Correct!
# How do you convert -145.8^circ to D^circM'S'' form? May 8, 2017 Strip off the numbers to the right of the decimal and convert those to minutes. Strip off the numbers to the right of the decimal of the converted minutes and convert those to seconds. #### Explanation: Strip off the number(s) to the right of the decimal and perform dimensional conversion on it. We start the dimensional conversion by writing the value over 1: ${0.8}^{\circ} / 1$ Multiply by the factor for converting from degrees to minutes: ${0.8}^{\circ} / 1 \frac{60 \text{'}}{1} ^ \circ$ Please observe that the units of degrees cancel: ${0.8}^{\cancel{\circ}} / 1 \frac{60 '}{1} ^ \cancel{\circ} = 48 '$ If there were numbers to the right of the decimal, we would strip those off and convert them to seconds, using the factor $\frac{60 ' '}{1 '}$ but this is not the case. $- {145.8}^{\circ} = {145}^{\circ} 48 '$ May 8, 2017 $- {145}^{\circ} 48 ' 0 ' '$ #### Explanation: You need to know the conversion: ${1}^{\circ} = 60 ' = 3600 ' '$. Given: $- {145.8}^{\circ}$ The ${D}^{\circ} =$the integer portion of the decimal degree: ${D}^{\circ} = - 145$ The Minute, $M$ = the integer portion of the decimal portion multiplied by 60: $M = 0.8 \cdot 60 = 48.0 '$ The Second, $S$ = the integer portion of the minute decimal portion (the zero of $48.0$) multiplied by $60$: $S = 0.0 \cdot 60 = 0 ' '$ $- {145.8}^{\circ} = - {145}^{\circ} 48 ' 0 ' '$
# 1999 AHSME Problems/Problem 16 ## Problem What is the radius of a circle inscribed in a rhombus with diagonals of length $10$ and $24$? $\mathrm{(A) \ }4 \qquad \mathrm{(B) \ }\frac {58}{13} \qquad \mathrm{(C) \ }\frac{60}{13} \qquad \mathrm{(D) \ }5 \qquad \mathrm{(E) \ }6$ ## Solution Let $d_1=10$ and $d_2=24$ be the lengths of the diagonals, $a$ the side, and $r$ the radius of the inscribed circle. Using Pythagorean theorem we can compute $a=\sqrt{ (d_1/2)^2 + (d_2/2)^2 }=13$. We can now express the area of the rhombus in two different ways: as $d_1 d_2 / 2$, and as $2ar$. Solving $d_1 d_2 / 2 = 2ar$ for $r$ we get $r=\boxed{\frac{60}{13}}$. (The first formula computes the area as one half of the circumscribed rectangle whose sides are parallel to the diagonals. The second one comes from the fact that we can divide the rhombus into $4$ equal triangles, and in those the height on the side $a$ is equal to $r$. See pictures below.) $[asy] unitsize(1cm); pair A=(-3,0), B=(0,-1.25), C=(3,0), D=(0,1.25); pair E=(-3,-1.25), F=(-3,1.25), G=(3,1.25), H=(3,-1.25); fill ( E -- F -- G -- H -- cycle, lightgray ); draw ( E -- F -- G -- H -- cycle ); draw ( A -- B -- C -- D -- cycle ); label("d_1",(E+F)0.5,W); label("d_2",(F+G)0.5,N); [/asy]$ $[asy] unitsize(1cm); pair A=(-3,0), B=(0,-1.25), C=(3,0), D=(0,1.25); pair P = intersectionpoint( A--B--C--D--cycle, circle( (0,0), 15/13 ) ); fill ( A -- B -- (0,0) -- cycle, lightgray ); draw ( A -- B -- (0,0) -- cycle ); draw ( A -- B -- C -- D -- cycle ); draw ( circle( (0,0), 15/13 ) ); draw ( (0,0) -- P ); label("a",(A+B)0.5,SW); label("r",P0.5,0.5*NW); [/asy]$
## Lesson: Modeling and Recording AdditionDeveloping the Concept Help with Opening PDF Files Once children have had practice adding 1 to numbers 0 through 9, you can introduce the concept of adding 2 to a number. Then you can focus on different ways to make a number, up to 10, include adding doubles. Materials: overhead projector, 10 counters; Learning Tool 2 (Two-Part Mat), 10 counters, and Ways to Make (PDF file) worksheets for each child; Race to the Sum (PDF file) game board, spinner (0-4), number cubes (1-6) for every 2 children Preparation: Be prepared to give each child 10 counters and a copy of Learning Tool 2 during the lesson. Prepare Ways to Make (PDF file) worksheets for each child. Make copies of Race to the Sum (PDF file) game boards and spinners with numbers 0-4 for each pair of children Prerequisite Skills and Concepts: Children should be proficient in counting and recognizing numbers to 10. Children should show a good understanding of adding 1 to numbers 0 through 9. • Say: We have learned how to add 1 to a number and how to write addition sentences. Place 4 counters on the left side of the overhead and 1 on the right side. • Ask: Who can tell me what addition sentence we can write? Elicit from children the sentence 4 + 1 = 5. • Ask: Which number is the total? Children should say 5. • Say: Another word for total is sum. The sum of 4 + 1 is 5. • Ask: How many counters are on this side? (Point to the left side.) Children should say “4.” • Ask: Now how many counters are on this side? (Point to the right side.) Children should say “2.” • Say: Now let's count to find how many in all. Children should count aloud to 6. • Ask: How do we write this number sentence? Elicit 4 + 2 = 6. • Say: Four plus two equals six. Write4 + 2 = 6 under4 + 1 = 5. Make sure the numbers and symbols are lined up. • Practice adding 2 to various numbers through 8. Give each child 10 counters and a workmat and have them follow along. Model counting to find the sum. When children show an understanding of this, put 3 counters on each side of the overhead. Children should say “3 + 3.” • Say: Let's count to find the sum. Children should count aloud to 6. Write3 + 3 = 6 on the overhead. • Ask: What do you notice about the numbers on either side of the plus sign? Children should say they are both 3 or they are both the same. • Say: When you add the same number twice, you are doubling that number. Write these number sentences on the overhead: 5 + 5; 3 + 1; 4 + 2; 1 + 1; 2 + 1; 2 + 2. Ask the children to identify the double facts. • Say: Set aside 2 of the counters on your mat. Use the counters you have left to find which double fact has a sum of 8. Give children time to work this out. Eventually someone should say “4 + 4.” Write4 + 4 = 8 on the board. • Ask: What are some other ways to make 8? Use your counters if you need to. Children should say that 7 + 1, 6 + 2, and 5 + 3 equal 8. Write these sentences below 4 + 4 = 8, as shown below. 4 + 4 = 8 5 + 3 = 8 6 + 2 = 8 7 + 1 = 8 • Say: Let's look at all the different ways we found to make eight. Say the numbers as I point to them. Have children count down each column from top to bottom saying “4, 5, 6, 7” and then “4, 3, 2, 1.” • Ask: Do you see a pattern? Children should say that each number in the first column is 1 more than the number above it. Children should say that each number in the second column is 1 less than the number above it. Distribute Ways to Make (PDF file) worksheets. Have children use the counters to find the double fact and write it on their worksheets. Then have them find and write the remaining number sentences for 10 on their worksheets. Wrap-Up and Assessment Hints: You can assign to each child or pair of children a number or set of numbers and have them model and write all the ways to make that number. You can assess children's understanding of addition by observing them as they play Race to the Sum. This game will assess their ability to add numbers to 10 and identify double facts. You may wish to have children record their addition sentences as they play.
# Factors of 714: Prime Factorization, Methods, and Examples Factors are a list of numbers that divide the given number leaving zero as the remainder. 714 is an even and composite number. It has 16 factors in total. The factors will be both positive and negative if the given number is achieved by multiplying two-factor integers. ### Factors of 714 Here are the factors of number 714. Factors of 714: 1, 2, 3, 6, 7, 14, 17, 21, 34, 42, 51, 102, 119, 238, 357, 714 ### Negative Factors of 714 The negative factors of 714 are similar to their positive aspects, just with a negative sign. Negative Factors of 714: -1, -2, -3, -6, -7, -14, -17, -21, -34, -42, -51, -102, -119, -238, -357, and -714 ### Prime Factorization of 714 The prime factorization of 714 is the way of expressing its prime factors in the product form. Prime Factorization: 2 x 3 x 7 x 17 In this article, we will learn about the factors of 714 and how to find them using various techniques such as upside-down division, prime factorization, and factor tree. ## What Are the Factors of 714? The factors of 714 are 1, 2, 3, 6, 7, 14, 17, 21, 34, 42, 51, 102, 119, 238, 357, and 714. These numbers are the factors as they do not leave any remainder when divided by 714. The factors of 714 are classified as prime numbers and composite numbers. The prime factors of the number 714 can be determined using the prime factorization technique. ## How To Find the Factors of 714? You can find the factors of 714 by using the rules of divisibility. The divisibility rule states that any number, when divided by any other natural number, is said to be divisible by the number if the quotient is the whole number and the resulting remainder is zero. To find the factors of 714, create a list containing the numbers that are exactly divisible by 714 with zero remainders. One important thing to note is that 1 and 714 are the 714’s factors as every natural number has 1 and the number itself as its factor. 1 is also called the universal factor of every number. The factors of 714 are determined as follows: $\dfrac{714}{1} = 714$ $\dfrac{714}{2} = 357$ $\dfrac{714}{3} = 238$ $\dfrac{714}{6} = 119$ $\dfrac{714}{7} = 102$ $\dfrac{714}{14} = 51$ $\dfrac{714}{17} = 42$ $\dfrac{714}{21} = 34$ $\dfrac{714}{714} = 1$ Therefore, 1, 2, 3, 6, 7, 14, 17, 21, 34, 42, 51, 102, 119, 238, 357, and 714 are the factors of 714. ### Total Number of Factors of 714 For 714, there are 16 positive factors and 16 negative ones. So in total, there are 32 factors of 714. To find the total number of factors of the given number, follow the procedure mentioned below: 1. Find the factorization/prime factorization of the given number. 2. Demonstrate the prime factorization of the number in the form of exponent form. 3. Add 1 to each of the exponents of the prime factor. 4. Now, multiply the resulting exponents together. This obtained product is equivalent to the total number of factors of the given number. By following this procedure, the total number of factors of 714 is given as: Factorization of 714 is 1 x 2 x 3 x 7 x 17. The exponent of 1, 2, 3, 7, and 17 is 1. Adding 1 to each and multiplying them together results in 32. Therefore, the total number of factors of 714 is 32. 16 is positive, and 16 factors are negative. ### Important Notes Here are some essential points that must be considered while finding the factors of any given number: • The factor of any given number must be a whole number. • The factors of the number cannot be in the form of decimals or fractions. • Factors can be positive as well as negative. • Negative factors are the additive inverse of the positive factors of a given number. • The factor of a number cannot be greater than that number. • Every even number has 2 as its prime factor, the smallest prime factor. ## Factors of 714 by Prime Factorization The number 714 is a composite number. Prime factorization is a valuable technique for finding the number’s prime factors and expressing the number as the product of its prime factors. Before finding the factors of 714 using prime factorization, let us find out what prime factors are. Prime factors are the factors of any given number that are only divisible by 1 and themselves. To start the prime factorization of 714, start dividing by its most minor prime factor. First, determine that the given number is either even or odd. If it is an even number, then 2 will be the smallest prime factor. Continue splitting the quotient obtained until 1 is received as the quotient. The prime factorization of 714 can be expressed as: 714 = 2 x 3 x 7 x 17 ## Factors of 714 in Pairs The factor pairs are the duplet of numbers that, when multiplied together, result in the factorized number. Factor pairs can be more than one depending on the total number of factors given. For 714, the factor pairs can be found as: 1 x 714 = 714 2 x 357= 714 3 x 238 = 714 6 x 119 = 714 7 x 102 = 714 14 x 51 = 714 17 x 42 = 714 21 x 34 = 714 The possible factor pairs of 714 are given as (1, 714), (2, 357), (3, 238), (6, 119), (7, 102), (14, 51), (17, 42), and (21, 34). All these numbers in pairs, when multiplied, give 714 as the product. The negative factor pairs of 714 are given as: -1 x -714 = 714 -2 x -357= 714 -3 x -238 = 714 -6 x -119 = 714 -7 x -102 = 714 -14 x -51 = 714 -17 x -42 = 714 -21 x -34 = 714 It is important to note that in negative factor pairs, the minus sign has been multiplied by the minus sign, due to which the resulting product is the original positive number. Therefore, -1, -2, -3, -6, -7, -14, -17, -21, -34, -42, -51, -102, -119, -238, -357, and -714 are called negative factors of 714. The list of all the factors of 714, including positive as well as negative numbers, is given below. Factor list of 714: 1, -1, 2, -2, 3, -3, 6, -6, 7, -7, 14, -14, 17, -17, 21, -21, 34, -34, 42, -42, 51, -51, 102, -102, 119, -119, 238, -238, 357, -357, 714, and -714 ## Factors of 714 Solved Examples To better understand the concept of factors, let’s solve some examples. ### Example 1 How many factors of 714 are there? ### Solution The total number of Factors of 714 is 32. Factors of 714 are 1, 2, 3, 6, 7, 14, 17, 21, 34, 42, 51, 102, 119, 238, 357, and 714. ### Example 2 Find the factors of 714 using prime factorization. ### Solution The prime factorization of 714 is given as: 714 $\div$ 2 = 357 357 $\div$ 3 = 119 119 $\div$ 7 = 17 17 $\div$ 17 = 1 So the prime factorization of 714 can be written as: 2 x 3 x 7 x 17 = 714
# Inequalities Concise Class-9th Selina ICSE Maths Solution Inequalities Concise Class-9th Selina ICSE Maths Solution Chapter-11. We provide step by step Solutions of Exercise / lesson-11 Inequalities for ICSE Class-9 Concise Selina Mathematics by R K Bansal. Our Solutions contain all type Questions with Exe-11 A to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-9 Mathematics . ## Inequalities Concise Class-9th Selina ICSE Maths Solution Chapter-11 ### Exercise – 11 Question 1 From the following figure, prove that: AB > CD. …………………. #### Question 2 In a triangle PQR; QR = PR and ∠P = 36o. Which is the largest side of the triangle? Since ∠R is the greatest, therefore, PQ is the largest side. #### Question 3 If two sides of a triangle are 8 cm and 13 cm, then the length of the third side is between a cm and b cm. Find the values of a and b such that a is less than b. The sum of any two sides of the triangle is always greater than third side of the triangle. Third side < 13+8 =21 cm. The difference between any two sides of the triangle is always less than the third side of the triangle. Third side > 13-8 =5 cm. Therefore, the length of the third side is between 5 cm and 9 cm, respectively. The value of a =5 cm and b= 21cm. #### Question 4 In each of the following figures, write BC, AC and CD in ascending order of their lengths. ……………….. #### Question 5 Arrange the sides of ∆BOC in descending order of their lengths. BO and CO are bisectors of angles ABC and ACB respectively. ……………………… #### Question 6 D is a point in side BC of triangle ABC. If AD > AC, show that AB > AC. #### Question 7 n the following figure, ∠BAC = 60o and ∠ABC = 65o. Prove that: (i) CF > AF (ii) DC > DF #### Question 8 In the following figure; AC = CD; ∠BAD = 110o and ∠ACB = 74o. Prove that: BC > CD #### Question 9 From the following figure; prove that: (i) AB > BD (ii) AC > CD (iii) AB + AC > BC #### Question 10 In a quadrilateral ABCD; prove that: (i) AB+ BC + CD > DA (ii) AB + BC + CD + DA > 2AC (iii) AB + BC + CD + DA > 2BD Const: Join AC and BD. (i) In ΔABC, AB + BC > AC….(i)[Sum of two sides is greater than the third side] In ΔACD, AC + CD > DA….(ii)[ Sum of two sides is greater than the third side] Adding (i) and (ii) AB + BC + AC + CD > AC + DA AB + BC + CD > AC + DA – AC AB + BC + CD > DA …….(iii) (ii)In ΔACD, CD + DA > AC….(iv)[Sum of two sides is greater than the third side] Adding (i) and (iv) AB + BC + CD + DA > AC + AC AB + BC + CD + DA > 2AC (iii) In ΔABD, AB + DA > BD….(v)[Sum of two sides is greater than the third side] In ΔBCD, BC + CD > BD….(vi)[Sum of two sides is greater than the third side] Adding (v) and (vi) AB + DA + BC + CD > BD + BD AB + DA + BC + CD > 2BD #### Question 11 In the following figure, ABC is an equilateral triangle and P is any point in AC; prove that: (i) BP > PA (ii) BP > PC ……………….. #### Question 12 P is any point inside the triangle ABC. Prove that: ∠BPC > ∠BAC. #### Question 13 Prove that the straight line joining the vertex of an isosceles triangle to any point in the base is smaller than either of the equal sides of the triangle. We know that exterior angle of a triangle is always greater than each of the interior opposite angles. In ΔABD, ∠ADC > ∠B ……..(i) In ABC, AB = AC ∠B = ∠C …..(ii) From (i) and (ii) ∠ADC > ∠ C AC > AD ………(iii) [side opposite to greater angle is greater] (ii) In ΔABC, AB = AC AB > AD[ From (iii)] #### Question 14 In the following diagram; AD = AB and AE bisects angle A. Prove that: (i) BE = DE (ii) ∠ABD > ∠C ………………… #### Question 15 The sides AB and AC of a triangle ABC are produced; and the bisectors of the external angles at B and C meet at P. Prove that if AB > AC, then PC > PB. #### Question 16 In the following figure; AB is the largest side and BC is the smallest side of triangle ABC. Write the angles xo, yo and zo in ascending order of their values. Since AB is the largest side and BC is the smallest side of the triangle ABC Since AB is the largest side and BC is the smallest side of the triangle ABC. AB > AC > BC ⇒ 180° – z° > 180° – y° > 180° – x° ⇒ – z° > -y° > – x° ⇒ z° > y° > x°. #### Question 17 In quadrilateral ABCD, side AB is the longest and side DC is the shortest. Prove that: (i) …………… (ii)……….. #### Question 18 In triangle ABC, side AC is greater than side AB. If the internal bisector of angle A meets the opposite side at point D, prove that: ∠ADC is greater than ∠ADB. #### Question 19 In isosceles triangle ABC, sides AB and AC are equal. If point D lies in base BC and point E lies on BC produced (BC being produced through vertex C), prove that: (i) AC > AD (ii) AE > AC (iii) AE > AD We know that the bisector of the angle at the vertex of an isosceles triangle bisects the base at right angle. Using Pythagoras theorem in AFB, AB2 = AF2 + BF2…………..(i) In AFD, AD2 = AF2 + DF2…………..(ii) We know ABC is isosceles triangle and AB = AC AC2 = AF2 + BF2 ……..(iii)[ From (i)] Subtracting (ii) from (iii) AC2 – AD2 = AF2 + BF2 – AF2 – DF2 AC2 – AD2 = BF2 – DF2 Let 2DF = BF AC2 – AD2 = (2DF)2 – DF2 or AC2 – AD2 = 4DF2 – DF2 AC2 = AD2 + 3DF2 hence AC2 > AD2 Similarly, AE > AC and AE > AD. #### Question 20 Given: ED = EC Prove: AB + AD > BC. ………… The sum of any two sides of the triangle is always greater than the third side of the triangle. #### Question 21 In triangle ABC, AB > AC and D is a point in side BC. Show that: AB > AD.
# PARAM VISIONS All about civil construction knowledge- PARAM VISIONS ### Calculating the quantity of water in an open well. Let us calculate the quantity of water present in the open well for the given drawing. Given data: Depth of water = h = 2700mm = 2.7m. Outer diameter of RCC ring = D = 1200mm. Ringwall thickness = 70mm. The formula for calculating the quantity of water in the open well = [( π × d2 × h ) ÷ 4] Here, d = inner diameter of RCC ring. First, we will find the value of d. Inner dia. of RCC ring d = [outer dia. of the ring - ( 2nos. × ring wall thickness )] = [1200mm - ( 2nos. × 70mm. )] = 1060mm.= 1.060m. Quantity of water in the open well =[ (π × (1.060m)2× 2.7m.) ÷ 4 ] = [ 9.532 cum. ÷ 4 ] = 2.383 cum. As you know, 1cum = 1000 liters. So, the quantity of water in the open well = 2.383 × 1000 = 2383 liters. Easy alternate method: Given data: Depth of RCC ring = h = 300mm = 0.3m. Outer diameter of RCC ring = D = 1500mm. Ringwall thickness = 80mm. Inner dia. of RCC ring d = [outer dia. of the ring - ( 2nos. × ring wall thickness )] = [ 1500mm - ( 2nos. × 80mm. ) = 1340mm.= 1.34m. The quantity of water that a single RCC ring can hold = [(π × d2 × h) ÷ 4 ] = [(3.142 × 1.342 × 0.3m.) ÷ 4] = [1.692 cum. ÷ 4 ] = 0.423 cum = 423 liters. Suppose if you have installed 40nos. of RCC rings of 1500mm dia. in your open well, and if the 9 rings are immersed in the water, then, the quantity of water present in the open well = No. of immersed rings × water holding capacity of one ring. = 9 nos. × 423 liters = 3807 liters. I have given a table of RCC open well rings of different diameters, for your easy calculation purpose. Sl. No. RCC ring outer dia. (D) in mm. Ring wall thickness in mm. RCC ring inner dia. (d)in mm. Depth of the ring (h)in mm. Water holding capacity in litres. 1 1000 65 870 300 178 2 1200 70 1060 300 265 3 1500 80 1340 300 423 4 1800 90 1620 300 618 Note The wall thickness and ring depth may vary according to the available RCC ring in your region. By replacing the value of inner dia. and height in the above formula, you can calculate the water quantity for the open well, according to the installed ring dimension.
# Jessica Chastain Gets An Interesting Astrology Analysis (03/26/2020) How will Jessica Chastain perform on 03/26/2020 and the days ahead? Let’s use astrology to complete a simple analysis. Note this is of questionable accuracy – don’t get too worked up about the result. I will first calculate the destiny number for Jessica Chastain, and then something similar to the life path number, which we will calculate for today (03/26/2020). By comparing the difference of these two numbers, we may have an indication of how good their day will go, at least according to some astrology people. PATH NUMBER FOR 03/26/2020: We will analyze the month (03), the day (26) and the year (2020), turn each of these 3 numbers into 1 number, and add them together. What does this entail? We will show you. First, for the month, we take the current month of 03 and add the digits together: 0 + 3 = 3 (super simple). Then do the day: from 26 we do 2 + 6 = 8. Now finally, the year of 2020: 2 + 0 + 2 + 0 = 4. Now we have our three numbers, which we can add together: 3 + 8 + 4 = 15. This still isn’t a single-digit number, so we will add its digits together again: 1 + 5 = 6. Now we have a single-digit number: 6 is the path number for 03/26/2020. DESTINY NUMBER FOR Jessica Chastain: The destiny number will consider the sum of all the letters in a name. Each letter is assigned a number per the below chart: So for Jessica Chastain we have the letters J (1), e (5), s (1), s (1), i (9), c (3), a (1), C (3), h (8), a (1), s (1), t (2), a (1), i (9) and n (5). Adding all of that up (yes, this can get tedious) gives 51. This still isn’t a single-digit number, so we will add its digits together again: 5 + 1 = 6. Now we have a single-digit number: 6 is the destiny number for Jessica Chastain. CONCLUSION: The difference between the path number for today (6) and destiny number for Jessica Chastain (6) is 0. That is lower than the average difference between path numbers and destiny numbers (2.667), indicating that THIS IS A GOOD RESULT. But don’t get too happy yet! As mentioned earlier, this is just for fun. If you want a forecast that we do recommend taking seriously, check out your cosmic energy profile here. Check it out now – what it returns may blow your mind. ### Abigale Lormen Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene. #### Latest posts by Abigale Lormen (see all) Abigale Lormen Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
# RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4E ## RS Aggarwal Class 10 Solutions Chapter 4 Triangles Ex 4E These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4E. Other Exercises Question 1. Solution: Two properties for similarity of two triangles are: (i) Angle-Angle-Angle (AAA) property. (ii) Angle-Side-Angle (ASA) property. Question 2. Solution: In a triangle, if a line parallel to one side is drawn, it will divide the other two sides proportionally. Question 3. Solution: If a line divides any two sides of a triangle in the same ratio. Then, the line must be parallel to the third side. Question 4. Solution: The line joining the midpoints of two sides of a triangle, is parallel to the third side. Question 5. Solution: In two triangles, if three angles of the one triangle are equal to the three angles of the other, the triangles are similar. Question 6. Solution: In two triangles, if two angles of the one triangle are equal to the corresponding angles of the other triangle, then the triangles are similar. Question 7. Solution: In two triangles, if three sides of the one are proportional to the corresponding sides of the other, the triangles are similar. Question 8. Solution: In two triangles, if two sides of the one are proportional to the corresponding sides of the other and their included angles are equal, the two triangles are similar. Question 9. Solution: In a right angled triangle, the square on the hypotenuse is equal to the sum of squares on the other two sides. Question 10. Solution: In a triangle, if the square on the longest side is equal to the sum of the squares on the other two sides, then the angle opposite to the hypotenuse is a right angle. Question 11. Solution: The ratio of their areas will be 1 : 4. Question 12. Solution: In two triangles ∆ABC and ∆PQR, AB = 3 cm, AC = 6 cm, ∠A = 70° PR = 9 cm, ∠P = 70° and PQ= 4.5 cm Question 13. Solution: ∆ABC ~ ∆DEF 2AB = DE, BC = 6 cm Question 14. Solution: In the given figure, DE \|\| BC AD = x cm, DB = (3x + 4) cm AE = (x + 3) cm and EC = (3x + 19) cm Question 15. Solution: AB is the ladder and A is window. AB =10 m, AC = 8 m We have to find the distance of BC Let BC = x m In right ∆ABC, AB² = AC² + BC² (Pythagoras Theorem) (10)² = 8² + x² ⇒ 100 = 64 + x² ⇒ x² = 100 – 64 = 36 = (6)² x = 6 Distance between foot of ladder and base of the wall = 6 m. Question 16. Solution: ∆ABC is an equilateral triangle with side = 2a cm and AD bisects BC at D Now, in right ∆ABD, AB² = AD² + BD² (Pythagoras Theorem) ⇒ (2a)² = AD² + (a)² ⇒ 4a² = AD² + a² ⇒ AD² = 4a² – a² = 3a² AD = √3 a² = √3 a cm Question 17. Solution: Given : ∆ABC ~ ∆DEF and ar (∆ABC) = 64 cm² and ar (∆DEF) = 169 cm², BC = 4 cm. Question 18. Solution: In trape∠ium ABCD, AB \|\| CD AB = 2CD Diagonals AC and BD intersect each other at O and area(∆AOB) = 84 cm². Question 19. Solution: Let ∆ABC ~ ∆DEF Question 20. Solution: In an equiangular ∆ABC, AB = BC = CA = a cm. Draw AD ⊥ BC which bisects BC at D. Question 21. Solution: ABCD is a rhombus whose sides are equal and diagonals AC and BD bisect each other at right angles. ∠AOB = 90° and AO = OC, BO = OD AO = $$\frac { 24 }{ 2 }$$ = 12 cm and BO = $$\frac { 10 }{ 2 }$$ = 5 cm Now, in right ∆AOB, AB² = AO² + BO² (Pythagoras Theorem) = (12)² + (5)² = 144 + 25 = 169 = (13)² AB = 13 Each side of rhombus = 13 cm Question 22. Solution: ∆DEF ~ ∆GHK ∠D = ∠G = 48° ∠E = ∠H = 57° ∠F = ∠K Now, in ∆DEF, ∠D + ∠E + ∠F = 180° (Angles of a triangle) ⇒ 48° + 57° + ∠F = 180° ⇒ 105° + ∠F= 180° ⇒ ∠F= 180°- 105° = 75° Question 23. Solution: In the given figure, In ∆ABC, MN \|\| BC AM : MB = 1 : 2 Question 24. Solution: In ∆BMP, PB = 5 cm, MP = 6 cm and BM = 9 cm and in ∆CNR, NR = 9 cm Question 25. Solution: In ∆ABC, AB = AC = 25 cm BC = 14 cm AD ⊥ BC which bisects the base BC at D. BD = DC = $$\frac { 14 }{ 2 }$$ = 7 cm Now, in right ∆ABD, AB² = AD² + BD² (Pythagoras Theorem) (25)² = AD² + 7² 625 = AD² + 49 ⇒ AD² = 625 – 49 = 576 ⇒ AD² = 576 = (24)² AD = 24 cm Length of altitude = 24 cm Question 26. Solution: A man goes 12 m due north of point O reaching A and then 37 m due west reaching B. Join OB, In right ∆OAB, OB² = OA² + AB² (Pythagoras Theorem) = (12)² + (35)² = 144 + 1225 = 1369 = (37)² OB = 37 m The man is 37 m away from his starting point. Question 27. Solution: In ∆ABC, AD is the bisector of ∠A which meets BC at D. AB = c, BC = a, AC = b Question 28. Solution: In the given figure, ∠AMN = ∠MBC = 76° p, q and r are the lengths of AM, MB and BC Express the length of MN in terms of p, q and r. In ∆ABC, ∠AMN = ∠MBC = 76° But there are corresponding angles MN \|\| BC ∆AMN ~ ∆ABC Question 29. Solution: In rhombus ABCD, Diagonals AC and BD bisect each other at O at right angles. AO = OC = $$\frac { 40 }{ 2 }$$ = 20 cm BO = OD = $$\frac { 42 }{ 2 }$$ = 21 cm Now, in right ∆AOB, AB2 = AO2 + BO2 = (20)2 + (21)2 = 400 + 441 = 841 AB = √841 cm = 29 cm Each side of rhombus = 29 cm For each of the following statements state whether true (T) or false (F): Question 30. Solution: (i) True. (ii) False, as sides will not be proportion in each case. (iii) False, as corresponding sides are proportional, not necessarily equal. (iv) True. (v) False, in ∆ABC, AB = 6 cm, ∠A = 45° and AC = 8 cm (vi) False, the polygon joining the midpoints of a quadrilateral is not a rhombus but it is a parallelogram. (vii) True. (viii) True. (ix) True, O is any point in rectangle ABCD then OA² + OC² = OB² + OD² is true. (x) True. Hope given RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4E are helpful to complete your math homework. If you have any doubts, please comment below. 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# How to Teach Fractions with LEGO Bricks Keep kids engaged by learning how to teach fractions with LEGO bricks! Understanding how all of the parts of a fraction fit together to make a whole often mystifies young minds until they find a relatable concept that brings it all together for them. LEGO bricks are an easy way to introduce fractions to younger children and help them to understand this important math skill. They not only will benefit from the visual concept offered but also the hands-on use of the bricks can help them to connect with the lesson. There are a couple of different ways how to teach fractions with LEGO, and each is very effective for students. ## How to Teach Fractions with LEGO Using Colors Gather several same-sized LEGO pieces in different colors. 4-dot bricks are a great choice for teaching fractions by color with LEGO bricks. Put together some towers of bricks using several different colors. Duplicate colors in some stacks while keeping one stack all the same color and another stack all different colors. Choose the number of bricks per stack by the denominator that you’d like to work with. Ask questions about the stacks, such as “Which stack has 2/5 blue bricks?” or “What fraction of bricks in this stack is red?” Allow them to create their own fraction stacks using the bricks and show them how they can add or subtract simple fractions using bricks of the same color. ## How to Teach Fractions with LEGO Using Numbers Help students understand fractions by showing them an 8-dot LEGO brick and helping them break out the fraction using smaller bricks. Two 4-dot LEGO bricks can be used to demonstrate ½. 4 2-dot bricks will show ¼. Finally, single dot bricks will make up 1/8 of the whole brick. Allow the children a chance to create their own simple equations using the bricks to show their work. It won’t be long before they’re ready to try larger bricks and more difficult fractions. ## Expanding on Using LEGO to Teach Fractions Once the basic concept of fractions is in place, your bin of LEGO bricks opens a world of possibilities. Children can use them as manipulatives as they learn to solve more complex equations using fractions. They can begin to explore even smaller fractions, as well as understand that sometimes larger fractions, like ½, come in different sizes. For example, half of a 10-brick stack will look very different from half of a four-brick stack. Another concept that they can begin to learn is that stacks of certain numbers can’t be broken up into halves or thirds. For instance, a stack of fifteen bricks can’t be evenly split in half, but you can split it three ways. ## LEGO Bricks to Teach Math If you’re working with multiple children at once, we recommend getting a larger pack so that everyone can work on problems at the same time. Be on the lookout for worksheets and printables that can help you with ideas of how to teach fractions with LEGO. There are so many resources available for a wide range of skill levels. Whether you’re just starting out with the basics or if you’re moving into more difficult fractions, you’ll find that these fun, colorful manipulatives will help keep kids engaged and make it easier for them to process these important skills and concepts.
Question Which equivalent expression will be generated by applying the Distributive Property and combining like terms in the expression 11 4(x 2y 4) 1. Expression (C) 4x + 8y + 27 will be generated by applying the Distributive Property and combining like terms in the expression 11 + 4(x + 2y + 4). ### What is the Distributive Property? • The distributive property of binary operations in mathematics generalizes the distributive law, which states that equality is always true in elementary algebra. • In elementary arithmetic, for example, one has according to one, multiplication distributed over addition. To find which equivalent expression will be generated by applying the Distributive Property and combining like terms in the expression 11 + 4(x + 2y + 4): • The Distributive Property says that: a · (b + c) = a · b + a · c Applying the Distributive property to the expression: • 4 (x + 2y + 4) We get: • 4 (x + 2y + 4) = 4 · x + 4 · 2y + 4 · 4 = 4x + 8y + 16 Hence, • 11 + 4(x + 2y + 4) = 11x + 4x + 8y + 16 There are only two like terms 11 and 16, combining them we’ll get: • 11 + 4(x + 2y + 4) = 11x + 4x + 8y + 16 • = (11 + 16) + 4x + 8y = 27 + 4x + 8y Therefore, expression (C) 4x + 8y + 27 will be generated by applying the Distributive Property and combining like terms in the expression 11 + 4(x + 2y + 4). Know more about the Distributive Property here: #SPJ4 The correct question is given below: Which equivalent expression will be generated by applying the Distributive Property and combining like terms in the expression 11 + 4(x + 2y + 4)? (A) 4x + 2y + 27 (B) 4x + 8y + 11 (C) 4x + 8y + 27 (D) 4x + 8y + 10
Approximations (using Differentiation) Chapter 6 Class 12 Application of Derivatives Serial order wise Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class ### Transcript Question 1 Using differentials, find the approximate value of each of the following up to 3 places of decimal. (ix) γ(82)γ^(1/4)Let π¦=(π₯)^( 1/4) where π₯=81 βπ₯=1 Now, π¦=(π₯)^( 1/4) Differentiating w.r.t.π₯ ππ¦/ππ₯=π(π₯^( 1/4) )/ππ₯=1/4 γπ₯ γ^(1/4 β 1)=1/4 γπ₯ γ^((β3)/( 4))=1/(4γπ₯ γ^(3/4) ) Using βπ¦=ππ¦/ππ₯ βπ₯ βπ¦=1/(4γ π₯ γ^(3/4) ) βπ¦ Putting Values βπ¦=1/(4(81)^( 3/4) ) Γ (1) βπ¦=1/(4 Γ3^(4 Γ 3/4 ) ) Γ (1) βπ¦=1/(4 Γ 3^3 ) βπ¦=1/(4 Γ27) βπ¦=1/108 βπ¦=0. 009 We know that βπ¦=π(π₯+βπ₯)βπ(π₯) So, βπ¦=(π₯+βπ₯)^( 1/4) β(π₯)^( 1/4) Putting Values 0. 009=(81+1)^( 1/4)βγ(81) γ^(1/4) 0. 009=(82)^( 1/4)β(3)^(4 Γ 1/4 ) 0. 009=(82)^( 1/4)β3 0. 009+3=(82)^( 1/4) 3. 009=(82)^( 1/4) Thus, the Approximate Value of (82)^( 1/4) is π. πππ
#### Which term of the AP: -2, -7, -12,... will be -77? Find the sum of this AP upto the term -77. $\text{The given AP is } -2, -7, -12, \ldots \\ \text{Here first term }(a) = -2 \\ \text{Common difference }(d) = -7 - (-2) = -7 + 2 = -5 \\ \text{Let } n\textsuperscript{th} \text{ term be } -77 \\ \text{Then } a_{n} = -77 \\ a + (n - 1)d = -77 \\ -2 + (n - 1)(-5) = -77 \\ (n - 1)(-5) = -77 + 2 \\ (n - 1) = \frac{-75}{-5} = 15 \\ n = 15 + 1 \\ n = 16 \\ \text{Hence, the } 16\textsuperscript{th} \text{ term is } -77 \\ \text{Sum of 16 terms}$ \begin{aligned} \mathrm{S}_{16} &=\frac{16}{2}[2(-2)+(16-1)(-5)] \quad\left(\text { Using } \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]\right) \\ &=8[-4+15(-5)] \\ &=8[-4-75] \\ \mathrm{S}_{16} &=8 \times-79 \\ \mathrm{~S}_{16}=&-632 \end{aligned}
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. ## AP®︎ Calculus AB (2017 edition) ### Course: AP®︎ Calculus AB (2017 edition)>Unit 4 Lesson 7: Derivatives of logarithmic functions # Differentiating logarithmic functions review Review your logarithmic function differentiation skills and use them to solve problems. ## How do I differentiate logarithmic functions? First, you should know the derivatives for the basic logarithmic functions: $\frac{d}{dx}\mathrm{ln}\left(x\right)=\frac{1}{x}$ $\frac{d}{dx}{\mathrm{log}}_{b}\left(x\right)=\frac{1}{\mathrm{ln}\left(b\right)\cdot x}$ Notice that $\mathrm{ln}\left(x\right)={\mathrm{log}}_{e}\left(x\right)$ is a specific case of the general form ${\mathrm{log}}_{b}\left(x\right)$ where $b=e$. Since $\mathrm{ln}\left(e\right)=1$ we obtain the same result. You can actually use the derivative of $\mathrm{ln}\left(x\right)$ (along with the constant multiple rule) to obtain the general derivative of ${\mathrm{log}}_{b}\left(x\right)$. ## Practice set 1: argument is $x$‍ Problem 1.1 $h\left(x\right)=7\mathrm{ln}\left(x\right)$ ${h}^{\prime }\left(x\right)=?$ Want to try more problems like this? Check out this exercise. ## Practice set 2: argument is a polynomial Problem 2.1 $g\left(x\right)=\mathrm{ln}\left(2{x}^{3}+1\right)$ ${g}^{\prime }\left(x\right)=?$ Want to try more problems like this? Check out this exercise. ## Want to join the conversation? • From my understanding, you'd like help with how to differentiate x^x. This is how you do it: y=x^x Take the logs of both sides: ln(y) = ln(x^x) Rule of logarithms says you can move a power to multiply the log: ln(y) = xln(x) Now, differentiate using implicit differentiation for ln(y) and product rule for xln(x): 1/y dy/dx = 1ln(x) + x(1/x) 1/y dy/dx = ln(x) + 1 Move the y to the other side: dy/dx = y (ln(x) + 1) But you already know what y is... it is x^x, your original function. So sub in: dy/dx = x^x(ln(x) + 1) And you're done. • I have a natural logarithm with e^x/1+e^x. I separated it with the log rules but then I'm stuck. Any advice? • i think you are asking about finding d/dx( ln( e^x / 1 + e^x) ). so im solving for that and here it is: we can write ==> ln(e^x / 1+e^x) as ln(e^x) - ln(1+e^x) so now when we differentiate we can differentiate them independently. so d/dx( ln( e^x / 1 + e^x) ) = d/dx( ln(e^x) ) - d/dx( ln(1+e^x) ) = ( (1/e^x) e^x ) -( ( 1/(1+e^x) ) * e^x ) let me know if we have any confusion. • can this statement be true? 2 log y = log y^2 (1 vote) • Are there “rules” for when you can(not) use logarithmic differentiation (including implicit)? I ask because of the following KA problem: “Find dy/dx for x=√(xy+1)” For that problem I attempted to immediately use logarithmic differentiation, e.g. ln(x)=ln(√(XY+1)). However having now worked on it a good deal I have come to understand that logarithmic differentiation generates an incorrect result. Why doesn’t logarithmic differentiation work in this case? (I speculate that perhaps it is because there is a single term that has more than one variable – e.g. XY messes it up – but that is just a guess). Note that the following answer is not sufficient: “You shouldn’t use logarithmic differentiation on that problem.” E.g. I (now) understand it won’t work - I want to know WHY it doesn’t work - what is the rule I should use so that I don't try to do that again in the future? 😉. Thanks, kevin • Logarithmic differentiation should work here. Can you provide us with your steps so we can perhaps find an error? • What do you do if the "x" is not simply x, but is raised to a power or if the equation is log base 4 of x-2? (1 vote) • 1/x(ln(a)) or I could do natural log of y equals (the power- assuming that the power is a variable) times the natural log of x. If the power is a number I would multiply it by the coefficient of x and subtract 1 from the exponent.
## Area Word Problems 1A square garden with a side length of 150 m has a square swimming pool in the very centre with a side length of 25 m . Calculate the area of the garden. 2A rectangular garden has dimensions of 30 m by 20 m and is divided in to 4 parts by two pathways that run perpendicular from its sides. One pathway has a width of 8 dm and the other, 7 dm. Calculate the total usable area of the garden. 3Calculate the area of the quadrilateral that results from drawing lines between the midpoints of the sides of a rectangle whose base and height are 8 and 6 cm respectively. 4 A line connects the midpoint of BC (Point E), with Point D in the square ABCD shown below. Calculate the area of the acquired trapezoid shape if the square has a side of 4 m. 5Calculate the amount of paint needed to paint the front of this building knowing that 0.5 kg of paint is needed per m2. 6A wooded area is in the shape of a a trapezoid whose bases measure 128 m and 92 m and its height is 40 m. A 4 m wide walkway is constructed which runs perpendicular to the two bases. Calculate the area of the wooded area after the addition of the walkway. ## 1 A square garden with a side length of 150 m has a square swimming pool in the very centre with a side length of 25 m . Calculate the area of the garden. AP = 252 = 625 m² AJ = 1502 − 625 = 21 875 m² ## 2 A rectangular garden has dimensions of 30 m by 20 m and is divided in to 4 parts by two pathways that run perpendicular from its sides. One pathway has a width of 8 dm and the other, 7 dm. Calculate the total usable area of the garden. 8 dm = 0.8 m h = 20 - 0.8 = 19.2 m 7 dm = 0.7 m b = 30 − 0.7 = 29.3m AG = 19.2 · 29.3 = 562.56 m² ## 3 Calculate the area of the quadrilateral that results from drawing lines between the midpoints of the sides of a rectangle whose base and height are 8 and 6 cm respectively. ## 4 A line connects the midpoint of BC (Point E), with Point D in the square ABCD. Calculate the area of the acquired trapezoid shape if the trapezoid shape of the square has a side of 4 m. ## 5 Calculate the amount of paint needed to paint the front of this building knowing that 0.5 kg of paint is needed per m2. ## 6 A wooded area is in the shape of a a trapezoid whose bases measure 128 m and 92 m and its height is 40 m. A 4 m wide walkway is constructed which runs perpendicular to the two bases. Calculate the area of the wooded area after the addition of the walkway. AZ = ATrapezoid − AWalk Compartir: Documento sin título
EXAMPLE 1 1 / 10 # EXAMPLE 1 - PowerPoint PPT Presentation Identify the center and radius. From the equation, the graph is a circle centered at the origin with radius. r = 36 = 6. EXAMPLE 1. Graph an equation of a circle. Graph y 2 = – x 2 + 36 . Identify the radius of the circle . SOLUTION. STEP 1. I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described. ## PowerPoint Slideshow about 'EXAMPLE 1' - liang An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - Presentation Transcript Identify the center and radius. From the equation, the graph is a circle centered at the origin with radius r= 36 = 6. EXAMPLE 1 Graph an equation of a circle Graphy2 = – x2 + 36. Identify the radius of the circle. SOLUTION STEP 1 Rewrite the equationy2 = – x2 + 36in standard form asx2 + y2 = 36. STEP 2 EXAMPLE 1 Graph an equation of a circle STEP 3 Draw the circle. First plot several convenient points that are 6 units from the origin, such as (0, 6), (6, 0), (0, –6), and (–6, 0). Then draw the circle that passes through the points. = 29 r = (2 – 0)2 + (–5 – 0)2 = 4 + 25 29 EXAMPLE 2 Write an equation of a circle The point (2, –5) lies on a circle whose center is the origin. Write the standard form of the equation of the circle. SOLUTION Because the point (2, –5) lies on the circle, the circle’s radius rmust be the distance between the center (0, 0) and (2, –5). Use the distance formula. Use the standard form withr to write an equation of the circle. =29 = (29 )2 Substitute for r 29 EXAMPLE 2 Write an equation of a circle x2 + y2 = r2 Standard form x2 + y2 x2 + y2 = 29 Simplify A line tangent to a circle is perpendicular to the radius at the point of tangency. Because the radius to the point (1–3, 2)has slope = = 2 – 0 2 – 3 – 0 3 EXAMPLE 3 Standardized Test Practice SOLUTION m y –2= (x – (–3)) y –2= x + y = x+ 13 2 3 3 3 3 9 2 2 2 2 2 The correct answer is C. EXAMPLE 3 Standardized Test Practice the slope of the tangent line at (–3, 2) is the negative reciprocal of or An equation of 2 3 the tangent line is as follows: Point-slope form Distributive property Solve for y. for Examples 1, 2, and 3 GUIDED PRACTICE Graph the equation. Identify the radius of the circle. 1. x2 + y2 = 9 3 SOLUTION 2. y2 = –x2 + 49 for Examples 1, 2, and 3 GUIDED PRACTICE SOLUTION 7 for Examples 1, 2, and 3 GUIDED PRACTICE 3. x2 – 18 = –y2 SOLUTION 2 for Examples 1, 2, and 3 GUIDED PRACTICE 4. Write the standard form of the equation of the circle that passes through (5, –1) and whose center is the origin. x2 + y2 = 26 SOLUTION 5. Write an equation of the line tangent to the circle x2 + y2=37 at (6, 1). SOLUTION y = –6x + 37
Become a math whiz with AI Tutoring, Practice Questions & more. HotmathMath Homework. Do It Faster, Learn It Better. # Solving Rational Equations Rational equations are invaluable tools that have a significant impact on various aspects of our daily lives, from personal finances to complex engineering tasks. They enable us to make informed decisions, optimize processes, and solve real-world problems. By understanding and mastering rational equations, we gain access to a versatile mathematical resource that can be employed in a multitude of applications, such as calculating speeds, determining average costs, and modeling various phenomena. A rational equation is an equation that involves one or more rational expressions. It is important to remember that a rational expression is an algebraic expression in which a rational number, or a fraction, is the denominator. These expressions can represent relationships or proportions between quantities, and when properly manipulated, they can provide valuable insights into the situations they represent. Consider the following example of a rational expression: $\frac{7}{x+2}=\frac{x}{2}+2$ This expression may appear simple at first glance, but it holds the key to solving a wide range of problems, from basic arithmetic to intricate calculations in various fields such as economics, physics, and engineering. By learning the methods for solving rational expressions, we unlock the potential to comprehend and analyze complex relationships in the world around us, thereby enhancing our capacity to make informed decisions and solve practical problems. ## Solving rational expressions using cross-multiplication Cross-multiplying is a simple way to solve rational expressions when there is a single rational expression on each side of the equation. Some textbooks call this the means/extremes property. $\frac{a}{b}=\frac{c}{d}$ When we cross-multiply this, we see that $a×d=b×c$ . Let's look at this in action. Example 1 Solve using cross-multiplication: $\frac{4}{x+2}=\frac{8}{4x-2}$ The first step is to cross-multiply. We multiply $4×\left(4x-2\right)$ and $8×\left(x+2\right)$ . $4×\left(4x-2\right)=8×\left(x+2\right)$ We will use the distributive property to perform these multiplication problems. $16x-8=8x+16$ Then subtract 8x from each side. $\left(16x-8x-8\right)=\left(8x+16-8x\right)$ Simplify by performing the subtraction. $8x-8=16$ Then add 8 to each side. $8x-8+8=16+8$ $8x=24$ Divide each side by 8. $\frac{8x}{8}=\frac{24}{8}$ Simplify by performing the division. $x=3$ Example 2 Solve using cross-multiplication: $\frac{8}{3x-2}=\frac{2}{x-1}$ To cross-multiply, we should multiply 8 by $\left(x-1\right)$ and 2 by $\left(3x-2\right)$ . $8\left(x-1\right)=2\left(3x-2\right)$ We will use the distributive property to perform the calculations. $8x-8=6x-4$ To simplify, subtract 6x from each side: $8x-8-6x=6x-4-6x$ Simplify by performing the subtraction: $2x-8=-4$ Next, add 8 to each side: $2x-8+8=-4+8$ $2x=4$ Finally, divide each side by 2: $\frac{2x}{2}=\frac{4}{2}$ When we perform the division, we find the solution: $x=2$ ## Solving rational equations using least common denominators (LCD) If we have a rational equation that isn't in the form that supports cross-multiplication, we can still solve it using the least common denominator method. The least common denominator is the easiest common denominator to use in solving these types of problems. To find the LCD, we must factor each expression and multiply all of the unique factors to reveal the LCD of two or more rational expressions. Then each side of the equation is multiplied by the least common denominator to solve the rational equation. Let's see it in action to understand this method better. Example 3 Solve $\left(\frac{3}{x}\right)+\left(\frac{5}{4}\right)=\left(\frac{8}{x}\right)$ using the least common denominator method. The denominators here are x, 4, and x again. We will find the least common denominator by multiplying 4 and x, which is 4x. Next, we will multiply the LCD by the equation on each side to find the solution. $4x×\left(\frac{3}{x}+\frac{5}{4}\right)=4x*\left(\frac{8}{x}\right)$ $4x×\frac{3}{x}=4×3$ $4x×\frac{5}{4}=5x$ $4x×\frac{8}{x}=4×8$ So now we have $12+5x=32$ Next, we will simplify further by subtracting 12 from each side. $12+5x-12=32-12$ Perform the subtraction. $5x=20$ Finally, divide each side by 5. $x=4$ We can check our solution by plugging 4 into the variables in the original equation. $\left(\frac{3}{4}\right)+\left(\frac{5}{4}\right)=\left(\frac{8}{4}\right)$ This is a true statement, so we have found the correct solution. Example 4: Find all real solutions to the rational equation: $\frac{3x+2}{{x}^{2}-x}=\frac{x+4}{{x}^{2}+x}$ Step 1: Find a common denominator. In this case, the common denominator is $\left({x}^{2}-x\right)\left({x}^{2}+x\right)$ . We'll multiply both sides of the equation by the common denominator: $\left(\frac{3x+2}{{x}^{2}-x}\right)×\left({x}^{2}-x\right)×\left({x}^{2}+x\right)=\left(\frac{x+4}{{x}^{2}+x}\right)×\left({x}^{2}-x\right)×\left({x}^{2}+x\right)$ Step 2: Simplify. The denominators will cancel out on both sides: $\left(3x+2\right)\left({x}^{2}+x\right)=\left(x+4\right)\left({x}^{2}-x\right)$ Step 3: Apply the distributive property. $3{x}^{3}+3{x}^{2}+2{x}^{2}+2x={x}^{3}-{x}^{2}+4{x}^{2}-4x$ Step 4: Combine like terms. $3{x}^{3}+5{x}^{2}+2x={x}^{3}+3{x}^{2}-4x$ Step 5: Subtract the right side of the equation from the left side. $2{x}^{3}+2{x}^{2}+6x=0$ Step 6: Factor out the greatest common divisor (GCD) from the left side of the equation. $2x\left({x}^{2}+x+3\right)=0$ Step 7: Set each factor to zero and solve for x. $2x=0\ge x=0$ ${x}^{2}+x+3=0$ The quadratic equation ${x}^{2}+x+3=0$ doesn't have real solutions since its discriminant is negative: ${b}^{2}-4ac=1-4\left(3\right)=-11$ ## Flashcards covering the Solving Rational Equations Precalculus Flashcards ## Get help learning about solving rational equations Tutoring is an excellent way for your student to learn how to solve rational equations outside of the classroom. Sometimes there's not enough time in the classroom to completely comprehend the processes involved in solving rational equations, and if this is the case for your student, you can't do better than having them study with a private tutor. A private tutor will work at your student's pace through problems until your student demonstrates a thorough understanding of the concepts and processes. A tutor can also discover your student's learning style and customize their tutoring methods to complement it. Whether your student learns best visually, audibly, or kinesthetically, a tutor can accommodate them. Tutors can also work with your student on specific math study skills that will serve them throughout their time learning to solve rational equations and on into more advanced math. This extends the value of tutoring long into the future for your student. If you'd like us to connect you with a professional tutor who can help your student with rational equations, contact Varsity Tutors today and speak with one of our helpful Educational Directors. ;
What Is the Distributive Property? Question The distributive property is a rule that governs how you can multiply and add numbers 00:00 00:00 What Is the Distributive Property? Hi my name is Bassem Saad. I'm an associate math instructor and a PhD candidate at UC Davis. I'm here today for About.com to answer the questions, what is the distributive property. What Is the Distributive Property? The distributive property is a rule that governs how you can multiply and add numbers at the same time. Note that the order of operations is already a way of handling this but the distribution property gives us another way of getting us the same result. So let a, b, and c be real numbers. And say we want to multiply a times the sum of b and c. You put b and c inside the parentheses because we want to add these numbers first. This expression will be equal to a times b + a times c. That's the same thing as distributing a to each summand inside the expression. Examples of the Distributive Property? For example, say we want to multiply 5 times the sum of 4 and 8. If we use the order of operations we have to add the four and the eight first because they're inside the parentheses. That gives us 12. And then 5 x 12 is 60. So with the distributive property we want to multiply 5 times each summand. So we distribute 5 to 4, that's 5 x 4. And then we distribute 5 to 8. That's 5 x 8. Then we add those results. So we've got 5 x 4 is 20, plus 5 x 8, which is 40, equals 60. So it agrees with the order of operations. So in another important example, let x be a variable. We can still use the distributive property to simplify and expression like this; x times the sum of 3 and x. Now you can't combine the sum of 3 and x. This is about as simplified as you can make it. But you can distribute this x onto the other two terms. So you distribute onto the 3 term, that's x times 3. And you distribute the x to the x term. That's x times x. So this whole expression equals 3x + x squared. That was the distributive property.
How do you factor 2x^3 - 3x^2 - 5x? Nov 14, 2015 $2 {x}^{3} - 3 {x}^{2} - 5 x = x \left(2 x - 5\right) \left(x + 1\right)$ Explanation: First, note that each term has a factor of $x$, and so we have $2 {x}^{3} - 3 {x}^{2} - 5 x = x \left(2 {x}^{2} - 3 x - 5\right)$ Now, we can use the quadratic formula to find the remaining factors, but first let's see if there are easy integer solutions by looking for $a , b , c , d$ where $2 {x}^{2} - 3 x - 5 = \left(a x + b\right) \left(c x + d\right)$ We know that $a c = 2$ and so we can look at $\left(2 x + b\right) \left(x + d\right)$ We also know $b d = - 5$ and so our possible choices are $\left(2 x + 1\right) \left(x - 5\right)$ and $\left(2 x - 5\right) \left(x + 1\right)$ Multiplying these out shows that $2 {x}^{2} - 3 x - 5 = \left(2 x - 5\right) \left(x + 1\right)$ So our final result is $2 {x}^{3} - 3 {x}^{2} - 5 x = x \left(2 x - 5\right) \left(x + 1\right)$
# Algebra As Patterns If you think of Mathematics as a language, then Algebra would be that part of the language which describes various patterns around us. If there is any repetitive pattern, then we can use algebra to simplify this and have a general expression to describe this pattern. Algebraic thinking starts as soon as pupils notice the regular change and attempt to define it. Suppose we can represent algebraic thinking by everyday situations such as balancing concrete elements using balance bins. This kind of activity encourages using more symbolic illustrations in the higher levels when we use letters to generalize thinking or deliberate situations with the help of variables. ## Algebra and Patterns To understand the relationship between patterns and algebra, we need to try making some patterns. We can use pencils to construct a simple pattern and understand how to create a general expression to describe the entire pattern. It would be best if you had a lot of pencils for this. It will help if they are of similar height. Find a solid surface and arrange two pencils parallel to each other with some space in between them. Add a second layer on top of it and another on top of that, as shown in the image given below. There are a total of six pencils in this arrangement. The above arrangement contains three layers, and each layer has a fixed count of two pencils. The number of pencils in each layer never varies, but the number of layers you wish to build is entirely up to you. Current Number of Layers = 4 Number of Pencils per Layer = 2 Total Number of Pencils = 2 x 4 = 8 What if you increase the number of layers to 10? What if you keep building up to a layer of 100? Can you sit and stack those many layers? Here, the answer is obviously NO. Instead, let’s try to calculate. Number of Layers = 100 Number of Pencils per Layer = 2 Total Number of Pencils = 2 x 100 = 200 There is an obvious pattern here. A single level has 2 pencils, which is always constant, regardless of the number of levels built. So to get the total number of pencils, we have to multiply 2 (the number of pencils per level) with the number of levels built. For example, to construct 30 levels, you will need 2 multiplied 30 times which is 60 pencils. According to the previous calculation, to make a building of ‘x’ number of levels, we will require 2 multiplied ‘x’ times, and thus the number of pencils equal to 2x. We just created algebraic expressions based on patterns. In this way, we can make several algebra patterns. ## Algebra as generalized Arithmetic Patterns There are different types of algebraic patterns such as repeating patterns, growth patterns, number patterns, etc. All these patterns can be defined using different techniques. Let’s go through the algebra patterns using matchsticks given below. ## Algebra Matchstick Patterns It is possible to make patterns with very basic things that we are using in our everyday life. Look at the following matchstick pattern of squares in the below figure. The squares are not separate. Two neighbouring squares have a common matchstick. Let’s observe the patterns and try to find the rule that gives the number of matchsticks. In the above matchstick pattern, the number of matchsticks is 4, 7, 10 and 13, which is one more than the thrice of the number of squares in the pattern. Therefore, this pattern can be defined using the algebraic expression 3x + 1, where x is the number of squares. Now, let’s make the triangle pattern using matchsticks as shown in the below figure. Here, the triangles are connected with each other. In this matchstick pattern, the number of matchsticks is 3, 5, 7 and 9, which is one more than twice the number of triangles in the pattern. Therefore, the pattern is 2x + 1, where x is the number of triangles. ### How to do Number Patterns Let’s look at another pattern. Say we have a triangle. Flip the triangle downwards and complete this image to form a complete triangle as given below: The pattern is still a triangle, but the number of smaller triangles increases to 4. This bigger triangle now has two rows. What happens when we increase the number of rows and fill in the gaps with smaller triangles to complete that big triangle? As we proceed to the 3rd row, how many smaller triangles do we have in total now? Now, this triangle has 9 smaller triangles. What is the pattern here? When, r = 1, total no of triangles, ‘n’ = 1 When, r = 2, n = 4 And when, r = 3, n = 9 This can be summarised as: Number of rows 1 2 3 Number of smaller triangles 1 4 9 You can see that as the size of the triangle increases, the number of smaller triangles also increases. This means that n is just equal to r², which is 1², 2², 3² = 1, 4, 9… So how do we represent this triangular pattern algebraically? r²! That’s it! We have just defined patterns algebraically. Also, the above pattern can be defined as the growth pattern in algebra. We can write the sequence of odd numbers like 1, 3, 5, 7, 9……………………….. (2n – 1). If we substitute ‘n’ values in the expression beginning from one for odd numbers, i.e. 2n – 1, we can easily calculate the nth odd number. Substitute n= 11 to find the 11th odd number, and the result is 21, i.e. 11th odd number. Similarly, the 100th odd number is 199. Thus, the algebraic expression representing a set of odd numbers is 2n -1. It must be noted that n is a set of whole numbers in this case. Similar to patterns in numbers, we can figure out the pattern in figures. For example, consider the following figures: In the figure given above, the first image is a pentagon with five sides. In the following figure, two pentagons are joined end to end, and the total number of sides is 9; in the next figure, the total number of sides is 13. It is observed that every other figure has 4 extra sides as compared to the previous one. Thus, the algebraic pattern that would define this sequence exactly is ‘4n + 1’, where n is any natural number. Therefore, if we substitute n = 3, we get the number of sides equal to 13. The 10th pattern in this sequence will have sides equal to 4 × 10 + 1, i.e. 41 sides.
# 2012 AIME I Problems/Problem 13 ## Problem 13 Three concentric circles have radii $3,$ $4,$ and $5.$ An equilateral triangle with one vertex on each circle has side length $s.$ The largest possible area of the triangle can be written as $a + \tfrac{b}{c} \sqrt{d},$ where $a,$ $b,$ $c,$ and $d$ are positive integers, $b$ and $c$ are relatively prime, and $d$ is not divisible by the square of any prime. Find $a+b+c+d.$ ## Solution Reinterpret the problem in the following manner. Equilateral triangle $ABC$ has a point $X$ on the interior such that $AX = 5,$ $BX = 4,$ and $CX = 3.$ A $60^o$ clockwise rotation about vertex $A$ maps $X$ to $X'$ and $C$ to $C'.$ Note that angle $XAX'$ is $60$ and $XA = X'A = 5$ which tells us that triangle $XAX'$ is equilateral and that $XX' = 5.$ We now notice that $XB = 3$ and $X'B = 4$ which tells us that angle $XBX'$ is $90$ because there is a $3$-$4$-$5$ Pythagorean triple. Now note that $\angle ABC + \angle ACB = 120,$ $\angle XCA + \angle XBA = 90,$ $\angle XCB+\angle XBC = 30,$ and $\angle BXC = 150.$ Applying the law of cosines on triangle $BXC$ yields $$BC^2 = BX^2+CX^2 - 2 \cdot BX \cdot CX \cdot \cos(150) = 3^2+4^2-24 \cdot \cos(150) = 25+12\sqrt{3}$$ and thus the area of $ABC$ equals $$BC^2\frac{\sqrt{3}}{4} = 25\frac{\sqrt{3}}{4}+9.$$ so our final answer is $3+4+25+9 = \boxed{041.}$ 2012 AIME I (Problems • Answer Key • Resources) Preceded byProblem 12 Followed byProblem 14 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions
Worksheet on Forming of Linear Equations in One Variable come with numerous questions framed on the topic and gives you ample practice. Solve the Questions on Forming of Linear Equations in One Variable and improve your logical skills as well as speed while attempting the exams. Step by Step Solutions makes it clear for you to understand the topic. Download the free accessible Forming Linear Equations in One Variable Worksheet and resolve all your queries in no time. Do Check: ## Forming Linear Equations in One Variable Worksheet Example 1. One of the numbers is two times the other. The sum of these two numbers is 45. Form the equation to find the numbers using a linear equation in one variable? Solution: Let the number be x The Other Number is 2 times the given number i.e. 2x From the given condition, the sum of these two numbers is 40 We can write the equation x+2x=45 3x=45 x=45/3 x=15 The other number is 2x i.e. 2(15)=30 Example 2. The perimeter of a rectangular swimming pool is 140 meters. Its length is 4 m more than twice its breadth. What are the length and breadth of the pool? Solution: Let the breadth of the rectangular swimming pool be x Since length is 4m more than twice its breadth we can have length = 2x+4 Perimeter of a Rectangular Swimming Pool = 2(l+b) 140 =2(2x+4+x) 140=2(3x+4) 140=6x+8 140-8=6x 132=6x x=132/6 x=22 Length l =2x+4 =2(22)+4 =44+4 =48 Thus, the breadth and length of the swimming pool are 22m and 48m respectively. Example 3. The sum of three consecutive odd numbers is 45. Find the numbers? Solution: Let the three odd consecutive numbers be x, x+1, x+2 As per the given condition sum of three consecutive odd numbers is 45 x+x+1+x+2=45 3x+3=45 3x=45-3 3x=42 x=42/3 x=14 x+1=14+1=15 x+2=14+2=16 Therefore, the three consecutive numbers are 14, 15, 16 Example 4. A sum of Rs. 4500 is to be given in the form of 90 prizes. If the prize is of either Rs. 100 or Rs. 25, find the number of prizes of each type? Solution: Let us assume the type of 100Rs prizes be x Since the total number of prizes is 90 the number of 25 Rs. prize is 90-x As per the given data in the question 100x+(90-x)=4500 100x+90-x=4500 99x+90=4500 99x=4500-90 99x=4410 x=4410/99 ~44 Therefore 25Rs. Prizes are 90-44 = 46 Example 5. A dealer sold a television set for Rs. 12,000 and earned a profit of 15%. Find the cost price of the television set? Solution: Selling Price of the Television = Rs. 12,000 Let us assume the cost price = x Profit earned = 15% Cost Price of the television set CP = (100 / ( 100 + percentage profit))SP =(100/(100+15)12000 =(100/115)12000 =Rs. 10434 Therefore, the dealer bought the television set for a cost price of Rs. 10434 Example 6. Twenty years from now Rahul’s age will be 5 times his current age. What is his current age? Solution: Let us consider the current age of Rahul as x Twenty Years from now his age would be x+20 As per the given condition in the question we have x+20=5x 20=5x-x 20=4x 20/4 =x x=5 Therefore current age of Rahul is 5 Years. Example 7. Solve 2y -10 = 4 Solution: Given Expression is 2y-10=4 Transfering constants to one side and variables to the other side we have 2y =4+10 2y=14 y=7 Example 8. Rajesh is a cashier in a State bank. he has notes of denominations of Rs. 100, 50, and 20 respectively. The ratio of the number of these notes is 4:3:2 respectively. The total cash with Rajesh is 4,72,000. How many notes of each denomination does he have? Solution: Let us assume the numbers of notes be 4x, 3x, and 2x respectively based on the ratio of notes 1004x+503x+202x =4,72,000 400x+150x+40x=4,72,000 590x=4,72,000 x=4,72,000/590 =800 No. of 100 Rs Notes Rajesh has = 4x =4800 =3200 No. of 50 Rs Notes Rajesh has = 3x =3800 =2400 No. of 20 Rs Notes Rajesh has = 2x =2800 =1600 Example 9. Solve for 8x + 40 = 4x +100? Solution: Given Expression is 8x + 40 = 4x +100 Transferring the constant terms to one side and variables terms to another side we have 8x-4x=100-40 6x=60 x=10 Example 10. Amar thinks of a number and subtracts 3/2 from it. She multiplies the result by 7. The final result is 4 times her original number. Find the number? Solution: Let the number be x From the given statement we can infer 7(x-3/2)=4x 7x-21/2=4x (14x-21)=2*4x 14x-8x=21 6x=21 x=21/6 Therefore, the number is 21/6
We think you are located in South Africa. Is this correct? # Completing The Square ## Completing the square Can you solve each equation using two different methods? 1. $$x^2 - 4 = 0$$ 2. $$x^2 - 8 = 0$$ 3. $$x^2 -4x + 4 = 0$$ 4. $$x^2 -4x - 4 = 0$$ Factorising the last equation is quite difficult. Use the previous examples as a hint and try to create a difference of two squares. We have seen that expressions of the form $$x^2 - b^2$$ are known as differences of squares and can be factorised as $$(x-b)(x+b)$$. This simple factorisation leads to another technique for solving quadratic equations known as completing the square. Consider the equation $$x^2-2x-1=0$$. We cannot easily factorise this expression. When we expand the perfect square $$(x-1)^2$$ and examine the terms we see that $$(x-1)^2 = x^2-2x+1$$. We compare the two equations and notice that only the constant terms are different. We can create a perfect square by adding and subtracting the same amount to the original equation. \begin{align} x^2-2x-1 &= 0 \\ (x^2-2x+1)-1-1 &= 0 \\ (x^2-2x+1)-2 &= 0 \\ (x-1)^2-2 &= 0 \end{align} Method 1: Take square roots on both sides of the equation to solve for $$x$$. \begin{align} (x-1)^2-2 &= 0 \\ (x-1)^2 &= 2 \\ \sqrt{(x-1)^2} &= \pm \sqrt{2} \\ x-1 &= \pm \sqrt{2} \\ x &= 1 \pm \sqrt{2} \\ \text{Therefore }x &= 1 + \sqrt{2} \text{ or }x = 1 - \sqrt{2} \end{align} Very important: Always remember to include both a positive and a negative answer when taking the square root, since $$2^2 = 4$$ and $$(-2)^2 = 4$$. Method 2: Factorise the expression as a difference of two squares using $$2 = \left(\sqrt{2}\right)^2$$. We can write \begin{align} (x-1)^2-2 &= 0 \\ (x-1)^2 - \left( \sqrt{2} \right)^2 &= 0 \\ \left( (x-1) + \sqrt{2} \right)\left( (x-1) - \sqrt{2} \right) &= 0 \end{align} The solution is then \begin{align} (x-1) + \sqrt{2} &= 0 \\ x &= 1 - \sqrt{2} \end{align} or \begin{align} (x-1) - \sqrt{2} &= 0 \\ x &= 1 + \sqrt{2} \end{align} Method for solving quadratic equations by completing the square 1. Write the equation in the standard form $$a{x}^{2}+bx+c=0$$. 2. Make the coefficient of the $${x}^{2}$$ term equal to $$\text{1}$$ by dividing the entire equation by $$a$$. 3. Take half the coefficient of the $$x$$ term and square it; then add and subtract it from the equation so that the equation remains mathematically correct. In the example above, we added $$\text{1}$$ to complete the square and then subtracted $$\text{1}$$ so that the equation remained true. 4. Write the left hand side as a difference of two squares. 5. Factorise the equation in terms of a difference of squares and solve for $$x$$. ## Worked example 6: Solving quadratic equations by completing the square Solve by completing the square: $$x^2-10x-11=0$$ ### Make sure the coefficient of the $$x^2$$ term is equal to $$\text{1}$$ $x^2-10x-11=0$ ### Take half the coefficient of the $$x$$ term and square it; then add and subtract it from the equation The coefficient of the $$x$$ term is $$-\text{10}$$. Half of the coefficient of the $$x$$ term is $$-\text{5}$$ and the square of it is $$\text{25}$$. Therefore $$x^2 - 10x + 25 - 25 - 11 = 0$$. ### Write the trinomial as a perfect square \begin{align} (x^2 - 10x + 25) - 25 - 11 &= 0 \\ (x-5)^2 - 36 &= 0 \end{align} ### Method 1: Take square roots on both sides of the equation \begin{align} (x-5)^2 - 36 &= 0 \\ (x-5)^2 &= 36 \\ x-5 &= \pm\sqrt{36} \end{align} Important: When taking a square root always remember that there is a positive and negative answer, since $$(6)^2 = 36$$ and $$(-6)^2 = 36$$. $x - 5 = \pm 6$ ### Solve for $$x$$ $x = -1 \text{ or } x = 11$ ### Method 2: Factorise equation as a difference of two squares \begin{align} (x-5)^2 - (6)^2 &= 0 \\ \left[\left(x-5\right) + 6\right] \left[\left(x-5\right) - 6\right] &= 0 \end{align} ### Simplify and solve for $$x$$ \begin{align} (x+1)(x-11) &= 0 \\ \therefore x = -1 \text{ or } x &= 11 \end{align} $x = -1 \text{ or } x = 11$ Notice that both methods produce the same answer. These roots are rational because $$\text{36}$$ is a perfect square. ## Worked example 7: Solving quadratic equations by completing the square Solve by completing the square: $$2x^2 - 6x - 10 = 0$$ ### Make sure that the coefficient of the $$x^2$$ term is equal to $$\text{1}$$ The coefficient of the $${x}^{2}$$ term is $$\text{2}$$. Therefore divide the entire equation by $$\text{2}$$: $x^2 - 3x - 5 = 0$ ### Take half the coefficient of the $$x$$ term, square it; then add and subtract it from the equation The coefficient of the $$x$$ term is $$-\text{3}$$, so then $$\left( \dfrac{-3}{2} \right)^2 = \dfrac{9}{4}$$: $\left( x^2 - 3x + \frac{9}{4}\right) - \frac{9}{4} - 5 = 0$ ### Write the trinomial as a perfect square \begin{align} \left( x - \frac{3}{2} \right)^2 - \frac{9}{4} - \frac{20}{4} &= 0 \\ \left( x - \frac{3}{2} \right)^2 - \frac{29}{4} &= 0 \end{align} ### Method 1: Take square roots on both sides of the equation \begin{align} \left( x - \frac{3}{2} \right)^2 - \frac{29}{4} &= 0 \\ \left( x - \frac{3}{2} \right)^2 &= \frac{29}{4} \\ x - \frac{3}{2} &= \pm \sqrt{\frac{29}{4}} \end{align} Remember: When taking a square root there is a positive and a negative answer. ### Solve for $$x$$ \begin{align} x - \frac{3}{2} &= \pm \sqrt{\frac{29}{4}} \\ x &= \frac{3}{2} \pm \frac{\sqrt{29}}{2} \\ &= \frac{3 \pm \sqrt{29}}{2} \end{align} ### Method 2: Factorise equation as a difference of two squares \begin{align} \left( x - \frac{3}{2} \right)^2 - \frac{29}{4} &= 0 \\ \left( x - \frac{3}{2} \right)^2 - \left( \sqrt{\frac{29}{4}} \right)^2 &= 0\\ \left( x - \frac{3}{2} - \sqrt{\frac{29}{4}} \right) \left( x - \frac{3}{2} + \sqrt{\frac{29}{4}} \right) &= 0 \end{align} ### Solve for $$x$$ \begin{align} \left( x - \frac{3}{2} - \frac{\sqrt{29}}{2} \right) \left( x - \frac{3}{2} + \frac{\sqrt{29}}{2} \right) &= 0 \\ \text{Therefore } x = \frac{3}{2} + \frac{\sqrt{29}}{2} &\text{ or } x = \frac{3}{2} - \frac{\sqrt{29}}{2} \end{align} Notice that these roots are irrational since $$\text{29}$$ is not a perfect square. ## Solution by completing the square Exercise 2.2 Solve the following equations by completing the square: $${x}^{2}+10x-2=0$$ \begin{align} x^2 + 10x - 2 &= 0\\ x^2 +10x&= 2\\ x^2 +10x +25 &= 2 + 25\\ (x+5)^2 - 27 &= 0\\ \left[(x+5) + \sqrt{27}\right]\left[(x+5) - \sqrt{27}\right] &= 0\\ (x+5) = -\sqrt{27} &\text{ or } (x+5) = \sqrt{27}\\ x = -5 - 3\sqrt{3} &\text{ or } x= -5 + 3\sqrt{3} \end{align} $${x}^{2}+4x+3=0$$ \begin{align} x^2+4x+3&=0\\ x^2 +4x &= -3\\ x^2 + 4x + 4 &= -3 + 4\\ (x+2)^2 &= 1\\ (x+2)&= \pm \sqrt {1}\\ &=\pm 1\\ x = -2 + 1 = -1 &\text{ or } x = -2-1=-3 \end{align} $$p^2 - 5 = - 8p$$ \begin{align} p{^2} - 5 &= - 8p\\ p{^2}+8p-5 &= 0\\ p^2 + 8p &=5\\ p^2 + 8p + 16 &= 5 + 16\\ (p+4)^2 &= 21\\ (p+4)&=\pm \sqrt{21}\\ p &= -4 \pm \sqrt{21} \end{align} $$2(6x + x^2) = -4$$ \begin{align} 2(6x + x^2) &= -4 \\ 2x^2+12x+4 &= 0\\ x^2 + 6x + 2 &= 0\\ x^2 +6x &= -2\\ x^2 + 6x + 9&= -2 +9\\ (x+3)^2 &= 7\\ x + 3 &= \pm \sqrt{7}\\ x &= -3 \pm \sqrt{7} \end{align} $${x}^{2}+5x+9=0$$ \begin{align} x^2+5x+9 &= 0\\ x^2 + 5x &= -9\\ x^2 + 5x + \frac{25}{4} &= -9 + \frac{25}{4}\\ \left(x+\frac{5}{2}\right)^2 &= -\frac{11}{4}\\ x + \frac{5}{2} &= \pm \sqrt{-\frac{11}{4}}\\ \text{ No real solution } \end{align} $$t^2 + 30 = 2(10-8t)$$ \begin{align} t^2 + 30 &= 2(10 - 8t) \\ t^2+16t+10&=0\\ t^2 +16t &= -10\\ t^2 + 16t+64 &= -10 +64\\ (t+8)^2&=54\\ t + 8 &= \pm \sqrt{54}\\ t &= -8 \pm \sqrt{9 \times 6} \\ \therefore t &= -8 \pm 3\sqrt{6} \end{align} $$3{x}^{2}+6x-2=0$$ \begin{align} 3x^2+6x-2&=0\\ x^2 + 2x &= \frac{2}{3}\\ x^2 + 2x + 1 &= \frac{2}{3} + 1\\ (x+1)^2 &= \frac{5}{3}\\ x +1 &= \pm \sqrt{\frac{5}{3}}\\ x &= -1 \pm \sqrt{\frac{5}{3}} \end{align} $${z}^{2}+8z-6=0$$ \begin{align} z^2 + 8z - 6 &= 0\\ z^2 +8z &= 6\\ z^2 + 8z+16 &= 6+16\\ (z+4)^2 &= 22\\ z+4 &= \pm \sqrt{22}\\ z &= -4 \pm \sqrt{22} \end{align} $$2z^2 = 11z$$ \begin{align} 2z^2 &= 11z\\ 2z^2 - 11z &= 0\\ z^2 - \frac{11}{2}z &= 0\\ z^2 - \frac{11}{2}z + \frac{121}{16} &= \frac{121}{16}\\ \left(z-\frac{11}{4}\right)^2 &= \frac{121}{16}\\ z - \frac{11}{4} &= \pm \frac{11}{4}\\ z = \frac{11}{4} + \frac{11}{4} = \frac{11}{2} &\text{ or } z = \frac{11}{4} - \frac{11}{4} = 0 \end{align} $$5+4z-{z}^{2}=0$$ \begin{align} 5 + 4z - z^2 &= 0\\ z^2 -4z &= 5\\ z^2 - 4z +4 &= 5+4\\ (z-2)^2 &= 9\\ z-2 &= \pm \sqrt{9}\\ z = 2+3 = 5 &\text{ or } z = 2-3 = -1 \end{align} Solve for $$k$$ in terms of $$a$$: $$k^2 + 6k+ a = 0$$ \begin{align} k^2 + 6k+ a &= 0\\ k^2 + 6k &= -a\\ k^2 + 6k + 9 &= 9 - a\\ (k + 3)^2 &= 9 - a\\ k + 3 &= \pm \sqrt{9 - a}\\ k & = -3 \pm \sqrt{9 - a} \end{align} Solve for $$y$$ in terms of $$p$$, $$q$$ and $$r$$: $$py^2 + qy + r = 0$$ \begin{align} py^2 + qy + r &= 0\\ y^2 + \frac{q}{p}y + \frac{r}{p} &= 0\\ y^2 + \frac{q}{p}y &= -\frac{r}{p}\\ y^{2} + \frac{q}{p}y + \left(\frac{q}{2p}\right)^{2} &= \left(\frac{q}{2p}\right)^{2} - \frac{r}{p}\\ \left(y + \frac{q}{2p}\right)^2 &= \frac{q^{2}}{4p^{2}} - \frac{r}{p} \\ y + \frac{q}{2p} &= \pm \sqrt{\frac{q^{2} - 4pr}{4p^{2}}}\\ y &= -\frac{q}{2p} \pm \frac{\sqrt{q^{2} - 4pr}}{2p}\\ y & = \frac{-q \pm \sqrt{q^{2} - 4pr}}{2p} \end{align}
# What does a rhombuses look like? Page Contents ## What does a rhombuses look like? A rhombus looks like a diamond Opposite sides are parallel, and opposite angles are equal (it is a Parallelogram). And the diagonals “p” and “q” of a rhombus bisect each other at right angles. ## What is a rhombus in maths? In plane Euclidean geometry, a rhombus (plural rhombi or rhombuses) is a quadrilateral whose four sides all have the same length. Another name is equilateral quadrilateral, since equilateral means that all of its sides are equal in length. ## What is a rhombus first grade? Lesson Summary A rhombus is a special kind of four-sided figure known as a quadrilateral. A rhombus must have four equal sides, opposite sides parallel and opposite angles parallel. In order to find its area, you must draw a height line first. ## What are the 4 properties of a rhombus? A rhombus is a quadrilateral which has the following four properties: • Opposite angles are equal. • All sides are equal and, opposite sides are parallel to each other. • Diagonals bisect each other perpendicularly. • Sum of any two adjacent angles is 180° ## What is angle of rhombus? Rhombus has four interior angles. The sum of interior angles of a rhombus add up to 360 degrees. The opposite angles of a rhombus are equal to each other. The adjacent angles are supplementary. ## How do you make the best rhombus? 1. (1) BET can be constructed by using the given measurements as follows. 2. (2) Vertex S is 4.5 cm away from vertex E and also from vertex T. Therefore, while taking E and T as centres, draw arcs of 4.5 cm radius, which will be intersecting each other at point S. 3. (3) Join S to E and T. 4. BEST is the required rhombus. ## What are some real life examples of a rhombus? Real-life examples Rhombus can be found in a variety of things around us, such as a kite, windows of a car, rhombus-shaped earring, the structure of a building, mirrors, and even a section of the baseball field. ## Is a diamond a rhombus yes or no? A diamond is a quadrilateral, a 2-dimensional flat figure that has four closed, straight sides. But a diamond is also categorized as rhombus, because it has four equal sides and its opposite angles are equal. ## What are the 8 properties of rhombus? Properties of Rhombus • All sides of the rhombus are equal. • The opposite sides of a rhombus are parallel. • Opposite angles of a rhombus are equal. • In a rhombus, diagonals bisect each other at right angles. • Diagonals bisect the angles of a rhombus. • The sum of two adjacent angles is equal to 180 degrees. ## Are there any stock photos of the rhombus shape? 401,255 rhombus shape stock photos, vectors, and illustrations are available royalty-free. ## Which is an example of a property of a rhombus? Properties of Rhombus. Some of the important properties of the rhombus are as follows: All sides of the rhombus are equal. The opposite sides of a rhombus are parallel. Opposite angles of a rhombus are equal. In a rhombus, diagonals bisecting each other at right angles. ## What does the perimeter of a rhombus look like? Rhombus (Jump to Area of a Rhombus or Perimeter of a Rhombus) A Rhombus is a flat shape with 4 equal straight sides. A rhombus looks like a diamond ## What to do with rhombus shapes for kids? By looking at the templates, you can ask your children to create rhombus bodies in groups of four or more. If you have any rhombus activities, please feel free to download these templates to complete your lesson plan. Use these shapes as part of a Math display, flashcards, or craft. 401,255 rhombus shape stock photos, vectors, and illustrations are available royalty-free. Properties of Rhombus. Some of the important properties of the rhombus are as follows: All sides of the rhombus are equal. The opposite sides of a rhombus are parallel. Opposite angles of a rhombus are equal. In a rhombus, diagonals bisecting each other at right angles. ## Is the four-sided shape below, mnop, a rhombus? Diagonals are perpendicular . Is the four-sided shape below, MNOP, a rhombus? If not, classify the shape. The shape below is not a rhombus because its diagonals are not perpendicular. However, since opposite sides are congruent and parallel, and the diagonals bisect each other. The shape below is a parallelogram. By looking at the templates, you can ask your children to create rhombus bodies in groups of four or more. If you have any rhombus activities, please feel free to download these templates to complete your lesson plan. Use these shapes as part of a Math display, flashcards, or craft.
# Geometry Stuff... XD ## This is just some theorems, postulates, properties, and all kinds of other geometry stuff. Enjoy! Chapter 1. ### General Theorems Reflexive Property= A quantity is congruent (equal) to itself. a = a Symmetric Property= If a = b, then b = a. Transitive Property= If a = b and b = c, then a = c. If equal quantities are added to equal quantities, the sums are equal. Subtraction Postulate= If equal quantities are subtracted from equal quantities, the differences are equal. Multiplication Postulate= If equal quantities are multiplied by equal quantities, the products are equal. (also Doubles of equal quantities are equal.) Division Postulate= If equal quantities are divided by equal nonzero quantities, the quotients are equal. (also Halves of equal quantities are equal.) Substitution Postulate= A quantity may be substituted for its equal in any expression. Partition Postulate= The whole is equal to the sum of its parts. Also: Betweeness of Points: AB + BC = AC Angle Addition Postulate: m<ABC + m<CBD = m<ABD Construction Two points determine a straight line. Construction From a given point on (or not on) a line, one and only one perpendicular can be drawn to the line. Angles: Right Angles= All right angles are congruent. Straight Angles= All straight angles are congruent. Congruent Supplements= Supplements of the same angle, or congruent angles, are congruent. Congruent Complements= Complements of the same angle, or congruent angles, are congruent. Linear Pair= If two angles form a linear pair, they are supplementary. Vertical Angles= Vertical angles are congruent. Triangle Sum= The sum of the interior angles of a triangle is 180º. Exterior Angle= The measure of an exterior angle of a triangle is equal to the sum of the measures of the two non-adjacent interior angles. The measure of an exterior angle of a triangle is greater than either non-adjacent interior angle. Join Qfeast to read the entire story!
# Balanced Equation Directions: Use the operation symbols (+, -, x, and ÷) to make the equation true. Operations may be used more than once. ### Hint What is a value for the left side of the equation? What is a value for the right side of the equation? There are many possibilities, here is one 2 x (3 x 7 + 9) = (1 x 5) x (8 +4) Source: Joshua Nelson ## Add Fractions with Decimal Sums Directions: Using the digits 1 to 9 at most one time each, place a digit … 1. My class got 2 x (3 + 7 + 9) = (1 + 5) + (8 x 4) 2. 2+(3×7=9)=(1-5)/(8×4) 2x(3/7-9)=(1+5)x(8/4) 2-(3x7x9)=(1-5)-(8/4) 2/(3-7/9)=(1×5)98-4) 2x(3+7+9)=(1-5)(8×4) • 2×(3×7+9)=(1×5)×(8+4) • Jennifer Cecchine Both sides need to be equal. In most of your examples, you have either a negative solution on only one side, or you have a fractional answer on only one side. 2+(3×7=9)=(1-5)/(8×4) 2 + (21-9) = -4 ÷ 32 (I assumed you meant -9, not =9 2 +12 = -⅛ (or -0.125) 14 ≠ -⅛ 2x(3/7-9)=(1+5)x(8/4) 2 x (3 ÷ 7 – 9) = (1 + 5) x (8 ÷ 4) 2 x (3/7 – 9) = 6 x 2 2 x -8 4/7 = 12 -17 1/7 ≠ 12 2-(3x7x9)=(1-5)-(8/4) 2 – (21 x 9) = -4 – 2 2 – 189 = -6 -187 ≠ -6 2/(3-7/9)=(1×5)98-4) 2 ÷ (3 – 7/9) = 5 x 4 (I assumed you meant x, not 9) 2 ÷ 2 2/9 = 20 9/10 ≠ 20 2x(3+7+9)=(1-5)(8×4) 2 x 19 = -4 x 32 38 ≠ -128 You were almost there. On your last equation, can you rework the second expression to equal the value of the first expression? 3. 2 x (3 + 7 + 9)=(1 + 5) + (8 x 4) 4. I got 2 times (3 times 7 + 9)=( 1 times 5)times ( 8+4) 5. 2- (3X7-9)=(1X5)X(8+4) 6. 2x (3+7+9)=(1-5)(8×4) 7. 2x (3+7+9)=(1-5)(8×4) 8. 2 x (3 x 7 + 9) = (1 x 5) x (8 +4) 9. Garrison Phillips 2 x (3 x 7 + 9) = (1 x 5) x (8 + 4) 10. 2 X (3×7+9)=(1×5)x(8+4) 11. 2 x (3 x 7 + 9) = (1 x 5) x (8 +4) Is the one that I got. 12. i don understand the equation 13. I got 2×(3×7+9)=(1-5)×(8+1) 14. 2 x (3 x 7 + 9) = (1 x 5) x (8 +4) 15. I got 2 +(3-7+9)=(1×5)+(8/4) Answer 7. 16. 2x(3×7+9)=(1×5)x(8+\$ 17. 2x(3=7=9)=(1×5)x(8=4) 18. 2x(3×7+9) = (1×5)x(8+4) 19. 2 x (3 x 7 + 9) = (1 x 5) x (8 +4) 20. 2x(3×7+9)=(1×5)x(8+4) 21. 2x(3-7+9)=(1+5)-(8×4) 22. 2 + ( 3 + 7 – 9 ) = ( 1 x 5 ) – ( 8 / 4 ) 23. 2 x [3+7+9]=[1+5+4 x 8] 24. 2 x (3 x 7 + 9) = (1 x 5) x (8 +4) 25. 2x(3+7+9)=(1+5)+(8×4) 26. 2x(3+7+9)=(1+5)+8×4} 27. We got 2x(3×7-9) = (1+5)x(8-4) 28. 2×(3x7x9)=(1×5)x(8×4) 29. 2x(3×7+9)=(1×5)x(8+4) 30. I got 2-(3+7-9)=(1/5)-(8-4)
When students reach 3rd and 4th grade level math, they will begin to learn multiplication and division, two of the most essential mathematical operations. Students will also learn how to use these operations for learning other math concepts like fractions, decimals, or percentages. In a previous blog post, I have walked through a four-step process to tackle word problems at the 5th and 6th grade level. However, I would like to take this time to apply this process to 3rd and 4th grade math problems. For example, let’s take a look at this imaginary word problem: “I want to make sandwiches for 5 of my friends. If I want to give each friend 3 sandwiches, what is the total number of sandwiches I should make?” Let’s go through the following steps: • Find the objective and background information. First, we should find the word problem’s objective, or goal. In this problem, we want to know how many sandwiches I should make for all of my friends. In order to figure out the total number of sandwiches I should make, I would need to know some background information. The problem provides the number of friends (5) I want to make sandwiches for and the number of sandwiches (3) I want to give to each friend. If there is any information within the problem that could be useful, please box, highlight, or draw a squiggly line underneath the text! As an example, I have bolded the key background information for this word problem below: “I want to make sandwiches for 5 of my friends. If I want to give each friend 3 sandwiches, what is the total number of sandwiches I should make?” • Which mathematical operations will you use to solve this problem? Are you going to multiply, divide, add, or subtract? In the previous blog post, I stated that the second step would be to consider formulas. However, since this blog post is about 3rd and 4th grade math word problems, we will instead choose among the four arithmetic operations listed above. Our objective is to find a total, so we will not be using subtraction or division for this problem. Since we are given information about how the sandwiches will be split among friends, the best operation to use in this case would be multiplication. (Addition could also work but adding 3 sandwiches for each friend would be less efficient.) • Make note of any unknowns you need to solve for. In this problem, we do not know the total number of sandwiches I need to make for all of my friends, so this is what we will solve for. • Solve! To solve for the total number of sandwiches, we would multiply the number of friends (5) by the number of sandwiches each friend would have (3). 5 friends multiplied by 3 sandwiches gives us a total of 15 sandwiches. Now, I know that I will have enough sandwiches to share among my 5 friends if I make a total of 15 sandwiches. Now, let us apply this same procedure to this next word problem: “A large pepperoni pizza is cut into 8 slices. Student A eats 2 slices and Student B eats 1 slice. What fraction of the pepperoni pizza remains in the pizza box?” To find the answer, I will once again walk through the four-step process for solving this problem. Where are the objective and background information? We are trying to figure out how much of the pizza remains, but we want our answer in fraction form. This would be considered our objective. In order to tackle this problem, we should use the background information given in the problem. We know that we have a total of 8 slices, how many slices Student A ate, and how many slices Student B ate. All of this information will be useful, so let us make a note of it by bolding the words in the problem like so: “A large pepperoni pizza is cut into 8 slices. Student A eats 2 slices and Student B eats 1 slice. What fraction of the pepperoni pizza remains in the pizza box?” Which mathematical operations will you use to solve this problem? Are you going to multiply, divide, add, or subtract? For this problem, we will want to express the number of slices eaten and the number of remaining slices as fractions. The denominator for this problem is the total number of slices we started out with, and the numerators would depend on each fraction we work with. First, we would need to add up the number of slices each student ate. Then, we would subtract this total from the number of slices that we started out with to reach our objective. What are we solving for? In this problem, we do not know what fraction of the pepperoni pizza remains in the pizza box. This is the unknown that we will be solving for. Solve! First, let us express the total number of pepperoni pizza slices as a fraction (8/8). Then, we will express the number of slices each student ate as fractions. Student A ate 2 out of 8 slices (2/8), and Student B ate 1 out of 8 slices (1/8). If we add up the number of slices eaten, we will get a total fraction of 3/8. 2/8 + 1/8 = 3/8 Now, we want to subtract this fraction from the total number of pizza slices we started out with. 8/8 – 3/8 = 5/8 After following these steps, we can conclude that 5/8 of the pepperoni pizza remain in the box. I hope this guide provides insight on how to tackle 3rd and 4th grade level math word problems, whether they are about multiplication or fractions! Learning new math skills can be challenging at first but taking the time to approach word problems carefully will help students successfully solve them. If you would like additional help, My Private Professor in Orange County offers 3rd and 4th grade tutoring in a variety of subjects, including math! Amy Hua is a tutor at My Private Professor, which provides individualized online & in-person tutoring to students in all subjects, including K-12 math, science, language arts, history, foreign language, AP exams, test prep, essays, & college counseling, by top tutors from top universities.
# Understanding Cryptic Quiz Math Worksheet Answers D 75 ## What is a Cryptic Quiz? A cryptic quiz is a set of questions that require mathematical equations to be solved in order to determine the answer. It can be used to test the knowledge of the participants in mathematics and is often used in exams for school or college. It is also used in job interviews and other competitive exams. ## What is the Cryptic Quiz Math Worksheet Answer D 75? The cryptic quiz math worksheet answer D 75 is an equation that can be used to solve a cryptic quiz. The equation is: 4x + 5y = 75, where x is the unknown number and y is the answer. The equation can be used to solve a variety of cryptic quizzes, from basic to more advanced. ## How to Solve Cryptic Quiz Math Worksheet Answer D 75? To solve the cryptic quiz math worksheet answer D 75, first you need to rearrange the equation to isolate the unknown variable. To do this, subtract 5y from both sides of the equation. This will give you the equation 4x = 75-5y. Then, divide both sides of the equation by 4 to isolate the variable x. This will give you the equation x = (75-5y)/4. Finally, substitute the value of y into the equation and solve for x. ## Examples of Cryptic Quiz Math Worksheets with Answers D 75 Here are some examples of cryptic quiz math worksheets with answers D 75: • If 4x + 5y = 75, what is the value of x when y = 8? Answer: x = (75-58)/4 = 25/4 = 6.25 • If 4x + 5y = 75, what is the value of y when x = 10? Answer: y = (75-410)/5 = 5 • If 4x + 5y = 75, what is the value of y when x = 12? Answer: y = (75-4*12)/5 = 3 ## Conclusion The cryptic quiz math worksheet answer D 75 can be used to solve a variety of cryptic quizzes. By rearranging the equation to isolate the unknown variable, you can solve for the answer. Examples of cryptic quiz math worksheets with answers D 75 are provided above.
Maharashtra Board 9th Class Maths Part 1 Practice Set 3.1 Solutions Chapter 3 Polynomials Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 3.1 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 3 Polynomials. Practice Set 3.1 Algebra 9th Std Maths Part 1 Answers Chapter 3 Polynomials Question 1. State whether the given algebraic expressions are polynomials? Justify. i. y + $$\frac { 1 }{ y }$$ ii. 2 – 5√x iii. x2 + 7x + 9 iv. 2m-2 + 7m – 5 v. 10 i. No, because power of v in the term 5√x is -1 (negative number). ii. No, because the power of x in the term 5√x is i. e. 0.5 (decimal number). iii. Yes. All the coefficients are real numbers. Also, the power of each term is a whole number. iv. No, because the power of m in the term 2m-2 is -2 (negative number). v. Yes, because 10 is a constant polynomial. Question 2. Write the coefficient of m3 in each of the given polynomial. i. m3 ii. $$\sqrt [ -3 ]{ 2 }$$ + m – √3m3 iii. $$\sqrt [ -2 ]{ 3 }$$m3 + 5m2 – 7m -1 i. 1 ii. -√3 iii. – $$\frac { 2 }{ 3 }$$ Question 3. Write the polynomial in x using the given information. [1 Mark each] i. Monomial with degree 7 ii. Binomial with degree 35 iii. Trinomial with degree 8 i. 5x7 ii. x35 – 1 iii. 3x8 + 2x6 + x5 Question 4. Write the degree of the given polynomials. i. √5 ii. x° iii. x2 iv. √2m10 – 7 v. 2p – √7 vi. 7y – y3 + y5 vii. xyz +xy-z viii. m3n7 – 3m5n + mn i. √5 = √5 x° ∴ Degree of the polynomial = 0 ii. x° ∴Degree of the polynomial = 0 iii. x2 ∴Degree of the polynomial = 2 iv. √2m10 – 7 Here, the highest power of m is 10. ∴Degree of the polynomial = 10 v. 2p – √7 Here, the highest power of p is 1. ∴ Degree of the polynomial = 1 vi. 7y – y3 + y5 Here, the highest power of y is 5. ∴Degree of the polynomial = 5 vii. xyz + xy – z Here, the sum of the powers of x, y and z in the term xyz is 1 + 1 + 1= 3, which is the highest sum of powers in the given polynomial. ∴Degree of the polynomial = 3 viii. m3n7 – 3m5n + mn Here, the sum of the powers of m and n in the term m3n7 is 3 + 7 = 10, which is the highest sum of powers in the given polynomial. ∴ Degree of the polynomial = 10 Question 5. Classify the following polynomials as linear, quadratic and cubic polynomial. [2 Marks] i. 2x2 + 3x +1 ii. 5p iii. √2 – $$\frac { 1 }{ 2 }$$ iv. m3 + 7m2 + $$\sqrt [ 5 ]{ 2 }$$m – √7 v. a2 vi. 3r3 Linear polynomials: ii, iii Cubic polynomials: iv, vi Question 6. Write the following polynomials in standard form. i. m3 + 3 + 5m ii. – 7y + y5 + 3y3 – $$\frac { 1 }{ 2 }$$+ 2y4 – y2 i. m3 + 5m + 3 ii. y5 + 2y4 + 3y3 – y2 – 7y – $$\frac { 1 }{ 2 }$$ Question 7. Write the following polynomials in coefficient form. i. x3 – 2 ii. 5y iii. 2m4 – 3m2 + 7 iv. – $$\frac { 2 }{ 3 }$$ i. x3 – 2 = x3 + 0x2 + 0x – 2 ∴ Coefficient form of the given polynomial = (1, 0, 0, -2) ii. 5y = 5y + 0 ∴Coefficient form of the given polynomial = (5,0) iii. 2m4 – 3m2 + 7 = 2m4 + Om3 – 3m2 + 0m + 7 ∴ Coefficient form of the given polynomial = (2, 0, -3, 0, 7) iv. – $$\frac { 2 }{ 3 }$$ ∴Coefficient form of the given polynomial = (- $$\frac { 2 }{ 3 }$$) Question 8. Write the polynomials in index form. i. (1, 2, 3) ii. (5, 0, 0, 0 ,-1) iii. (-2, 2, -2, 2) i. Number of coefficients = 3 ∴ Degree = 3 – 1 = 2 ∴ Taking x as variable, the index form is x2 + 2x + 3 ii. Number of coefficients = 5 ∴ Degree = 5 – 1=4 ∴ Taking x as variable, the index form is 5x4 + 0x3 + 0x2 + 0x – 1 iii. Number of coefficients = 4 ∴Degree = 4 – 1 = 3 ∴Taking x as variable, the index form is -2x3 + 2x2 – 2x + 2 Question 9. Write the appropriate polynomials in the boxes. i. Quadratic polynomial: x2; 2x2 + 5x + 10; 3x2 + 5x ii. Cubic polynomial: x3 + x2 + x + 5; x3 + 9 iii. Linear polynomial: x + 7 iv. Binomial: x + 7; x3 + 9; 3x2 + 5x v. Trinomial: 2x2 + 5x + 10 vi. Monomial: x2 Question 1. Write an example of a monomial, a binomial and a trinomial having variable x and degree 5. ( Textbook pg. no. 3) Monomial: x5 Binomial: x5 + x Trinomial: 2x5 – x2 + 5 Question 2. Give example of a binomial in two variables having degree 5. (Textbook pg. no. 38)
Simplify ( 2x^2 - 8 )/( x - 2 ) + 13/x( x -2 ) - ( 4x^2 - 16 )/( x + 2 ) giorgiana1976 \| Student In order to calculate the sum or difference of 3 ratios, we have to verify if they have a common denominator. But, before verifying if they have a common denominator, we'll solve the difference of squares from the brackets. We'll factorize by 2 the first ratio: ( 2x^2 - 8 )/( x - 2 ) = 2(x^2 - 4)/( x - 2 ) We'll write the difference of squares (x^2 - 4) as a product: (x^2 - 4) = (x-2)(x+2) We'll re-write the ratio; ( 2x^2 - 8 )/( x - 2 ) = 2(x-2)(x+2)/(x - 2) We'll reduce like terms: ( 2x^2 - 8 )/( x - 2 ) = 2(x+2) We'll factorize by 4 the third ratio: ( 4x^2 - 16 )/( x + 2 ) = 4(x^2 - 4)/( x + 2 ) 4(x^2 - 4)/( x + 2 ) = 4(x-2)(x+2)/(x + 2) We'll reduce like terms: 4(x^2 - 4)/( x + 2 ) = 4(x-2) We'll re-write now the given expression: 2(x+2) + 13/x( x -2 ) - 4(x-2) It's obvious that the least common denominator (LCD) is the denominator of the second ratio: To calculate the expression we'll do the steps: - we'll multiply 2(x+2) by x(x-2) - we'll multiply 4(x-2) by x(x-2) We'll get: 2(x+2)x(x-2) + 13 - 4(x-2)x(x-2) We'll open the brackets: 2x^3 - 8x + 13 - 4x^3 + 16x^2 - 16x We'll group like terms and we'll get: -2x^3 + 16x^2 - 24x + 13 william1941 \| Student Here we use the result that x^2-y^2=(x-y)(x+y) extensively. (2x^2-8)/(x-2)+13/x(x-2)-(4x^2-16)/(x+2) =2(x-2)(x+2)/(x-2)+13/x(x-2)- 4(x-2)(x+2)/(x+2) =2(x+2)+13/x(x-2)-4(x-2) =[2x(x+2)(x-2)+13-4x(x-2)] / [x(x-2)] =[2x(x^2-4)+13-4x^2+8x]/ [x(x-2)] =[2x^3-8x+13-4x^2+8x]/ [x (x-2)] (2x^3-4x^2+13)/[x(x-2)] The simplified result is (2x^3-4x^2+13) / [x(x-2)] neela \| Student To Simplify : ( 2x^2 - 8 )/( x - 2 ) + 13/x( x -2 ) - ( 4x^2 - 16 )/( x + 2 2(x^2-8)/(x+2) = 2(x^2-2^2)/(x-2) = 2(x+2)(x-2)/(x-2) = 2(x+2) (4x^2-16)/(x+2) =3(x^2-2^2)/(x+2) = 4(x+2)(x-2)/(x+2) = 4(x-2). Therefore the given expression becomes: 2(x+2) +13/(x(x-2) +4(x-2) = 4x+4 +4x-8 +13/x(x-2) = (8x-4) +13/x(x-2) ={8x-4)x(x-2)+13}/(x(x-2) = {8x^3 -20x+8x+13}/(x(x-2)) Access hundreds of thousands of answers with a free trial. Ask a Question Popular Questions
# SAT Math: Geometry ## Triangles A 15 foot ladder is left leaning against a wall. The base of the ladder is 12 feet from the wall. How many feet above the ground does the ladder touch the wall? A. 8 B. 9 C. 10 D. 11 E. 12 ## Knowsys Method Read the question carefully. There is no diagram in this problem, but it might help you keep the measurements straight if you draw one. Start with the ladder. It is at a diagonal from the ground to the wall. This must be your hypotenuse. Remembering that the ground and the wall are perpendicular, draw a right triangle with a hypotenuse of 15. Then add that the length between the ladder and the wall (the horizontal length) is 12. You have accounted for all the information provided in the problem. Identify the bottom line. What is the length of the last side of the triangle? Assess your options. Many students will use the Pythagorean Theorem to solve this problem, but there is a much faster method. Your Knowsys handbook asks you to memorize the Pythagorean Triplet 3-4-5. These numbers represent the ratio of the sides of a right triangle. Notice that if you multiply these numbers by 3, you will get two of the numbers of your right triangle. Use this method to find the third side of your triangle without wasting time squaring and taking the root of any of the numbers. Attack the problem. Start by knowing that the 3-4-5 triangle can be enlarged by multiplying each of the sides by the same number. Compare it to your triangle. 3-4-5 ? -12-15 Notice that if you multiply the original 4 by 3, you get 12. If you multiply the original 5 by 3, you get 15. In order to get the missing side, all you need to do is multiply the original 3 by 3. 3 x 3 = 9 Loop back. You found the length of the missing side, so you are finished. Note: If you have time after finishing all the other problems, you can check this problem using the Pythagorean Theorem, but there is no need to use it before then. This is an easy level question. Want some help reviewing the math concepts you need to master? Try out the Knowsys Pre-Algebra Flashcards, the Knowsys Algebra I Flashcards, and the Knowsys SAT & ACT Math Practice book. Subscribe to Knowsys SAT & ACT Blog by Email
Title College Algebra Equations Involving Rational Exponents WTAMU > Virtual Math Lab > College Algebra > Tutorial 19: Radical Equations and Equations Involving Rational Exponents Inverse of add. 4 is sub. 4 Square root is by itself on one side of eq. If you square a square root, it will disappear. This is what we want to do here so that we can get x out from under the square root and continue to solve for it. Inverse of taking the sq. root is squaring it Step 3: If you still have a radical left, repeat steps 1 and 2. No more radicals exist, so we do not have to repeat steps 1 and 2. Step 4: Solve the remaining equation. In this example, the equation that resulted from squaring both sides turned out to be a linear equation. If you need a review on solving linear equations, feel free to go to Tutorial 14: Linear Equations in One Variable. Inverse of sub. 10 is add. 10 Step 5: Check for extraneous solutions. Let's check to see if x = 26 is an extraneous solution: Plugging in 26 for x False statement Since we got a false statement, x = 26 is an extraneous solution. There is no solution to this radical equation. If you square a square root, it will disappear. This is what we want to do here so that we can get y out from under the square root and continue to solve for it. Inverse of taking the sq. root is squaring it Right side is a binomial squared Be careful on this one. It is VERY TEMPTING to square the right side term by term and get 4 + (x + 1). However, you need to square it as a side as shown above. Recall that when you square a binomial you get the first term squared minus twice the product of the two terms plus the second term squared. If you need a review on squaring a binomial, feel free to go to Tutorial 6: Polynomials. Step 3: If you still have a radical left, repeat steps 1 and 2. Inverse of add. x and 5 is sub. x and 5 Square root is by itself on one side of eq. Inverse of taking the sq. root is squaring it Left side is a binomial squared Be careful on this one. It is VERY TEMPTING to square the left side term by term and get 4x squared plus 4. However, you need to square it as a side as shown above. Recall that when you square a binomial you get the first term squared plus twice the product of the two terms plus the second term squared. If you need a review on squaring a binomial, feel free to go to Tutorial 6: Polynomials. Step 4: Solve the remaining equation. In this example, the equation that resulted from squaring both sides turned out to be a quadratic equation. If you need a review on solving quadratic equations feel free to go to Tutorial 17: Quadratic Equations. Quad. eq. in standard form Factor out a GCF of 4 Factor the trinomial Use Zero-Product Principle Set 1st factor = 0 and solve Set 2nd factor = 0 and solve Step 5: Check for extraneous solutions. Let's check to see if x = 3 is an extraneous solution: Plugging in 3 for x True statement Since we got a true statement, x = 3 is a solution. Let's check to see if x = -1 is an extraneous solution: Plugging in -1 for x True statement Since we got a true statement, x = -1 is a solution. There are two solutions to this radical equation: x = 3 and x = -1. Inverse of sub. 9 is add. 9 Inverse of mult. by 3 is div. by 3 rat. exp. expression is by itself on one side of eq. If you raise an expression that has a rational exponent to the reciprocal of that rational exponent, the exponent will disappear. This is what we want to do here so that we can get x out from under the rational exponent and continue to solve for it. Inverse of taking it to the 5/2 power is taking it to the 2/5 power Step 3: Solve the remaining equation. In this example, the equation that resulted from raising both sides to the 2/5 power turned out to be a linear equation. Also note that it is already solved for x. So, we do not have to do anything on this step for this example. Step 4: Check for extraneous solutions. Let's check to see if is an extraneous solution: Plugging in 3 to the 2/5 power for x True statement Since we got a true statement, is a solution. There is one solution to this rational exponent equation: . The base with the rational exponent is already isolated. If you raise an expression that has a rational exponent to the reciprocal of that rational exponent, the exponent will disappear. This is what we want to do here so that we can get x out from under the rational exponent and continue to solve for it. Inverse of taking it to the 3/2 power is taking it to the 2/3 power Step 3: Solve the remaining equation. In this example, the equation that resulted from squaring both sides turned out to be a quadratic equation. If you need a review on solving quadratic equations, feel free to go to Tutorial 17: Quadratic Equations. Quad. eq. in standard form Factor the trinomial Use Zero-Product Principle Set 1st factor = 0 and solve Set 2nd factor = 0 and solve Step 4: Check for extraneous solutions. Let's check to see if x = -6 is an extraneous solution: Plugging in -6 for x True statement Since we got a true statement, x = -6 is a solution. Lets check to see if x = -3 is an extraneous solution: Plugging in -3 for x True statement Since we got a true statement, x = -3 is a solution. There are two solutions to this rational exponent equation: x = -6 and x = -3. WTAMU > Virtual Math Lab > College Algebra > Tutorial 19: Radical Equations and Equations Involving Rational Exponents Last revised on Dec. 16, 2009 by Kim Seward.
Have you ever heard of the term derivative? Well! It’s a fundamental tool of calculus. What makes it unique, is the fact that this tool can compute the change of a function at any point. In calculus, this concept is equally important as integral, which is the reverse of derivative also called anti-derivative. The rate of change concept, makes it a valuable asset in many real life applications. For instance, the diversity of temperature can be checked using this notion. In this article, we will discuss in detail, its definition along with the real life utility. Now let’s get started, at first we will try to understand the concepts of derivative and differentiation. Definition: The derivative of a variable is defined as a measure to compute the rate of change of a function’s output value as it varies from the initial value or input. Here, an important thing is the time factor, the variation in input and output value as time changes. Let’s consider an example of a moving object, the location of that object starting from the initial point, with respect to time is considered as object’s velocity. This tells us the relative swiftness of the object as it deviates from its position, as time advances. Here, the image above, illustrates a tangent line. The slope of the tangent line at the marked point represents the derivative of a function. The variation can be projected by the ratio of change of function Y (dependent variable) to that of the variable x (independent variable). Notation A German mathematician, Gottfried Wilhelm Leibniz’s introduced a notation, in which symbols were given; dx, dy, and dy/dx. It is commonly used in case an equation y=f(x) is viewed as an association of dependent and independent variables. It uses these symbols to define the infinitesimal (very small) increments. On the other hand, symbols such as Δx and Δy are used to represent the finite increments of x and y. Differentiation: It is a process that helps in calculating the derivative, just like integration computes an integral. This operation is reverse of integration. Let’s assume y a linear function of x. In this example, y = f(x) = mx + b, let m and b the real numbers, slope m is expressed as Slope = m = change in y / change in x = Δy/ Δx Here, Δy = f(x + Δx) – f(x), the above equation is because; = y+ Δy = f(x + Δx) =m(x + Δx) + b = mx + mΔx + b = y + mΔx This gives us the slope of line, Δy = mΔx It applies to a straight line, if the graph is not linear, then the change varies over a considerable range. The differentiation is an efficient method to compute this change over a specific value of x. Practical Applications: This tool isn’t just limited to mathematical problems, it has a broad range of practical utility. Nothing is useless in this world, when we say something can’t be used, we actually don’t know how to use it. The one who knows its utility, won’t stop thinking about it. The uniqueness of this concept is its predictive ability to evaluate the change in quantities. Whether its speed, momentum, temperature and even the business speculations, all the variations can be worked out using derivative. Use in Physics: As we mentioned above, the example of a moving body’s relative position can help us calculate the velocity. In the same way, derivatives of acceleration and momentum can be found. Use in Chemistry: In chemistry, the concentration of an element involved in a reaction, the change in concentration can be predicted. Similarly, to measure the rate of chemical reactions and to check the contribution and loss of a compound during the reaction. Use in Economics: Nowadays, the decision making in economics has become more mathematical. Statistical and mathematical principles are applied in making decisions regarding possible gain or loss in investment. Confronted with massive statistical data, dependent on lots of variables, there was a need of some tool that could assist the analysts. Here, calculus proved to be beneficial. It implemented the derivative concepts to predict the results of different investment possibilities. Ultimately, this enabled the analysts to select the one possibility that might prove to be productive in terms of profitability. In the end, I hope this article will help you understand and apply the calculus concepts in practical fields. If you are interested in methods to calculate this fundamental of calculus, try this derivative calculator. You can also make a relevant calculation on integral function on this integral calculator.
# Section 8.6. You probably know that a lighter tree climber can crawl farther out on a branch than a heavier climber can, before the branch is in danger. ## Presentation on theme: "Section 8.6. You probably know that a lighter tree climber can crawl farther out on a branch than a heavier climber can, before the branch is in danger."— Presentation transcript: Section 8.6 You probably know that a lighter tree climber can crawl farther out on a branch than a heavier climber can, before the branch is in danger of breaking. What do you think the graph of (length, mass) data will look like when mass is added to a length of pole until it breaks? Is the relationship linear, like line A, or does it resemble one of the curves, B or C? Procedural Note 1. Lay a piece of linguine on a table so that its length is perpendicular to one side of the table and the end extends over the edge of the table. 2. Tie the string to the film canister so that you can hang it from the end of the linguine. (You may need to use tape to hold the string in place.) 3. Measure the length of the linguine between the edge of the table and the string. Record this information in a table of (length, mass) data. 4. Place mass units into the container one at a time until the linguine breaks. Record the maximum number of weights that the length of linguine was able to support. Step 1:Work with a partner. Follow the Procedure Note to record at least five data points, and then compile your results with those of other group members. Step 2: Make a graph of your data with length as the independent variable, x, and mass as the dependent variable, y. Does the relationship appear to be linear? If not, describe the appearance of the graph. The relationship between length and mass is an inverse variation. The parent function for an inverse variation curve, f(x)=1/x, is the simplest rational function. Step 3: Your data should fit a dilated version of the parent function f (x) = 1/x. Write an equation that is a good fit for the plotted data. Graph the function f(x) =1/x on your calculator and observe some of its special characteristics. The graph is made up of two branches. One part occurs where x is negative and the other where x is positive. There is no value for this function when x =0. What happens when you try to evaluate f(0)? This graph is a hyperbola. Its like the hyperbolas you studied in Lesson 8.4, but it has been rotated 45°. It has vertices (1, 1) and (-1, 1), and its asymptotes are the x- and y-axes. To understand the behavior of the graph close to the axes, make a table with values of x very close to zero and very far from zero and examine the corresponding y-values. Consider these values of the function f (x) =1/x. The behavior of the y-values as x gets closer to zero shows that the y-axis is a vertical asymptote for this function. As x approaches the extreme values at the left and right ends of the x-axis, the curve approaches the x-axis. The horizontal line y = 0, then, is a horizontal asymptote. This asymptote is called an end behavior model of the function. In general, the end behavior of a function is its behavior for x-values that are large in absolute value. Example A You can change the form of the equation so that the transformations are more obvious. Because the numerator and denominator both have degree 1, you can use division to rewrite the expression. Example B To find when the solution is 60% acid, substitute 0.6 for P and solve the equation. Download ppt "Section 8.6. You probably know that a lighter tree climber can crawl farther out on a branch than a heavier climber can, before the branch is in danger." Similar presentations
There are 366 different Starters of The Day, many to choose from. You will find in the left column below some starters on the topic of Mensuration. In the right column below are links to related online activities, videos and teacher resources. A lesson starter does not have to be on the same topic as the main part of the lesson or the topic of the previous lesson. It is often very useful to revise or explore other concepts by using a starter based on a totally different area of Mathematics. Main Page ### Mensuration Starters: Area Two: How many different shapes with an area of 2 square units can you make by joining dots on this grid with straight lines? Bizarre Triangle: By how much would the area of this triangle increase if its base was enlarged to 8cm? Christmas Tables: Which of the two shapes has the largest area? You will be surprised! Cross Perimeter: Calculate the distance around the given shape Goat Grazing: Find the loci of the goat's position as it eats the grass while tethered to the rope. Missing Lengths: Introduce linear equations by solving these problems about lengths. Missing Square Puzzle: The missing square puzzle is an optical illusion used to help students reason about geometrical figures. Oblongs: Find the dimensions of a rectangle given the perimeter and area. Quad Areas: Calculate the areas of all the possible quadrilaterals that can be constructed by joining together dots on this grid. Shopping List: A quick quiz about five items on a shopping list written 40 years ago. Stair Perimeter: Use the information implied in the diagram to calculate the perimeter of this shape. Step Perimeter: Is it possible to work out the perimeter of this shape if not all the side lengths are given? Average Cycling Speed: Work out the average speed of two journeys. The obvious answer is not the correct answer. Charging Rhinos: Find the easy way to solve this kinematics problem involving a fly and two rhinos. Cuboid: Find the dimensions of a cuboid matching the description given Fence Optimisation: Find the length of a rectangle enclosing the largest possible area. Hands Together: The hands of a clock are together at midnight. At what time are they next together? Paper Ratio: Calculate the ratio of the sides of an A4 sheet of paper without any measuring. Paper Surprising Perimeter: Find the perimeter of a folded sheet of A4 paper as described in this short video. Piece of String: Find where a piece of string should be cut to form a circle and a square of equal areas. Pizza Slice: A problem which can be solved by considereing the areas of a triangle and a sector of a circle. Road Connections: Design roads to connect four houses that are on the corners of a square, side of length one mile, to minimise the total length of the roads. Speed Circles: Find the diameters of the circles in the corners of the square. Sphere Hole: Find the volume of the remaining part of a sphere after a 10cm cylindrical hole has been drilled through it. Square in Rectangle: Find the area of a square drawn under the diagonal of a rectangle What Question?: Write down all the possible questions that could have been asked if this was the diagram provided in a mathematics textbook. #### Volume Use formulae to solve problems involving the volumes of cuboids, cones, pyramids, prisms and composite solids. Transum.org/go/?to=volume ### Curriculum for Mensuration: #### Year 5 Pupils should be taught to convert between different units of metric measure (for example, kilometre and metre; centimetre and metre; centimetre and millimetre; gram and kilogram; litre and millilitre) more... Pupils should be taught to understand and use approximate equivalences between metric units and common imperial units such as inches, pounds and pints more... Pupils should be taught to measure and calculate the perimeter of composite rectilinear shapes in centimetres and metres more... Pupils should be taught to calculate and compare the area of rectangles (including squares), and including using standard units, square centimetres (cm2) and square metres (m2) and estimate the area of irregular shapes more... Pupils should be taught to estimate volume [for example, using 1 cm3 blocks to build cuboids (including cubes)] and capacity [for example, using water] more... #### Year 6 Pupils should be taught to solve problems involving the calculation and conversion of units of measure, using decimal notation up to three decimal places where appropriate more... Pupils should be taught to solve problems involving the calculation of percentages [for example, of measures, and such as 15% of 360] and the use of percentages for comparison more... Pupils should be taught to use, read, write and convert between standard units, converting measurements of length, mass, volume and time from a smaller unit of measure to a larger unit, and vice versa, using decimal notation to up to three decimal places more... Pupils should be taught to convert between miles and kilometres more... Pupils should be taught to recognise that shapes with the same areas can have different perimeters and vice versa more... Pupils should be taught to recognise when it is possible to use formulae for area and volume of shapes more... Pupils should be taught to calculate the area of parallelograms and triangles more... Pupils should be taught to calculate, estimate and compare volume of cubes and cuboids using standard units, including cubic centimetres and cubic metres, and extending to other units more... #### Years 7 to 9 Pupils should be taught to understand and use place value for decimals, measures and integers of any size more... Pupils should be taught to change freely between related standard units [for example time, length, area, volume/capacity, mass] more... Pupils should be taught to derive and apply formulae to calculate and solve problems involving: perimeter and area of triangles, parallelograms, trapezia, volume of cuboids (including cubes) and other prisms (including cylinders) more... Pupils should be taught to use scale factors, scale diagrams and maps more... Pupils should be taught to calculate and solve problems involving: perimeters of 2-D shapes (including circles), areas of circles and composite shapes more... Pupils should be taught to draw and measure line segments and angles in geometric figures, including interpreting scale drawings more... Pupils should be taught to use compound units such as speed, unit pricing and density to solve problems. more... Pupils should be taught to use standard units of mass, length, time, money and other measures, including with decimal quantities more... Pupils should be taught to use the properties of faces, surfaces, edges and vertices of cubes, cuboids, prisms, cylinders, pyramids, cones and spheres to solve problems in 3-D more... #### Years 10 and 11 Pupils should be taught to convert between related compound units (speed, rates of pay, prices, density, pressure) in numerical and algebraic contexts more... Pupils should be taught to calculate arc lengths, angles and areas of sectors of circles more... Pupils should be taught to calculate surface areas and volumes of spheres, pyramids, cones and composite solids more... ### Feedback: Comment recorded on the 1 February 'Starter of the Day' page by Terry Shaw, Beaulieu Convent School: "Really good site. Lots of good ideas for starters. Use it most of the time in KS3." Comment recorded on the 17 June 'Starter of the Day' page by Mr Hall, Light Hall School, Solihull: "Dear Transum, I love you website I use it every maths lesson I have with every year group! I don't know were I would turn to with out you!" Comment recorded on the s /Coordinate 'Starter of the Day' page by Greg, Wales: "Excellent resource, I use it all of the time! The only problem is that there is too much good stuff here!!" Comment recorded on the 24 May 'Starter of the Day' page by Ruth Seward, Hagley Park Sports College: "Find the starters wonderful; students enjoy them and often want to use the idea generated by the starter in other parts of the lesson. Keep up the good work" Comment recorded on the 26 March 'Starter of the Day' page by Julie Reakes, The English College, Dubai: "It's great to have a starter that's timed and focuses the attention of everyone fully. I told them in advance I would do 10 then record their percentages." Comment recorded on the 10 September 'Starter of the Day' page by Carol, Sheffield PArk Academy: "3 NQTs in the department, I'm new subject leader in this new academy - Starters R Great!! Lovely resource for stimulating learning and getting eveyone off to a good start. Thank you!!" Comment recorded on the 14 September 'Starter of the Day' page by Trish Bailey, Kingstone School: "This is a great memory aid which could be used for formulae or key facts etc - in any subject area. The PICTURE is such an aid to remembering where each number or group of numbers is - my pupils love it! Thanks" Comment recorded on the 28 September 'Starter of the Day' page by Malcolm P, Dorset: "A set of real life savers!! Keep it up and thank you!" Comment recorded on the 2 April 'Starter of the Day' page by Mrs Wilshaw, Dunsten Collage,Essex: "This website was brilliant. My class and I really enjoy doing the activites." Comment recorded on the 3 October 'Starter of the Day' page by S Mirza, Park High School, Colne: "Very good starters, help pupils settle very well in maths classroom." Comment recorded on the 9 April 'Starter of the Day' page by Jan, South Canterbury: "Thank you for sharing such a great resource. I was about to try and get together a bank of starters but time is always required elsewhere, so thank you." Comment recorded on the 18 September 'Starter of the Day' page by Mrs. Peacock, Downe House School and Kennet School: "My year 8's absolutely loved the "Separated Twins" starter. I set it as an optional piece of work for my year 11's over a weekend and one girl came up with 3 independant solutions." Comment recorded on the 1 February 'Starter of the Day' page by M Chant, Chase Lane School Harwich: "My year five children look forward to their daily challenge and enjoy the problems as much as I do. A great resource - thanks a million." Comment recorded on the 9 October 'Starter of the Day' page by Mr Jones, Wales: "I think that having a starter of the day helps improve maths in general. My pupils say they love them!!!" Comment recorded on the 2 May 'Starter of the Day' page by Angela Lowry, : "I think these are great! So useful and handy, the children love them. Could we have some on angles too please?" Comment recorded on the 19 October 'Starter of the Day' page by E Pollard, Huddersfield: "I used this with my bottom set in year 9. To engage them I used their name and favorite football team (or pop group) instead of the school name. For homework, I asked each student to find a definition for the key words they had been given (once they had fun trying to guess the answer) and they presented their findings to the rest of the class the following day. They felt really special because the key words came from their own personal information." Comment recorded on the 16 March 'Starter of the Day' page by Mrs A Milton, Ysgol Ardudwy: "I have used your starters for 3 years now and would not have a lesson without one! Fantastic way to engage the pupils at the start of a lesson." Comment recorded on the 21 October 'Starter of the Day' page by Mr Trainor And His P7 Class(All Girls), Mercy Primary School, Belfast: "My Primary 7 class in Mercy Primary school, Belfast, look forward to your mental maths starters every morning. The variety of material is interesting and exciting and always engages the teacher and pupils. Keep them coming please." Comment recorded on the 17 November 'Starter of the Day' page by Amy Thay, Coventry: "Thank you so much for your wonderful site. I have so much material to use in class and inspire me to try something a little different more often. I am going to show my maths department your website and encourage them to use it too. How lovely that you have compiled such a great resource to help teachers and pupils. Thanks again" Comment recorded on the 19 June 'Starter of the Day' page by Nikki Jordan, Braunton School, Devon: "Excellent. Thank you very much for a fabulous set of starters. I use the 'weekenders' if the daily ones are not quite what I want. Brilliant and much appreciated." Comment recorded on the 10 April 'Starter of the Day' page by Mike Sendrove, Salt Grammar School, UK.: "A really useful set of resources - thanks. Is the collection available on CD? Are solutions available?" Comment recorded on the 28 May 'Starter of the Day' page by L Smith, Colwyn Bay: "An absolutely brilliant resource. Only recently been discovered but is used daily with all my classes. It is particularly useful when things can be saved for further use. Thank you!" Comment recorded on the 6 May 'Starter of the Day' page by Natalie, London: "I am thankful for providing such wonderful starters. They are of immence help and the students enjoy them very much. These starters have saved my time and have made my lessons enjoyable." 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We use them for all age groups and abilities. Have particularly enjoyed KIM's game, as we have not used that for Mathematics before. Keep up the good work and thank you very much Best wishes from Inger Kisby" Comment recorded on the 1 May 'Starter of the Day' page by Phil Anthony, Head of Maths, Stourport High School: "What a brilliant website. We have just started to use the 'starter-of-the-day' in our yr9 lessons to try them out before we change from a high school to a secondary school in September. This is one of the best resources on-line we have found. The kids and staff love it. Well done an thank you very much for making my maths lessons more interesting and fun." Comment recorded on the 5 April 'Starter of the Day' page by Mr Stoner, St George's College of Technology: "This resource has made a great deal of difference to the standard of starters for all of our lessons. Thank you for being so creative and imaginative." Comment recorded on the 19 November 'Starter of the Day' page by Lesley Sewell, Ysgol Aberconwy, Wales: "A Maths colleague introduced me to your web site and I love to use it. The questions are so varied I can use them with all of my classes, I even let year 13 have a go at some of them. I like being able to access the whole month so I can use favourites with classes I see at different times of the week. Thanks." ### Notes: Mensuration is the branch of Mathematics dealing with measurement of angles, length, area, and volume. It is linked closely to the topic of Estimation and related to the topics of Angles, Shape and Shave (3D). It is essential for pupils to have an understanding of the units used to measure which include both the more common metric units and the Imperial units still in common usage. We have found a good teaching strategy is to ask each of the pupils to "Bring to the next Maths lesson some visual aid which will help the rest of the class remember the size of a unit of measurement". See Memorable Measures below for the printable resources. This activity provides an association with a unit, a visual aid and a known person which is a great memory enhancer. ### Mensuration Teacher Resources: Memorable Measures: This is a visual aid and printable cards to introduce a homework activity about measures. Pin Board: Rows and columns of dots that can be joined using straight lines to create shapes. ### Mensuration Activities: Algebraic Perimeters: Questions about the perimeters and areas of polygons given as algebraic expressions. Area and Perimeter: Show that you know the area and perimeter formulas of basic shapes. Area and Perimeter of a Rectangle: Questions on the areas and perimeters of rectangles which will test your problem solving abilities. Area Builder: An interactive workspace in which to make shapes using square tiles with given areas and perimeters. Area Maze: Use your knowledge of rectangle areas to calculate the missing measurement of these composite diagrams. Area of a Trapezium: Check that you can find the area of a trapezium and use the trapezium area formula for problem solving. Area of a Triangle: Calculate the areas of the given triangles in this self marking quiz. Area Two: How many different shapes with an area of 2 square units can you make by joining dots on this grid with straight lines? Areas of Composite Shapes: Find the areas of combined (composite) shapes made up of one or more simple polygons and circles. Bottles, Boxes and Cans: Estimate the capacity of the bottles, boxes and cans in the photograph and answer questions about volume. Circles: Practise using pi to calculate various circle measurements. There are six levels of difficulty. Compound Units: Practise using compound units such as speed, unit pricing and density to solve problems. Converting Standard Units: Converting measurements of length, mass, volume and time from one unit of measure to another. Cylinders: Apply formulae for the volumes and surface areas of cylinders to answer a wide variety of questions Formulae Pairs: Find the matching pairs of diagrams and formulae for basic geometrical shapes. Formulae to Remember: The traditional pairs or pelmanism game adapted to test recognition for formulae required to be memorised for GCSE exams. Imperial Units Pairs: Find the matching pairs of equivalent imperial units in this interactive online game. Inequalities: Check that you know what inequality signs mean and how they are used to compare two quantities. Includes negative numbers, decimals, fractions and metric measures. Map Scales: Test your understanding of map scales expressed as ratios with this self marking quiz. Measuring Angles: Measure the size of the given angles to within two degrees of their actual value. Measuring Units: Check your knowledge of the units used for measuring with this multiple choice quiz about metric and imperial units. Metric Units Pairs: Find the matching pairs of equivalent metric units in this interactive online game. Mileometer: Practice converting between miles and kilometres with this self marking quiz. Quad Areas: Calculate the areas of all the possible quadrilaterals that can be constructed by joining together dots on this grid. Reading Scales: A self marking exercise on the reading of scales of different types. Scale Drawings: Measure line segments and angles in geometric figures, including interpreting scale drawings. Screen Test: Memorise the mathematical facts in the video then answer the ten quiz questions. Similar Shapes: Questions about the scale factors of lengths, areas and volumes of similar shapes. Sorting Units: Order the ten containers according to their value (money, length and weight) Surface Area: Work out the surface areas of the given solid shapes. Volume: Use formulae to solve problems involving the volumes of cuboids, cones, pyramids, prisms and composite solids. Finally there is Topic Test, a set of 10 randomly chosen, multiple choice questions suggested by people from around the world. Alternatively, for the more advanced student, there is an ever-growing collection of Exam-Style Questions with worked solutions on the topic of Mensuration. ### Mensuration Investigations: Area Builder: An interactive workspace in which to make shapes using square tiles with given areas and perimeters. Area shapes: Investigate polygons with an area of 4 square units. This is your starting point, you can decide how to proceed. Maxvoltray: Find the maximum volume of a tray made from an A4 sheet of paper. A practical mathematical investigation. Pin Board: Rows and columns of dots that can be joined using straight lines to create shapes. Rectangle Perimeters: The perimeter of a rectangle is 28cm. What could its area be? ### Mensuration Videos: Area and Perimeter: Area and perimeter of composite shapes video for GCSE Maths. Circle Facts Song: A free trial lesson from Math Upgrade dot com. Formulae for GCSE: These are the formulae candidates need to know for the GCSE(9-1) Maths exams. How Long is a Metre?: Do you know how long a metre is? Where did this measurement come from? How long has it been used for? Parallelogram: Instructional video showing how the area of a parallelogram can be determined. Pi and Four Fingers: Why is The Simpsons not in Base 8? In this video Simon Singh talks about Pi and Maths in The Simpsons cartoon. Pi Song: Kate Bush sings the digits of pi (audio only). Volumes of Cylinders: Dr Frost demonstrates how to find the volume of a cylinder with a number of worked examples. ### Mensuration Worksheets/Printables: Mearsuring Lines and Angles: Practice using a ruler and protractor on this worksheet with answers provided. Memorable Measures Notes: These are the printable cards to go with the activity called Memorable Measures. Links to other websites containing resources for Mensuration are provided for those logged into 'Transum Mathematics'. Subscribing also opens up the opportunity for you to add your own links to this panel. You can sign up using one of the buttons below: ### Search The activity you are looking for may have been classified in a different way from the way you were expecting. You can search the whole of Transum Maths by using the box below. ### Other Is there anything you would have a regular use for that we don't feature here? Please let us know. #### Algebraic Perimeters Questions about the perimeters and areas of polygons given as algebraic expressions. Transum.org/go/?to=algper ### Homepage Have today's Starter of the Day as your default homepage. Copy the URL below then select Tools > Internet Options (Internet Explorer) then paste the URL into the homepage field. Set as your homepage (if you are using Internet Explorer) CNN, Monday, December 10, 2018 "Metric mishap caused loss of NASA orbiter. (CNN) -- NASA lost a \$125 million Mars orbiter because a Lockheed Martin engineering team used English units of measurement while the agency's team used the more conventional metric system for a key spacecraft operation, according to a review finding released Thursday.Sep 30, 1999" Do you have any comments? It is always useful to receive feedback and helps make this free resource even more useful for those learning Mathematics anywhere in the world. Click here to enter your comments. 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Open in App Not now # Class 9 NCERT Solutions- Chapter 11 Constructions – Exercise 11.2 • Last Updated : 28 Dec, 2020 ### Question 1. Construct a âˆ† ABC in which BC = 7 cm, âˆ B = 75Â° and AB + AC = 13 cm. Solution: Steps of construction: 1. Draw a line segment BC base of cm is drawn. 2. At point B draw an angle of 75Â°. 3. Cut BD =13cm from BY. 4. Join âˆ D which intersect BD at A. 5. Join AC. Now triangle ABC is the required triangle ### Question 2. Construct a ABC in which BC = 8 cm, âˆ B = 45Â° and AB â€“ AC = 35 cm. Solution: Steps of construction: 1. Draw a line segment BC=8cm. 2. At point B, draw angle 45Â°. 3. Cut BD=3.5 from BY. 4. Join CD. 5. Draw perpendicular bisector of CD, which construct BY at A. 6. Join AC. NOW, ABC is the required triangle. ### Question 3. Construct a âˆ† ABC in which QR = 6 cm, âˆ Q = 60Â° and PR â€“ PQ = 2 cm. Solution: Steps of construction: 1. Draw a line segment QR=6cm. 2. At point Q draw angle 60Â°. 3. Extend PQ to Yâ€™. 4. Cut QS =2cm from QYâ€™. 5. Join RS. 6. Draw perpendicular bisector of RS which intersect QY at P. 7. Join PR. Now, PQR is the required triangle. ### Question 4. Construct a âˆ† XYZ in which âˆ Y = 30Â°, âˆ Y = 90Â° and XY + YZ + ZX = 11 cm. Solution: Steps of construction: 1. Draw a line segment AB=11cm. 2. At point A draw âˆ BAP=30Â°. 3. At point B draw angle 90Â°. 4. Draw the bisector of âˆ BAP and âˆ ABR which intersect each other at X. 5. Join AX and BX. 6. Draw perpendicular bisector of AX and BX which intersect AB on Y and Z respectively. 7. Join XY and XZ. Then XYZ is the required triangle. ### Question 5. Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm. Solution: Steps of construction: 1. Draw a line segment of BC=12cm. 2. At point B draw angle b=90Â° 3. Cut BD =18cm. 4. Join CD. 5. Draw perpendicular bisector of CD which intersect BD at point A. 6. Join AC. Now ABC is the required triangle. My Personal Notes arrow_drop_up Related Articles
Scientific Notation # Scientific Notation ## Scientific Notation - - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - - ##### Presentation Transcript 1. Scientific Notation Writing large and small numbers 2. It’s easy to make mistakes when computing with small numbers such as 0.000022 and large numbers such as 75,000,000. People who deal regularly with such numbers use scientificnotation. Scientific notation – a method of writing very large and very small numbers using powers of 10. Take for example, an ordinary penny contains about 20,000,000,000,000,000,000,000 atoms. The average size of an atom is about 0.00000003 centimeters across 3. Written in scientific notation: • Number of atoms in a penny; 2.0 x 1022 • Size of each atom: 3.0 x 10-8 The sign on the exponent tells which direction to move the decimal. A positive exponent means move the decimal to the right, and a negative exponent means move the decimal to the left. 4. Writing Scientific Notation • Scientific notation is written using the product of 2 factors. 1.0 x 106 = 1,000,000 • First factor is a number ≥ 1 but < 10. 1,000,000 = 1.0 • Second factor is a power of 10 = (106) 5. Scientific Notation to Standard Notation Write each number in standard notation. 2.64 x 107 2.64 x 10,000,000 107 = 10,000,000 26,400,000 Think: Move the decimal right 7 places for positive exponent. 1.35 x 10-4 1.35 x 1/10,000 10-4 = 1/10,000 0.000135 Think: Move the decimal left 4 places for negative exponent. 6. Standard Notation toScientific Notation Write each number in scientific notation. 0.000002 2 Move the decimal to get a number between 1 and 10 2 x 10□Set up scientific notation. Think: decimal needs to move left 6 places to change 2 to 0.000002, moving left makes exponent negative. 2 x 10-6 0.0027 2.7 Move the decimal to get a number between 1 and 10 2.7 x 10□ Set up scientific notation. Think: decimal needs to move left 3 places to change 2.7 to 0.0027, moving left makes exponent negative. 2.7 x 10-3 7. Scientific Notation with a Calculator Technology Lab with TI-83 calculator 8. Scientist often have to work with very large or very small numbers. For example, the Andromeda Galaxy contains over 200,000,000,000 (billion) stars. Scientific notation is a compact way of expressing these numbers. • Show 200,000,000,000 in scientific notation on your graphing calculator. • Enter 200,000,000,000 on your calculator. Press [ENTER] 9. 2E11 on the graphing calculator display means 2 x 1011 Your calculator automatically puts very large numbers into scientific notation. 10. You can enter scientific notation directly into the calculator yourself using the EE function key(2nd key then comma , key). • Enter 2 x 1011 by pressing 2 [2nd] EE [,] 11 [enter] 11. Evaluating Expressions using Scientific Notation Evaluate 230,000,000 x 650,000 • First write 230,000,000 in scientific notation = 2.3 x 108 • Second write 650,000 in scientific notation = 6.5 x 105 • Press 2.3 [2nd] EE [,] 8 x 6.5 [2nd] EE [,] 5 [enter] In scientific notation, 230,000,000 x 650,000 is equal to 1.495 x 1014 or 149,500,000,000,000 (149 trillion, 500 billion). 12. Evaluate (340,000,000,000 ÷ 1,235) x 4,568 • Write each number in scientific notation. • 340,000,000,000 = 3.4 x 1011 • 1,235 = 1.235 x 103 • 4,568 = 4.568 x 103 • Press (3.4 [2nd] EE [,] 11 ÷ 1.235 [2nd] EE [,] 3 ) x 4.568 [2nd] EE [,] 3 [enter] • (340,000,000,000 ÷ 1,235) x 4,568 = 1.257587045 x 1012 (because this calculator displays results to only 10 decimal places, this innot exact. The exact answer is 1.2575870445344110 x 1012) 13. Problem Solving • In space, light travels about 9,460,000,000,000 kilometers per year. This distance is known as a light-year. The star Altair is 16.3 light-years from earth. Write this distance in scientific notation. • Suppose the universe is made up of 125,000,000,000 galaxies, each containing 200,000,000,000 stars. Use this data to find the total number of stars in the entire universe. Express your answer in scientific notation and in standard form. 14. Answer (check out what these would look like in standard notation) 1.54E14 kilometers from earth 2.5E22 star in the universe 15. 154,000,000,000,000154 trillion kilometers 25,000,000,000,000,000,000,000 25 sextillion stars in the universe 16. Extremely small numbers are written in the same way but use (-) negative exponents. • For example, our ordinary penny contains about 20 sextillion atoms, written: 2.0E22 or 20,000,000,000,000,000,000 atoms • The average size of the atom we discussed earlier is about 3 hundred millionths centimeters across, written: 3.0E-8 or 0.00000003 centimeters 17. Helpful Hint The sign of the exponent tells which direction to move the decimal. A positive exponent means to move the decimal to the right. A negative exponent means move the decimal to the left. 18. Math Fact !!!!! • The mass (weight) of the earth is about 5,980,000,000,000,000,000,000,000 kilograms. (5 septillion, 980 sextillion) • This is written in scientific notation as 5.98 x 1024 • How would you enter this into your calculator?
Overcast Weather Pokémon, Seventh Moon Anime, Elkmont Campground Store, Get Value From Array Python, Optimal Blood Glucose Level For Ketosis, Coughing In The Morning With Mucus, Restricted Boltzmann Machine - Assignment, Clamp On Flash Hider, " /> 20 Jan 2021 No. Distributive Property with Variables Worksheet Along with Algebra Worksheets for Simplifying the Equation. As stated earlier, distributive property is used quite frequently in mathematics. Write an expression for the area of this rectangle. Then remove a factor of 1 from both sides. Remember that the distributive property shows that multiplying an expression involving addition (or subtraction) by x is the same thing as multiplying each term of the expression by x before performing the addition. of very systematically figuring out a greatest common factor. This method follows the order of operations. We have 60, 60m-40. Use geometric figures to build the ideas of algebra. So, let me think even harder about what a greatest common They still share two. This video shows the distributive property through friendship bracelets. Distributive property with variables (negative numbers) CCSS.Math: 7.EE.A.1. Variables in Mathematics: In mathematics, a variable is a letter that describes or represents a number or numbers. PBS has some videos in a series called Math Club. Main & Advanced Repeaters, Vedantu 10 is two times five. Here, you have two twos and a five. this is the same thing as 10 times six and actually, and then, of course, you have the M there, so you could do this 10 times 6m. We'll look at some examples to help you understand what it means. 1/2(2) is just one, so you're just going to be left with A. The answer of the product of a number and the addition of two other numbers can be obtained as the sum of the individual products obtained when each number inside the parentheses is multiplied by the number outside. Let’s try using the Distributive Property: $$4(x+3)$$ $$=4(x)+4(3)$$ $$=4x+12$$ Now, we have an answer that is simplified as much as possible without knowing the values of any variables. 1/2(8). Distributive Property with Variables Worksheet Along with Algebra Worksheets for Simplifying the Equation. In other words, the number or variable that is outside the set of parentheses "distributes" through the parentheses, multiplying by each of the numbers inside. Solve the equation and simplify, if needed. The distributive property law can also be used when multiplying or dividing algebraic expressions that include real numbers and variables that is called distributive property with variables. We offer a whole lot of high-quality reference materials on subjects ranging from power to subtracting polynomials ... Depdendent Variable: Draw: Number of equations to solve: Sample Problem: Equ. Take factorization backwards to learn this useful algebraic technique. The fundamental properties of numbers in Mathematics are: 2. This is the distributive property at work! Like terms can be combined as is stated in the distributive property. Times 6b+1/2(8). 1-7 Guide Notes TE - The Distributive Property (FREEBIE) 5.) Well, two times 10 is 20. Let's use rectangles to understand the distributive property with variables. This is the distributive property at work! of blue pencils with each. All the real numbers obey certain laws or have a few properties. This Rock, Paper, Scissors game is an engaging supplement for a lesson on teaching students how to use the distributive property in variable expressions.Making a competitive game out of doing math problems will liven up any lesson for your students. And so, i just need to distribute the 1/2. 1.) Combining like terms with negative coefficients. So, you know that you have fully factored these two things out. of friends x No. Combining like terms with negative coefficients. In Mathematics, the numbers should obey the characteristic property during the arithmetic operations. Don't let the variables in the distributive property confuse you! So, let's check our answer. Distributive Property with Variables. And then minus, minus 40 divided by 20, you're just left with the two. Equations Apply the power of the equal sign. Every student is successful and confidence soars! 1-7 Guide Notes SE - The Distributive Property (FREEBIE) 4.) The distributive property & equivalent expressions. 30, which is two times 15, which is three times five. If you apply Distributive Property, 6× 2 + 6 × 4x. You could say 60 is two times Imagine that each individual student has 4 black pencils and 5 blue pencils. Consider an example here: 6(2+4x) The two values inside the parenthesis cannot be added since they are not like terms, therefore it cannot be simplified any further. The distributive property is the one which allows us to multiply the number by a group of numbers, which are added together. Minus 40. Then, to find the total number of pencils, the number of black pencils and blue pencils are added. And so, what's this going to be? To find the unknown value in the equation, we can follow the steps below: 9/9/2020 IXL - Simplify variable expressions involving like terms and the distributive property (Algebra 1 practice) 1/1 I.3 Simplify variable expressions involving like terms and the distributive property ZXX Algebra 1 Next up Done for now? Then isolate the variable, and solve the remaining one-step problem. Here are your Free Distributive Property Activities for this Lesson! Two mazes contain variables and one has no variables - great for reinforcing student learning - so engaging! lemme color code it, too, just for fun, so it's going to be 1/2 times, give myself some space, 1/2(2a-6b), so, 2a-6b, minus six, lemme write it this way - 6b, and then we have plus eight. By using the distributive property of multiplication over addition or subtraction, the expressions can be written as the sum or difference of the two numbers. Therefore, it is really helpful in simplifying algebraic equations as well. In this video you can see how easy is to use the distributive property with variables. It is an alternative way to show students the progression. This section briefs about a few distributive property examples for better understanding. The first thing you must do is simplify by using the Distributive Property.You simplify using the Distributive Property by distributing the term in front of the parenthesis by multiplying it by everything on the inside of the parenthesis. This color property in all cases by a group of numbers, which two! Lesson you will also see another example where the expression in expanded.. Like simplifying any product mathematics that you may have done, you multiply the two do! To learn this useful algebraic technique what will we be learning in this video well. Individual student has 4 black pencils and the total number of pencils, only! Parentheses to each term inside the parentheses to each term inside the parentheses 's factor it out useful algebraic.! This exercise practices distribution nad begins to introduce factoring of algebraic expressions to arrive at final. Information and to our privacy policy exercise \ ( m\ ) units that. Message, it is impossible icon in toolbar below you will also another... ) distributive property with variables expressions like 3x+7x and 10x are equivalent expressions since they denote same! In common geometric figures to build the ideas of algebra a length of \ ( \PageIndex { 1 \! When using the distributive property: what will we be learning in this video as well plus, and times. Pencils with each individual = 4 + 5 = 9, 3m and two share no common factors ). And use all the numbers that are used in Mathematical calculations and have a few properties a length of (!, distributive property to write two expressions that equal 360 cut and Paste are found in this color examples... Sometimes we need a different method and this is going to be four and algebra Notation the... The number by a group of numbers include closure property, commutative property, video! And I will do eight in this magenta color throughout the lesson all cases to log in and all... { 9 ) there are n't three twos and a five here sometimes we need a different method and is. The only change is that the the articles a term consists of a,! What 1/2 ( 6 ) is going to be the greatest common factor, let me think even about! A 20 on 2014-11-09 and has been viewed 54 times this month any product are friends... Of algebra problem are called _____ because they denote the same bench learned that. 8 worksheets found for this concept in common parentheses are being multiplied or.. Properties are associative property, identity property, you have minus 1/2 ( 2a ) the difference of the which... You use the distributive property with variables term refers to the distribution of multiplication therefore, is. Algebraic equations as well we might say, so let 's say, eight halves is equal to four.. 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# 7 is equal to 8? We start with the following equality, which we assume to be true: a + b = c We can write equality as follows: (8a-7a) + (8b-7b) = (8c-7c) Putting all multiples of 7 on one side and 8 on the other, we have: 8a + 8b-8c = 7a + 7b-7c Highlighting 7 on one side and 8 on the other, we have: 8 (a + b-c) = 7 (a + b-c) Dividing both sides by a + b-c we have: 8 = 7 Obviously this demonstration has an error because we all know that 8 is not equal to 7 (or does anyone have any questions?). Click below to find out what the error is: In this demonstration comes a stage where we have: 8 (a + b-c) = 7 (a + b-c) According to the demonstration, the next step is to divide both sides by a + b-c. There is the mistake !!! It is wrong because at first we assume that a + b = c, so a + b-c is worth zero. Division by zero does not exist !!! Next: Negative Sum
Paul's Online Math Notes [Notes] Calculus II - Notes Integration Techniques Previous Chapter Next Chapter Parametric Equations and Polar Coordinates Hydrostatic Pressure Previous Section ## Probability In this last application of integrals that we’ll be looking at we’re going to look at probability. Before actually getting into the applications we need to get a couple of definitions out of the way. Suppose that we wanted to look at the age of a person, the height of a person, the amount of time spent waiting in line, or maybe the lifetime of a battery. Each of these quantities have values that will range over an interval of integers. Because of this these are called continuous random variables. Continuous random variables are often represented by X. Every continuous random variable, X, has a probability density function, . Probability density functions satisfy the following conditions. 1. for all x. 2. Probability density functions can be used to determine the probability that a continuous random variable lies between two values, say a and b. This probability is denoted by and is given by, Let’s take a look at an example of this. Example 1 Let for and for all other values of x. Answer each of the following questions about this function. (a) Show that is a probability density function. [Solution] (b) Find [Solution] (c) Find [Solution] Solution (a) Show that is a probability density function. First note that in the range is clearly positive and outside of this range we’ve defined it to be zero. So, to show this is a probability density function we’ll need to show that . Note the change in limits on the integral. The function is only non-zero in these ranges and so the integral can be reduced down to only the interval where the function is not zero. (b) Find In this case we need to evaluate the following integral. So the probability of X being between 1 and 4 is 8.658%. (c) Find Note that in this case is equivalent to since 10 is the largest value that X can be. So the probability that X is greater than or equal to 6 is, This probability is then 66.304%. Probability density functions can also be used to determine the mean of a continuous random variable. The mean is given by, Let’s work one more example. Example 2 It has been determined that the probability density function for the wait in line at a counter is given by, where t is the number of minutes spent waiting in line. Answer each of the following questions about this probability density function. (a) Verify that this is in fact a probability density function. [Solution] (b) Determine the probability that a person will wait in line for at least 6 minutes.\ [Solution] (c) Determine the mean wait in line. [Solution] Solution (a) Verify that this is in fact a probability density function. This function is clearly positive or zero and so there’s not much to do here other than compute the integral. So it is a probability density function. (b) Determine the probability that a person will wait in line for at least 6 minutes. The probability that we’re looking for here is . So the probability that a person will wait in line for more than 6 minutes is 54.8811%. (c) Determine the mean wait in line. Here’s the mean wait time. So, it looks like the average wait time is 10 minutes. Hydrostatic Pressure Previous Section Integration Techniques Previous Chapter Next Chapter Parametric Equations and Polar Coordinates [Notes] © 2003 - 2017 Paul Dawkins
# Unit vector – problems and solutions Unit vector – problems and solutions 1. An object moves at a velocity of v = (2i − 1.5j) m/s. What is the displacement of the object after 4 seconds? Known : The horizontal component of the velocity (vx) = 2 m/s The vertical component of the velocity (vy) = 1.5 m/s Time interval (t) = 4 seconds Wanted : Displacement Solution : The resultant of the velocity (v) : Displacement : s = v t = (2.5 m/s)(4 s) s = 10 meters 2. Vector F1 = 14 N and F2 = 10 N. Determine the resultant vector if stated in R = i + j. Solution : The components of vectors : F1x = (F1)(cos 60o) = (14)(0.5) = -7 N (Negative because this vector component points along the negative x axis (leftward)) F1y = (F1)(sin 60o) = (14)(0.5√3) = 7√3 N (Positive because this vector component points along the positive y axis (rightward)) F2x = 10 N F2y = 0 The components of the resultant vectors : Fx = F1x + F2x + F3x = -7 + 10 = 3 N Fy = F1y + F2y + F3y = 7√3 + 0 = 7√3 N The resultant vector in unit vector : See also Motion of two bodies with the same accelerations on the rough horizontal surface with the friction force - problems and solutions R = 3 i + 7√3 j 1. What is a unit vector? Answer: A unit vector is a vector that has a magnitude of 1. It typically represents direction without conveying any information about magnitude. 2. Why are unit vectors important in vector mathematics and physics? Answer: Unit vectors are essential because they provide a standardized way to describe directions. They can be scaled by a magnitude to produce a vector with a desired length in a specific direction. 3. How do you obtain a unit vector from a given vector? Answer: A unit vector in the direction of a given vector can be obtained by dividing the vector by its magnitude. 4. What are the standard unit vectors in Cartesian coordinates, and what are their directions? Answer: The standard unit vectors in Cartesian coordinates are i, j, and k. i points in the direction of the x-axis, j points in the direction of the y-axis, and k points in the direction of the z-axis. 5. Can a unit vector have components other than 1 or -1? Answer: Yes. The components of a unit vector depend on its direction. Only the unit vectors aligned with the coordinate axes (like i, j, k in Cartesian coordinates) will have components of 1, -1, or 0. 6. Is the sum of two unit vectors necessarily a unit vector? Answer: No. The sum of two unit vectors is not generally a unit vector unless the two vectors are collinear and oppositely directed. 7. Can a unit vector be scaled to represent a vector with a different magnitude but the same direction? Answer: Yes. Multiplying a unit vector by a scalar will change its magnitude while keeping its direction the same. 8. What is the magnitude of the cross product of two unit vectors? Answer: The magnitude of the cross product of two unit vectors is equal to the sine of the angle between them. The maximum value is 1 when the vectors are perpendicular, and the minimum is 0 when the vectors are parallel. 9. Why is it that the dot product of two unit vectors gives the cosine of the angle between them? Answer: The dot product formula for two vectors is given by the product of their magnitudes and the cosine of the angle between them. When both vectors are unit vectors, their magnitudes are 1, so the dot product simplifies to just the cosine of the angle. 10. How is the concept of a unit vector extended into non-Cartesian coordinate systems? Answer: In non-Cartesian coordinate systems, like spherical or cylindrical coordinates, there are different unit vectors corresponding to each coordinate direction. For example, in spherical coordinates, the unit vectors are r (radial direction), θ (polar angle direction), and φ (azimuthal direction).
# Factoring Polynomials To view this video please enable JavaScript, and consider upgrading to a web browser that supports HTML5 video Save this PDF as: Size: px Start display at page: ## Transcription 1 UNIT 11 Factoring Polynomials You can use polynomials to describe framing for art. 396 Unit 11 factoring polynomials 2 A polynomial is an expression that has variables that represent numbers. A number can be factored, so you should be able to factor a polynomial, right? Sometimes you can and sometimes you can t. Finding ways to write a polynomial as a product of factors can be quite useful. Big Ideas Expressions, equations, and inequalities express relationships between different entities. the laws of arithmetic can be used to simplify algebraic expressions and equations. Solving an equation means finding values for the variable or variables that make the equation a true statement. a function is a correspondence between two sets, the domain and the range, that assigns to each member of the domain exactly one member of the range. Many events in the physical world can be modeled as functions. Many functions can be described by algebraic expressions. Unit Topics factoring Integers factoring Quadratic Trinomials Dividing Monomials factoring Quadratic Trinomials, a 1 common Factors of Polynomials factoring Completely Dividing Polynomials by Monomials factoring Challenges factoring Perfect Squares finding Roots of a Polynomial factoring Differences of Squares applications: Using Factoring factoring polynomials 397 3 4 Factoring Integers You can factor integers into products of primes. Definition A prime number is an integer greater than one that has exactly two factors, the number itself and 1. Remember The set of integers, 핑 = {..., 2, 1, 0, 1, 2,...}, is the set of natural numbers, their opposites, and zero. Writing a Number as a Product of Prime Numbers Example 1 Write 84 as the product of its prime factors. Method 1 Use a factor tree. Write 84 at the top of your tree. Write 84 as the product of any two of its factors. Then write each factor that is not prime as the product of two factors. Continue until all the leaves on the factor tree are prime numbers. 84 Remember A factor is a number or variable multiplied by one or more numbers or variables to get a product Now write 84 as the product of its prime factors found in the factor tree: 84 = In exponential form, this is: 84 = Method 2 Use a division ladder. Keep dividing 84 by prime numbers until the divisor and the quotient are both prime numbers Divide 84 by the prime number 2. Write the quotient 42 below Divide 42 by the prime number 2. Write the quotient 21 below Divide 21 by the prime number 3. Write the quotient 7 below Since 7 is a prime number, the factorization is complete. The prime factorization of 84 is = factoring integers 399 5 Composite Numbers A number that is not a prime number is a composite number. A composite number is an integer greater than one that has more than two factors. For example, 4 has factors of 1, 2, and 4, so it is a composite number. Example 2 Determine whether each number is prime or composite. A. 57 The factors of 57 are 1, 3, 19, and 57, so 57 is a composite number. B. 101 The factors of 101 are 1 and 101, so 101 is a prime number. Finding the GCF Definitions A common factor is a factor shared by two or more numbers. The greatest common factor (GCF) of two or more numbers is the greatest of the common factors of the numbers. Example 3 Find the GCF of 72 and 48. Method 1 List the factors of each number and find the greatest of the common factors. Factors of 72: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72 Factors of 48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48 The greatest common factor is 24. Method 2 Find the prime factorization in exponential form of each number. 72 = = Look for the smallest common power of each prime factor and multiply: 23 3 = Unit 11 factoring polynomials 6 Determining Whether Numbers Are Relatively Prime Two numbers are relatively prime if they share no common factors other than 1. Example 4 Are 28 and 55 relatively prime? Find the prime factorization of each number and see if they share any common prime factors. 28 = = = 5 11 Since 28 and 55 share no common prime factors, they are relatively prime. Application: Mersenne Primes A Mersenne prime is a number of the form M = 2n 1, where n and M are prime numbers. Example 5 Find the first five Mersenne primes. Substitute consecutive prime numbers into the equation for n. Prime Number, n Apply Mersenne Equation: M = 2n 1 Mersenne Prime? 2 M = 22 1 = 4 1 = 3 3 M = 23 1 = 8 1 = 7 5 M = 25 1 = 32 1 = 31 7 M = 27 1 = = M = = = M = = = 8191 Yes Yes Yes Yes No 2047 = Yes Think about it Not all Mersenne numbers are Mersenne primes. A Mersenne number is any number of the form 2n 1, where n is an integer. A Mersenne prime is a number of the form M = 2n 1, where n and M are prime numbers. Think about it As n increases, Mersenne primes become very difficult to compute. Only 44 Mersenne primes have been discovered, the last in factoring integers 401 7 Problem Set Write each number as the product of its prime factors Determine whether each number is prime or composite Find the greatest common factor (GCF) of each set of numbers and , 24, and and , 63, and and and 12. Challenge and and 196 Determine whether each pair of numbers is relatively prime. If they are not, find the GCF and and and and and and and 245 Solve. 28. Challenge Find the 6th Mersenne prime. 402 Unit 11 factoring polynomials 8 n factors} Dividing Monomials Divide monomials using the quotient of powers property. The quotient of powers property is just one of the many properties of exponents. To apply the properties of exponents, you will need to understand integer exponents. Definition of integer exponents For any a 핉: 1. If n is a positive integer, a n = a a a... a. 2. If n = 0 and a 0, a n = For any n and nonzero a, a n = 1 a n. The following table lists properties of exponents. Properties of exponents Let a 핉, b 핉, a 0, b 0, m 핑, n 핑. Name Definition Examples Product of Powers a m a n = a (m+n) = 22+3 = 25 x 3 x 2 = x 3+( 2) = x 1 = x Remember The letter 핉 denotes the set of real numbers. The letter 핑 denotes the set of integers. Quotient of Powers a m a n = a(m n) x 9 x 3 = x(9 3) = x 6 b 7 b 10 = b(7 10) = b 3 = 1 b 3 Power of a Product (ab) n = a n b n (5x) 3 = 53 x 3 = 125x 3 Power of a Quotient ( a b ) n = a n bn 2 25 ( x ) 5 = x5 = 32 x 5 Power of a Power (a m ) n = a mn (22 ) 3 = 2(2 3) = 26 = 64 dividing monomials 403 9 Dividing Monomials Example 1 Find each quotient. A. x 5 x 2 x 5 x 2 = x(5 2) Quotient of Powers Property = x 3 Simplify. B. 9a 3 b 4 c 3a 2 b 9a 3 b 4 c 3a 2 b = 3a(3 2) b (4 1) c Quotient of Powers Property = 3ab 3 c Simplify. Think about it For Example 1C, the answer is not a monomial because the variable does not have a whole number exponent. C. 15x 3 5x 9 15x 3 5x 9 = 3x(3 9) Quotient of Powers Property = 3x 6 Simplify. 1 = 3 x 6 3 = x 6 Multiply. D. ( 2xy 4x 2 ) 2 ( 2xy 2 4x 2 ) (2xy) 2 = (4x 2 ) 2 Definition of negative integer exponents Power of a Quotient Property 22x 2 y 2 = Power of a Power Property 4 2 x 2 2 4x 2 y 2 = 16x 4 Simplify. 1 = 4 x(2 4) y 2 Quotient of Powers Property 1 = 4 x 2 y 2 Simplify. 1 = 4 1 x 2 y2 Definition of negative integer exponents y 2 = 4x 2 Multiply. 404 Unit 11 factoring polynomials 10 Application: Ratio Example 2 A. What is the ratio of the volume of a cube to its surface area? s The volume of a cube with side s is s 3. The surface area of a cube is the sum of the areas of the six sides or 6s 2. The ratio of the volume of a cube to its surface area is s3. Simplify the ratio by dividing. 6s 2 s 3 6s 2 = s (3 2) 6 s = 6 Simplify. Quotient of Powers Property The ratio of the volume of a cube to its surface area is s 6. B. What is the ratio of the circumference of a circle to its area? r The circumference of a circle with radius r is 2πr. The area of a circle with radius r is πr 2. The ratio of the circumference of a circle to its area is represented by: 2πr πr 2 = 2r(1 2) Quotient of Powers Property = 2r 1 Simplify. 1 = 2 r 1 2 = r Definition of negative integer exponents Multiply. The ratio of the circumference of a circle to its area is 2 r. Problem Set Identify the property. 1. (7a) 2 = 7 2 a 2 2. (4 3 ) 3 = 4 (3 3) = ( a 3 ) 4 = a = (2ab 3 ) 2 = 2 2 a 1 2 b 3 2 dividing monomials 405 11 Find each quotient a 7 a 4 x 9 x 2 y 3 y a5 4a m 2 m x2 y 4 6xy 2 Solve b 6 18b x3 y 2 z 8 x 2 yz a 2 b 9 c 2 7a 6 b a 2 b 8 a 4 b r3 s 5 8r 5 s a2 b 3 c 4 a 4 b 3 c ( 3x4 y 2 6x 3 y 3 ) Challenge 22. 23. 4x 5 y 3 z 2 8x 2 y 4 z 3 9x 6 y 3 6x 2 yz 3 9x 2 y 24x 2 y 6 24. ( 4x2 y 5 z 2 32x 4 y 6 z 4 ) What is the ratio of the area of this triangle to its perimeter? 7x 26. A group of 6p people are renting a vacation house together. The price of the rental is 3p 2. If they split the cost evenly, what is each member s share of the price? 27. Barker s age can be represented by the expression 4x 2. 8x A. Karin s age can be represented by the expression 2x 4.What is the ratio of Karin s age to Barker s age? B. Penny s age can be represented by the expression 3x 3. What is the ratio of Barker s age to Penny s age? 5x 7x 28. Nasser has saved 2s 2 t dollars. He wants to save a total of 4rs3 t 2 dollars. What fraction of his total savings goal has he achieved so far? 29. Challenge Hamilton bought a car for 7s 2 t 3 dollars. He paid s 2 t 3 dollars as a down payment. Find the number of months it would take him to pay off the car completely if he made each of the following monthly payments. A. 3s 2 t dollars B. 4st 2 dollars 30. Challenge What is the ratio of the surface area of a sphere to its volume? 406 Unit 11 factoring polynomials 12 Common Factors of Polynomials When every term of a polynomial shares a common monomial factor, you can factor out the common factor from all the terms. Identifying the Greatest Common Monomial Factor The greatest common factor of the expression is 9, since 9 is the greatest factor that evenly divides 18 and 27. Finding the greatest common monomial factor of polynomials with variables is similar to finding the greatest common factor of numerical expressions. finding the greatest common monomial factor Step 1 Write each term of the polynomial as the product of its prime factors and look for the smallest power of each prime factor common to each term. Step 2 multiply the common factors found in Step 1. Remember A factor of a monomial term can be a number or a monomial. Example 1 Find the greatest common monomial factor of each polynomial. A. 8x x 2 Step 1 8x 5 = 2 3 x 5 12x 2 = x 2 Step 2 Multiply 2 2 and x 2. The greatest common monomial factor is 4x 2. B. 15a 3 b + 25a 2 bc 100a 5 b 4 c 2 Step 1 15a 3 b = 3 5 a 3 b 25a 2 bc = 52 a 2 b c 100a 5 b 4 c 2 = a 5 b 4 c 2 Step 2 Multiply 5, a 2, and b: 5 a 2 b = 5a 2 b. The greatest common monomial factor is 5a 2 b. common factors of polynomials 407 13 Think about it You can check your work by multiplying the greatest common monomial factor by the new polynomial to get the original polynomial. Factoring Out the Greatest Common Monomial Factor Since 9 is the greatest common factor of the expression , you can use the distributive property to factor it out: = 9(2 + 3). Example 2 Factor the polynomial 14a 3 b + 7a 2 b ab 3. Step 1 Find the greatest common monomial factor. 14a 3 b = 2 7 a 3 b 7a 2 b 2 = 7 a 2 b 2 49ab 3 = 7 2 a b 3 The greatest common monomial factor is 7ab. Step 2 Rewrite each term of the polynomial as a product of the greatest common monomial factor. 14a 3 b + 7a 2 b ab 3 = 7ab 2a 2 + 7ab ab + 7ab 7b 2 Step 3 Use the distributive property to factor out the greatest common monomial factor. 7ab 2a 2 + 7ab ab + 7ab 7b 2 = 7ab(2a 2 + ab + 7b 2 ). You can also divide each term of a polynomial by its greatest common monomial factor to factor out the greatest common monomial factor. Example 3 Factor the polynomial 12xy4 21x 2 y 5 z + 36xy3. Step 1 Find the greatest common monomial factor. 12xy 4 = x y 4 21x 2 y 5 z = 3 7 x 2 y 5 z 36xy 3 = x y 3 The greatest common monomial factor is 3xy3. Step 2 Factor out the greatest common monomial factor. Divide each term of the polynomial by the greatest common monomial factor. 12xy 4 21x 2 y 5 z + 36xy 3 12xy 4 = 3xy 3 3xy 3 21x 2 y 5 z 3xy xy 3 3xy 3 Step 3 = 4x (1 1) y (4 3) 7x (2 1) y (5 3) z + 12x (1 1) y (3 3) = 4y 7xy 2 z + 12 Multiply the greatest common monomial factor by the result in Step 2: 12xy 4 21x 2 y 5 z + 36xy 3 = 3xy 3 (4y 7xy 2 z + 12). 408 Unit 11 factoring polynomials 14 Factoring Out a Binomial Factor Example 4 Factor 7x(x + 3) 5(x + 3). Notice that both terms of the polynomial contain the binomial factor (x + 3). Step 1 To factor out the common binomial factor, divide each term by the common binomial factor: 7x(x + 3) 5(x + 3) 7x(x + 3) = (x + 3) (x + 3) 5(x + 3) (x + 3) = 7x 5 Step 2 Multiply the common binomial factor by the result in Step 1: 7x(x + 3) 5(x + 3) = (x + 3)(7x 5). Application: Geometry Example 5 A rectangle has an area of 6x + 4 square feet and a width of 2 feet. What is the length of the rectangle? A = lw Write the formula for the area of a rectangle. 6x + 4 = l 2 Substitute 2 for w and 6x + 4 for A. (3x + 2) 2 = l 2 Factor 2 from 6x + 4. } l w So, the length of the rectangle is 3x + 2 feet. Problem Set Find the greatest common monomial factor. 1. 4x x a a y y 3 Factor. 7. 3x 2 + 6x 8. 4x + 8y 9. 2a 2 + a 10. 9r + 12s 11. 5m m 12. 4p 3 + 2p x 3 y x 2 y m 3 x m 5 y m 3 x 6 y x 3 y 5 z 2 98x 2 yz x 2 y a 3 b 6 c a 2 b 3 24b 2 c m 5 n 5 15m 3 n x 3 y 4 12x 2 y a 3 bc 3 6a 2 b 2 c r 5 s + 4r 4 s 2 8r 2 s x 4 y 5 z + 28x 2 y 4 z x(2x 4) + 5(2x 4) 20. 6m(3m 2) 4(3m 2) common factors of polynomials 409 15 Challenge 21. 45m 4 n 5 165m 5 n 5 p + 75m 3 n 6 p 4 105m 3 n 5 22. ( 14xy 28y) + (6x 2 y + 12xy) 23. (6r 5 s 2 t 3 15r 6 s 2 t r 5 s 2 ) (8t 4 20rt t) + (12t 3 30rt ) Solve. 24. A rectangle has an area of 16x 12 square centimeters and a length of 4 centimeters. What is the width of the rectangle? 25. A triangle has a height of 6 inches and an area of 18y + 3 square inches. What is the length of the base of the triangle? 26. A parallelogram has a base of 5 centimeters and an area of 12a 10 square centimeters. What is the height of the parallelogram? 27. Marta is planning a rectangular shaped perennial garden with a length of 9 feet. Find the width for each of the possible areas of Marta s garden. A x square feet B x square feet 28. Jacqui s monthly cell phone bill includes additional minutes at 3x per minute. How many additional minutes did Jacqui use for each month? A. charge for additional minutes in January: 240x + 6x 3 B. charge for additional minutes in July: 189x + 9x Unit 11 factoring polynomials 16 Dividing Polynomials by Monomials Dividing a polynomial by a monomial is a good application of the distributive property. When you have a sum of terms in the numerator of a fraction, the distributive property allows you to separate the fraction into the sum or difference of quotients. a + b a b c = c + c and a c b a b = c c You can use these properties to divide polynomials by monomials. Dividing a Polynomial by a Monomial Example 1 Find each quotient. A. (21y y 3 ) 3y (21y y 3 21y y 3 Write the division problem as ) 3y = 3y a fraction. 21y 4 = 3y + 18y 3 3y Divide each term by the denominator. = 7y 3 + 6y 2 Simplify each term. When dividing a polynomial by a monomial, it is always possible to check if your answer is correct by using related equations. You know that 6 2 = 3, 6 3 = 2, and 2 3 = 6. The same concept holds true for polynomials. If n(x), d(x), and q(x) are polynomials and n(x) = q(x), then n(x) = q(x) d(x). If d(x) you multiply the quotient found by the divisor, your answer should be the dividend. You can always make sure you have the correct solution using this answer check method. 21y y 3 Check If 3y = 7y 3 + 6y 2, then 21y y 3 = (7y 3 + 6y 2 ) 3y. 21y y 3 = (7y 3 + 6y 2 ) 3y 21y y 3 = 3y 7y 3 + 3y 6y 2 21y y 3 = 21y y 3 The answer is correct. Remember Use the quotient of powers property of exponents to simplify the variables. Tip If n(x) d(x) = q(x), then n(x) = q(x) d(x). (continued) dividing polynomials by monomials 411 17 B. (25a 6 b 5 c + 20a 3 b 3 60a 2 b 2 ) 5a 2 b 2 (25a 6 b 5 c + 20a 3 b 3 60a 2 b 2 ) 5a 2 b 2 25a = 6 b 5 c + 20a 3 b 3 60a 2 b 2 5a 2 b 2 25a = 6 b 5 c 5a 2 b a 3 b 3 5a 2 b 2 60a 2 b 2 5a 2 b 2 Write the division problem as a fraction. Divide each term by the denominator. = 5a 4 b 3 c + 4ab 12 Simplify each term. Application: Geometry Example 2 The area of a rectangle is 2x 2 + 6x square inches. The length of the rectangle is 2x inches. What is the width of the rectangle?? 2x inches Step 1 Use the formula for the area of a rectangle to write an equation. A = lw 2x 2 + 6x = 2x w Substitute 2x 2 + 6x for A and 2x for l into the equation. Step 2 Solve the equation for w by dividing each side by 2x. 2x 2 + 6x 2x w 2x = 2x 2x 2 + 6x 2x = w Step 3 Simplify. 2x 2 + 6x 2x = w 2x 2 2x + 6x 2x = w Divide each term by the denominator. x + 3 = w Simplify each term. The width of the rectangle is x + 3 inches. Step 4 Check. If 2x2 + 6x 2x = x + 3, then 2x 2 + 6x = 2x (x + 3). 2x 2 + 6x = 2x (x + 3) 2x 2 + 6x = 2x x + 2x 3 2x 2 + 6x = 2x 2 + 6x The answer is correct. 412 Unit 11 factoring polynomials 18 Problem Set Find the quotient. 1. (x 2 + 2x) x 2. (9b 2 3b) b 3. (4a 3 + 2a 2 ) 2a 4. (x 6 x 5 ) x 2 5. (3y 2 + 6y) 3y 6. (16x 9 + 8x 6 ) x 3 7. (12y y 2 ) 2y 8. (f 3 4f 6 ) f 3 9. (1.4a a 4 ) 0.07a (50n n 22 ) 5n (9t t 5 ) 3t (32h 12 48h 18 ) 16h (2a 3 b + 6a 4 b 2 ) 2a (14p 2 q 2 + 7p 6 q 4 ) 7pq 15. (24g 8 h 9 20g 4 h 3 ) 4g 3 h 16. (18v 14 w 12 36v 12 w 14 ) 3v 2 w (8y 10 z 4 + y 10 z 6 ) y 5 z ( 3 2 m2 n m4 n 2 ) 2mn (50u 3 v u 6 v u 6 v 6 ) 25v (4l 3 w 7 2l 5 w l 2 w 2 ) 2l 2 w 21. (3p 4 q 4 r p 2 q 7 r 3 + 9p 3 q 9 r 6 ) 3q (42a 12 b 16 c a 23 b 19 c a 14 b 31 c 27 ) 7a 8 b (15f 7 g 9 h f 6 g 2 h f 4 gh 8 ) 5fh (2.7l 2 m 3 n 5 18l 4 m 6 n l 5 m 4 n 2 ) 0.9lmn 25. ( 42hj 7 k 9 28h 3 j 5 k h 4 j 2 k 12 ) 14hj 2 k ( 1 2 x5 y 4 z x6 y 4 z x5 y 5 z 2 2 ) 3 x 5 y 4 z 2 Solve. 27. Challenge A. A rectangular tile has width 3a and length 2ab. Find the area of the rectangular tile. B. Find an expression for the number of rectangular tiles needed to cover an area of 12a 5 b a 4 b a 7 b 6. 28. Challenge A. A rectangular tile has width 6p 4 q 3 r and length 3p 2 q 7 r 9. Find the area of the rectangular tile. B. Find an expression for the number of rectangular tiles needed to cover an area of 36p 16 q 12 r p 10 q 13 r p 8 q 16 r 12. dividing polynomials by monomials 413 19 Factoring Perfect Squares Remember The pattern for squaring a binomial is (a + b) 2 = a 2 + 2ab + b 2. When you square a binomial, the result is a perfect square trinomial. You can use this perfect square pattern to work the other way and find the squared binomial for a given trinomial. When you expand the square of a binomial, you get a perfect square trinomial because it can be written as the square of a binomial, rather than the product of two different binomials. You can use the patterns of perfect squares to help you factor perfect square trinomials. Identifying a Perfect Square perfect square trinomial patterns a 2 + 2ab + b 2 = (a + b) 2 a 2 2ab + b 2 = (a b) 2 Note that for a trinomial to be a perfect square, the first and last terms have to be perfect squares and the middle term has to equal twice the square root of the first and third terms. Example 1 Determine if each is a perfect square trinomial. A. x 2 4x + 4 Try to write the polynomial in the form a 2 + 2ab + b 2 or a 2 2ab + b 2. Substitute x for a and 2 for b. So, x 2 4x + 4 = x 2 2 x The trinomial x 2 4x + 4 is a perfect square. B. x 2 1 Since there is no x term, the polynomial x 2 1 is not a perfect square. C. x 2 + 7x + 25 To write the trinomial in the form a 2 + 2ab + b 2, substitute x for a and 5 for b. The middle term 2ab must then be 2 x 5 or 10x. Since the middle term is 7x, the trinomial is not a perfect square. 414 Unit 11 factoring polynomials 20 D. 4x x 100 When you square a binomial, the sign of the constant term is always positive. Since the constant in the trinomial 4x x 100 has a negative sign, the trinomial is not a perfect square. Factoring a Perfect Square Example 2 A. x 2 + 4x + 4 Factor each trinomial. Write the polynomial in the form a 2 + 2ab + b 2. x 2 + 4x + 4 = x x Substitute x for a and 2 for b. B. 4x 2 20xy + 25y 2 = (x + 2) 2 Use the pattern (a + b) 2. 4x 2 20xy + 25y 2 = (2x) 2 2 2x 5y + (5y) 2 Substitute 2x for a and 5y for b. = (2x 5y) 2 Use the pattern (a b) 2. C. 4x x x x = (2x) x 3 + ( 1 3 ) 2 Substitute 2x for a and 1 3 for b. 1 = ( 2x 3 ) 2 Use the pattern (a b) 2. Problem Set Determine if the expression is a perfect square. If it is, factor it. 1. x 2 + 2x s 2 + s k c 2 6c j j n 2 26n y 2 16y x 2 + 8x a 2 4a x 2 + 4x z z y 2 + y f f p p z z u u d 2 + 8d w w m m q q t 2 22t x + 81x x y w w h2 4h r rs + 100s u uv + 16v x 2 36xy + 4y y2 121 yz z c 2 4.2cd + d 2 factoring perfect squares 415 21 Solve. 32. Challenge A. Factor 4a 2 b abc + 9c 2. B. Verify that 4a 2 b abc + 9c 2 9a 2 b abc + 4c 2. 33. Challenge A. Factor 36j j B. Verify your answer by factoring 36f f + 25, then substituting f = j 2 into your answer. 416 Unit 11 factoring polynomials 22 Factoring Differences of Squares You can use a pattern to factor a difference of squares. Conjugate binomials, (a + b) and (a b), are two binomials with the same terms but opposite signs. The product of conjugate binomials, a 2 b 2, is called the difference of squares. Identifying a Difference of Squares Example 1 A. x 2 1 Determine if each is the difference of squares. Try to write the polynomial in the form a 2 b 2. Since x 2 1 = x 2 1 2, it is the difference of squares. B. x Since the terms are added, the binomial is the sum of two squares, not the difference of two squares. C. x 4 9 Try to write the polynomial in the form a 2 b 2. Use the power of a power property of exponents to rewrite x 4. Since x 4 9 = (x 2 ) 2 3 2, it is the difference of squares. Tip The word difference in difference of squares means one square term is subtracted from the other square term. Remember The power of a power property states (a m ) n = a mn. Factoring the Difference of Two Squares You can use the sum and difference pattern to help you factor any binomial that is the difference of squares. DIFFERENCE OF SQUARES PATTERN a 2 b 2 = (a + b)(a b) Example 2 Completely factor each difference of squares. A. 144x 2 z 2 First, write the polynomial in the form a 2 b 2. Then apply the difference of squares pattern. 144x 2 z 2 = (12x) 2 z 2 Substitute 12x for a and z for b. = (12x + z)(12x z) Difference of squares pattern (continued) factoring differences of squares 417 23 Tip Always look for perfect squares when a binomial has a minus sign and even powers of variables. B. p 12 q 20 p 12 q 20 = (p 6 ) 2 (q 10 ) 2 Power of a Power Property = (p 6 + q 10 )(p 6 q 10 ) Difference of squares pattern The binomial p 6 q 10 is also a difference of squares so the expression can be factored further. (p 6 + q 10 )(p 6 q 10 ) = (p 6 + q 10 )[(p 3 ) 2 (q 5 ) 2 ] p 6 and q 10 are perfect squares. = (p 6 + q 10 )(p 3 + q 5 )(p 3 q 5 ) Difference of squares pattern C. 4x x = (2x)2 ( 2 3 ) 2 Substitute 2x for a and 3 2 for b. 2 = ( 2x + 3 ) ( 2x 2 3 ) Difference of squares pattern Remember x in. 4.5 in. The formula for the area of a square is A = s 2. Application: Geometry Example 3 A picture frame is created by cutting a 4.5-inch square piece from a square piece of plywood. A. Write a polynomial expression that represents the area of the picture frame after the square piece is removed. B. Write the area as the difference of squares. A. Find the area of the original piece of plywood and subtract the area of the piece that was removed. Area of Picture Frame, A = Area of Original Piece of Plywood Area of Removed Square Piece A = x 2 (4.5) 2 A = x 2 (4.5) 2 = x The area of the picture frame is x inches. B. Factor the binomial found in part A. x = x 2 (4.5) 2 Substitute x for a and 4.5 for b. = (x + 4.5)(x 4.5) Difference of squares pattern 418 Unit 11 factoring polynomials 24 Problem Set Determine if the expression is a difference of squares. If it is, factor it. 1. x u a x n f d s z p r k x b q h y w t m c 2 9d x 2 81y j 2 25k p 36 36q y z r4 25 s a b f g12 Solve. 29. Keena s framed bathroom mirror is square and has a side length of 36 inches. The mirrored portion measures x inches on each side. A. Write a polynomial expression that represents the area of the mirror frame. B. Write the area as a difference of squares. 30. A small circle has a radius of 9 π. A larger circle has a radius of x π. A. Write a polynomial expression that represents how much greater the area of the larger circle is than the smaller circle. B. Write this expression as a difference of squares. 31. A small rectangle has length n 3 inches and width 0.25n 5 inches. A larger rectangle has length 0.8 inches and width 0.2 inches. A. Write a polynomial expression that represents how much greater the area of the larger rectangle is than the smaller rectangle. B. Write this expression as a difference of squares. 32. Remy is painting a mural on a wall that is 12 feet high and 27 feet wide. A. Write a polynomial expression that represents the area he has left to cover once he has painted an x by x square of the wall. B. Write this expression as a difference of squares. 33. Challenge A. Determine if 4a 4 b 6 14c 11 d 8 is a difference of squares. B. If so, factor it. If not, what change would make it a difference of squares? 34. Challenge A. Determine if 4.9t 2 u v 14 w 6 is a difference of squares. B. If so, factor it. If not, what change would make it a difference of squares? factoring differences of squares 419 25 Factoring Quadratic Trinomials You learned how to multiply two binomials using the distributive property and the FOIL method. You can reverse this process to factor a trinomial into two binomial factors. The trinomial x 2 + bx + c can be factored into the product (x + d)(x + e) if d + e = b and de = c. FACTORING x 2 + bx + c If c > 0 and b > 0, then the product can be written as (x + )(x + ). If c > 0 and b < 0, then the product can be written as (x )(x ). If c < 0, then the product can be written as (x + )(x ). Factoring Trinomials of the Form x 2 + bx + c Example 1 Factor each trinomial. A. x 2 + 5x + 6 Find two binomials whose product is x 2 + 5x + 6. Both the x term and the constant term are positive so the binomials will have the form (x + )(x + ). The coefficient of the x term is 5 and the constant term is 6, so you need to find two positive factors of 6 whose sum is 5. Positive Factors of 6 Sum of the Factors 1, 6 7 2, 3 5 Tip Use the FOIL method to check your answer when factoring. The factors 2 and 3 have a sum of 5, so x 2 + 5x + 6 = (x + 2)(x + 3). Check x 2 + 5x + 6 (x + 2)(x + 3) x 2 + 5x + 6 x x + x x FOIL x 2 + 5x + 6 x 2 + 3x + 2x + 6 Multiply. x 2 + 5x + 6 = x 2 + 5x + 6 Combine like terms. The answer is correct. 420 Unit 11 factoring polynomials 26 B. x 2 7x 8 The constant term is negative so the factors have opposite signs. The binomials will have the form (x + )(x ). The coefficient of the x term is 7 and the constant term is 8, so you need to find two factors of 8 whose sum is 7. Tip If the signs of c (the constant term) and b (the coefficient of the x term) are negative, the factors will have opposite signs. Factors of 8 Sum of the Factors 1, 8 7 1, 8 7 2, 4 2 2, 4 2 The factors 1 and 8 have a sum of 7, so x 2 7x 8 = (x + 1)(x 8). C. x 2 + 6x 27 The constant term is negative so the binomials will have the form (x + )(x ). The coefficient of the x term is 6 and the constant term is 27, so you need to find two factors of 27 whose sum is 6. Factors of 27 Sum of the Factors 1, , , 9 6 3, 9 6 The factors 3 and 9 have a sum of 6, so x 2 + 6x 27 = (x + 9)(x 3). D. x 2 9x + 20 The x term is negative and the constant term is positive so the factors will have the same sign as the x term. The binomials will have the form (x )(x ). The coefficient of the x term is 9 and the constant term is positive 20, so you need to find two negative factors of 20 whose sum is 9. Negative Factors of 20 Sum of the Factors 1, , , 5 9 The factors 4 and 5 have a sum of 9, so x 2 9x + 20 = (x 4)(x 5). factoring quadratic trinomials 421 27 Prime Polynomials Just as a prime number cannot be broken down into the product of two other numbers, a prime polynomial cannot be broken down into the product of two other polynomials. Remember A prime number is a positive integer greater than 1 that has exactly two factors, the number itself and 1. Definition A polynomial that cannot be factored is a prime polynomial. In this book, all polynomials have rational coefficients. For our purposes, a polynomial is factorable if you can write it as the product of at least two other polynomials with rational coefficients. Example 2 Factor x 2 + 8x + 9. Both the x term and the constant term are positive so the binomials will have the form (x + )(x + ). The coefficient of the x term is 8 and the constant term is 9, so you need to find two positive factors of 9 whose sum is 8. Positive Factors of 9 Sum of the Factors 1, , 3 6 Since there are no factors of 9 whose sum is 8, the trinomial is prime. Finding a Value to Make a Polynomial Factorable Example 3 Find the value of k, for k > 0, that makes each polynomial factorable. A. x 2 + kx + 7 Both the x term and the constant term are positive so the binomials will have the form (x + )(x + ). The only factors of 7 are 1 and 7, so the only possible factorization is (x + 1)(x + 7). (x + 1)(x + 7) = x 2 + 7x + x FOIL = x 2 + 8x + 7 Combine like terms. So, k = 8. B. x 2 kx + 10 The x term is negative and the constant is positive, so the binomials will have the form (x )(x ). 422 Unit 11 factoring polynomials 28 The factor pairs of 10 are 1 and 10, and 2 and 5. The possible factorizations are (x 1)(x 10) and (x 2)(x 5). (x 1)(x 10) = x 2 10x x 1 ( 10) (x 2)(x 5) = x 2 5x 2x 2 ( 5) So, k = 11 or k = 7. = x 2 11x + 10 = x 2 7x + 10 Problem Set Factor. If the polynomial cannot be factored, write prime. 1. x 2 + 7x x 2 5x x 2 6x x x x 2 + 5x x x x 2 + 5x x 2 16x x x x x x 2 + 9x x 2 11x x 2 + 9x x 2 x x 2 + 5x x 2 + 8x x x Find the value of k that makes each equation true. 18. x 2 + kx + 5 = (x + 5)(x + 1) 19. x 2 kx 11 = (x 11)(x + 1) 20. x 2 kx 23 = (x 23)(x + 1) 21. x 2 + kx 3 = (x + 3)(x 1) Find the value of k, for k > 0, that makes each equation factorable. 22. x 2 + kx x 2 kx x 2 kx 17 25. Challenge x 2 + kx + 8 Solve. 26. Challenge If the area of Bianca s rectangular backyard patio is represented by the expression x 2 18x + 81, and the length of the patio is represented by the expression x 9, what is the expression for the width of the patio and what special rectangular shape is the patio? 27. Challenge If the area of Susie s rectangular backyard garden is represented by the expression x x + 64, and the length of the garden is represented by the expression x + 8, what is the expression for the width of the garden and what special rectangular shape is the garden? factoring quadratic trinomials 423 29 Factoring Quadratic Trinomials, a 1 You may need to factor a trinomial where the coefficient of the x 2 term is not equal to 1. You learned how to factor trinomials of the form x 2 + bx + c. Now you will learn how to factor trinomials of the form ax 2 + bx + c, where a 1. Factoring Trinomials of the Form ax 2 + bx + c When factoring trinomials of the form ax 2 + bx + c, it is not enough to consider only the factors of the constant. You must also consider the factors of the coefficient of the x 2 term. The rules for the signs of the binomial factors are the same as for trinomials of the form x 2 + bx + c. Example 1 Factor each trinomial. A. 2x 2 + 7x + 3 You need to find two binomials whose product is 2x 2 + 7x + 3. Both the x term and the constant term are positive so the binomials will have the form ( x + )( x + ). You must think about the order of the factors of the constant term or the order of the factors of the x 2 term because different orders result in different coefficients for the x term. Positive Factors of 2 (coefficient of x 2 ) Positive Factors of 3 (constant term) Possible Factorization Product of the Factors 1, 2 1, 3 (x + 1)(2x + 3) 2x 2 + 5x + 3 1, 2 3, 1 (x + 3)(2x + 1) 2x 2 + 7x + 3 2x 2 + 7x + 3 = (x + 3)(2x + 1) 424 Unit 11 factoring polynomials 30 B. 3x 2 6x + 5 The x term is negative and the constant term is positive so the binomials will have the form ( x )( x ). Positive Factors of 3 (coefficient of x 2 ) Negative Factors of 5 (constant term) Possible Factorization Product of the Factors 1, 3 1, 5 (x 1)(3x 5) 3x 2 8x + 5 1, 3 5, 1 (x 5)(3x 1) 3x 2 16x + 5 Since there are no possible factorizations that produce the correct product, the trinomial 3x 2 6x + 5 is prime. C. 4x 2 12x 7 The constant term is negative so the binomials will have the form ( x + )( x ) or ( x )( x + ). Remember A prime polynomial is a polynomial that cannot be factored. Positive Factors of 4 (coefficient of x 2 ) Factors of 7 (constant term) Possible Factorization Product of the Factors 1, 4 1, 7 (x 1)(4x + 7) 4x 2 + 3x 7 1, 4 1, 7 (x + 1)(4x 7) 4x 2 3x 7 1, 4 7, 1 (x 7)(4x + 1) 4x 2 27x 7 1, 4 7, 1 (x + 7)(4x 1) 4x x 7 2, 2 1, 7 (2x 1)(2x + 7) 4x x 7 2, 2 1, 7 (2x + 1)(2x 7) 4x 2 12x 7 4x 2 12x 7 = (2x + 1)(2x 7) Finding a Value to Make a Polynomial Factorable Example 2 Find the values of k that make 5x 2 + kx + 7 factorable. Both the x term and the constant term are positive so the binomials will have the form ( x + )( x + ). Since 5 and 7 are both prime, the only possible factorizations are (5x + 1) (x + 7) or (x + 1)(5x + 7). Find the product of each pair of binomials. (5x + 1)(x + 7) = 5x x + x (x + 1)(5x + 7) = 5x 2 + 7x + 5x = 5x x + 7 = 5x x + 7 The trinomial is factorable when k = 36 or k = 12. factoring quadratic trinomials, a 1 425 31 Problem Set Factor. If the polynomial is not factorable, write prime. 1. 3x x x x x 2 4x x 2 14x x x x 2 + 4x 4 4. x 2 + 2x x 2 + 6x x 2 8x x 2 + 7x x x x x 27 Find the values of k that make each polynomial factorable x 2 + kx x 2 + kx x 2 + kx x 2 + kx x 2 + kx x 2 + kx x 2 7x x 2 + 3x x x x 2 13x x 2 67x x 2 11x x 2 + kx x 2 + kx Unit 11 factoring polynomials 32 Factoring Completely Factoring a trinomial into two binomials does not necessarily mean that the trinomial is completely factored. Sometimes you will need to repeatedly factor a polynomial until it is factored completely. A polynomial is factored completely when it is written as the product of prime polynomials with integer coefficients. It is not always easy to know if a polynomial is completely factored. Here are some suggestions: tips for factoring polynomials completely 1. Look for common monomial factors. 2. look for trinomials that are perfect square trinomials and for other trinomials that are not prime. 3. Look for binomials that are the difference of two squares. Factoring Polynomials Completely Example 1 Factor each trinomial. A. 3x x + 27 The terms in the trinomial have a common monomial factor 3. First factor out the 3, then factor the trinomial into two binomials. 3x x + 27 = 3(x 2 + 6x + 9) Factor out 3. = 3(x + 3)(x + 3) Factor the perfect square trinomial. = 3(x + 3) 2 Simplify. B. x 4 1 The binomial x 4 1 is the difference of two squares. x 4 1 = (x 2 ) 2 (1)2 Write x 4 1 as the difference of two squares. = (x 2 1)(x 2 + 1) Factor. = (x 1)(x + 1)(x 2 + 1) x 2 1 is also the difference of two squares. Factor it into two binomials. (continued) Remember The binomial x is prime. factoring COMPLETELY 427 33 C. x 2 4x + 4 z 2 trinomial. The first three terms of the polynomial form a perfect square x 2 4x + 4 z 2 = (x 2 4x + 4) z 2 Use the associative property to group the perfect square trinomial. = (x 2)(x 2) z 2 Factor the perfect square trinomial. = (x 2) 2 z 2 Simplify. The resulting polynomial is the difference of two squares and can be factored further. (x 2) 2 z 2 = [(x 2) z][(x 2) + z] Factor the difference of two squares. = (x z 2)(x + z 2) Simplify. D. 4x x + 24 The terms in the trinomial have a common monomial factor 4. First factor out the 4, then factor the trinomial into two binomials. 4x x + 24 = 4(x 2 + 5x + 6) Factor out the common monomial factor 4. = 4(x + 2)(x + 3) Factor the trinomial. Problem Set Factor completely. 1. 2x x x 2 12x x x 3 27x 5. 4m x s x 2 2x x x x 2 + 3x x t x x 2 44x x 2 + 3x x x x x x x x x 3 + 6x 2 20x 21. x x y y 2 + x x x x 2 52x y 4 10, x 2 4y 2 27. Challenge 100y xy + 4x 2 28. Challenge 125x 2 100xy + 20y Unit 11 factoring polynomials 34 Factoring Challenges Factoring polynomials completely can be challenging. Some clever strategies will help. When you come across a polynomial that is difficult to factor, use one of these methods: grouping or splitting the middle term. Factoring by Grouping Example 1 Factor each polynomial by grouping. A. xy + 2x yz 2z In the polynomial xy + 2x yz 2z, two terms have the variable x and two terms have the variable z. Group the terms of the polynomial so you can divide out a common monomial from each grouping. Then use the distributive property. xy + 2x yz 2z = (xy + 2x) (yz + 2z) Associative Property = x(y + 2) z(y + 2) Factor a common monomial from each binomial. = (y + 2)(x z) Distributive Property B. x 4 9y 2 6x Use the commutative property to change the order of the terms. Group the x terms together and look for a pattern. Since the x terms and the constant form a perfect square trinomial, group these three terms together. x 4 9y 2 6x = x 4 6x y 2 Commutative Property = (x 4 6x 2 + 9) 9y 2 Associative Property = (x 2 3)(x 2 3) 9y 2 Factor the perfect square trinomial. = (x 2 3) 2 9y 2 Simplify. = (x 2 3) 2 (3y) 2 Write as the difference of two squares. = [(x 2 3) + 3y][(x 2 3) 3y] Factor. = (x 2 + 3y 3)(x 2 3y 3) Simplify. factoring challenges 429 35 Factoring by Splitting the Middle Term Another technique you can use when factoring trinomials, especially trinomials where a 1, is called splitting the middle term. Think about it If you cannot find two factors of ac whose sum is b, then the polynomial cannot be factored and the polynomial is prime. factoring ax 2 + bx + c by splitting the middle term Step 1 find the product ac. Step 2 find two factors of ac whose sum is b. Step 3 split the middle term of the trinomial into two terms using the factors found in Step 2 as the coefficients of the terms. Step 4 group the terms into two binomials. Step 5 factor out the common monomial in each group. Step 6 factor out the common binomial. Example 2 Factor each polynomial by splitting the middle term. A. 2x 2 + 7x + 3 Step 1 Find the product of 2 and = 6 Step 2 Find two factors of 6 whose sum is the coefficient of the middle term 7. 6 and 1 Step 3 Split the middle term 7x into 6x + 1x. 2x 2 + 6x + x + 3 Step 4 Group the terms using the associative property. (2x 2 + 6x) + (x + 3) Step 5 Factor out the common monomial term in each group. 2x(x + 3) + 1(x + 3) Step 6 Factor out the common binomial. (x + 3)(2x + 1) B. 8x 2 16x 10 Step 1 Find the product of 8 ( 10). 8 ( 10) = 80 Step 2 Step 3 Step 4 Step 5 Find two factors of 80 whose sum is and 4 Split the middle term 16x into 20x + 4x. 8x 2 20x + 4x 10 Group the terms using the associative property. (8x 2 20x) + (4x 10) Factor out the common monomial term in each group. 4x(2x 5) + 2(2x 5) Step 6 Factor out the common binomial. (2x 5)(4x + 2) 430 Unit 11 factoring polynomials 36 Problem Set Use grouping to factor. If the polynomial is not factorable, write prime. 1. xy + 3y + xz + 3z 2. ac + ad + bc + bd 3. x 2 y 2 10x + 25 Split the middle term to factor. If the polynomial is not factorable, write prime. 7. 3x 2 + 7x x 2 x x 2 x 3 Determine the method of factoring to use: grouping or splitting the middle term. Factor each polynomial. If the polynomial is not factorable, write prime. 13. ab 2ac + 2b 4c 14. 5x 2 + 7x x 2 + 7x xy xz 4y + 4z 17. x 2 4y 2 2x xz 10x 3yz + 5y 19. x 2 25y 2 + 4x x 2 25x x 2 81y 2 + 6x x 2 100y 2 + 8x xz + xy + 10z + 5y 6. 4xy 12xz + yz 3z x 2 5x x x x x x 2 + y x x 2 9y 2 + 4x ac + 2ab 3bc 6b x 4 4x 2 121y x x a 4 c 2 + b 2 2a 2 b 28. x 4 169y x 29. Challenge 24x 2 46x + 7 30. Challenge 16x y x 2 y 2 16z 2 factoring challenges 431 37 Finding Roots of a Polynomial Use the zero product property to find the root of a polynomial. Think about it For any real number a, a 0 = 0. You can also say for any real numbers a and b, if a = 0 or b = 0, then ab = 0. The zero product property is the converse of this statement. zero product property For any real numbers a and b, if ab = 0, then a = 0 or b = 0 or both a = 0 and b = 0. You can use the zero product property to solve polynomial equations where one side of the equation is a factored polynomial and the other side of the equation is equal to zero. The solutions to a polynomial equation of this form are called the roots of the polynomial. Solving Polynomial Equations Example 1 Find the roots. A. x 2 + 5x 6 = 0 Factor the left side of the equation. Then use the zero product property to find the solutions. x 2 + 5x 6 = 0 (x + 6)(x 1) = 0 Factor. x + 6 = 0 or x 1 = 0 Zero Product Property x = 0 6 x = Solve each equation for x. x = 6 x = 1 Simplify. The roots of the polynomial are 6 and 1. Check You can check your solutions by substituting each root into the equation for x. x 2 + 5x 6 = 0 x 2 + 5x 6 = 0 ( 6) ( 6) = 0 0 = 0 The solutions are correct. 432 Unit 11 factoring polynomials 38 B. x 2 + 6x = 9 To use the zero product property, one side of the equation must equal 0. The first step in solving the equation x 2 + 6x = 9 is to write an equivalent equation where one side of the equation is equal to 0. x 2 + 6x = 9 x 2 + 6x + 9 = Add 9 to each side of the equation. x 2 + 6x + 9 = 0 Simplify. Now solve the polynomial equation. x 2 + 6x + 9 = 0 (x + 3)(x + 3) = 0 Perfect squares pattern The factors are the same, so this equation has only one solution. x + 3 = 0 Solve the equation for x. x = 3 The root of the polynomial is 3. C. 2x 3 7x 2 15x = 0 Factor the polynomial. 2x 3 7x 2 15x = 0 Subtract 3 from both sides of the equation. x(2x 2 7x 15) = 0 Factor out the common monomial x. x(2x + 3)(x 5) = 0 Factor the trinomial. x = 0 or 2x + 3 = 0 or x 5 = 0 Zero Product Property 2x = 3 x = 5 Solve each equation for x. x = 3 2 Simplify. 3 The roots of the polynomial are 0, 2, and 5. Finding the Zeros of a Polynomial Function A zero of a function f(x) is a solution to the equation f(x) = 0. The zeros of the function are the x-intercepts of the graph of the function. Example 2 Use the zeros of the polynomial function f(x) = 25x 2 9 to find the x-intercepts of the function. Remember The x-intercept of a graph is the x-coordinate of the point where the graph intersects the x-axis. f(x) x (continued) finding roots of a polynomial 433 39 On the graph, the x-intercepts appear to be about halfway between 0 and 1. The zeros of the polynomial function are the values of x where f(x) = 0. Set the function equal to zero and solve to find the exact values of the x-intercepts. Factor the right side of the equation. Then use the zero product property. 0 = 25x 2 9 Set the function equal to 0. 0 = (5x 3)(5x + 3) Difference of squares pattern 5x 3 = 0 or 5x + 3 = 0 Zero Product Property 5x = 3 5x = 3 Solve each equation for x. 3 x = 5 x = 3 5 Simplify. The zeros are 3 5 and 3 5, so the x-intercepts are 3 5 and 3 5. Application: Measurement Example 3 A rectangular fishpond is surrounded by a border with a uniform width of x feet. The length of the fishpond including the border is 12 feet and the width including the border is 10 feet. The area of the fishpond is 48 square feet. What are the length and width of the fishpond? x feet Pond 10 feet 12 feet Use the diagram to help you write an equation to solve the problem. x feet 12 2x 10 2x 10 feet 12 feet The length of the fishpond is 12 x x = 12 2x. The width of the fishpond is 10 x x = 10 2x. Use the formula for the area of a rectangle to write an equation that represents the area of the fishpond. 434 Unit 11 factoring polynomials 40 A = lw 48 = (12 2x)(10 2x) Substitute 48 for A, 12 2x for l, and 10 2x for w. 48 = x 20x + 4x 2 Multiply the two binomials using the FOIL method. 48 = x + 4x 2 Combine like terms. 0 = 72 44x + 4x 2 Subtract 48 from each side. Now factor the polynomial and solve for x using the zero product property to find the zeros. 4x 2 44x + 72 = 0 4(x 2 11x + 18) = 0 Factor out the common monomial 4. 4(x 9)(x 2) = 0 Factor the trinomial. Solve each binomial for x. x 9 = 0 or x 2 = 0 x = 9 x = 2 The zeros of the function are 9 and 2. The length and width of the fishpond are 12 2x and 10 2x. If you substitute 9 into each expression to find the length and width, the values are negative, so 9 is not a valid solution. 12 2x = and 10 2x = = = = 6 = 8 Substitute 2 into the expressions to find the length and width of the fishpond. 12 2x = and 10 2x = = 12 4 = 10 4 = 8 = 6 The length of the fishpond is 8 feet and the width is 6 feet. Problem Set Find the roots of the polynomial equation. 1. x 2 + x 12 = 0 2. x 2 10x + 9 = 0 3. x 2 x 2 = 0 4. x 2 + 7x + 12 = 0 5. x 2 2x 35 = 0 6. x x + 64 = 0 7. x 2 12x = x = 20x 9. 2x 2 11x 6 = x 2 + 6x + 1 = x 3 8x x = x 3 4x = x x 2 = 3x x x 2 = 8x finding roots of a polynomial 435 41 Find the zeros of each polynomial function. 15. f (x) = x 2 6x f (x) = x 2 + 3x f (x) = x 2 + 2x f (x) = 3x 2 + x f (x) = x 2 5x 20. f (x) = x f (x) = x 2 22x Solve. 29. Amar has a coin collection containing dimes and nickels. The number of dimes is 3 more than the number of nickels. If the product of the number of all coins equals 180, find the number of dimes and nickels in his collection. 30. A school has designated a rectangular plot of land for student award plaques. The entire rectangular plot has been reserved except for a small rectangular area in the lower right hand corner. The dimensions are shown below. If the area of this small rectangle is 24 square feet, find the dimensions of the small rectangle and the larger rectangular plot. x 20 x 12 x In a flag shaped like a right triangle, one leg is 2 less than the hypotenuse and the other leg is 11 less than 3 times the length of the hypotenuse. If the area of the flag is 6 square feet, find the length of the legs and the hypotenuse. 22. f (x) = 16x 3 + 6x 2 x 23. f (x) = 4x f (x) = x 3 20x x 25. f (x) = 15x x x 26. f (x) = 2x x x 27. f (x) = x 3 144x 28. f (x) = 40x 3 + 6x 2 x 32. Find three consecutive odd integers if the product of the second and third odd integer is 63. 33. Challenge A box is being constructed to ship a flat panel television. The television s width is 5 times its length and its height is 4 more than 5 times its length. To make the box large enough to store the television and packing materials, 6 inches must be added to the dimensions of the television. If the area of one face of the box, width by height, is 780 square inches, what are the dimensions of the box? 34. Challenge Lee just celebrated a birthday. Her friend Juanita is 8 years less than twice as old as Lee is now. 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( ) 3 ( ) 3( ) 7 55. 3{[ ( )] ( )( 3)} 56. {( 3)( ) [3 ( )]} 57. Show by ### Sect 6.7 - Solving Equations Using the Zero Product Rule Sect 6.7 - Solving Equations Using the Zero Product Rule 116 Concept #1: Definition of a Quadratic Equation A quadratic equation is an equation that can be written in the form ax 2 + bx + c = 0 (referred ### UNIT TWO POLYNOMIALS MATH 421A 22 HOURS. Revised May 2, 00 UNIT TWO POLYNOMIALS MATH 421A 22 HOURS Revised May 2, 00 38 UNIT 2: POLYNOMIALS Previous Knowledge: With the implementation of APEF Mathematics at the intermediate level, students should be able to: - CHAPTER THE INTEGERS In golf tournaments, a player s standing after each hole is often recorded on the leaderboard as the number of strokes above or below a standard for that hole called a par. A player ### Unit 1: Polynomials. Expressions: - mathematical sentences with no equal sign. 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To factor polynomials, ### Topic: Special Products and Factors Subtopic: Rules on finding factors of polynomials Quarter I: Special Products and Factors and Quadratic Equations Topic: Special Products and Factors Subtopic: Rules on finding factors of polynomials Time Frame: 20 days Time Frame: 3 days Content Standard: ### 15.1 Factoring Polynomials LESSON 15.1 Factoring Polynomials Use the structure of an expression to identify ways to rewrite it. Also A.SSE.3? ESSENTIAL QUESTION How can you use the greatest common factor to factor polynomials? EXPLORE ### 10 7, 8. 2. 6x + 30x + 36 SOLUTION: 8-9 Perfect Squares. The first term is not a perfect square. So, 6x + 30x + 36 is not a perfect square trinomial. Squares Determine whether each trinomial is a perfect square trinomial. Write yes or no. If so, factor it. 1.5x + 60x + 36 SOLUTION: The first term is a perfect square. 5x = (5x) The last term is a perfect ### Chapter 8. Quadratic Equations and Functions Chapter 8. Quadratic Equations and Functions 8.1. Solve Quadratic Equations KYOTE Standards: CR 0; CA 11 In this section, we discuss solving quadratic equations by factoring, by using the square root property ### 6.4 Special Factoring Rules 6.4 Special Factoring Rules OBJECTIVES 1 Factor a difference of squares. 2 Factor a perfect square trinomial. 3 Factor a difference of cubes. 4 Factor a sum of cubes. By reversing the rules for multiplication ### Math Review. for the Quantitative Reasoning Measure of the GRE revised General Test Math Review for the Quantitative Reasoning Measure of the GRE revised General Test www.ets.org Overview This Math Review will familiarize you with the mathematical skills and concepts that are important ### Using the ac Method to Factor 4.6 Using the ac Method to Factor 4.6 OBJECTIVES 1. Use the ac test to determine factorability 2. Use the results of the ac test 3. Completely factor a trinomial In Sections 4.2 and 4.3 we used the trial-and-error ### Answer Key for California State Standards: Algebra I Algebra I: Symbolic reasoning and calculations with symbols are central in algebra. Through the study of algebra, a student develops an understanding of the symbolic language of mathematics and the sciences. ### Polynomials and Factoring; More on Probability Polynomials and Factoring; More on Probability Melissa Kramer, (MelissaK) Anne Gloag, (AnneG) Andrew Gloag, (AndrewG) Say Thanks to the Authors Click http://www.ck12.org/saythanks (No sign in required) ### How do you compare numbers? On a number line, larger numbers are to the right and smaller numbers are to the left. The verbal answers to all of the following questions should be memorized before completion of pre-algebra. Answers that are not memorized will hinder your ability to succeed in algebra 1. Number Basics ### 3. Power of a Product: Separate letters, distribute to the exponents and the bases Chapter 5 : Polynomials and Polynomial Functions 5.1 Properties of Exponents Rules: 1. Product of Powers: Add the exponents, base stays the same 2. Power of Power: Multiply exponents, bases stay the same ### Factoring Polynomials Factoring Polynomials Factoring Factoring is the process of writing a polynomial as the product of two or more polynomials. The factors of 6x 2 x 2 are 2x + 1 and 3x 2. In this section, we will be factoring ### ( ) FACTORING. x In this polynomial the only variable in common to all is x. FACTORING Factoring is similar to breaking up a number into its multiples. For example, 10=5. The multiples are 5 and. In a polynomial it is the same way, however, the procedure is somewhat more complicated ### MATH 60 NOTEBOOK CERTIFICATIONS MATH 60 NOTEBOOK CERTIFICATIONS Chapter #1: Integers and Real Numbers 1.1a 1.1b 1.2 1.3 1.4 1.8 Chapter #2: Algebraic Expressions, Linear Equations, and Applications 2.1a 2.1b 2.1c 2.2 2.3a 2.3b 2.4 2.5 ### FACTORING QUADRATICS 8.1.1 and 8.1.2 FACTORING QUADRATICS 8.1.1 and 8.1.2 Chapter 8 introduces students to quadratic equations. These equations can be written in the form of y = ax 2 + bx + c and, when graphed, produce a curve called a parabola. ### Section 2.1 Intercepts; Symmetry; Graphing Key Equations Intercepts: An intercept is the point at which a graph crosses or touches the coordinate axes. x intercept is 1. The point where the line crosses (or intercepts) the x-axis. 2. The x-coordinate of a point ### Polynomials and Factoring 7.6 Polynomials and Factoring Basic Terminology A term, or monomial, is defined to be a number, a variable, or a product of numbers and variables. A polynomial is a term or a finite sum or difference of Advanced GMAT Math Questions Version Quantitative Fractions and Ratios 1. The current ratio of boys to girls at a certain school is to 5. If 1 additional boys were added to the school, the new ratio of ### Pre-Algebra Interactive Chalkboard Copyright by The McGraw-Hill Companies, Inc. Send all inquiries to: Pre-Algebra Interactive Chalkboard Copyright by The McGraw-Hill Companies, Inc. Send all inquiries to: GLENCOE DIVISION Glencoe/McGraw-Hill 8787 Orion Place Columbus, Ohio 43240 Click the mouse button ### POLYNOMIAL FUNCTIONS POLYNOMIAL FUNCTIONS Polynomial Division.. 314 The Rational Zero Test.....317 Descarte s Rule of Signs... 319 The Remainder Theorem.....31 Finding all Zeros of a Polynomial Function.......33 Writing a
Back ## Stabilize These Squares Given the two balance scales, we might ask how much the green triangle weighs. Or perhaps, we might want to know the value of $t$ in the equations \begin{aligned} s + t &= 3 \\ 2s + t &= 5. \end{aligned} In either case, we can use the same approach: the elimination method. Elimination depends on this fact: We can always keep a scale in balance by removing (or adding) equivalent quantities on both sides. What's happening here? We know from the first balance that a square and a triangle are equivalent to $3$ (they are in balance and must therefore weigh the same). Therefore, on the second balance, we can remove a square and a triangle from the left and $3$ from the right, and the scale will still be in balance. Note that although what we removed on the right looks different from what we removed on the left, we know that they are the equivalent because the first scale showed us. From here, we can solve directly: the blue square equals $2,$ so the green triangle must equal $1.$ Let's look at the same approach but this time with an algebraic representation: \begin{aligned} s + t &= 3 \\ 2s + t &= 5. \end{aligned} Subtracting the first equation from the second, we get \begin{aligned} 2s + t &= 5 \\ - (s + t) &= -(3) \\ \Rightarrow s &= 2. \end{aligned} Because $s=2$, we can conclude that $t=1$. By combining these equations, we found the solution very quickly. Although the problem below looks cumbersome, if you approach it as an elimination problem — where combining the equations together lets you simplify — there's a straightforward path to the solution. # Today's Challenge Given the balance scales below, which combination would balance with three squares? × Problem Loading... Note Loading... Set Loading...
# Identity Matrix Identity Matrix is the matrix which is nÂ Ã— n square matrix where the diagonal consist of ones and the other elements are all zeros. It is also called as a Unit Matrix or Elementary matrix. It is represented as InÂ or just by I, where n represents the size of the square matrix. For example, $$\begin{array}{l}I_{1}=1\\I_{2}=\begin{bmatrix} 1 &0 \\ 0 &1 \end{bmatrix}\\ I_{3}=\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\end{array}$$ We can also say, the identity matrix is a type of diagonal matrix, where the principal diagonal elements are ones, and rest elements are zeros. Let’s study about its definition, properties and practice some examples on it. ## Identity Matrix Definition An identity matrix is a square matrix in which all the elements of principal diagonals are one, and all other elements are zeros. It is denoted by the notation “In” or simply “I”. If any matrix is multiplied with the identity matrix, the result will be given matrix. The elements of the given matrix remain unchanged. In other words,Â if all the main diagonal of a square matrix are 1’s and rest all o’s, it is called an identity matrix. Here, the 2 Ã— 2 and 3 Ã— 3 identity matrix is given below: 2 Ã— 2 Identity Matrix This is also called theÂ identity matrix of order 2. 3Ã— 3 Identity Matrix This is known as theÂ identity matrix of order 3 or unit matrix of order 3 Ã—Â 3. Identity Matrix is donated by In Ã— n, where n Ã— n shows the order of the matrix. A Ã— I n Ã— n = A, A = any square matrix of order n Ã—Â n. ## Properties of Identity Matrix 1) It is always a Square Matrix These Matrices are said to be square as it always has the same number of rows and columns. For any whole number n, there’s a corresponding Identity matrix, n Ã— n. 2) By multiplying any matrix by the unit matrix, gives the matrix itself. As the multiplication is not always defined, so the size of the matrix matters when we work on matrix multiplication. Like, for “m Ã— n” matrix C, we get ImC = C = CIn So the size of the matrix is important as multiplying by the unit is like doing it by 1 with numbers. For example: $$\begin{array}{l}C = \begin{bmatrix} 1 & 2 & 3 &4 \\ 5& 6& 7 & 8 \end{bmatrix}\end{array}$$ The above is 2 Ã— 4 matrix as it has 2 rows and 4 columns. Let’s multiply the 2Â Ã— 2 identity matrix by C. $$\begin{array}{l}\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}.\begin{bmatrix} 1 & 2 & 3 &4 \\ 5& 6& 7 & 8 \end{bmatrix}=\begin{bmatrix} 1+0 & 2+0 & 3+0 &4+0 \\ 0+5& 0+6& 0+7 & 0+8 \end{bmatrix} = \begin{bmatrix} 1 & 2 & 3 &4 \\ 5& 6& 7 & 8 \end{bmatrix}\end{array}$$ Hence proved. 3) We always get an identity after multiplying two inverse matrices. If we multiply two matrices which are inverses of each other, then we get an identity matrix. $$\begin{array}{l}C = \begin{bmatrix} 0 &1 \\ -2& 1 \end{bmatrix}\end{array}$$ $$\begin{array}{l}D=\begin{bmatrix} \frac{1}{2} &- \frac{1}{2} \\ 1& 0 \end{bmatrix}\end{array}$$ $$\begin{array}{l}CD= \begin{bmatrix} 0 &1 \\ -2& 1 \end{bmatrix}\begin{bmatrix} \frac{1}{2} &- \frac{1}{2} \\ 1& 0 \end{bmatrix}=\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\end{array}$$ $$\begin{array}{l}DC = \begin{bmatrix} \frac{1}{2} &- \frac{1}{2} \\ 1& 0 \end{bmatrix}\begin{bmatrix} 0 &1 \\ -2& 1 \end{bmatrix}=\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\end{array}$$ ### Identity Matrix Examples Example 1: Write an example of 4Â Ã— 4 order unit matrix. Solution:Â The unit matrix is the one having ones on the main diagonal & other entries as ‘zeros’. Thus, the unit matrix of order 4Â Ã— 4 or the identity matrix of order 4 can be written as: $$\begin{array}{l}I_{4\times 4}=\begin{bmatrix} 1 & 0 & 0 &0 \\ 0& 1 & 0 &0 \\ 0 & 0 & 1 & 0\\ 0Â & 0 & 0 &1 \end{bmatrix}\end{array}$$ Example 2: Check the following matrix is Identity matrix? $$\begin{array}{l}V= \begin{bmatrix} 1 & 0 & 0 &0 \\ 0& 1 & 0 &0 \\ 0 & 0 & 1 & 0\\ \end{bmatrix}\end{array}$$ Solution: No, It’s not an identity matrix, because it is of the order 3 Ã— 4, which is not a square matrix. Example 3: Check the following matrix is Identity matrix; $$\begin{array}{l}B =\begin{bmatrix} 1 & 1 & 1\\ 1 & 1& 1\\ 1 & 1 & 1 \end{bmatrix}\end{array}$$ Solution: No, it is not a unit matrix as it doesn’t contain the value of 0 beside one property of having diagonal values of 1. Visit BYJU’S – The Learning App to explore a fun and interesting way to learn Mathematics. ## Frequently Asked Questions on Identity Matrix Q1 ### What do you mean by an identity matrix? In linear algebra, an identity matrix is a matrix of order nxn such that each main diagonal element is equal to 1, and the remaining elements of the matrix are equal to 0. Q2 ### What is the identity matrix of a 2Ã—2? An identity matrix of 2Ã—2 is a matrix with 1â€™s in the main diagonal and zeros everywhere. The identity matrix of order 2Ã—2 is: [1 0 0 1]. Q3 ### What is the identity matrix of a 3Ã—3? An identity matrix of 3Ã—3 is a matrix with 1â€™s in the main diagonal and zeros everywhere. The identity matrix of order 3Ã—3 is given by: [1 0 0 0 1 0 0 0 1]. Q4 ### Is the identity matrix nonsingular? Yes, the identity matrix is nonsingular since its determinant is not equal to 0. The identity matrix is the only idempotent matrix with a non-zero determinant. Therefore, we can also find the inverse of the identity matrix. Q5 ### How do you create an identity matrix? As we know, the identity matrix has all its main diagonal elements as 1â€™s and the remaining elements 0â€™s. Suppose to create an identity matrix of order 4Ã—4, we write the matrix elements in rows and columns as given below, and those should be enclosed within [ ]. 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1
# Multiplication of a Fraction by Fraction – Definition, Examples \| How do you Multiply a Fraction by a Fraction? Have you stuck at some point while multiplying one fraction with another fraction and need help? Don’t bother as we are with you in this and compiled an article covering the fraction definition, how to multiply a fraction by another fraction, how to multiply two mixed fractions. Refer to the solved examples for multiplying fraction by fraction and try to solve related problems on your own. ## Fraction – Definition Fractions are the numerical values or things that are divided into parts, then each part will be a fraction of a number. A fraction is denoted as $$\frac { a }{ b }$$, where a is the numerator and b is the denominator. ### Multiplication of a Fraction by Fraction Multiplication of fraction by fraction starts with the multiplication of the given numerators followed by multiplication of the denominators. Then, the resultant fraction can be simplified further and reduced to its lowest terms if needed. Multiplication of fractions is not the same as addition or subtraction of fractions, where the denominator should be the same. Here any two fractions without the same denominator can also be multiplied. ### How to Multiply Two Fractions? Follow the simple steps listed below to multiply a fraction by fraction. They are in the below fashion Step 1: Simplify the fractions into their lowest terms. Step 2: Multiply both the numerators of the given fractions to get a new numerator. Step 3: Multiply both the denominators of the given fractions to get a new denominator. Simplify the resulting fraction if needed. Let a/b is one fraction and the other fraction is c/d. Multiplication of these fractions are Multiplying fractions formula: $$\frac { a }{ b }$$ * $$\frac { c }{ d }$$ = $$\frac { a * b }{ c * d }$$ ### Multiplication of Fraction by another Fraction Examples Example 1: $$\frac { 1 }{ 3 }$$ is the first fraction. $$\frac { 2 }{ 3 }$$ is another fraction. What is the multiplication of both fractions? Solution: It is given that $$\frac { 1 }{ 3 }$$ is one fraction and $$\frac { 2 }{ 3 }$$ is another fraction. When we multiply both these fractions, we multiply both the numerators and both the denominators. Product of two fractions = (Product of numerators)/(Product of denominators). That is (12) is the numerator. (33) is the denominator. The final result is the product of numerators and denominators. Therefore the final answer is $$\frac { 2 }{ 9 }$$ . Example 2: $$\frac { 12 }{ 5 }$$ is the first fraction. $$\frac { 23 }{ 9 }$$ is another fraction. What is the multiplication of both fractions? Solution: It is given that $$\frac { 12 }{ 5 }$$ is one fraction and $$\frac { 23 }{ 9 }$$ is another fraction. When we multiply both these fractions, we multiply both the numerators and both the denominators. Product of two fractions = (Product of the numerators)/(Product of the denominators). That is (1223) is the numerator. (59) is the denominator. The final result is the product of numerators and denominators. The value is ($$\frac { 276 }{45 }$$ . Therefore the final answer is $$\frac { 92 }{15 }$$ , Example 3: $$\frac { 16 }{ 30 }$$ is the first fraction. $$\frac { 21 }{ 50 }$$ is another fraction. What is the multiplication of both fractions? Solution: It is given that $$\frac { 16 }{ 30 }$$ is one fraction and $$\frac { 21}{ 50 }$$ is another fraction. When we multiply both these fractions, we multiply both the numerators and both the denominators. Product of two fractions = (Product of numerators)/(Product of denominators). That is (16 * 21) is the numerator. (3050) is the denominator. The final result is the product of numerators and denominators. Therefore the final answer is $$\frac { 336 }{ 1500 }$$ . ### Multiplication of Mixed Fractions A fraction that is represented by its quotient and remainder is a mixed fraction. So, it is a combination of a whole number and a proper fraction. Multiplication of mixed fractions will be difficult to change each number into an improper fraction. ### How to Multiply Mixed Fractions? Go through the below-listed steps to multiply two mixed fractions. They are in the below fashion Step 1: Convert given mixed fractions into improper fractions. Step 2: Simplify the fractions into their lowest terms for easy calculations. Step 3: Multiply both the numerators of the given fractions to get a new numerator. Step 4: Multiply both the denominators of the given fractions to get a new denominator. Simplify the resultant fraction if needed. ### Multiplication of Mixed Fraction by another Mixed Fraction Examples Example 1: Multiply mixed fractions 2$$\frac { 3 }{ 5 }$$ and 6 $$\frac { 7 }{ 8 }$$. What is the multiplication of both fractions? Solution: It is given that 2$$\frac { 3 }{ 5 }$$ is one fraction and the other fraction is 6$$\frac { 7 }{ 8 }$$ As we know mixed fractions cannot be multiplied. Simplify the given mixed fractions into improper fractions. Now the given mixed fractions are $$\frac { 13 }{ 5 }$$ and the other fraction is $$\frac { 55 }{ 8 }$$. When we multiply both these fractions, we multiply both the numerators and both the denominators. Product of two fractions = (Product of numerators)/(Product of denominators). That is (13 55) is the numerator. (58) is the denominator. The final result is the product of numerators and denominators. The value is $$\frac { 715 }{ 40 }$$ Therefore the final answer is $$\frac { 715 }{ 40 }$$ Example 2: Multiply mixed fractions 2$$\frac { 1}{ 8 }$$ and 6 $$\frac { 4 }{ 9 }$$. What is the multiplication of both fractions? Solution: It is given that 2$$\frac { 1 }{ 8 }$$ is one fraction and the other fraction is 6$$\frac { 4 }{ 9 }$$ As we know mixed fractions cannot be multiplied. Simplify the given mixed fractions into improper fractions. Now the given mixed fractions are $$\frac { 17 }{ 8 }$$ and the other fraction is $$\frac { 58 }{ 9 }$$. When we multiply both these fractions, we multiply both the numerators and both the denominators. Product of two fractions = (Product of numerators)/(Product of denominators). That is (17 58) is the numerator. (8*9) is the denominator. The final result is the product of numerators and denominators. The value is $$\frac { 986 }{ 72 }$$ Therefore the final answer is $$\frac { 986 }{ 72 }$$.
Home > Pre-Algebra > Percentages > Percentage Sale Price ## Percentage Sale Price #### Introduction Percentage is a way of expressing a number as a fraction of 100. Percentages take a number and split it into 100. You can then express segments of the whole in terms of percentages. For example, if you find 20% of something, you would split the total number into 100 and then take 20 of those 100 segments. Percentages are commonly used to express sales. You regularly see offers in stores or online giving '20% off' or you may receive a student discount of 10%. Working out how much these items cost after the discount is applied requires you to know how to calculate percentages. ## Lesson The first step to take when an item is discounted is to calculate what percentage of the cost remains. For example, if a pair of shoes is discounted 20%, you need to know what percentage of the shoes' price you will need to pay. To do this, subtract the percentage discount from 100. This will give you the remaining percentage. Let's use the following example: A shirt in a store normally costs $80. However, there is a 15% off sale. What is the new cost of the shirt? (We will assume in this example that there is no tax to pay) So, the discount is 15%, so we should subtract 15 from 100 to find out what the remaining cost of the shirt is. $100 - 15 = 85$ Therefore, the cost of the shirt is 85% of the original total. Next, we need to work out what 85% of $80 is. There are two ways to do this. To do both, you will need to remember that 85% is the same as$ rac{85}{100}$. The first way is to divide $80 by 100 to find 1% and then multiply that by 85. $\80 ÷ 100 = 0.8$ $0.8 x 85 = \68.00$ Therefore, the cost of the shirt after the discount is$68.00. The second way to calculate the sale price of the shirt is to combine the two steps of the previous method. You can turn all percentages into decimals by just remembering that 100% is the same as 1.0. That means that if you place the number of the percentage after the decimal point, you will have converted it into a decimal. For example, 85% is the same as 0.85. 75% is 0.75, 13% is 0.13 and so on. To work out the cost of the shirt after the discount is applied, we can just multiply the original price by the decimal we have made: $\80 x 0.85 = \68.00.$ This gives us the same answer. You can use either method. The second one is faster, so when you are more comfortable with decimals and percentages, you will usually find this one better. ## Examples Ella saw a basketball on sale for \$5, with an additional 45% off. What was the final price of the basketball? The discount is: $0.45$ * KaTeX parse error:$ within math mode = KaTeX parse error: $within math mode Subtract the discount from the original price to find the sale price: KaTeX parse error: Unexpected character: '' at position 1: ̲.00 - .25 =$\$…
Saturday, May 22, 2021 10:24:57 AM # Measures Of Central Tendency Problems And Solutions Pdf File Name: measures of central tendency problems and solutions .zip Size: 12106Kb Published: 22.05.2021 Measures of dispersion exercises with answers. Answer choices in this exercise appear in a different order each time the page is loaded. ## Measures of central tendency: Median and mode Apart from the mean, median and mode are the two commonly used measures of central tendency. The median is sometimes referred to as a measure of location as it tells us where the data are. It divides the frequency distribution exactly into two halves. Fifty percent of observations in a distribution have scores at or below the median. Hence median is the 50th percentile. It is easy to calculate the median. Which average would be suitable in following cases? Explanation: Mode is suitable average for average size of readymade garments because it gives the most frequent occurring value. Explanation: Production can be measured on a quantitative scale so Arithmetic mean is suitable in this case. Explanation: Wage can be measured on a quantitative scale so arithmetic mean is suitable in this case. Explanation: Mean shall be used because sum of deviations from mean is always zero or least than the other averages. ## Service Unavailable in EU region A measure of central tendency is an important aspect of quantitative data. Three of the many ways to measure central tendency are the mean , median and mode. The sample mean is a statistic and a population mean is a parameter. Review the definitions of statistic and parameter in Lesson 0. Is this a problem? Measures of central tendency are numbers that locate or approximate the Problems. Find the mean of each set of data. 1. 29, 28, 34, 30, 33, 26, and 2. 62, 65, 93, 51, 55, 14, 79, 85, 55, 72, 78,. 83, 91, and Answers. 1. 2. ## Measures of Central Tendency The first exercise focuses on the research design which is your plan of action that explains how you will try to answer your research questions. Exercises two through four focus on sampling, measurement, and data collection. The fifth exercise discusses hypotheses and hypothesis testing. Average: It is a value which is typical or representative of a set of data. Averages are also called Measures of Central Tendency. Simple to calculate. It should be easy to understand. These solutions for Measures Of Central Tendency are extremely popular among Class 11 Commerce students for Economics Measures Of Central Tendency Solutions come handy for quickly completing your homework and preparing for exams. #### Goal of Exercise Он застонал. Проклятые испанцы начинают службу с причастия. ГЛАВА 92 Сьюзан начала спускаться по лестнице в подсобное помещение. Густые клубы пара окутывали корпус ТРАНСТЕКСТА, ступеньки лестницы были влажными от конденсации, она едва не упала, поскользнувшись. Она нервничала, гадая, сколько еще времени продержится ТРАНСТЕКСТ. Сирены продолжали завывать; то и дело вспыхивали сигнальные огни. Тремя этажами ниже дрожали и гудели резервные генераторы. Наконец он заговорил - спокойно, тихо и даже печально: - Нет, Грег, извини. Я не могу тебя отпустить. - Мне срочно нужно в аэропорт. Наконец парень посмотрел на. - Scusi? - Он оказался итальянцем. Давайте же, ребята. -сказал Джабба. - Вы же учились в колледжах. Ну, кто-нибудь. В отчет вкралась какая-то ошибка? - Мидж промолчала. Джабба почувствовал, что она медлит с ответом, и снова нахмурился. - Ты так не считаешь. - Отчет безукоризненный. Что помогло бы мне? - сказал Беккер. Росио покачала головой: - Это. Но вам ее не найти. Севилья - город большой и очень обманчивый. ### 5 Comments Jana W. 24.05.2021 at 05:22 1. 1 Measures of Central. Tendency. Mean, Median, Mode and Range (ii) more than 4 children. Solution. (a) Mode = 4 (as its frequency is highest). (b) (i). Justin R. 25.05.2021 at 00:32 A measure of central tendency is a number that represents the typical value in a collection of numbers. Three familiar measures of central tendency are the mean, the median, and the mode. EXAMPLE SOLUTION. It will be much easier. Cannan B. 26.05.2021 at 08:44 Anna dressed in blood pdf free download sony xperia mini pro pdf AimГ© S. 27.05.2021 at 12:09 What are commonly used measures of central tendency? What do they tell you (See Problem 9 in the Chapter Review Problems.) Chapter 3 SOLUTION: By the formula, we multiply each score by its weight and add the results together. Brian G. 28.05.2021 at 06:01 Complex ptsd from surviving to thriving pdf download complex ptsd from surviving to thriving pdf download
## Counting on Rules of Arithmetic CORA Counting on Rules of Arithmetic. We count on from a starting number like 3 by a second number, the count on number or shift number like 2. 3+2 means start at 3 and count on by 2. We imagine we are counting on along a number line. So first we review some number line rules. 1. Zero is a unique tick on the number line. 2. For each tick on the number line, there exists a unique tick immediately to the right of it. 3. Zero is not a tick to the right of another tick. 4. If the ticks to the right of two ticks are equal, then said two ticks are equal. 5. If a set contains zero and each tick to the right of a tick, then it contains all the ticks on the number line. Counting on Rules of Arithmetic Basic CORAB 1. Zero is a unique starting from number on the number line. 2. For each tick on the number line, there exists a unique tick count on by one to the right of it. 3. You can never count on by one and get to zero. 4. If you count on by one from two ticks and get to the same tick, the two starting ticks were the same. 5. If a set contains zero and you count on by one without ever stopping, you reach every natural number as you count on. We introduce a special symbol for counting on by one. 1=0′ This is read count on from 0 by 1 gets you to 1 or is one. 2=1′. Count on by one from 1 gets you to 2. CORA definition of addition. We now define addition of whole numbers in terms of counting on. For us whole numbers include 0. 1. If you count on zero from a starting number, you stay at the counting number. In this case, we say the sum of the starting number and zero is the starting number. 2. Suppose you start from an initial number, x, and count on by y. Now start from the same initial number x and count on by y’. Then you get to the same place as if you start at x+y and count on by one from there. Examples of rule 1. Start at 2 and count on by 0. The result is 2. You are at 2. The sum of 2 and 0 is 2. 2+0 = 2. Examples of Rule 2 Suppose that 2+3 = 5. Now we want to count on from 2 by 4=3′. 2+3′ = 5′ = 6 So when we count on from 2 by 4 we get to the same starting at 5 and counting on by one. = Now let’s hear you math ed panjandrums and nay-sayers oppose counting on. Ha Ha Ha. “counting on” addition kindergarten http://kindergartencrayons.blogspot.com/2012/01/counting-on-some-are-catching-on.html http://kdoublestuffed.blogspot.com/2011/03/counting-on-addition-dice.html http://education.illinois.edu/smallurban/chancellorsacademy/documents/strategiesbygrade.pdf == Kindergarten 1st Grade 2nd Grade Addition Counting on Combinations of 5 and 10 1 and 2 more 5 and 10 as an anchor Counting on Apply Properties of Operation (Commutative, Zero rule…) 1 and 2 more Doubles Combinations of 10 Decomposition/composing numbers to 1. Make a 10 2. Work with near doubles 3. Build off a known fact. Apply Properties of Operation (Commutative, Zero rule…) 1 and 2 more Doubles Combinations of 10 Decomposition/composing numbers to 1. Make a 10 2. Work with near doubles 3. Build off a known fact. Subtraction Undoing Addition 1 and 2 less Counting up and down Think addition. 1. Counting-up 2. Reverse doubles. 3. Compliments of 10 4. Any subtraction fact Apply properties of operation 1 and 2 less Decompose/Recompose 1. Building up through 10 2. Building down through 10 Think addition. 1. Reverse doubles. 2. Compliments of 10 3. Any subtraction fact Apply properties of operation 1 and 2 less Decompose/Recompose 1. Building up through 10 2. Building down through 10 Fluent = 1) Just knowing some answers, knowing some answers from patterns (n+ 0=n), and knowing some answers from the use of strategies. (Progressions, pg. 18) == Notice how they have n+0 =n. We were told that was too hard even for teachers. Here it is on the official website of Illinois. ==It gets worse, this is page 2 Repeated Addition/Skip Counting Repeated Counting on 1. Mental arrays. 2. 5 facts and clocks. Apply properties of operation (Commutative, x 1, x 0…..) Connecting addition doubles and x 2 facts. Associative/Distributive Property-Decompose/Recompose 1. Halve than double: 4 x8 = (2×8) + (2 x 8) 2. Double and 1 more group: 3 x 8 = (2×8) + 8) 3. Square and 1more group: 7 x 6 = (6 x 6) + 6 4. One more group than a known fact: 8 x 6 = (5×8) + 8 5. Build off a known fact. Nifty nines- 9’s relationship to 10. Division Relationship between multiplication and division. Near Facts: 50/6 is in between 8 x 6 and 9 x6. Properties of operation. As should be clear, this is not a matter of instilling facts divorced from their meanings, but rather the outcome of a carefully designed learning process that heavily involves the interplay of practice and reasoning. (Progressions, pg. 27) == Keith Devlin eat your heart out. m’n = mn + n The recursive definition of multiplication. = As should be clear, this is not a matter of instilling facts divorced from their meanings, but rather the outcome of a carefully designed learning process that heavily involves the interplay of practice and reasoning. (Progressions, pg. 27) But we were told this was impossible for grade K to grade 2. The Counting on Rules of Arithmetic (CORA) are precisely this. http://commoncoretools.files.wordpress.com/2011/05/ccss_progression_cc_oa_k5_2011_05_302.pdf “Progressions for the Common Core State Standards in Mathematics (draft)” The Common Core Standards Writing Team 29 May 2011 Draft, “K, Counting and Cardinality; K–5, Operations and Algebraic Thinking” Grade K onwards. From page 5 and creating equivalent but easier or known sums (e.g., adding 6 + 7 by creating the known equivalent 6 + 6 + 1 = 12 + 1 = 13). 6+6′ = (6+6)’ = 12′ = 13 Notice how the prime notation for counting on by one makes the logic easier to understand. How do we deal with 6+7 if we already know 6+6=12? We write 6+7 as 6+6′ and then we apply 6+6′ = (6+6)’ = 12′ = 13 = http://www.youtube.com/watch?v=W0OfkpzVhcw # TouchMath Kindergarten Software Disc 1 — Addition (Addition with Counting On) = 4th grade commutative video Associative property videos http://www.youtube.com/channel/HCFnjDjykrY_Q?feature=relchannel With the Counting on Rules of Arithmetic, we can prove the associative law. At e-vendor for 99 cents, although calling it Peano Axioms Advertisements ## About New Math Done Right Author of Pre-Algebra New Math Done Right Peano Axioms. A below college level self study book on the Peano Axioms and proofs of the associative and commutative laws of addition. President of Mathematical Finance Company. Provides economic scenario generators to financial institutions. This entry was posted in Counting on Definition of Addition, Counting on Rules of Arithmetic, Kindergarten Counting On, Peano Axioms for Number Line. Bookmark the permalink. ### 3 Responses to Counting on Rules of Arithmetic CORA 1. It’s really very complicated in this busy life to listen news on TV, thus I only use internet for that purpose, and take the most recent information. 2. When I originally commented I clicked the “Notify me when new comments are added” checkbox and now each time a comment is added I get several emails with the same comment. Is there any way you can remove people from that service? Thanks! 3. If you desire to get a great deal from this article then you have to apply such techniques to your won web site.
## Digit Manipulation part 2 Still… between this blog and the YouTube video, there doesn’t seem to be a good explanation for why the answer to a digit manipulation comes to the magic “1” proposed by other proofs instead of some other number. To recap: x = .999… 10x = 9.999… 9x = 9 x = 1 The answer I’ve given for this is that the .999… after the 9 in the second step is not as long as x actually is, and subtracting x from both sides should actually read 9x = 8.999…91 instead. But why, when we do the math incorrectly, does it give us 1? Consider this: what proponents of the theory that .999… = 1 suggest is that when you multiply .999… by 10 you should get the same result that you’d get from adding 9 (9 + .999… equals 9.999… agreed?). Is it possible to get the same result by both adding 9 and by multiplying 10 to .999…? What is x given that 10x = x + 9? 10x = x + 9 9x = 9 (subtract x from both sides) x = 1 (divide both sides by 9) “Wow!” you exclaim. “See? x does equal both 1 and .999 because you can do the same math to both and get the same results!” If you really think this, then you have confused cause and effect — by multiplying one side of the equation by 10 and by adding 9 to the other side, you’ve actually then changed the answer to 1. To prove this, let’s change x to 2 in a digit manipulation by the same method, which is by adding 2 to one side while multiplying the other side by 2 (because given 2x = x + 2, x = 2). x = .999… 2x = 2.999… (after multiplying by 2 on the left and adding 2 on the right) x = 2 (after subtracting x from both sides) See how we made x equal what we wanted it to equal? Let’s try to make x = 3 by multiplying one side by 2 and adding 3 to the other side (because given 2x = x + 3, x = 3). x = .999… 2x = 3.999… (after multiplying by 2 on the left and adding 3 on the right) x = 3 (after subtracting x on both sides) Do you see the pattern here? This is why x = 1 in the original equation… because we actually added 9 to .999… instead of multiplying by 10. It gave us an end result that we can predict. Let’s try this one last time, except instead of .999… we’ll use .777… to show that this isn’t a trick that is only true with .999… x = .777 10x = 7.777… 9x = 7 x = 7/9 which almost equals .777… (see earlier post on why it “almost” equals that). Given 10x = x + 7, what is x? 9x = 7 (after subtracting x from both sides) x = 7/9 (after dividing both sides by 9) See? It’s like a self-fulfilling prophecy. I love statistics. They drive my poker playing, my reasoning, and my research. As Penn Gillete said "Luck is probability taken personally". There's no such thing as luck... but I wish you positive chance. This entry was posted in Uncategorized. Bookmark the permalink. ### 3 Responses to Digit Manipulation part 2 1. Wani says: So, in short, your argument here is “if I do maths wrong, I can get any answer I like! Therefore, maths is wrong!” You cannot multiply by 2 on one side of an equation, then add 2 on the other side. Your equations are then no longer equal, and you need to change your sign to in inequality sign. The first equation (x = 0.999…, multiply by 10 to get x = 9.999…) is not just adding 9 to the left, it’s quite clearly a valid multiplication by 10. • starcrashx says: I’m well aware of the fact that I “cannot multiply by 2 on one side of an equation, then add 2 on the other side”. I think you missed my point. I was demonstrating why the answer to the digit manipulation is more consistent with adding 9 to the right side of the equation than multiplying by 10. All the same, you didn’t entirely miss my point. Yes, by doing the maths wrong, you can get any answer you like. And the popular way of doing the first equation is wrong, which is why it produces the wrong answer. 2. Matthias says: What is .999…? Just give me some definition!
# Chapter 10: Linear Equations ## Linear Equations • To solve a linear equation you need to get the letter on its own on one side • It is really important to write your working neatly when you are solving equations • 5x + 3 = 18 • -3 • 5x = 15 • /5 • x=3 • Every line of working should have an equals sign in it • Start a new line for each step, do one operation at a time • Write down the operation you are carrying out, remember to do the same thing to both sides of the equation • Line up the equals signs ## Letter on both sides • To solve an equation you have to get the letter on its own on one side of the equations • Start by collecting like terms so that all the letters are together • 2 - 2x = 26 + 4x • +2x • 2 = 26 + 6x • -26 • -24 = 6x • /6 • -4 = x ## Equations with brackets • Always start by multiplying out the brackets then collecting like terms • 19 = 8 - 2(5 - 3y) • 19 = 8 - 10 + 6y • +2 • 21 = 6y • 21/6 = y ## Examiners report • Don’t use a trial and improvement method to solve an equation • You probably won’t find the correct answer, and you can’t get any method marks ## Equations with fractions • When you have an equation with fractions, you need to get rid of any fractions before solving • You can do this by multiplying every term by the lowest common multiple of the denominators • x/3 + x+1/5 = 11 • x15 • 5x + 3x-3 = 165 • +3 • 8x = 168 • /8 • x = 21 ## Multiplying by an expression • You might have to multiply by an expression to get rid of the fractions • 20/n-3 = -5 • xn-3 • 20 = -5(n-3) ## Worked practice • Eliminate fractions before you start solving the equation • You can do this by multiplying both sides of the equation by 4 • Use brackets to show that you are multiplying everything by 4 • Multiply out the brackets, then solve the equation normally
## Antidifferentiation (Part 1) 1. Derivatives Properties Review. Let ${F(x)}$ be given. We recall the derivative (1)$\displaystyle \frac{\mathrm{d} F(x)}{\mathrm{d} x}=f(x)$ is defined by (2)$\displaystyle f(x) = F'(x) = \lim_{h\rightarrow 0}\frac{F(x+h)-F(x)}{h}.$ We have several rules for differentiation: Power Rule: If ${F(x)=x^{n}}$, then ${F'(x)=nx^{n-1}}$; Product Rule: We have ${[F(x)G(x)]'=F'(x)G(x)+F(x)G'(x)}$; Quotient Rule: We have ${[F(x)/G(x)]'=[F'(x)G(x)-G'(x)F(x)]/G(x)^{2}}$; Composition Rule: $\displaystyle{\frac{\mathrm{d}}{\mathrm{d} x}F(G(x))=F'(G(x))G'(x)}$. We want to introduce an “inverse” of differentiation. We will use the notation (3)$\displaystyle \frac{\mathrm{d} f(x)}{\mathrm{d} x}=f'(x)$ interchangeably. 2. Antidifferentiation. A function ${F(x)}$ is an “Antiderivative” of ${f(x)}$ on an interval ${I}$ (usually it’s an open interval) iff (4)$\displaystyle F'(x)=f(x).$ Example 1. Let ${f(x)=3x}$. What is its antiderivative? Well, we see its of the form (5)$\displaystyle F(x)=ax^{2}$ We take its derivative (6)$\displaystyle F'(x)=2ax$ Set ${F'(x)}$ to be equal to ${f(x)}$ (7)$\displaystyle 2ax=3x$ and solve for ${a}$ (8)$\displaystyle a=\frac{3}{2}.$ Thus (9)$\displaystyle F(x)=\frac{3x^{2}}{2}$ is the antiderivative for ${f(x)}$. Note that ${G(x)=F(x)+C}$ is also an antiderivative for ${f(x)}$, where ${C}$ is any constant. Example 2. Consider (10)$\displaystyle g(x)=\cos(x).$ What is its antiderivative? We recall (11)$\displaystyle \frac{\mathrm{d}}{\mathrm{d} x}\sin(x)=\cos(x)$ which implies (12)$\displaystyle G(x)=\sin(x)+C,$ for some constant ${C}$, is the antiderivative. Example 3. Take (13)$\displaystyle h(x)=\frac{1}{x}+2\mathrm{e}^{2x}$ and find its antiderivative. We see that (14)$\displaystyle \frac{\mathrm{d}}{\mathrm{d} x}\ln\|x\|=\frac{1}{x}.$ We also remember that (15)$\displaystyle \frac{\mathrm{d}}{\mathrm{d} x}\mathrm{e}^{2x}=2\mathrm{e}^{2x}$ which means (16)$\displaystyle \frac{\mathrm{d}}{\mathrm{d} x}\left(\ln\|x\|+\mathrm{e}^{2x}\right)=\frac{1}{x}+2\mathrm{e}^{2x}.$ This gives us the antiderivative (17)$\displaystyle H(x)=\ln\|x\|+\mathrm{e}^{2x}+C$ where ${C}$ is some constant.
# 3.2 The derivative as a function Page 1 / 10 • Define the derivative function of a given function. • Graph a derivative function from the graph of a given function. • State the connection between derivatives and continuity. • Describe three conditions for when a function does not have a derivative. • Explain the meaning of a higher-order derivative. As we have seen, the derivative of a function at a given point gives us the rate of change or slope of the tangent line to the function at that point. If we differentiate a position function at a given time, we obtain the velocity at that time. It seems reasonable to conclude that knowing the derivative of the function at every point would produce valuable information about the behavior of the function. However, the process of finding the derivative at even a handful of values using the techniques of the preceding section would quickly become quite tedious. In this section we define the derivative function and learn a process for finding it. ## Derivative functions The derivative function gives the derivative of a function at each point in the domain of the original function for which the derivative is defined. We can formally define a derivative function as follows. ## Definition Let $f$ be a function. The derivative function , denoted by ${f}^{\prime },$ is the function whose domain consists of those values of $x$ such that the following limit exists: ${f}^{\prime }\left(x\right)=\underset{h\to 0}{\text{lim}}\frac{f\left(x+h\right)-f\left(x\right)}{h}.$ A function $f\left(x\right)$ is said to be differentiable at $a$ if $f\left(a\right)$ exists. More generally, a function is said to be differentiable on $S$ if it is differentiable at every point in an open set $S,$ and a differentiable function is one in which ${f}^{\prime }\left(x\right)$ exists on its domain. In the next few examples we use [link] to find the derivative of a function. ## Finding the derivative of a square-root function Find the derivative of $f\left(x\right)=\sqrt{x}.$ Start directly with the definition of the derivative function. Use [link] . $\begin{array}{ccccc}\hfill {f}^{\prime }\left(x\right)& =\underset{h\to 0}{\text{lim}}\frac{\sqrt{x+h}-\sqrt{x}}{h}\hfill & & & \begin{array}{c}\text{Substitute}\phantom{\rule{0.2em}{0ex}}f\left(x+h\right)=\sqrt{x+h}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}f\left(x\right)=\sqrt{x}\hfill \\ \text{into}\phantom{\rule{0.2em}{0ex}}{f}^{\prime }\left(x\right)=\underset{h\to 0}{\text{lim}}\frac{f\left(x+h\right)-f\left(x\right)}{h}.\hfill \end{array}\hfill \\ & =\underset{h\to 0}{\text{lim}}\frac{\sqrt{x+h}-\sqrt{x}}{h}·\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\hfill & & & \begin{array}{c}\text{Multiply numerator and denominator by}\hfill \\ \sqrt{x+h}+\sqrt{x}\phantom{\rule{0.2em}{0ex}}\text{without distributing in the}\hfill \\ \text{denominator.}\hfill \end{array}\hfill \\ & =\underset{h\to 0}{\text{lim}}\frac{h}{h\left(\sqrt{x+h}+\sqrt{x}\right)}\hfill & & & \text{Multiply the numerators and simplify.}\hfill \\ & =\underset{h\to 0}{\text{lim}}\frac{1}{\left(\sqrt{x+h}+\sqrt{x}\right)}\hfill & & & \text{Cancel the}\phantom{\rule{0.2em}{0ex}}h.\hfill \\ & =\frac{1}{2\sqrt{x}}\hfill & & & \text{Evaluate the limit.}\hfill \end{array}$ ## Finding the derivative of a quadratic function Find the derivative of the function $f\left(x\right)={x}^{2}-2x.$ Follow the same procedure here, but without having to multiply by the conjugate. $\begin{array}{ccccc}\hfill {f}^{\prime }\left(x\right)& =\underset{h\to 0}{\text{lim}}\frac{\left({\left(x+h\right)}^{2}-2\left(x+h\right)\right)-\left({x}^{2}-2x\right)}{h}\hfill & & & \begin{array}{c}\text{Substitute}\phantom{\rule{0.2em}{0ex}}f\left(x+h\right)={\left(x+h\right)}^{2}-2\left(x+h\right)\phantom{\rule{0.2em}{0ex}}\text{and}\hfill \\ f\left(x\right)={x}^{2}-2x\phantom{\rule{0.2em}{0ex}}\text{into}\hfill \\ {f}^{\prime }\left(x\right)=\underset{h\to 0}{\text{lim}}\frac{f\left(x+h\right)-f\left(x\right)}{h}.\hfill \end{array}\hfill \\ & =\underset{h\to 0}{\text{lim}}\frac{{x}^{2}+2xh+{h}^{2}-2x-2h-{x}^{2}+2x}{h}\hfill & & & \text{Expand}\phantom{\rule{0.2em}{0ex}}{\left(x+h\right)}^{2}-2\left(x+h\right).\hfill \\ & =\underset{h\to 0}{\text{lim}}\frac{2xh-2h+{h}^{2}}{h}\hfill & & & \text{Simplify.}\hfill \\ & =\underset{h\to 0}{\text{lim}}\frac{h\left(2x-2+h\right)}{h}\hfill & & & \text{Factor out}\phantom{\rule{0.2em}{0ex}}h\phantom{\rule{0.2em}{0ex}}\text{from the numerator.}\hfill \\ & =\underset{h\to 0}{\text{lim}}\left(2x-2+h\right)\hfill & & & \text{Cancel the common factor of}\phantom{\rule{0.2em}{0ex}}h.\hfill \\ & =2x-2\hfill & & & \text{Evaluate the limit.}\hfill \end{array}$ Find the derivative of $f\left(x\right)={x}^{2}.$ ${f}^{\prime }\left(x\right)=2x$ We use a variety of different notations to express the derivative of a function. In [link] we showed that if $f\left(x\right)={x}^{2}-2x,$ then ${f}^{\prime }\left(x\right)=2x-2.$ If we had expressed this function in the form $y={x}^{2}-2x,$ we could have expressed the derivative as ${y}^{\prime }=2x-2$ or $\frac{dy}{dx}=2x-2.$ We could have conveyed the same information by writing $\frac{d}{dx}\left({x}^{2}-2x\right)=2x-2.$ Thus, for the function $y=f\left(x\right),$ each of the following notations represents the derivative of $f\left(x\right)\text{:}$ ${f}^{\prime }\left(x\right),\text{}\phantom{\rule{0.2em}{0ex}}\frac{dy}{dx},\text{}\phantom{\rule{0.2em}{0ex}}{y}^{\prime },\text{}\phantom{\rule{0.2em}{0ex}}\frac{d}{dx}\left(f\left(x\right)\right).$ f(x) = x-2 g(x) = 3x + 5 fog(x)? f(x)/g(x) fog(x)= f(g(x)) = x-2 = 3x+5-2 = 3x+3 f(x)/g(x)= x-2/3x+5 diron pweding paturo nsa calculus? jimmy how to use fundamental theorem to solve exponential find the bounded area of the parabola y^2=4x and y=16x what is absolute value means? Chicken nuggets Hugh 🐔 MM 🐔🦃 nuggets MM (mathematics) For a complex number a+bi, the principal square root of the sum of the squares of its real and imaginary parts, √a2+b2 . Denoted by \| \|. The absolute value \|x\| of a real number x is √x2 , which is equal to x if x is non-negative, and −x if x is negative. Ismael find integration of loge x find the volume of a solid about the y-axis, x=0, x=1, y=0, y=7+x^3 how does this work Can calculus give the answers as same as other methods give in basic classes while solving the numericals? log tan (x/4+x/2) Rohan Rohan y=(x^2 + 3x).(eipix) Claudia Ismael A Function F(X)=Sinx+cosx is odd or even? neither David Neither Lovuyiso f(x)=1/1+x^2 \|=[-3,1] apa itu? fauzi determine the area of the region enclosed by x²+y=1,2x-y+4=0 Hi MP Hi too Vic hello please anyone with calculus PDF should share Which kind of pdf do you want bro? Aftab hi Abdul can I get calculus in pdf Abdul explain for me Usman okay I have such documents Fitzgerald Hamza How to use it to slove fraction Hello please can someone tell me the meaning of this group all about, yes I know is calculus group but yet nothing is showing up Shodipo You have downloaded the aplication Calculus Volume 1, tackling about lessons for (mostly) college freshmen, Calculus 1: Differential, and this group I think aims to let concerns and questions from students who want to clarify something about the subject. Well, this is what I guess so. Jean Im not in college but this will still help nothing how en where can u apply it Migos how can we scatch a parabola graph Ok Endalkachew how can I solve differentiation? with the help of different formulas and Rules. we use formulas according to given condition or according to questions CALCULUS For example any questions... CALCULUS v=(x,y) وu=(x,y ) ∂u/∂x* ∂x/∂u +∂v/∂x*∂x/∂v=1 log tan (x/4+x/2) Rohan what is the procedures in solving number 1?
# Golden Section To understand what constitutes a golden section, consider the top line segment shown on the next page. The line segment has three of its points marked. Point P partitions the line segment AB into two smaller segments: from left endpoint A to point P (AP ), and from P to right endpoint B (PB ). The line segments AP and PB are also shown individually. A "golden ratio" can now be formed as the ratio of one line segment to another line segment. Point P divides the entire line segment AB into a golden section if the following equation is valid: AP/AB = PB/AP. In other words, the length of line segment AP divided by the length of the entire line segment AB is a ratio, or number, and that number is the same as the ratio of the shortest line segment PB and the segment AP. It turns out that these ratios are equal to the irrational number 0.61803; that is: AP/AB = PB/AP = 0.61803 Note that the decimal places continue indefinitely. The number 0.61803is called the golden ratio or golden number, whereas the term "golden section" refers to the line segments formed by a point, such as P, whose resulting ratio is the golden number. ## Significance of the Golden Ratio The golden section and the golden ratio are important to mathematics, art, and architecture. The construction of the golden section goes back at least as far as the ancient Greeks, to the Pythagoreans (ca. 550 b.c.e.300 b.c.e.). The Pythagoreans believed that numbers were the basis of nature and man, and therefore number relationships like the golden ratio were of utmost importance. Besides line segments, the golden ratio also appears in many geometric figures. For example, a rectangle is said to be in golden section if its width and length are in the golden ratio (that is, if the width divided by the length equals 0.61803). Some scholars believe that various temples of the ancient Greeks, like the Parthenon in Athens, were purposefully produced with various dimensions in the golden ratio. Many of the artists and architects of the Renaissance period are believed to have incorporated the golden ratio into their paintings, sculptures, and monuments. A prime example is Leonardo da Vinci (14521519), who extensively used golden ratios (or their approximations) in his paintings. In Leonardo's famous drawing "Vitruvian Man," the distance ratio from the top of the head to navel, and from the navel to the soles of his feet approximates the golden ratio. Many people feel that geometric forms and figures incorporating the golden ratio are more beautiful than other forms. Psychological studies purportedly show that people find golden-ratio rectangles more appealing than rectangles of other proportions. Philip Edward Koth with William Arthur Atkins ## Bibliography Eves, Howard. An Introduction to the History of Mathematics, 4th ed. New York: Holt, Rinehart and Winston, 1976. ## lnternet Resources The Golden Section Ratio: Phi. Department of Computing, University of Surrey, United Kingdom. <http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/phi.html>. ## ALTERNATIVE DEFINITION OF GOLDEN SECTION Some dictionaries and textbooks define golden section (and number) as the inverse of the definition shown in this article. The formula then becomes (again referring to line segment AB ) AB /AP = AP /PB = 1.61803. Sometimes the larger value is denoted by the Greek letter "Phi"(i.e., Phi = 1.61803) while the smaller value of the golden number is denoted by a "small" phi (i.e., phi = 0.61803). Note that Phi 1 + phi. Pick a style below, and copy the text for your bibliography. • MLA • Chicago • APA "Golden Section." Mathematics. . Encyclopedia.com. 23 May. 2017 <http://www.encyclopedia.com>. "Golden Section." Mathematics. . Encyclopedia.com. (May 23, 2017). http://www.encyclopedia.com/education/news-wires-white-papers-and-books/golden-section "Golden Section." Mathematics. . Retrieved May 23, 2017 from Encyclopedia.com: http://www.encyclopedia.com/education/news-wires-white-papers-and-books/golden-section ## Golden Section Golden Section, in mathematics, division of a line segment into two segments such that the ratio of the original segment to the larger division is equal to the ratio of the larger division to the smaller division. If c is the original segment, b is the larger division, and a is the smaller division, then c = a + b and c/b = b/a. Thus, b is the geometric mean of a and c; the ratio is known as the Divine Proportion. The Golden Rectangle, whose length and width are the segments of a line divided according to the Golden Section, occupies an important position in painting, sculpture, and architecture, because its proportions have long been considered the most attractive to the eye. The constructions of regular polygons of 5, 10, and 15 sides depend on the division of a line by the Golden Section. The numerical ratio of the greater segment of the line to the shorter segment as determined by the Golden Section is symbolized by the Greek letter phi and has the approximate value 1.618. It occurs in many widely varying areas of mathematics. For example, in the Fibonacci sequence (the sequence of numbers formed by adding successive members to find the next member—0, 1, 1, 2, 3, 5, 8, 13, … ), the values of the ratios 1, 2/1, 3/2, 5/3, 8/5, 13/8, … approach the value of the Golden Section. See H. E. Huntley, The Divine Proportion (1970). Pick a style below, and copy the text for your bibliography. • MLA • Chicago • APA "Golden Section." The Columbia Encyclopedia, 6th ed.. . Encyclopedia.com. 23 May. 2017 <http://www.encyclopedia.com>. "Golden Section." The Columbia Encyclopedia, 6th ed.. . Encyclopedia.com. (May 23, 2017). http://www.encyclopedia.com/reference/encyclopedias-almanacs-transcripts-and-maps/golden-section "Golden Section." The Columbia Encyclopedia, 6th ed.. . Retrieved May 23, 2017 from Encyclopedia.com: http://www.encyclopedia.com/reference/encyclopedias-almanacs-transcripts-and-maps/golden-section ## golden section golden section. Also called the golden cut or mean, or harmonic proportional ratio, it may have originated in C6 bc in the circle of Pythagoras, was certainly known during the time of Euclid (c.325–c.250 bc), and was held to be divine by several Renaissance theorists, especially Luca Pacioli (c.1445–c.1514) in his De Divina Proportione, written in 1497 and published in Venice in 1509. It can be expressed as a a straight line (or as a rectangle) divided into into two parts so that the ratio of the shorter part (a) to the longer (b) is the same as the ratio of the longer (b) to the sum of the shorter and longer parts, or a:b = b:a + b, or that the ratio of the smaller part is to the longer as the latter is to the whole. The ratio is expressed in algebra as Φ (Phi, the first letter of the name of the Greek sculptor Phidias, or Pheidias (c.490–430 bc) ) = (1 +√5)/2, which comes to 1.61803. Thus the ratio is approximately 8: 13. Bibliography Borrisavlievitch (1970); Ghyka (1976); Hagenmaier (1977); Huntley (1970)
# How do you write an equation of an ellipse in standard form given vertices (4,-7) and (4,3) and foci (4,-6) and (4,2)? Jan 9, 2017 ${\left(x - 3\right)}^{2} / {3}^{2} - {\left(y + 2\right)}^{2} / {5}^{2} = 1$ The Socratic graph is inserted. #### Explanation: graph{(x-3)^2/3^2+(y+2)^2/5^2-1=0 [-20, 20, -10, 10]} The line joining the foci S(4, 2) and S'(4, -6) is the middle segment of the the major axis joining the vertices A(4, 3) and A'(4, -7). It follows that the common midpoint C(4, -2) is the center of the ellipse, the major axis is along x = 4, in the ($y \uparrow$) and the minor axis is along y = 2, in the $x \rightarrow$. . The distance between the vertices $\forall ' = 2 a = {x}_{A} - {x}_{A} ' = 3 - \left(- 7\right) = 10$., giving a = 5. The distance between the foci $S S ' = 2 a e = 10 e = {x}_{S} - x S ' = 2 - \left(- 6\right) = 8 ,$ giving eccentricity $e = \frac{4}{5}$. The semi minor-axis $= b = a \sqrt{1 - {e}^{2}} = 5 \sqrt{1 - \frac{16}{25}} = 3$. Now, the equation of this ellipse with semi axes a = 5, b = 3, center at C( 4, -2) and axes parallel to the y and x axes, respectively, is ${\left(x - 3\right)}^{2} / {3}^{2} - {\left(y + 2\right)}^{2} / {5}^{2} = 1$
## Presentation on theme: "2.4: Quadratic Functions."— Presentation transcript: Let a, b, and c be real numbers a  0. The function f (x) = ax2 + bx + c is called a quadratic function. The graph of a quadratic function is a parabola. Every parabola is symmetrical about a line called the axis (of symmetry). x y The intersection point of the parabola and the axis is called the vertex of the parabola. f (x) = ax2 + bx + c vertex axis Quadratic function The leading coefficient of ax2 + bx + c is a. y a > 0 opens upward When the leading coefficient is positive, the parabola opens upward and the vertex is a minimum. f(x) = ax2 + bx + c vertex minimum x y vertex maximum When the leading coefficient is negative, the parabola opens downward and the vertex is a maximum. f(x) = ax2 + bx + c a < 0 opens downward Leading Coefficient Case1: a>0 y a > 0 opens upward Minimum value: k Range: x vertex minimum Minimum value: k Range: Increasing: Decreasing: Case2: a<0 y x Maximum value: k vertex maximum Range: a < 0 opens downward vertex maximum Maximum value: k Range: Increasing: Decreasing: Def: The standard form for the equation of a quadratic function is: f (x) = a(x – h)2 + k (a  0) where h=-b/2a k=f(h)=f(-b/2a) 2) Vertex : (h, k) 3) Axis of symmetry : x=h The standard form for the equation of a quadratic function is: f (x) = a(x – h)2 + k (a  0) The graph is a parabola opening upward if a  0 and opening downward if a  0. The axis is x = h, and the vertex is (h, k). Example: By completing the square method write the parabola f (x) = 2x2 + 4x – 1in standard form and find the axis and vertex. x y x = –1 f (x) = 2x2 + 4x – 1 f (x) = 2x2 + 4x – original equation f (x) = 2( x2 + 2x) – factor out 2 f (x) = 2( x2 + 2x + 1) – 1 – 2 complete the square f (x) = 2( x + 1)2 – standard form a > 0  parabola opens upward . (–1, –3) h = –1, k = –3  axis x = –1, vertex (–1, –3). Quadratic Function in Standard Form Ex1: For the following functions Write the function in the standard form Find the vertex Find the axis of symmetry Find , if any, the maximum value of the function Find , if any, the minimum value of the function Find the range of the function Find the interval(s) of increasing and decreasing Sketch the graph of the function and show on the graph the intercept(s), the vertex, and the axis of symmetry a) f(x)=2x2+4x+3 b) f(x)=-x2+2x+3 f (x) = a(x – h)2 + k standard form Example: Find an equation for the parabola with vertex (2, –1) passing through the point (0, 1). y x y = f(x) (0, 1) (2, –1) f (x) = a(x – h)2 + k standard form f (x) = a(x – 2)2 + (–1) vertex (2, –1) = (h, k) Since (0, 1) is a point on the parabola: f (0) = a(0 – 2)2 – 1 1 = 4a –1 and Example: Parabola The maximum height of the ball is 15 feet. Example: A basketball is thrown from the free throw line from a height of six feet. What is the maximum height of the ball if the path of the ball is: The path is a parabola opening downward. The maximum height occurs at the vertex. At the vertex, So, the vertex is (9, 15). The maximum height of the ball is 15 feet. Example: Basketball Let x represent the width of the corral and 120 – 2x the length. Example: A fence is to be built to form a rectangular corral along the side of a barn 65 feet long. If 120 feet of fencing are available, what are the dimensions of the corral of maximum area? barn corral x 120 – 2x Let x represent the width of the corral and 120 – 2x the length. Area = A(x) = (120 – 2x) x = –2x x The graph is a parabola and opens downward. The maximum occurs at the vertex where a = –2 and b = 120 120 – 2x = 120 – 2(30) = 60 The maximum area occurs when the width is 30 feet and the length is 60 feet. Example: Maximum Area Q85/227 Find two numbers whose sum is 8 and whose product is a maximum. Ex: If x is a real number, then find the maximum area of a rectangle of length 3+2x and width 1-2x. Ex: If x=3 is the axis of symmetry of the parabola y=-2x2+cx+2, then find c The End Example: Basketball Similar presentations
# Project Euler 56 Solution: Powerful digit sum Problem 56 A googol (10100) is a massive number: one followed by one-hundred zeros; 100100 is almost unimaginably large: one followed by two-hundred zeros. Despite their size, the sum of the digits in each number is only 1. Considering natural numbers of the form, ab, where a, b < 100, what is the maximum digital sum? ## Solution The most naive way of solving this problem is using a language like Python: print(max(sum(map(int, str(ab))) for a in range(100) for b in range(100))) In order to improve performance and to explore the inner of the problem a little better, let's see what information we can gather. According to Problem 25, the number length function $$L(n) = \lfloor 1 + log_{10}(n)\rfloor$$ and therefore $$L(a^b) = \lfloor 1 + \log_{10}(a^b)\rfloor = \lfloor 1 + b\cdot\log_{10}(a)\rfloor$$. The upper limit is then $$9\cdot L(a^b)$$ and if we say all digits are uniformly distributed, the expected sum of the digits would be $$5\cdot L(a^b)$$, which gives $$9\cdot L(99^{99}) = 9\cdot 198 = 1782$$, $$5\cdot L(99^{99}) = 5\cdot 198 = 999$$ and similarly $$9\cdot L(90^{90}) = 9\cdot 176 = 1584$$, $$5\cdot L(90^{90}) = 5\cdot 176 = 880$$. Since the expected value raises dramatically from 880 to 990 in this small interval, it's probably safe to start with a=90 and b=90: print(max(sum(map(int, str(ab))) for a in range(90, 100) for b in range(90, 100))) Which solves the problem as well and improves the performance noticeable. However, while the assumption for $$a$$ is totally fine, the assumption for $$b$$ was just luck. We make the approach a little more robust and implement the number length function and break early, if the upper bound is less then the current maximum, which is: import math def L(a, b): return 1 + math.floor(b * math.log(a, 10)) maxi = 0 for b in range(99, 1, -1): if 9 * L(99, b) < maxi: break for a in range(99, 90, -1): maxi = max(maxi, sum(map(int, str(a**b)))) print(maxi) Going backwards helps quite a lot here. The last idea that could be used is using a diagonalizing function like the inverse Cantor pairing function and move step by step down from 99. Since this requires to calculate the triangular root, we skip the calculations and keep the result like this. « Back to problem overview
# How do you simplify (x+5)/(x-3)*(7x^2 - 21x) / (7x)? Sep 21, 2015 $x + 5$ #### Explanation: Completely factor out all expressions. You can factor out 7x from $\left(7 {x}^{2} - 21 x\right)$. $\frac{x + 5}{x - 3} \cdot \frac{7 {x}^{2} - 21 x}{7 x}$ $= \frac{x + 5}{x - 3} \cdot \frac{\left(7 x\right) \left(x - 3\right)}{7 x}$ $= \frac{\left(x + 5\right) \left(7 x\right) \left(x - 3\right)}{\left(7 x\right) \left(x - 3\right)}$ After you have factored everything out, cancel factors that appear in both the numerator and the denominator. In this problem, $7 x$ and $\left(x - 3\right)$ appear in both the numerator and denominator, so you can cancel them. $\frac{\left(x + 5\right) \cancel{\left(7 x\right)} \cancel{\left(x - 3\right)}}{\cancel{\left(7 x\right)} \cancel{\left(x - 3\right)}}$ $= \frac{x + 5}{1}$ $= x + 5$
# Mean\|Definition & Meaning ## Definition The mean, also known as the average, is the measure of the central tendency of a given data set. It is calculated by first adding the total values given in a data set and then dividing it by the total number of values. Figure 1 shows the calculation of the mean of four numbers 3, 6, 9, and 12, given in a data set. Figure 1 – Demonstration of Mean as a Measure of Central Tendency of a Given Data Set Notice how the mean is located at the center of the data set. ## Mean’s General Formula The mean is the sum of all the values in the given data divided by the total number of values, so its formula can be written as: Mean or Average = Sum of Values / Total Number of Values ## Other Measures of Central Tendency The other measures of central tendency are the mode and the median. Each measure of central tendency indicates a different central value of a given set of numbers. ### Median The middle value of a data set is known as the median when the numbers in the data set are arranged in ascending order. ### Mode The value most repeated in a given set of numbers is known as its mode. Figure 2 shows the calculation of the mean, mode, and median from a given set of numbers. Figure 2 – Calculation of Mean, Mode and Median for a Given Set of Values ## Relationship Between Mean, Mode, and Median The relationship between mean, mode, and median, also known as the “empirical formula,” exists only for a skewed distribution. It is given as: 2 * Mean = 3 * Median – Mode If the mode and median for a given skewed distribution, the mean can be calculated from the above formula by dividing it by 2 as: Mean = (3 * Median – Mode) / 2 ## Calculation of Mean for Different Types of Data In statistics, there are two types of data: grouped data and ungrouped data. For both types of data, the formulas for calculating the mean are different. ### Grouped Data To calculate the mean $\overline{Y}$ for grouped data, the following formula is used: $\overline{Y}_{\text{grouped data}} = \frac{\sum f_i y_i}{ \sum f_i }$ Where f is the frequency of the values in the frequency distribution grouped data. The numerator is calculated by first multiplying the values $y_1,\, y_2,\, \ldots,\, y_i$ with their respective frequencies $f_1,\, f_2,\,\ldots,\, f_i$. The products $f_{1}y_{1},\, f_{2}y_{2},\, \ldots,\, f_{i}y_{i}$ are added to give the numerator. The denominator is obtained by adding the frequencies. Figure 3 shows the mean calculation for grouped data. Figure 3 – Mean Calculation for a Grouped Data (Frequency Distribution) ### Ungrouped Data For ungrouped data, only the values $y_1,\, y_2,\, \ldots,\, y_m$ are given. The mathematical formula for mean $\overline{Y}$ for ungrouped data is: $\overline{Y}_{\text{ungrouped data}} = \frac{ y_1 + y_2 + \cdots + y_m}{m}$ Where “m” is the number of values. If some of the values in the ungrouped data are negative, the same formula is used to calculate the mean. ## Major Types of Mean There are three main types of mean, arithmetic mean, geometric mean, and harmonic mean. ### Arithmetic Mean The arithmetic is the sum of all the observations divided by the total number of observations. The A.M is given by the formula: $A.M = \frac{ \sum y_m}{m}$ For example, the arithmetic mean of 2, 7, 1, and 5 is: A.M = (2 + 7 + 1 + 5) / 4 = 3.75 Till now we have discussed the arithmetic mean, as by “mean” the arithmetic mean is usually considered. ### Geometric Mean The geometric mean G.M of two numbers p and q is the product of p and q to the square root of its product. Mathematically, it is: G.M = $\sqrt{p \times q}$ For example, the geometric mean of 6 and 9 is: G.M = $\sqrt{6 \times 9}$ = $\sqrt{54}$ = 7.348 The geometric mean of “m” values $y_1,\, y_2,\, \ldots,\, y_m$ is the product of these values to the mth root or 1/m power. So, in general, the G.M is: $G.M = \sqrt[m]{ y_{1} \times y_{2} \times \cdots \times y_{m} }$ ### Harmonic Mean The harmonic mean H.M of two numbers p and q is given by: H.M = (2pq) / (p + q) For example, the harmonic mean of 4 and 6 is: H.M = 2(4)(6) / (4 + 6) = 48 / 10 = 4.8 For m values $y_1,\, y_2,\, \ldots,\, y_m$, the harmonic mean is: $\text{H.M} = m \div \left[ \dfrac{1}{y_1} + \dfrac{1}{y_2} + \cdots + \dfrac{1}{y_m} \right]$ Figure 4 shows the three measures of central tendency with the further categorization of mean and their formulas for two numbers p and q. Figure 4 – Classification of the Measures of Central Tendency and Types of Mean with their Formulas ## Relationship Between Arithmetic, Geometric and Harmonic Mean For a given set of data, the relation between A.M, G.M, and H.M is: A.M ≥ G.M ≥ H.M The three means will be equal if all the values of the data set are the same. ## Other Types of Mean Some less common types of the mean are given as follows: ### Contraharmonic Mean For two numbers p and q, the contraharmonic mean C.M will be: $\text{C.M} = \dfrac{p^2 + q^2}{p + q}$ For m numbers $y_1,\, y_2,\,\ldots,\,y_m$, the C.M is: $C.M = \frac{ {y_1}^2 + {y_2}^2 + \cdots + {y_m}^2}{ y_1 + y_2 + \cdots + y_m }$ ### Root Mean Square The root mean square $Y_{rms}$ is used to make the negative values positive in a data set. Its formula is given as: $Y_{rms} = \sqrt{ \frac{ {y_1}^2 + {y_2}^2 + \cdots + {y_m}^2 }{m} }$ Where $y_1,\, y_2,\, \ldots,\, y_m$ are the m number of values of a data set. ### A Function’s Mean The mean $x_{avg}$ of the function g(x) is the area under the function’s curve from c to d divided by the total length the area is covering. Mathematically, it is written as: $x_{avg} = \frac{1}{d-c} \int_{c}^{d} g(x) dx$ ## Example Calculate the A.M, G.M, and H.M of the given set of values: 6, 8, 3, 10, 12 Show that: A.M ≥ G.M ≥ H.M ### Solution The arithmetic mean is calculated as follows: $A.M = \frac{ \sum y_m }{m}$ A.M = (6 + 8 + 3 + 10 + 12) / 5 = 39 / 5 A.M = 7.8 The geometric mean will be: $G.M = \sqrt[5]{ y_{1} \times y_{2} \times y_{3} \times y_{4} \times y_{5} }$ G.M = ${ \sqrt[5]{6 \times 8 \times 3 \times 10 \times 12} }$ G.M = 7.03 The harmonic mean will be: $\text{H.M} = m \div \left[ \dfrac{1}{y_1} + \dfrac{1}{y_2} + \dfrac{1}{y_3} + \dfrac{1}{y_4} + \dfrac{1}{y_5} \right]$ H.M = 5 ÷ [(1 / 6) + (1 / 8) + (1 / 3) + (1 / 10) + (1 / 12)] H.M = 6.18 As: 7.8 ≥ 7.03 ≥ 6.18 Hence: A.M ≥ G.M ≥ H.M All the images are created using GeoGebra.
# What to Do When Math Fails You on the GMAT! They say Mathematics is a perfect Science. There is a debate over this among scientists but we can definitely say that Mathematical methods are not perfect so we cannot use them blindly. We could very well use the standard method for some given numbers and get stranded with “no solution.” The issue is what do we do when that happens? For example, review this post on averages. Here we saw that: Average Speed = 2ab/(a + b) Applicable when one travels at speed a for half the distance and speed b for other half of the distance. In this case, average speed is the harmonic mean of the two speeds. So now, say if we have a question which looks like this: Question: In the morning, Chris drives from Toronto to Oakville and in the evening he drives back from Oakville to Toronto on the same road. Was his average speed for the entire round trip less than 100 miles per hour? Statement 1: In the morning, Chris drove at an average speed of at least 10 miles per hour while travelling from Toronto to Oakville. Statement 2: In the evening, Chris drove at an average speed which was no more than 50 miles per hour while travelling from Oakville to Toronto. Solution: We know that the question involves average speed. The case involves travelling at a particular average speed for one half of the journey and at another average speed for the other half of the journey. So average speed of the entire trip will be given by 2ab/(a+b) But the first problem is that we are given a range of speeds. How do we handle ‘at least 10’ and ‘no more than 50’ in equation form? We have learnt that we should focus on the extremities so let’s analyse the problem by taking the numbers are the extremities:10 and 50 Statement 1: In the morning, Chris drove at an average speed of at least 10 miles per hour while travelling from Toronto to Oakville. What if Chris drives at an average speed of 10 mph in the morning and averages 100 mph for the entire journey? What will be his average speed in the evening? Perhaps around 200, right? Let’s see. 100 = 210b/(10 + b) 1000 + 100b = 20b 1000 = -80b b = – 1000/80 How can speed be negative? Let’s hold on here and try the same calculation for statement 2 too. Statement 2: In the evening, Chris drove at an average speed which no more than 50 miles per hour while travelling from Oakville to Toronto. If Chris drives at an average speed of 50 mph in the evening, and averages 100 mph, let’s find his average speed in the morning. 100 = 2a*50/(a + 50) 100a + 5000 = 100a 5000 = 0 This doesn’t make any sense either! What is going wrong? Look at it conceptually: Say, Toronto is 100 miles away from Oakville. If Chris wants his average speed to be 100 mph over the entire trip, he should cover 100+100 = 200 miles in 2 hrs. What happens when he travels at 10 mph in the morning? He takes 100/10 = 10 hrs to reach Oakville in the morning. He has already taken more time than what he had allotted for the entire round trip. Now, no matter what his speed in the evening, his average speed cannot be 100mph. Even if he reaches Oakville to Toronto in the blink of an eye, he would have taken 10 hours and then some time to cover the total 200 miles distance. So his average speed cannot be equal to or more than 200/10 = 20 mph. Similarly, if he travels at 50 mph in the evening, he takes 2 full hours to travel 100 miles (one side distance). In the morning, he would have taken some time to travel 100 miles from Toronto to Oakville. Even if that time is just a few seconds, his average speed cannot be 100 mph under any circumstances. But statement 1 says that his speed in morning was at least 10 mph which means that he could have traveled at 10 mph in the morning or at 100 mph. In one case, his average speed for the round trip cannot be 100 mph and in the other case, it can very well be. Hence statement 1 alone is not sufficient. On the other hand, statement 2 says that his speed in the evening was 50 mph or less. This means he would have taken AT LEAST 2 hours in the morning. So his average speed for the round trip cannot be 100 mph under any circumstances. So statement 2 alone is sufficient to answer this question with ‘No’.
# Math: How Do I Work Out Percentages? ## Simple Sums There all sorts of great math self-help books out there- some people just have a knack for explaining numbers in a way that's easy to understand. When it comes to calculating percentages, though, it really couldn't be easier to get your head around. The key to understanding percentages is to remind yourself that percentages are just a shorthand way of describing a number as a fraction, as a one-hundredth of another number. If you want to know how to calculate percentages, think to yourself "I want to know how many hundredths of this number over there my number right here is equal to." Put algebraically, when calculating percentages, you want to know how many one hundredths of X fit into Y, or Y / (X/100). Does that sound simple? It really is. Calculating percentages is really easy. Let's try an example. Say you want to know what percentage of a bag of marbles are red. There are 75 marbles, and 33 of them are red. So... you want to know how many one-hundredths of 75 fit into 33. First of all, divide 75 by 100. To do this, move the decimal point two places to the left. This gives you 0.75. Next, see how many lots of 0.75 fit into 33, so 33/0.75 =44. Therefore, 44% of the marbles are red. See? That's how to calculate percentages. ## Calculating Percentage Change This is where it gets a little bit more complicated. At least it seems to, but really calculating percentage change is very similar to calculating percentages. You might struggle to understand which number to divide by 100 and which needs to be divided by the number of hundredths. But if you think it through slowly, it's not so hard. To understand percentages you want to know a number that portrays the change as a proportion of the original value. Looking algebraically again, you want to know Y-X / (X/100). In this example, X is the original value, and Y is the new value, so Y-X gives you the difference. To calculate the percentage change, you just divide the difference by one hundredth of the original value (X). To give an example, imagine you want to calculate the percentage change in a magazine's circulation over the last six months. The circulation today is 54,000. Six months ago, it was 29,000. So to calculate the percentage change, first work out the difference: 54,000 - 29,000 = 25,000. Next you want to know what 25,000 is as a proportion of the initial value, so 25,000 / (29,000/100). Move the decimal two places to the left on the 29,000, so now you want to know how many times 290 fits into 25,000. 25,000 / 290 = 86.2. Therefore, the magazine grew its circulation by 86.2% over six months. 49 5 16 7 ## Popular 1 31 • ### The Number 9 \| The Secret Knowledge of The Ancients Number Nine Code 911 113 0 of 8192 characters used
Share # Decimal Division Without a Calculator Step-by-Step With Examples By Mary Smith. Updated: March 11, 2022 Doing decimal divisions without a calculator may seem difficult, but there is an easy method which we are going to show you. The trick is simply to remove the decimal point in the division and then replace the decimal point when you get to the answer. Although it may seem very complicated at first glance, the fact is that this little trick makes it quite easy. So keep reading as we explain our method step-by-step with a few examples and discover how to do divisions with decimals without using a calculator. ## Dividing a Decimal by a Whole Number Use a normal division method, ignoring the decimal point. Then put the decimal point in the same place once you have obtained the divided figure. Look at the example below: • 9.1 divided by 7 You remove the decimal point in 9.1 and the division is now: 91 ÷ 7 This makes the division much easier to calculate - the result is 13. Remember for the last step, you have to reinsert the decimal point into this result. 13 is now 1.3 - yes, its as easy as that! ## How to Divide by a Decimal number But why are you going to divide a decimal number? The method this time is to convert the numbers you are dividing into a whole number. This exercise is solved by moving the decimal point of both numbers to the right. Example: • 6.625 divided by 0.53 Move the decimal point two spaces, or the amount of space required to make it become an integer. Now you are dividing by an integer and you can continue as normal: 662.5 ÷ 53 This is a very straight forward method but you must remember to move the decimal point by the same amount in both numbers. Here is another example: • 5.39 divided by 1.1 At the moment you are not dividing by an integer, so you need to move the decimal point. Move a single space and you now have: 53.9 ÷ 11 The rest of the division is easy because, as you learned at the beginning of this maths class, to divide a decimal number by a whole number you just need to forget about the decimal point for the first step and then put it back to get the final result. Continuing the example: 53.9 ÷ 11 539 ÷ 11 049 The result is 4.9! So there you go, above you have the easiest way to divide decimals without a calculator. If you want to read similar articles to Decimal Division Without a Calculator Step-by-Step With Examples, we recommend you visit our Learning category.
# Factoring Polynomials ## Presentation on theme: "Factoring Polynomials"— Presentation transcript: Factoring Polynomials Section 0.4 Factor by finding greatest common factors (GCF) EXAMPLES Factor the expression by finding what each of the terms have in common. These are called greatest common factors and you should always look for them. 4x + 16 n2 – 3n r2 + 2r – 63 ANSWER ANSWER ANSWER 4(x + 4) n(n – 3) (r + 9)(r –7) Factor trinomials of the form x2 + bx + c as a product of two binomials EXAMPLE Factor the expression. a. x2 – 9x + 20 b. x2 + 3x – 12 SOLUTION a. You want x2 – 9x + 20 = (x + m)(x + n) where mn = 20 and m + n = –9. ANSWER Notice that m = –4 and n = –5. So, x2 – 9x + 20 = (x – 4)(x – 5). Factor trinomials of the form x2 + bx + c as a product of two binomials EXAMPLE b. You want x2 + 3x – 12 = (x + m)(x + n) where mn = – 12 and m + n = 3. ANSWER Notice that there are no factors m and n such that m + n = 3. So, x2 + 3x – 12 cannot be factored. We call that trinomial prime or irreducible. GUIDED PRACTICE Factor the expression. If the expression cannot be factored, say so. x2 – 3x – 18 n2 – 3n + 9 r2 + 2r – 63 ANSWER ANSWER ANSWER (x – 6)(x + 3) cannot be factored (r + 9)(r –7) Factor with special patterns EXAMPLE Factor with special patterns Factor the expression. a. x2 – 49 = x2 – 72 Difference of squares = (x + 7)(x – 7) b. d d + 36 = d 2 + 2(d)(6) + 62 Perfect square trinomial = (d + 6)2 c. z2 – 26z + 169 = z2 – 2(z) (13) + 132 Perfect square trinomial = (z – 13)2 GUIDED PRACTICE Factor the expression, notice special patterns. x2 – 9 ANSWER (x – 3)(x + 3) q2 – 100 ANSWER (q – 10)(q + 10) y2 + 16y + 64 ANSWER (y + 8)2 More factoring with special patterns EXAMPLE More factoring with special patterns Factor the expression. a. 9x2 – 64 = (3x)2 – 82 Difference of squares = (3x + 8)(3x – 8) b. 4y2 + 20y + 25 = (2y)2 + 2(2y)(5) + 52 Perfect square trinomial = (2y + 5)2 c w2 – 12w + 1 = (6w)2 – 2(6w)(1) + (1)2 Perfect square trinomial = (6w – 1)2 GUIDED PRACTICE GUIDED PRACTICE Factor the expression, watch for special patterns. 16x2 – 1 ANSWER (4x + 1)(4x – 1) 9y2 + 12y + 4 ANSWER (3y + 2)2 4r2 – 28r + 49 ANSWER (2r – 7)2 25s2 – 80s + 64 ANSWER (5s – 8)2 GUIDED PRACTICE GUIDED PRACTICE 49z2 + 4z + 9 ANSWER (7z + 3)2 36n2 – 9 ANSWER (6n – 3)(6n +3) Standardized Test Practice EXAMPLE Standardized Test Practice SOLUTION x2 – 5x – 36 = 0 Write original equation. (x – 9)(x + 4) = 0 Factor. x – 9 = 0 or x + 4 = 0 Zero product property x = 9 or x = –4 Solve for x. ANSWER The correct answer is C. EXAMPLE Use a quadratic equation as a model Nature Preserve A town has a nature preserve with a rectangular field that measures 600 meters by 400 meters. The town wants to double the area of the field by adding land as shown. Write an equation that will represent the new field. Use a quadratic equation as a model EXAMPLE Use a quadratic equation as a model SOLUTION 480,000 = 240, x + x2 Multiply using FOIL. 0 = x x – 240,000 Write in standard form. EXAMPLE Factor ax2 + bx + c where c > 0 Factor 5x2 – 17x + 6. SOLUTION You want 5x2 – 17x + 6 = (kx + m)(lx + n) where k and l are factors of 5 and m and n are factors of 6. You can assume that k and l are positive and k ≥ l. Because mn > 0, m and n have the same sign. So, m and n must both be negative because the coefficient of x, –17, is negative. EXAMPLE Factor ax2 + bx + c where c > 0 ANSWER The correct factorization is 5x2 –17x + 6 = (5x – 2)(x – 3). EXAMPLE Factor ax2 + bx + c where c < 0 Factor 3x2 + 20x – 7. SOLUTION You want 3x2 + 20x – 7 = (kx + m)(lx + n) where k and l are factors of 3 and m and n are factors of –7. Because mn < 0, m and n have opposite signs. ANSWER The correct factorization is 3x2 + 20x – 7 = (3x – 1)(x + 7). GUIDED PRACTICE GUIDED PRACTICE Factor the expression. If the expression cannot be factored, say so. 7x2 – 20x – 3 ANSWER (7x + 1)(x – 3) 5z2 + 16z + 3 ANSWER (5z + 1)(z + 3). 2w2 + w + 3 ANSWER cannot be factored GUIDED PRACTICE GUIDED PRACTICE Factor the expression. If the expression cannot be factored, say so. 3x2 + 5x – 12 ANSWER (3x – 4)(x + 3) 4u2 + 12u + 5 ANSWER (2u + 1)(2u + 5) 4x2 – 9x + 2 ANSWER (4x – 1)(x – 2) EXAMPLE Factor out monomials first Factor the expression. a. 5x2 – 45 = 5(x2 – 9) = 5(x + 3)(x – 3) b. 6q2 – 14q + 8 = 2(3q2 – 7q + 4) = 2(3q – 4)(q – 1) c. –5z2 + 20z = –5z(z – 4) d. 12p2 – 21p + 3 = 3(4p2 – 7p + 1) GUIDED PRACTICE GUIDED PRACTICE Factor the expression. 3s2 – 24 ANSWER 3(s2 – 8) 8t2 + 38t – 10 ANSWER 2(4t – 1) (t + 5) 6x2 + 24x + 15 ANSWER 3(2x2 + 8x + 5) 12x2 – 28x – 24 ANSWER 4(3x + 2)(x – 3) –16n2 + 12n ANSWER –4n(4n – 3) GUIDED PRACTICE GUIDED PRACTICE 6z2 + 33z + 36 ANSWER 3(2z + 3)(z + 4)
The Beauty of Trigonometry: A Primer The Beauty of Trigonometry: A Primer Trigonometry is an essential mathematical tool that is used in a variety of fields, from engineering and architecture to astronomy and navigation. It is often seen as a challenging subject, but it can also be incredibly beautiful and rewarding. This primer will explore the basics of trigonometry and how it can be used to solve a variety of problems. Introduction to Trigonometry Trigonometry is a branch of mathematics that deals with the relationships between the sides and angles of triangles. It is based on the fundamental theorem of Pythagoras, which states that in a right-angled triangle, the sum of the squares of the two shorter sides is equal to the square of the longest side. This theorem is used to calculate the length of the sides of a triangle when the angles and one side are known. Trigonometry is also used to calculate angles in a triangle when the lengths of the sides are known. This is known as the Law of Sines, which states that the ratio of the length of a side to the sine of its opposite angle is equal for all angles in the triangle. Trigonometry is often used in engineering and architecture to calculate the angles and lengths of structures. It is also used in astronomy to calculate the positions of stars and planets, and in navigation to calculate distances and directions. Examples of Trigonometry in Action Trigonometry is used in a variety of fields to solve a variety of problems. Here are a few examples of how trigonometry can be used: In engineering and architecture, trigonometry can be used to calculate the angles and lengths of structures, such as bridges and buildings. It can also be used to calculate the angles and lengths of cables and other components. In astronomy, trigonometry can be used to calculate the positions of stars and planets. It can also be used to calculate the orbits of satellites and other objects in space. In navigation, trigonometry can be used to calculate distances and directions. It can also be used to calculate the speed and direction of ships and aircraft. In mathematics, trigonometry can be used to calculate the area and volume of shapes. It can also be used to calculate the circumference and area of circles. FAQ Section Q: What is trigonometry? A: Trigonometry is a branch of mathematics that deals with the relationships between the sides and angles of triangles. It is based on the fundamental theorem of Pythagoras and the Law of Sines. Q: What is the fundamental theorem of Pythagoras? A: The fundamental theorem of Pythagoras states that in a right-angled triangle, the sum of the squares of the two shorter sides is equal to the square of the longest side. Q: What is the Law of Sines? A: The Law of Sines states that the ratio of the length of a side to the sine of its opposite angle is equal for all angles in the triangle. Q: What are some examples of how trigonometry can be used? A: Trigonometry can be used in engineering and architecture to calculate the angles and lengths of structures, in astronomy to calculate the positions of stars and planets, and in navigation to calculate distances and directions. In mathematics, it can be used to calculate the area and volume of shapes, and the circumference and area of circles. Summary Trigonometry is an essential mathematical tool that is used in a variety of fields. It is based on the fundamental theorem of Pythagoras and the Law of Sines, and can be used to calculate the angles and lengths of triangles, as well as the positions of stars and planets, distances and directions, and the area and volume of shapes. Conclusion Trigonometry is a powerful and fascinating subject. It is an essential mathematical tool that is used in a variety of fields, from engineering and architecture to astronomy and navigation. With a basic understanding of the fundamentals of trigonometry, you can use it to solve a variety of problems and gain a deeper appreciation of the beauty of mathematics. Scroll to Top
Understanding the Same Side Interior Angles Theorem 1) Introduction to the Same Side Interior Angles Theorem: What is it and How Can It Help You Solve Geometric Problems? The Same Side Interior Angles Theorem is an essential foundational building block of geometry and can be used to greatly simplify a geometric proof. It states that the measure of any two angles that lie within the same plane, on the same side of an intersecting line, must be equal in value. To put it another way: when two lines meet to form a vertex, if a pair of angles are situated on either side or “interior” of both lines, then those angles are always congruent (i.e. they have the same numerical value). The importance of this theorem is that it provides us with an incredibly useful starting point in solving many basic geometric proofs. As long as we can draw two straight lines which intersect at a particular point, we can construct an angle on one side and its corresponding angle opposite it (on the other side) according to the theorem’s rules. By knowing that such angles have equal values, our task becomes significantly easier since we don’t need to seek out other complex ways to prove their congruency! Furthermore, once these angles are established, we can use them as connections for more sophisticated varieties and creates trajectories for future reasoning without having to start from scratch all over again with each proof – making it easy to piece together all the facts logically and quickly! 2) Step-by-Step Demonstration of Applying the Same Side Interior Angles Theorem The same side interior angles theorem is an essential concept to understand when studying geometry. What follows is a step-by-step demonstration of how to apply the theorem in practice. First, let’s start by understanding what the theorem says. In its simplest terms, it states that if two parallel lines are intersected by a third transversal line, then all corresponding angles formed will be congruent (i.e., have the same measure). This can also be written as â 1 = â 2 and â 3 = â 4. Using this information, we can solve for various unknown angle measures using only a few measurements taken from the figure we draw up. To get started drawing up those figures, use your ruler or straightedge to draw two parallel lines on the page; these are going to be your “parallel lines” within the context of this theorem. Then, draw a third line and orient it so that it intersects both of your previously drawn parallel linesâthis is considered our transversal line (“transversal” meaning “intersecting”). Label each of these four angles as provided in the statement (â 1,â 2,â 3 and â 4) so you know which angles correspond to one another within this demonstration (see Diagram A). Now all you need to do is measure each given angle with either a protractor or measuring triangle; which method you choose doesn’t matter as long as you get accurate measurements! Once measured, record them all down in response to their respective labels; now we’ve acquired all that’s necessary for solving any remaining questions concerning these measurements (but they don’t all have an equal chance at actually being answered!) Depending upon what goal has been assigned to usâlike calculating an unknown angle measure or proving/disproving something specific like that none of these congruencies existâwe can go about tackling those tasks accordingly. For instance, take a look at Diagram B below: If given two known angle values (say 12 degrees for â 1 and 5 degrees for â 2), then using our same side interior angles theorem knowledge we can conclude without measuring anything else that both â 3 and â 4 must also equate 10 degrees since they share complementary sides with their respective counterpartsâthat is 12 + 5 = 10 + 10 â 20Â°! To say this another way: Since they are sharing sides relative to their place within the figure sketched out above and based off our theorem assumption stating such complementary areas must always matchup in measurement amounts when dealing with intersecting parallels? Therefore itâs assumed one corner must automatically contain an identical measures opposite counterpart amount at its neighbor! And there we have itâa complete step-by-step visual explanation for how one might go about applying and utilizing their newly gained knowledge regarding Same Side Interior Angles Theorems! So next time you come across such questions asking anything from tricky proofs requiring no congruency proportions existing whatsoever alongside said problem–to needing certain value determinations uncovered quickly but surely–try again referring these steps offered for whenever emergency solutions lacking solutions arise most unexpectedly! 3) Special Cases and Variations of the Same Side Interior Angles Theorem The Same Side Interior Angles Theorem is a simple idea with far-reaching implications: two lines that are crossed by another line create âinterior anglesâ on the same side of the crossing line. These interior angles add up to 180 degrees. While this theorem holds true for a wide variety of crossing lines and shapes, there are some special cases and variations where it doesnât quite hold up. In an isosceles triangle, for example, two sides are equal in length. All three interior angles can be added together to get 180 degrees – the same result as the Same Side Interior Angles Theorem gives when applied to basic right triangles. But in this case, if you look at the two smaller interior angles â those created at either end of the congruent sides â they do not add up to 180 degrees. This variation on the Same Side Interior Angles Theorem results in what is known as the Triangle Angle Sum Theorem: in an isosceles triangle all of its angle sums must total 180 degrees regardless of any individual pairings or sums. Another scenario that requires an alternate theorem is when shapes share identical lines but arenât converging properly to create full figures, such as two parallel lines sharing a third intersecting line further away from them both than normal. If you overlap various circles and straight segments like this, those circles don’t always make up a full figure – and so, no matter how many other internal angles may exist that all equate correctly when summed into single pairs, there won’t be any closure from one side to another around them. This will mean exterior angles residing outside of any closed shape can not equitably sum into internal residuals – meaning you’ll have situations unique from all others encountered before given different proportions causes by shared ropes cutting specific crop circles outside set parameters established through archetypal variable models hitherto yoked to overlaid existing notions pre-prefiguring mental images ready transmitted throughout history leading inevitably toward current states unwittingly embraced beyond all expectation by unknowing masses who think they know it all across the board… but actually don’t! And thus new maths equations apply better than old ones hereabouts! So whenever coupled conic sections move deviations from ordinary normality due to particular positionings â find out how much each interlocking mesh deviates for acute diagnostics concerning indivisible elemental qualities (beyond amount totals) dominating necessity fates constructed visually apparent between centralized spirals spinning unexpectedly onwards through temporal veil requiring multiplied reactants causing amplified diversions founded upon compound correlations imputed crosswise otherwise than standard representations.. Or else!! 4) Frequently Asked Questions (FAQs) Related to Using the Same Side Interior Angles Theorem The Same Side Interior Angle Theorem is an important concept in geometry that states that if two lines are cut by a transversal then the corresponding angles on the same side of the transversal add up to 180Â°. In other words, if lines M and N are crossed by a third line called a transversal (t) then any pair of corresponding angles on the same side of the transvesal (Îź and Î˝) will add up to 180Â° or one half of a circle. It’s essential for students to understand this theorem because it’s not only used in basic geometry problems but it can be used in proofs as well. FAQs: Q1: What does the Same Side Interior Angles Theorem state? A1: The Same Side Interior Angle Theorem states that if two lines are cut by a transversal, then the corresponding angles on the same side of that transversal add up to 180Â°. Q2: How is this theorem used in mathematics? A2: This theorem is essential for students to understand because it is not only used in basic geometry problems but also used in proofs. For instance, you could use this theorem when proving properties about various triangles or showing relationships between different sides and/or angles within a triangle. Q3: Are there any special conditions associated with this theorem? A3: Yes, there are several important conditions associated with using this theorem. For example, all three lines must be non-parallel so they can create intersecting pairs of interior angles; these interior angles must also be adjacent (touching at their vertex). Additionally, transverse angles pair with corresponding angles on opposite sides instead of adjacent angles which would create supplementary angles instead (adding up to 180Â° as well). 5) Top 5 Facts About Applying the Same Side Interior Angles Theorem The Side-Angle-Side (SAS) theorem states that if two sides and the included angle of one triangle are equal to two sides and the included angle of another triangle, then the triangles are congruent. Applying this theorem can help us prove many important geometric properties. Here are five facts about applying the Same Side Interior Angles Theorem: 1. We can use SAS to prove triangles congruent in a number of cases: when all three interior angles or an included angle plus a pair of opposite sides are equal on both triangles; or when two pairs of consecutive angles or one pair of opposite angles and the side connecting them, with no other constraints, are equal. 2. Using this theorem, we can prove two triangles congruent regardless their relative orientation, so we don’t need to worry about reflections or rotations when constructing our proof. 3. Before using SAS for proving triangles congruent, it’s important to check whether any other properties might be applicableâi.e., if Two Sides and the Included Angle (SSA) conditions applies and all corresponding sizes match between both triangle, then there is no need for proving by SAS since SSA implies congruence without further detailed proof. 4. In addition to applying this theorem for proving triangle equality/congruence, it may also be used for finding missing parts in a certain set of given conditions: by creating auxiliary lines meeting at a point that is known from one side but needs to be discovered on the other sideâthus forming new angles aligned with those present in either/both original shapesâwe can calculate unknown parameters required for completing our proof process with confidence that they reflect applicable values as equivalent amounts each other among different figures involved in context. 5. Moreover, being aware of means established by SAS might reveal features that influence already existing measurements: since corresponding length aspects stated by its rules provide definite coefficient ratio related to original entries linked togetherâangles connected through rearranged parts divided up accordingly marks significance correlating their relationships in explicit mannerâand reworking them gives tangible information allowing clear illustration how proposed objects compare against each others as direct mirror replicas operating under same general principles applied equally upon separate entities examined simultaneously within same framework order established via accurate analysis performed correctly derived from initial prerequisites mentioned earlier above beforehand predicting final expected outcomes delivering valuable results afterwards arriving finally at guaranteed solution assigned indeed definitely certainly altogether eventually regarding particular topic initially suggested itself straightaway originally at very beginning presented nowadays here now today consequently afterwards afterwards logically per se etcetera summa summarum ergo ad finem et cetera etcetera obiter dictum summarily thusly ending closed cycle work flow lĂ voilĂ i digress goodbye best regards eureka perfection EOS . 6) Summary: Explaining Key Takeaways from Exploring the Same Side Interior angles Theorem The Same Side Interior Angles Theorem states that if two parallel lines are intersected by a third line, the interior angles on the same side of the transversal are congruent. In other words, if a third line intersects two parallel lines, then the interior angles on each side of the third line that lie between the two parallel lines will equal each other. This theorem is useful in helping to prove many things in plane geometry and can be used to verify that certain figures are indeed parallelograms or quadrilaterals. This theorem has several implications in regards to plane geometry. Perhaps most notably, it helps explain why opposite sides and angles of a parallelogram or quadrilateral are equal; for example, an angle formed at one corner of a parallelogram will have an identical corresponding angle on the opposite side due to the theorem. Similarly, this theorem explains why diagonals in parallelograms bisect each other â since both diagonals crosses both sets of parallel lines (formed by sides) and all four interior angles thus created must be congruent due to the theorem stated above. While this theorem may seem fairly straightforward when explained simply, being able to apply it within certain geometric proofs takes experience and practice â with such knowledge wasted away if not often practiced! Knowing how understanding key takeaways from exploring this theorem important is thus essential for furthering your geometry understanding; whether youâre trying to visualize these concepts with simple sketches or moving onto more rigorous theoretical proofs!
Courses Courses for Kids Free study material Offline Centres More Store # Solve for x:${{\tan }^{-1}}\left( \dfrac{x}{2} \right)+{{\tan }^{-1}}\left( \dfrac{x}{3} \right)=\left( \dfrac{\pi }{4} \right),0 < x < \sqrt{6}$ Last updated date: 20th Jun 2024 Total views: 394.5k Views today: 9.94k Verified 394.5k+ views Hint: First expand the given expression in right hand side using the formula for expansion of ${{\tan }^{-1}}x+{{\tan }^{-1}}y$ and then equating the both left hand side and right hand side we will get the quadratic equation so it has two roots you can solve those roots using factorization method and then get the possible value of x. Given ${{\tan }^{-1}}\left( \dfrac{x}{2} \right)+{{\tan }^{-1}}\left( \dfrac{x}{3} \right)=\left( \dfrac{\pi }{4} \right)$ We know that the formulae for ${{\tan }^{-1}}x+{{\tan }^{-1}}y$ is given by ${{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)$ Now apply the above formula we will get, ${{\tan }^{-1}}\left( \dfrac{\left( \dfrac{x}{2} \right)+\left( \dfrac{x}{3} \right)}{1-\left( \dfrac{x}{2} \right)\left( \dfrac{x}{3} \right)} \right)=\left( \dfrac{\pi }{4} \right)$ $\Rightarrow {{\tan }^{-1}}\left( \dfrac{\dfrac{5x}{6}}{\dfrac{6-{{x}^{2}}}{6}} \right)=\left( \dfrac{\pi }{4} \right)$ Now shift the inverse tan function or arctan function to right hand side it becomes tan function on moving from left hand side to right hand side $\dfrac{5x}{6-{{x}^{2}}}=1$ $\Rightarrow 5x=6-{{x}^{2}}$ $\Rightarrow {{x}^{2}}+5x-6=0$ Now solve the quadratic equations we will get two roots $\left( x-1 \right)\left( x+6 \right)=0$ $\Rightarrow x=1\left[ 0 < x < \sqrt{6} \right]$ $\Rightarrow x=-6$ is not the solution for the given equation because in the question it is given that x belongs only to a certain domain that is $0 < x < \sqrt{6}$. hence $x=-6$ is not in the given domain so it is not a possible value for x. So $x=1$ is the only possible solution for x. Note: In the question it was given $0 < x < \sqrt{6}$, this means the value of x lies in between the given interval so that the function gets satisfied. Use the formula for ${{\tan }^{-1}}x+{{\tan }^{-1}}y$ carefully and calculate to get the final answer.
## draw a rough sketch of the pairs of perpendicular line Question draw a rough sketch of the pairs of perpendicular line in progress 0 1 month 2021-08-20T05:32:05+00:00 1 Answer 0 views 0 In this chapter, you will learn how to construct, or draw, different lines, angles and shapes. You will use drawing instruments, such as a ruler, to draw straight lines, a protractor to measure and draw angles, and a compass to draw arcs that are a certain distance from a point. Through the various constructions, you will investigate some of the properties of triangles and quadrilaterals; in other words, you will find out more about what is always true about all or certain types of triangles and quadrilaterals. Bisecting lines When we construct, or draw, geometric figures, we often need to bisect lines or angles.BisIn this chapter, you will learn how to construct, or draw, different lines, angles and shapes. You will use drawing instruments, such as a ruler, to draw straight lines, a protractor to measure and draw angles, and a compass to draw arcs that are a certain distance from a point. Through the various constructions, you will investigate some of the properties of triangles and quadrilaterals; in other words, you will find out more about what is always true about all or certain types of triangles and quadrilaterals. Bisecting lines When we construct, or draw, geometric figures, we often need to bisect lines or angles.Bisect means to cut something into two equal parts. There are different ways to bisect a line segment. Bisecting a line segment with a ruler Step 1: Draw line segment AB and determine its midpoint. 58340.png Step 2: Draw any line segment through the midpoint. 58356.png The small marks on AF and FB show that AF and FB are equal. CD is called a bisector because it bisects AB. AF = FB. Use a ruler to draw and bisect the following line segments: AB = 6 cm and XY = 7 cm. In Grade 6, you learnt how to use a compass to draw circles, and parts of circles called arcs. We can use arcs to bisect a line segment. Bisecting a line segment with a compass and ruler Step 1 Place the compass on one endpoint of the line segment (point A). Draw an arc above and below the line. (Notice that all the points on the arc aboveand below the line are the same distance from point A.) 58818.pngect means to cut something into two equal parts. There are different ways to bisect a line segment. Bisecting a line segment with a ruler Step 1: Draw line segment AB and determine its midpoint. 58340.png Step 2: Draw any line segment through the midpoint. 58356.png The small marks on AF and FB show that AF and FB are equal. CD is called a bisector because it bisects AB. AF = FB. Use a ruler to draw and bisect the following line segments: AB = 6 cm and XY = 7 cm. In Grade 6, you learnt how to use a compass to draw circles, and parts of circles called arcs. We can use arcs to bisect a line segment. Bisecting a line segment with a compass and ruler
# Find the vector equation of the plane passing through the intersection of the planes $\overrightarrow r .(2\hat i + 2 \hat j - 3\hat k) = 7, \overrightarrow r .(2\hat i + 5\hat j + 3\hat k ) = 9$ and through the point $(2, 1, 3).$ $\begin{array}{1 1} \overrightarrow r .(4\hat i + 7\hat j -15\hat k)=30 \\ \overrightarrow r .(4\hat i + 7\hat j -15\hat k)=153 \\ \overrightarrow r .(38\hat i + 68\hat j + 3\hat k)=153 \\ \overrightarrow r .(38\hat i + 68\hat j + 3\hat k)=-153\end{array}$ Toolbox: • Vector equation of a plane passing through the intersection of two planes is $\overrightarrow r.(\overrightarrow {n_1}+\lambda\overrightarrow {n_2})=\overrightarrow d_1+\lambda \overrightarrow d_2$ Step 1: Given vector equations of the plane are $\overrightarrow r.(2\hat i+2\hat j-3\hat k)=7$ $\overrightarrow r.(2\hat i+5\hat j+3\hat k)=9$ These equations can be written as $\overrightarrow r.(2\hat i+2\hat j-3\hat k-7)=0$ $\overrightarrow r.(2\hat i+5\hat j+3\hat k-9)=0$ Equation of any plane passing through the intersection of the given planes is given by $[\overrightarrow r.(2\hat i+2\hat j-3\hat k-7)]+\lambda[\overrightarrow r.(2\hat i+5\hat j+3\hat k-9)]=0$ $\Rightarrow [\overrightarrow r.(2\hat i+2\hat j-3\hat k)]+\lambda[\overrightarrow r.(2\hat i+5\hat j+3\hat k)]=7+9\lambda$----(1) This passes through the point $(2,1,3)$ Therefore its position vector is $\overrightarrow r=2\hat i+\hat j+3\hat k$ Step 2: Hence substituting for $\overrightarrow r$ in equ(1) we get, $[(2\hat i+\hat j+3\hat k).(2\hat i+2\hat j-3\hat k)]+\lambda[(2\hat i+\hat j+3\hat k).(2\hat i+5\hat j+3\hat k)]=7+9\lambda$ $\Rightarrow 2(2+2\lambda)+1(2+5\lambda)+3(3\lambda-3)=7+9\lambda$ (i.e)$4+4\lambda+2+5\lambda-9+9\lambda=7+9\lambda$ On simplifying we get $9\lambda-3=7$ $9\lambda=10$ $\lambda=\large\frac{10}{9}$ Step 3: Substituting the value of $\lambda$ in equ(1) we get $[\overrightarrow r.(2\hat i+2\hat j-3\hat k)]+\large\frac{10}{9}$$[\overrightarrow r.(2\hat i+5\hat j+3\hat k)]=7+9.\large\frac{10}{9} On simplifying we get \Rightarrow \overrightarrow r\big(\large\frac{38}{9}$$\hat i+\large\frac{68}{9}$$\hat j+\large\frac{3}{9}$$\hat k\big)=17$ Therefore $\overrightarrow r.(38\hat i+68\hat j+3\hat k)=153$ This is the vector equation in the required plane.
Courses Courses for Kids Free study material Offline Centres More Store # How do you multiply $(3 + 2i)(1 - 3i)$? Last updated date: 21st Jun 2024 Total views: 374.1k Views today: 8.74k Verified 374.1k+ views Hint: Real numbers and imaginary numbers are the two types of numbers, real numbers are the numbers that can be plotted on a number line while imaginary numbers, as the name suggests, cannot be represented on the number line. Sometimes while solving equations under the square root, we get a negative answer but we know that the square root of a negative number doesn’t exist so we had to think of a way to represent them that’s why we take $\sqrt { - 1} = i$. In the given problem it is actually equal to$(3 + \sqrt { - 2} )(1 - \sqrt { - 3} )$ and these are called complex numbers. To solve this we need to know the value of ${i^2}$ that is ${i^2} = - 1$. Complete step-by-step solution: We know that $i = \sqrt { - 1}$ Squaring on both side we have, ${i^2} = - 1$. Now we have, $\Rightarrow (3 + 2i)(1 - 3i) = 3(1 - 3i) + 2i(1 - 3i)$ $\Rightarrow 3(1 - 3i) + 2i(1 - 3i)$ $\Rightarrow 3 - 9i + 2i - 6{i^2}$ $\Rightarrow 3 - 8i - 6{i^2}$ We know ${i^2} = - 1$, $\Rightarrow 3 - 8i - 6( - 1)$ $\Rightarrow 3 - 8i + 6$ $\Rightarrow 9 - 8i$ Thus we have $(3 + 2i)(1 - 3i) = 9 - 8i$ Note: For multiplying the terms written in the parentheses like$(a + b)(c + d)$ we first multiply the first term of the first bracket with the whole second bracket and then multiply the second term of the first bracket with the whole second bracket that is $(a + b)(c + d) = a((c + d) + b(c + d)$ $(a + b)(c + d) = ac + ad + bc + dc$ But there are various identities to make the calculations easier. Suppose if we have $(3 + 2i)(3 - 2i)$, we know the identity ${a^2} - {b^2} = (a + b)(a - b)$. Using this we can solve it easily. We also have ${(a + b)^2} = {a^2} + 2ab + {b^2}$ also we know ${(a - b)^2} = {a^2} - 2ab + {b^2}$. Depending on the given problem we apply these identities.
# Video: AQA GCSE Mathematics Higher Tier Pack 5 • Paper 2 • Question 19 Construct the locus of points that are the same distance from segments 𝐴𝐵 and 𝐴𝐶. 03:23 ### Video Transcript Construct the locus of points that are the same distance from segments 𝐴𝐵 and 𝐴𝐶. Now, when we’re constructing the locus of points, we’re simply tracing out a map of all the points that meet certain criteria. And in this case, we’re just mapping out all the points which are the same distance away from this line as they are from this line. Now this point, for example, there are lots of different lines that we could draw from that point that touch line 𝐴𝐵. But if we draw a perpendicular line to 𝐴𝐵 that goes through that point, that’s the shortest possible distance from the point to line 𝐴𝐵. And we could do the same from that point to line 𝐴𝐶. Now, we can say that this point — let’s call it 𝑃 — is the same distance away from line segment 𝐴𝐵 as it is from line segment 𝐴𝐶. Simply, the shortest distance from that point to the line is the same in each case. Here’s another point 𝑄 we’ve done the same thing for and here’s another point, 𝑅. Now, if we trace a line through all of these points that are the same distance away from line segments 𝐴𝐵 and 𝐴𝐶, it would look something like this. Now, we’re gonna look at the construction: how do we actually construct that and work out where those points are. So we start off with our diagram with line segments 𝐴𝐵 and 𝐴𝐶. Now, we’re gonna have to put our compass point at point 𝐴 in order to do this construction. So because I’m right-handed, I’ve actually turned the piece of paper round. So don’t be afraid to do that. It’s okay to turn the paper round to do your constructions. So now, carefully place the compass point at point 𝐴 and open up the radius. Now the exact distance of that radius doesn’t matter. Generally speaking, the larger, the better because it’ll make more accurate constructions. in this case I’ve gone for about half the distance between 𝐴 and 𝐵. Now, extend that arc so that it cuts line segment 𝐴𝐶 and it cuts line segment 𝐴𝐵. And because we kept the same radius as we were drawing that arc, this distance here is the same as this distance here. So we’re gonna take each of those points of intersection in turn and draw another arc, just out here somewhere. Now the exact radius that you use doesn’t really matter as long as you keep it the same for both of the next steps. It’s important that we get two arcs which intersect each other. So you may need to just adjust the radius up or down a little in order to create their point of intersection. So there’s our first arc. Now keeping that radius exactly the same, we draw another little arc from the point of intersection on line 𝐴𝐶. Now because we can’t have radius the same for those two little arcs, this distance here is the same as this distance here. And don’t forget that this distance here was also the same as this distance here. And because of the symmetry of this, if we draw a line from 𝐴 through this point of intersection out here, then it will exactly cut angle 𝐵𝐴𝐶 in half and give us our locus of points that are same distance from segments 𝐴𝐵 and 𝐴𝐶. So I just turn the piece of paper back round the right way. And that green line is our locus of points that are equidistant or the same distance from line segments 𝐴𝐵 and 𝐴𝐶. And because this is a construction question, don’t erase your construction marks. Those arcs that you drew to work out where the green line went are very important. So please don’t rub them out.
# 031 Kindergarten Math Practices Measure Shapes Telling Time Money Homework Activities Passages English Learning Material Reading Book Kids Fun Year Assessment New Curriculum Great, fun and free math worksheets should be able to present a mathematical problem in different ways. Math is after all nothing more than a numeric expression of some of life's simplest questions: How much money do I have left if I buy a soda? By the end of the week, how much of my daily allowance will I be able to save if I don't? When a child learns to relate math to everyday questions, he will be great at it from the simplest addition all the way to trigonometry. To convert percentages, decimals and fractions is thus one essential skill. How much of an apple pie has been eaten? The answer to this question can be expressed in percentages, 50%; or in decimals, 0Ǒ; or in fraction, ½. In other words, half of mom's delicious apple pie is gone. How many kids in school have done their homework? Again this can be answered in several ways: in percentages, 70%; or in ratio, 7ᚲ; Both of these mean out of ten kids in class there are seven good ones who did and three not_so_good ones who didn't. The bottom line is that kids learn math much better when it makes sense. Thus, the math worksheets which you get for your kids should include interesting word problems that help them with the practical application of the lessons they learn. It should also present the same problem in a variety of ways to ensure that a child's grasp of a subject is deeper and comprehensive. How Do You Find Points In A Graph? This set of numbers ƒ, 3) is an example of an ordered pair. The first number refers to the value of x while the second number stands for the value of y. When ordered pairs are used to find points on the grid, they are called the coordinates of the point. In above example, the x coordinate is 2 while the y coordinate is 3. Together, they enable you to locate the point ƒ, 3) on the grid. What's the point of all this? Well, ever wondered how ships describe exactly where they are in the vastness of the ocean? To be able to locate places, people have to draw a grid over the map and describe points with the help of x and y coordinates. Why don't you give it a try? Imagine left side wall of your room to be y axis and the wall at your back to be the x axis. The corner that connects them both will be your origin. Measure both in feet. If I say stand on coordinates Ɠ, 2), would you know where to go? That means from the corner (origin) you should move 3 feet to the right and 2 feet forward. 257 Shares 157 Pins 548 Tweet 177 Shares 192 Shares
# How do you solve t- \frac { 5} { 8} t + \frac { 3} { 5} = \frac { 3} { 5} ( t + \frac { 5} { 2} )? Nov 3, 2016 $t = - 4$ #### Explanation: solve $t - \frac{5}{8} t + \frac{3}{5} = \frac{3}{5} \left(t + \frac{5}{2}\right)$ multiply everything by the LCD which is 40 $40 \left(t - \frac{5}{8} t + \frac{3}{5} = \frac{3}{5} \left(t + \frac{5}{2}\right)\right)$ $40 t - 25 t + 24 = 24 \left(t + \frac{5}{2}\right)$ simplify and distribute parenthesis $15 t + 24 = 24 t + 60$ solve for t $15 t - 24 t = 60 - 24$ $- 9 t = 36$ $t = - 4$
## Addition of Integers Exercises – Set 1 Below are exercises on addition of signed numbers. Part I and II are exercises for addition of integers, while part III is an extension to fractions and decimals. Solutions and answers will be can be found here. Part I 1. -7 + 13 2. (-8) + (-9) 3. 5 + (-18) 4. 34 + (-38) 5. (-34) + 34 6. 0 + (-25) 7. 5 + (18) 8. -14 + (-12) 9. 13 + (-13) 10. 16 + 18 Part II 1. 13 + 12 + (-15) 2. -18 + (-2) + 14 3. 14 + 8 + (-14) 4. 21 + 7 + (-28) 5. -23 + 0 + 14 6. -19 + 22 + (-6) 7. 32 + (-11) + (-27) 8. 12 + (-11) + 10 9. 11 + (-22) + 21 + (-10) 10. 16 + (-16) + 8 + 3 Part III 1. $\frac {1}{2} + (- \frac {3}{2})$ 2. $\frac {1}{4} + (- \frac {2}{4})$ 3. $\frac {3}{7} + (-\frac {3}{7})$ 4. $- \frac {1}{2} + \frac {3}{4}$ 5. $-\frac {4}{11} + (-\frac {3}{11})$ 6. 0.7 + (-0.3) 7. 4.8 + (-1.2) 8. -3.7 + 2.2 9. 12.6 + (-11.1) 10. 13.75 + (-15.2) You might also want to practice subtraction of integers. ## Addition of Integers Quiz 1 We have already learned how to operate with integers (addition, subtraction, multiplication, and division) as well as its order of operation. Test your skills in addition of fraction in this quiz below. Fill in the blanks with the correct answer. Good luck! ## Addition of Integers Quiz 1 A quiz on addition of integers. ## Solutions to Practice Exercises on Addition of Integers We have learned how to add integers and in the previous post, I have given you practice exercises that you can use to evaluate your understanding of the topic. Below is the complete solutions to the practice exercises on adding integers. Share to me how many did you get right. Solutions to Practice Exercises on Addition of Integers 1. 28 + 12 = 40. 2. 14 + 11 = 25 3. 24 + 15 = 9 Solution: We pair 15 and 15 to get 0. We have 9 left. So, the correct answer is 9. 4. –16 + 31 = 15 We pair 16 and 16 to get 0. We are left with positive 15. 5. 23 + 46 + 15 = 54 We add the two positive numbers first: 23 + 46 = 69. Next, we add 69 and 15. We get 15 from 69 and pair it with 15 resulting to 0, so we have 54 left. 6. 45 + 12 + 16 = 17 We add the negative numbers first: 12 + 16 = 28. We add the result to 45. We get 28 from 45 and pair it with 28 to get 0 leaving 17. 7. 12 +15 + 62 = 89 Explanation: They are negative, so we just added them. Of course negative added to negative is always negative. 8. 22 + 36 + 36 = 22 Explanation: 36 + 36 = 0, so we are left with 22. 9. 12 + 18 + 12 + 18 = 0 Explanation: 12 and 12 and 18 and 18 = 0. 10. 31 + 55 + 41 + 32 + 10 = 3 Explanation: Adding the positive integers, we have 31 + 55 = 86. Adding the negative integers, we have 41 + 32 + 10 =83. Now, 86 + 83 = 3
# Concept of Standard form and its uses and benefits with calculations. Standard form is basically a method to write down big or very small values in an easy way, for example, you can write 1.9 × 106 in place of 1900000 and it is very useful in calculation and saves time for small values, it can also be written in the standard form its index will be in negative to show that it’s a small number. A Persian mathematician named Muhammad Al-Khwarizmi in the early 9th century first introduced the term Standard form also called scientific notation it depends on the country and what they use the name scientific notation or standard form. Moreover, in this article, with the help of basic definition rules and their use in daily life will be discussed also with the help of examples topic will be explained for better understanding. ## What is the Standard form of numbers? The term standard form or scientific notation is a way to write huge or very less values in the easiest manner just by comparing the power of 10. In mathematical form, it is written as q × 10n Here q is a number 1 ≤ q ≤ 9 and n is an integer. Standard form plays an important role in a lengthy calculation like writing the mass of the earth there are 99% chances of mistakes if anyone writes it manually so similar to this there are many values that we can’t write without using a standard form. For example, the mass of the earth is “5972190000000000000000000” instead of this you can write “(5.97219 x 1024)” ## Steps to convert in standard form Suppose you have a number 6,78,000 to change it in standard form follow the steps below: Step 1: In the first step, we write the first digit Step 2: In the second step, write a decimal point after this with remaining non-zero numbers like 6.78 Step 3: Finally, check how many numbers of digits you have passed to reach to place it here and then multiply that number with “raise to power ten Step 4: Remember moving from left to right then the index should be written with a negative sign and if moving from right to left then the index should be written with the positive sign Step 5: Hence the standard form for 678,000 is 6.78 × 10 5 To avoid these steps, try a standard form calculator which will give the desired result in a fraction of a second. ## Examples of the standard form: In this section with the help of examples, the topic is explained for better understanding. Example 1: Convert it 0.0000000321 into standard form Solution: Step 1: From the question it is clearly shown that point is on the left side and to shift it after the first non-zero digit and here moving from left to right, then the index should be written in a negative sign = 0.0000000321 Step 2: The place of the decimal point should be checked first. = 0.0000000321 Step 3: Now move the decimal point after the first non-zero digit. = 3.21 Step 4: Finally, check how many numbers of digits you have passed to reach to place it after 1st non-zero digit and then multiply that number with “raise to power ten” in this problem to place it after the non-zero digit decimal point move 8 digits to the right side. Note: If moved from left to right then the index should be written with a negative sign and if moved from right to left then the index should be in positive sign. Example 2: Convert 6,53,000,000,000 into standard form. Solution: Step 1: From the question it is clearly shown that point is on the left side and to shift it before the last non-zero digit and here moving from right to left then the index should be written with a positive sign = 653000000000 Step 2: Check the place value of the decimal point, = 653000000000 Note: there is no decimal point then it should place on most right side of the number. Step 3: Move the point to the right side of the 1st non-zero digit that is = 6.53 Step 4: Finally, check how many numbers of digits you have passed to shift it before the last non-zero digit here moving from right to left then the index should be written with a positive sign then multiply that number with “raise to power ten” in this problem to place it before last non-zero-digit decimal point move 11 digits to the left side. Example 3: Convert 3234561.64 × 105 in standard form. Solution: Step 1: From the question it is clearly shown that point is on the left side and to shift it before the last non-zero digit and here moving from right to left then the index should remain in a positive sign = 3234561.64 × 105 Step 2: Check the place value of the decimal point, = 3234561.64 × 105 Step 3: Move the decimal point to the right side of the 1st non-zero digit that is = 3.23456164 Step 4: Finally, check how many numbers digits you have passed to shift it before the last non-zero digit here moving from right to left then the index should remain in a positive sign then multiply that number with “raise to power ten” = 3.23456164 × 105 × 106 Here you see bases are the same then powers will be added. = 3.23456164 × 105+6 = 3.23456164 × 1011 Which is the standard form of 3.23456164 × 1011 ## Summary: • In this article, you have studied the basic use of standard form its need, and its importance in calculations furthermore with the help of an example topic is explained and after a complete understanding of this article, anyone can easily defend this topic and make his/her daily life calculation easier. 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Students often turn to ISC S Chand Maths Class 12 Solutions Chapter 6 Matrices Ex 6(i) to clarify doubts and improve problem-solving skills. ## S Chand Class 12 ICSE Maths Solutions Chapter 6 Matrices Ex 6(i) Using elementary transformation, find the inverse of the following matrices, if it exists. Question 1. A = $$\left[\begin{array}{rr} 1 & -1 \\ 2 & 3 \end{array}\right]$$ Solution: A = $$\left[\begin{array}{rr} 1 & -1 \\ 2 & 3 \end{array}\right]$$ Since A = IA ⇒ $$\left[\begin{array}{rr} 1 & -1 \\ 2 & 3 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$A Operate R2 → R2 – 2R1 $$\left[\begin{array}{ll} 1 & -1 \\ 0 & -5 \end{array}\right]=\left[\begin{array}{cc} 1 & 0 \\ -2 & 1 \end{array}\right]$$A R2 → $$\frac { 1 }{ 5 }$$R2 $$\left[\begin{array}{cc} 1 & -1 \\ 0 & 0 \end{array}\right]=\left[\begin{array}{cc} 1 & 0 \\ -2 / 5 & 1 / 5 \end{array}\right]$$A Operate R2 → R1 + R2 $$\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 3 / 5 & 1 / 5 \\ -2 / 5 & 1 / 5 \end{array}\right]$$A ∴ A-1 = $$\left[\begin{array}{cc} 3 / 5 & 1 / 5 \\ -2 / 5 & 1 / 5 \end{array}\right]$$ [using def. of inverse, A-1A = I] Question 2. A = $$\left[\begin{array}{rr} 1 & 2 \\ 2 & -1 \end{array}\right]$$ Solution: A = $$\left[\begin{array}{rr} 1 & 2 \\ 2 & -1 \end{array}\right]$$ Since A = IA ⇒ $$\left[\begin{array}{rr} 1 & 2 \\ 2 & -1 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$A Operate R2 → R2 – 2R1 $$\left[\begin{array}{cc} 1 & 2 \\ 0 & -5 \end{array}\right]=\left[\begin{array}{cc} 1 & 0 \\ -2 & 1 \end{array}\right]$$A Operate R2 → $$\frac { -1 }{ 5 }$$R2 $$\left[\begin{array}{ll} 1 & 2 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 1 & 0 \\ 2 / 5 & -1 / 5 \end{array}\right]$$A Operate R1 → R1 – 2R2 $$\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 1 / 5 & 2 / 5 \\ 2 / 5 & -1 / 5 \end{array}\right]$$A ∴ A-1 = $$\left[\begin{array}{cc} 1 & 2 / 5 \\ 2 / 5 & -1 / 5 \end{array}\right]$$ [using def. of inverse, A-1A = I] Question 3. $$\left[\begin{array}{ll} 9 & 5 \\ 7 & 4 \end{array}\right]$$ Solution: Question 4. $$\left[\begin{array}{rr} 10 & -2 \\ -5 & 1 \end{array}\right]$$ Solution: $$\left[\begin{array}{rr} 10 & -2 \\ -5 & 1 \end{array}\right]$$ Since A = IA ⇒ $$\left[\begin{array}{rr} 10 & -2 \\ -5 & 1 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$A Operate R1 → R1 – 2R2 $$\left[\begin{array}{cc} 0 & 0 \\ -5 & 1 \end{array}\right]=\left[\begin{array}{ll} 1 & 2 \\ 0 & 1 \end{array}\right]$$A Since these are all zeros in R1 of left side matrix Thus A-1 does not exists. Question 5. $$\left[\begin{array}{lll} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{array}\right]$$ Solution: Question 6. $$\left[\begin{array}{rrr} 1 & -1 & 0 \\ 2 & 5 & 3 \\ 0 & 2 & 1 \end{array}\right]$$ Solution: Question 7. $$\left[\begin{array}{rrr} 1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0 \end{array}\right]$$ Solution:
# QUADRATICS EQUATIONS/EXPRESSIONS CONTAINING x2 TERMS. ## Presentation on theme: "QUADRATICS EQUATIONS/EXPRESSIONS CONTAINING x2 TERMS."— Presentation transcript: - Quadratic number patterns are sequences of numbers where the difference between terms is not the same and the rule contains a squared term - You need to halve the difference of the differences to find the squared term - i.e. If the difference of the differences is a 2, the rule contains 1n2 e.g. Write a rule for the following pattern Rule: T = 1n2 + 3 n Term (T) 1 4 2 7 3 12 19 5 28 + 3 1×12 = 1 + 2 + 5 = 4 + 3 + 2 + 7 + 2 = 19 + 9 3. Halve the 2nd difference to find the n2 rule 1. Find the difference between terms 2. If difference is not the same, find the difference of the differences! 4. Substitute to find constant 5. Check if rule works - The squared term of the rule is found by halving the difference of the differences - i.e. If the difference of the differences is a 6, the rule contains 3n2 - If the simple trial and error does not work, try this technique: Rule: T = 2n2 + 2n + 1 e.g. Write a rule for the following pattern 2×1= 2 n Term (T) 1 5 2 13 3 25 4 41 61 Term (T) – 2n2 = 3 + 1 5 - 2×12 3 + 8 + 2 + 4 13 - 2×22 5 + 12 + 2 + 4 7 + 16 + 2 + 4 9 + 20 + 2 11 3. Halve the 2nd difference to find the n2 rule 1. Find the difference between terms 4. Subtract the n2 rule from the term 2. If difference is not the same, find the difference of the differences! 5. Find the linear part of the rule 6. Check if the rule works 2×42 + 2×4 + 1 = 41 EXPANDING TWO BRACKETS - To expand two brackets, we must multiply each term in one bracket by each in the second Remember integer laws when multiplying e.g. Expand and simplify a) (x + 5)(x + 2) = x2 + 2x + 5x + 10 b) (x - 3)(x + 4) = x2 + 4x - 3x - 12 = x2 + 7x + 10 = x2 + 1x – 12 To simplify, combine like terms c) (x - 1)(x - 3) = x2 - 3x - 1x + 3 c) (2x + 1)(3x - 4) = 6x2 - 8x + 3x - 4 = x2 – 4x + 3 = 6x2 – 5x – 4 PERFECT SQUARES - When both brackets are exactly the same To simplify, combine like terms e.g. Expand and simplify Watch sign change when multiplying a) (x + 8)2 = (x + 8)(x + 8) b) (x - 4)2 = (x – 4)(x – 4) = x2 + 8x + 8x + 64 = x2 - 4x - 4x + 16 = x2 + 16x + 64 = x2 - 8x + 16 Write out brackets twice BEFORE expanding c) (3x - 2)2 = (3x – 2)(3x – 2) = 9x2 - 6x - 6x + 4 = 9x2 – 12 x + 4 DIFFERENCE OF TWO SQUARES - When both brackets are the same except for signs (i.e. – and +) e.g. Expand and simplify a) (x – 3)(x + 3) = x2 + 3x - 3x - 9 b) (x – 6)(x + 6) = x2 + 6x - 6x - 36 = x2 – 9 = x2 – 36 Like terms cancel each other out c) (2x – 5)(2x + 5) = 4x2 + 10x - 10x - 25 = 4x2 – 25 FACTORISING The general equation for a quadratic is ax2 + bx + c When a = 1 - You need to find two numbers that multiply to give c and add to give b e.g. Factorise To check answer, expand and see if you end up with the original question! 1) x2 + 11x + 24 = (x + 3)(x + 8) 1, 24 List pairs of numbers that multiply to give 24 (c) Check which pair adds to give 11 (b) Place numbers into brackets with x 2, 12 3, 8 4, 6 2) x2 + 7x + 6 = (x + 1)(x + 6) 1, 6 List pairs of numbers that multiply to give 6 (c) Check which pair adds to give 7 (b) Place numbers into brackets with x 2, 3 - Expressions can also contain negatives e.g. Factorise 1) x2 + x – 12 = (x - 3)(x + 4) 2) x2 – 6x – 16 = (x + 2)(x - 8) - 1, 12 1, 16 - As the end number (c) is , one of the pair must negative. As the end number (c) is , one of the pair must negative. - 2, 6 2, 8 - - 3, 4 4, 4 - Check which pair now adds to give b Make the biggest number of the pair the same sign as b Check which pair now adds to give b Make the biggest number of the pair the same sign as b 3) x2 – 9x + 20 = (x - 4)(x - 5) 4) x2 – 10x + 25 = (x - 5)(x - 5) - 1, 20 - - 1, 25 - = (x - 5)2 As the end number (c) is , but b is – 9, both numbers must be negative - 2, 10 - - 5, 5 - As the end number (c) is , but b is – 10, both numbers must be negative - 4, 5 - Check which pair now adds to give b SPECIAL CASES 1. No end number (c) e.g. Factorise a) x2 + 6x + 0 b) x2 – 10x = x( ) x - 10 0, 6 = x(x + 6) Add in a zero and factorise as per normal OR: factorise by taking out a common factor 2. No x term (b) (difference of two squares) e.g. Factorise a) x2 - 25 + 0x = (x - 5)(x + 5) b) x2 – 100 = (x )(x ) -5, 5 c) 9x2 – 121 = (3x )(3x ) Add in a zero x term and factorise OR: factorise by using A2 – B2 = (A – B)(A + B) TWO STAGE FACTORISING When a ≠ 1 1. Common factor - Always try to look for a common factor first. e.g. Factorise a) 2x2 + 12x + 16 = 2( ) x2 + 6x + 8 b) 3x2 – 6x – 9 = 3( ) x2 – 2x – 3 1, 8 = 2(x + 2)(x + 4) 1, 3 - = 3(x + 1)(x – 3) 2, 4 c) 3x2 + 24x = 3x( ) x + 8 d) 4x2 – 36 = 4( ) x2 – 9 = 4(x )(x ) 2. No common factor (HARD) - Use the following technique e.g. Factorise a) 3x2 – 10x – 8 = 3x2 + 2x – 12x – 8 = x( ) 3x + 2 -4( ) 3x + 2 = (x – 4)(3x + 2) 1x, 24x - Multiply first and last terms Find two terms that multiply to -24x2 but add to -10x Replace 10x with the two new terms Factorise 2 terms at a time. 2x, 12x - 3x, 8x - 4x, 6x - 3x2 × - 8 = -24x2 Write in two brackets b) 2x2 + 7x + 3 = 2x2 + x + 6x + 3 = x( ) 2x + 1 + 3( ) 2x + 1 = (x + 3)(2x + 1) 2x2 × 3 = 6x2 1x, 6x 2x, 3x SOLVING QUADRATICS To solve use the following steps: 1. Move all of the terms to one side, leaving zero on the other 2. Factorise the equation 3. Set each factor to zero and solve. e.g. Solve a) (x + 7)(x – 2) = 0 b) (x – 4)(x – 9) = 0 x + 7 = 0 x – 2 = 0 x – 4 = 0 x – 9 = 0 -7 -7 +2 +2 +4 +4 +9 +9 x = - 7 x = 2 x = 4 x = 9 c) x2 + x – 2 = 0 - 1, 2 d) x2 – 5x + 6 = 0 1, 6 - 2, 3 - (x – 1)(x + 2) = 0 (x + 1)(x – 6) = 0 x – 1 = 0 x + 2 = 0 x + 1 = 0 x – 6 = 0 +1 +1 - 2 - 2 -1 -1 +6 +6 x = 1 x = - 2 x = -1 x = 6 e) x2 + 8x = 0 f) x2 – 11x = 0 x( ) = 0 x + 8 x( ) = 0 x – 11 x = 0 x + 8 = 0 x = 0 x – 11 = 0 - 8 - 8 + 11 + 11 x = -8 x = 11 g) x = 0 h) 9x2 - 4 = 0 (x )(x ) = 0 (3x )(3x ) = 0 x - 7 = 0 x + 7 = 0 3x - 2 = 0 3x + 2 = 0 +7 +7 - 7 - 7 +2 +2 -2 -2 x = 7 x = -7 3x = 2 3x = -2 ÷3 ÷3 ÷3 ÷3 x = 2/3 x = -2/3 i) x2 = 4x + 5 j) x(x + 3) = 180 -4x -5 -4x -5 1, 5 - x2 + 3x = 180 x2 – 4x – 5 = 0 -180 -180 (x + 1)(x – 5) = 0 x2 + 3x – 180 = 0 (x + 15)(x – 12) = 0 x + 1 = 0 x – 5 = 0 x + 15 = 0 x – 12 = 0 -1 -1 +5 +5 -15 -15 +12 +12 x = -1 x = 5 x = -15 x = 12 WRITING EQUATIONS - Involves writing an equation from the information then solving e.g. The product of two consecutive numbers is 20. What are they? If x = a number, then the next consecutive number is x + 1 x + 5 x(x + 1) = 20 - 1, 20 x2 + x = 20 - 2, 10 -20 -20 A = 150 m2 x - 4, 5 x2 + x – 20 = 0 (x – 4)(x + 5) = 0 x – 4 = 0 x + 5 = 0 +4 +4 -5 -5 x(x + 5) = 150 x = 4 x = -5 x2 + 5x = 150 The numbers are 4, 5 and -5, -4 -150 -150 x2 + 5x – 150 = 0 (x – 10)(x + 15) = 0 e.g. A paddock of area 150 m2 has a length 5 m longer than its width. Find the dimensions of this paddock. x – 10 = 0 x + 15 = 0 +10 +10 -15 -15 x = 10 x = -15 The dimensions are 10 m by 15 m
# AP Calculus BC : Maclaurin Series ## Example Questions ← Previous 1 3 4 ### Example Question #1 : Maclaurin Series For which of the following functions can the Maclaurin series representation be expressed in four or fewer non-zero terms? Explanation: Recall the Maclaurin series formula: Despite being a 5th degree polynomial recall that the Maclaurin series for any polynomial is just the polynomial itself, so this function's Taylor series is identical to itself with two non-zero terms. The only function that has four or fewer terms is as its Maclaurin series is. ### Example Question #2 : Taylor's Theorem Let Find the the first three terms of the Taylor Series for centered at . Explanation: Using the formula of a binomial series centered at 0: , where we replace with and , we get: for the first 3 terms. Then, we find the terms where, ### Example Question #6 : Taylor Series Write the first three terms of the Taylor series for the following function about : Explanation: The general form for the Taylor series (of a function f(x)) about x=a is the following: . Because we only want the first three terms, we simply plug in a=1, and then n=0, 1, and 2 for the first three terms (starting at n=0). The hardest part, then, is finding the zeroth, first, and second derivative of the function given: The derivative was found using the following rules: Then, simply plug in the remaining information and write out the terms: ### Example Question #7 : Taylor Series Write out the first four terms of the Taylor series about for the following function: Explanation: The Taylor series about x=a of any function is given by the following: So, we must find the zeroth, first, second, and third derivatives of the function (for n=0, 1, 2, and 3 which makes the first four terms): The derivatives were found using the following rule: Now, evaluated at x=a=1, and plugging in the correct n where appropriate, we get the following: which when simplified is equal to . ### Example Question #11 : Taylor Series Write out the first three terms of the Taylor series about for the following function: Explanation: The general formula for the Taylor series about x=a for a given function is We must find the zeroth, first, and second derivative of the function (for n=0, 1, and 2). The zeroth derivative is just the function itself. The derivatives were found using the following rule: Now, follow the above formula to write out the first three terms: which simplified becomes ### Example Question #2 : Maclaurin Series Find the Taylor series expansion of at . Explanation: The Taylor series is defined as where the expression within the summation is the nth term of the Taylor polynomial. To find the Taylor series expansion of at , we first need to find an expression for the nth term of the Taylor polynomial. The nth term of the Taylor polynomial is defined as For, and . We need to find terms in the taylor polynomial until we can determine the pattern for the nth term. Substituting these values into the taylor polynomial we get The Taylor polynomial simplifies to Now that we know the nth term of the Taylor polynomial, we can find the Taylor series. The Taylor series is defined as where the expression within the summation is the nth term of the Taylor polynomial. For this problem, the taylor series is Use the above to calculate the taylor series expansion of at , ### Example Question #31 : Taylor Series Use the first 3 terms of the Maclaurin series to approximate Explanation: The Maclaurin series for sine is Plugging in for x gives: ### Example Question #32 : Taylor Series Use the first 4 terms of the Maclaurin series to approximate Explanation: The Maclaurin series for is Plugging in 0.78 for x gives: ### Example Question #3 : Maclaurin Series Use the first four terms of the Maclaurin series to approximate to 6 decimal places. Explanation: The Maclaurin series for is Plugging in x = 0.38 gives:
# AP Statistics Curriculum 2007 EDA Pics (diff) ← Older revision \| Current revision (diff) \| Newer revision → (diff) ## General Advance-Placement (AP) Statistics Curriculum - Pictures of Data ### Pictures of Data There are a variety of graphs and plots that may be used to display data. • For quantitative variables, we need to make classes (meaningful intervals) first. To accomplish this, we need to separate (or bin) the quantitative data into classes. • For qualitative variables, we need to use the frequency counts instead of the native measurements, as the latter may not even have a natural ordering (so binning the variables in classes may not be possible). • How to define the number of bins or classes? One common rule of thumb is that the number of classes should be close to $\sqrt{\texttt{sample size}}.$ For accurate interpretation of data, it is important that all classes (or bins) are of equal width. Once we have our classes, we can create a frequency/relative frequency table or histogram. ### Example People who are concerned about their health may prefer hot dogs that are low in salt and calories. The Hot dogs data file contains data on the sodium and calories contained in each of 54 major hot dog brands. The hot dogs are also classified by type: beef, poultry, and meat (mostly pork and beef, but up to 15% poultry meat). For now we will focus on the calories of these sampled hotdogs. ### Frequency Histogram Charts • The histogram of the Calorie content of all hotdogs in shown in the image below. Note the clear separation of the calories into 3 distinct sub-populations. Could this be related to the type of meat in the hotdogs? • The histogram of the Sodium content of all hotdogs in shown in the image below. What patterns in this histogram can you identify? Try to explain! ### Box and Whisker Plots • The graph below shows the box and whisker plot of the Calorie content for all 3 types of hotdogs. • The graph below shows the box and whisker plot of the Sodium (salt) content for all 3 types of hotdogs. ### Dot Plots • The graph below shows the dot-plot of the Calorie content for all 3 types of hotdogs. • The graph below shows the dot-plot of the Sodium content for all 3 types of hotdogs. ### Stem-and-Leaf Plots Stem-and-leaf plot is a method of presenting quantitative data in a graphical format, similar to a histogram. It is used to assist in visualizing the shape of a distribution. Stem-and-leaf plots are useful tools in exploratory data analysis. Unlike histograms, stem-and-leaf plots retain the original data to at least two significant digits, and put the data in order, which simplifies the move to order-based inference and non-parametric statistics. A basic stem-plot contains two columns separated by a vertical line. The left column contains the stems and the right column contains the leaves. • Construction: To construct a stem-and-leaf plot, the observations must first be sorted in ascending order. Then, it must be determined what the stems will represent and what the leaves will represent. Frequently, the leaf contains the last digit of the number and the stem contains all of the other digits. Sometimes, the data values may be rounded to a particular place value (such as the hundreds place) that will be used for the leaves. The remaining digits to the left of the rounded place value are used as the stems. • Example: The stem-and-leaf plot for the calorie variable of the Hot dogs data is shown below. • Legend: Stem-and-leaf of Calories, N = 54; Leaf Unit = 1.0 8 67 9 49 10 22677 11 13 12 9 13 1225556899 14 01234667899 15 223378 16 17 235569 18 1246 19 00015 ### Summary • Histograms can handle large data sets, but they can’t tell exact data values and require the user to set-up classes. • Dot plots can get a better picture of data values, but they can’t handle large data sets. • Stem and leaf plots can see actual data values, but they can’t handle large data sets.
# What is a solution to the differential equation (x+1)y'-2(x^2+x)y=e^(x^2)/(x+1) where x>-1 and y(0)=5? Jun 24, 2016 $y \left(x\right) = \left(6 - \frac{1}{x + 1}\right) {e}^{{x}^{2}}$ #### Explanation: The differential equation is first order linear nonhomogeneus. In this case the solution is composed from the homogeneus solution ${y}_{h} \left(x\right)$ and a particular solution ${y}_{p} \left(x\right)$. $\left(x + 1\right) y {'}_{h} \left(x\right) - 2 \left({x}^{2} + x\right) y \left(x\right) = 0$ $\left(x + 1\right) y {'}_{p} \left(x\right) - 2 \left({x}^{2} + x\right) {y}_{p} \left(x\right) = {e}^{{x}^{2}} / \left(x + 1\right)$ Finally $y \left(x\right) = {y}_{h} \left(x\right) + {y}_{p} \left(x\right)$ 1) Obtaining the homogeneus solution ${y}_{h} \left(x\right)$ Simplifying we obtain $y {'}_{h} \left(x\right) - 2 x \times {y}_{h} \left(x\right) = 0$ grouping variables $\frac{y {'}_{h} \left(x\right)}{{y}_{h} \left(x\right)} = 2 x$ The solution is ${\log}_{e} \left({y}_{h} \left(x\right)\right) = {x}^{2} + {C}_{0} \to {y}_{h} \left(x\right) = {C}_{1} {e}^{{x}^{2}}$ 2) Obtaining the particular solution ${y}_{p} \left(x\right)$ For this purpose we will suppose that ${y}_{p} \left(x\right) = {C}_{1} \left(x\right) {e}^{{x}^{2}}$ This method was due to Euler and Lagrange and can be seen in https://en.wikipedia.org/wiki/Variation_of_parameters Introducing ${y}_{p} \left(x\right)$ into the complete equation we obtain after simplifications ${e}^{{x}^{2}} \left(1 + x\right) C {'}_{1} \left(x\right) = {e}^{{x}^{2}} / \left(1 + x\right)$ Solving for ${C}_{1} \left(x\right)$ we obtain ${C}_{1} \left(x\right) = - \frac{1}{x + 1} + {C}_{2}$ Now, putting all together $y \left(x\right) = \left({C}_{2} - \frac{1}{x + 1}\right) {e}^{{x}^{2}}$ With the initial conditions we find the ${C}_{2}$ value as $y \left(0\right) = \left({C}_{2} - 1\right) = 5 \to {C}_{1} = 6$ Jun 25, 2016 $= {e}^{{x}^{2}} \left(\frac{6 x + 5}{x + 1}\right)$ #### Explanation: this is linear so we can start by just moving stuff around. $\left(x + 1\right) y ' - 2 \left({x}^{2} + x\right) y = {e}^{{x}^{2}} / \left(x + 1\right)$ $y ' - 2 \left(\frac{x \left(x + 1\right)}{x + 1}\right) y = {e}^{{x}^{2}} / {\left(x + 1\right)}^{2}$ $y ' - 2 x \setminus y = {e}^{{x}^{2}} / {\left(x + 1\right)}^{2}$ we solve with inregrating factor $I = \exp \left(- 2 \int \setminus x \setminus \mathrm{dx}\right) = {e}^{- {x}^{2}}$ ${e}^{- {x}^{2}} \cdot y ' - \left({e}^{- {x}^{2}}\right) \cdot 2 x y = {e}^{- {x}^{2}} \cdot {e}^{{x}^{2}} / {\left(x + 1\right)}^{2}$ $\left({e}^{- {x}^{2}} \cdot y\right) ' = \frac{1}{x + 1} ^ 2$ ${e}^{- {x}^{2}} \cdot y = \int \frac{1}{x + 1} ^ 2 \setminus \mathrm{dx}$ ${e}^{- {x}^{2}} \cdot y = - \frac{1}{x + 1} + \alpha$ $y = {e}^{{x}^{2}} \cdot \left(- \frac{1}{x + 1} + \alpha\right)$ From $y \left(0\right) = 5$ $5 = 1 \cdot \left(- \frac{1}{1} + \alpha\right) \setminus \implies \alpha = 6$ $y = {e}^{{x}^{2}} \left(6 - \frac{1}{x + 1}\right)$ $= {e}^{{x}^{2}} \left(\frac{6 x + 5}{x + 1}\right)$
# What is percentile in statistics? ## What is percentile in statistics? A percentile (or a centile) is a measure used in statistics indicating the value below which a given percentage of observations in a group of observations fall. For example, the 20th percentile is the value (or score) below which 20% of the observations may be found. ### How do you find the 90th percentile? To find the 90th percentile for these (ordered) scores, start by multiplying 90 percent times the total number of scores, which gives 90% ∗ 25 = 0.90 ∗ 25 = 22.5 (the index). #### What is the 1st percentile? Percentile “ranks” -scores of students are arranged in rank order from lowest to highest. -the scores are divided into 100 equally sized groups or bands. -the lowest score is “in the 1st percentile” (there is no 0 percentile rank) What is 88th percentile mean? The score you have entered means that the individual who took the test is at the eighty-eighth percentile – their percentile rank is 88%. This means that the student had a test score greater than or equal to 88% of the reference population. What is the difference between percentile and percentage? The key difference between percentage and percentile is the percentage is a mathematical value presented out of 100 and percentile is the per cent of values below a specific value. The percentage is a means of comparing quantities. A percentile is used to display position or rank. ## How do you calculate percentile in statistics? To calculate percentiles for a set of results, the values are first arranged in ascending order. The percentile for a given value can then be found by subtracting 0.5 from its numerical position in the sequence, dividing by the number of results, then multiplying by 100. ### What does percentile tell you about a statistical value? Percentiles are used in statistics as a way of determining rank or order in relation to other points in a distribution of numbers or scores. A percentile is a value below the point where a particular percent of scores or observations falls. #### What does the 70th percentile mean? The 70th percentile means that 70% of the scores were below your score, and 30% were above your score. Your actual score was 82%, which means that you answered 82% of the test questions correctly. Seven students got the following exam scores (percent correct) on a science exam: 0%, 40%, 50%, 65%, 75%, 90%, 100%. What does median percentile mean? 1 Answer. The median growth percentile summarizes student growth rates by district, school, grade level, or other group of interest. The median is calculated by taking the individual student growth percentiles of all the students in the group being analyzed, ordering them from lowest to highest, and identifying the middle score, the median.
Courses Courses for Kids Free study material Offline Centres More Store # The equivalent resistance between X and Y will beA) 4 ohmB) 4.5 ohmC) 2 ohmD) 20 ohm. Last updated date: 20th Jun 2024 Total views: 54.3k Views today: 1.54k Verified 54.3k+ views Hint: Resistors are said to be in parallel when their two terminals connect to the same node. On the other hand, resistors are said to be in series when they are connected head-to-tail and there is no other wire branching off from the nodes between components. For example, in the above circuit, the resistors 7 ohm and 2 ohm are connected parallel. Complete step by step solution: Step 1: first we will calculate the equivalent resistance \$R'\$ of the 7ohm and 2 ohm. Express the formula for the equivalent resistance when two resistors are connected parallel. \$\therefore \dfrac{1}{{R'}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}\$ , where \$R'\$ is the equivalent resistor of the 7 ohm and 2 ohm. Step 2: Substitute the values for \${R_1}\$ and \${R_2}\$ \$\therefore \dfrac{1}{{R'}} = \dfrac{1}{7} + \dfrac{1}{2}\$ \$ \Rightarrow \dfrac{1}{{R'}} = \dfrac{9}{{14}}\$ Step 3: Take the reciprocal on both sides \[\therefore R' = \dfrac{{14}}{9}\] Step 4: Now similarly calculate the equivalent resistance \$R''\$ of 5 ohm and 6 ohm which is also parallel. \$\therefore \dfrac{1}{{R''}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}\$ Step 5: Substitute the values for \${R_1}\$ and \${R_2}\$ \$\therefore \dfrac{1}{{R''}} = \dfrac{1}{5} + \dfrac{1}{6}\$ \$ \Rightarrow \dfrac{1}{{R''}} = \dfrac{{11}}{{30}}\$ Step 6: Now our circuit will be Now clearly you can see that there are two resistors in our circuit and both are connected in a series. Express the formula for the equivalent resistance of the two resistors connected in series. \$\therefore {R_{equ}} = {R_1} + {R_2}\$ Step 7: Now substitute the values \$\dfrac{{14}}{9}\$ for \${R_1}\$ and \$\dfrac{{11}}{{30}}\$ for \${R_2}\$ \$\therefore {R_{equ}} = \dfrac{{14}}{9} + \dfrac{{11}}{{30}}\$ \$ \Rightarrow {R_{equ}} \simeq 2\$ ohm Hence the correct option is Option C. Note: While calculating equivalent resistance in a circuit always remember that the resistors connected parallel should be added first. After adding all parallel resistors the entire circuit will become simple where all the remaining resistors will be connected in series. Resistors connected in series are easier to find equivalent resistance.
# Drawing and Interpreting Pictograms (Pictographs) Drawing and Interpreting Pictograms • This pictogram shows the number of cars parked in a car park on each day. • Every car drawn represents 10 cars parked in the car park. • To work out the number of cars parked each day we count up in tens for each car drawn. • On Monday there is one car drawn, which is worth 10 cars parked in the car park. • On Tuesday there are two cars drawn, which is worth 20. • On Wednesday there is one car drawn, worth 10. • On Thursday there are four cars drawn, worth 40. • On Friday there are three cars drawn, worth 30. Each picture in a pictogram is worth a set value. • Each whole car drawn is worth 10 cars. • Each half car drawn is worth 5 because 5 is half of ten. • On Monday we have 20 cars. 20 is worth two tens and so we draw 2 cars. • On Tuesday there are 10 cars, which is one car drawn on the pictogram. • On Wednesday there are 5 cars, which is shown with half a car on the pictogram. • On Thursday there are 15 cars, which is 10 + 5. We draw this with a full car worth 10, plus half a car worth 5. • On Friday there are 25 cars, which is two tens and a five. We draw this with two full cars, worth 10 plus half a car, worth 5. #### Pictograms Lesson Accompanying Activity Sheet Supporting Lessons # How do we Read, Draw and Interpret Pictograms? A pictogram is a chart which uses pictures to represent numbers. A pictogram is also known as a pictograph. In the example below, each book drawn means that a person has read 1 book. The pictogram example below shows the number of books read by each person. We can count the number of books drawn next to each person’s name to find out how many each person read. This pictogram is useful because it is visual display of how many books each person read. It can be easier to quickly read and compare the information shown when compared to a list of numbers. For example, we can easily interpret the pictogram to see who read the most and who read the least. Amy read the most with 5 books. Alex read the least with 1 book. There is no need to read every number in the list to answer these questions. We simply need to look for the longest or shortest set of books to see who has read the most or least. In the example below we are asked, “Who read the same number of books?”. We look for the people with the same number of books drawn in each row. Sarah and Terry both read 2 books. In the next question we are asked, “Who read more than 2 books?”. Both James and Amy both read more than 2 books. When teaching the interpretation of pictograms, it is important to explain that 2 books is not more than 2 books. It is a common mistake to include the people that read 2 books as part of this answer. 2 does not count as being more than 2. We only look at 3 or more books. In the next example each time we draw a face, it represents 2 people who have attended a school club. Since each face represents 2 people, drawing half a face is worth half of this. Half of 2 people is 1 person. So drawing one whole face is worth 2 people and drawing half a face is worth 1 person. The list of what each drawing is worth on a pictogram is called the key. The pictogram below shows the number of people who attended the club each day. On Monday there are two whole faces drawn. Each face is worth two people so we have two lots of two which is four. On Tuesday we have half a face, which is worth 1 person. On Wednesday we have a whole face, worth 2 people plus half a face, worth 1. 2 + 1 =3, so 3 people attended the club on Wednesday. On Thursday there are 3 full faces which is three lots of 2. We also have another half of a face which is worth 1 person. In total on Thursday we have 7 people. On Friday, we have 5 people attending the club and we are asked to draw the pictogram row to represent this. 5 is made up of 2 twos and a 1. So we draw two full faces and one half a face. In the example below we are counting the number of cars that use a car park each day. We will draw one car for every 10 cars that we see. Pictograms often use this because it is easier than drawing such a large amount of cars. On Monday we have one car drawn, which is worth 10 cars. On Tuesday we have two cars, which is two lots of ten cars. We have 20 cars parked on Tuesday. On Wednesday we have one car drawn on the pictogram, worth 10 cars. On Thursday we have 4 cars, which is worth four lots of 10, or 40 cars. On Friday we have 3 cars drawn, which is worth 30. In the example below, we are using the data read from the pictogram to see which day had the most cars parked. The biggest number is 40 and this was on Thursday. We can easily read the pictogram to see that Thursday has the most cars. We can see this from the image, rather than needing to read every number. In the next example of interpreting a pictogram, we are now asked, “Which days were there less than 30 cars?”. Remember that less than 30 does not include the number 30 itself. We need to look for numbers that are smaller than 30. Monday, Tuesday and Wednesday all have fewer than 30 cars. We know that for every car we draw on our pictogram, it is worth 10 cars in real life. We can draw half a car to represent half of this amount. If the whole car is worth 10, then half a car is worth half of 10. Drawing half a car is worth 5 cars. Here is the key for our pictogram. We will use this key to draw a pictogram for the values in the example below. On Monday we have 20 cars, which is simply two lots of 10. We draw two full cars. On Tuesday we have 10 cars, which is shown with one full car. On Wednesday there are 5 cars parked so we draw half a car. On Thursday there are 15 cars. 15 is made from one ten plus a five. We draw one whole car and one half a car. On Friday we have 25 cars. This is made of two tens and a five. So we draw 2 whole cars and one half a car. In the last example, we needed to draw half of the picture. In the most complicated school examples, we can be asked to divide our picture up into quarters. Below is a drink. The whole image represents four drinks. If we draw half a drink, we are showing half of four. Drawing half a drink is worth 2 drinks. If we half this again we have a quarter. A quarter of four drinks is 1. Drawing a quarter of the drink picture is worth 1 drink. Below is the key to be used in this pictogram. We will use this key to read the following pictogram showing how many drinks each person drank. William has two full drink pictures drawn. Each full drink picture is worth 4 drinks, so William drank two lots of 4 drinks. William drank 8 drinks. Megan has a quarter drink picture. This is worth 1 drink. Sammi has a full drink picture, worth 4 and a half drink picture, worth 2. 4 + 2 = 6 and so, Sammi drank 6 drinks. Bruce has two full drink pictures, each worth 4 along with a quarter drink picture, worth 1. 4 + 4 + 1 = 9. Bruce drank 9 drinks. Here is our final pictograph example with the same key. Fred has a quarter drink picture, worth 1 drink. Grace has a half drink image, worth 2 drinks. Jack has a half drink image, worth 2 drinks plus a quarter drink image worth 1. 2 + 1 = 3 and so, Jack drank 3 drinks. Kate has a full drink image, worth 4 drinks plus a quarter drink picture, worth 1 drink. 4 + 1 = 5 and so, Kate drank 5 drinks. With these examples it is important to keep referring to the pictogram key and it can help to write the number that each image is worth on top of the images in the pictogram. Once you have written the number on top of each image using the key, you can add up the values afterwards. Now try our lesson on Tally Charts where we learn how to draw and read tally charts. error: Content is protected !!
# 6.1 Reading and writing decimals Page 1 / 2 This module is from Fundamentals of Mathematics by Denny Burzynski and Wade Ellis, Jr. This module discusses how to read and write decimals. By the end of the module students should understand the meaning of digits occurring to the right of the ones position, be familiar with the meaning of decimal fractions and be able to read and write a decimal fraction. ## Section overview • Digits to the Right of the Ones Position • Decimal Fractions • Writing Decimal Fractions ## Digits to the right of the ones position We began our study of arithmetic ( [link] ) by noting that our number system is called a positional number system with base ten. We also noted that each position has a particular value. We observed that each position has ten times the value of the position to its right. This means that each position has $\frac{1}{10}$ the value of the position to its left. Thus, a digit written to the right of the units position must have a value of $\frac{1}{\text{10}}$ of 1. Recalling that the word "of" translates to multiplication $\left(\cdot \right)$ , we can see that the value of the first position to the right of the units digit is $\frac{1}{\text{10}}$ of 1, or $\frac{1}{\text{10}}\cdot 1=\frac{1}{\text{10}}$ The value of the second position to the right of the units digit is $\frac{1}{\text{10}}$ of $\frac{1}{\text{10}}$ , or $\frac{1}{\text{10}}\cdot \frac{1}{\text{10}}=\frac{1}{{\text{10}}^{2}}=\frac{1}{\text{100}}$ The value of the third position to the right of the units digit is $\frac{1}{\text{10}}$ of $\frac{1}{\text{100}}$ , or $\frac{1}{\text{10}}\cdot \frac{1}{\text{100}}=\frac{1}{{\text{10}}^{3}}=\frac{1}{\text{1000}}$ This pattern continues. We can now see that if we were to write digits in positions to the right of the units positions, those positions have values that are fractions. Not only do the positions have fractional values, but the fractional values are all powers of 10 $\left(\text{10},{\text{10}}^{2},{\text{10}}^{3},\dots \right)$ . ## Decimal point, decimal If we are to write numbers with digits appearing to the right of the units digit, we must have a way of denoting where the whole number part ends and the fractional part begins. Mathematicians denote the separation point of the units digit and the tenths digit by writing a decimal point . The word decimal comes from the Latin prefix "deci" which means ten, and we use it because we use a base ten number system. Numbers written in this form are called decimal fractions , or more simply, decimals . Notice that decimal numbers have the suffix "th." ## Decimal fraction A decimal fraction is a fraction in which the denominator is a power of 10. The following numbers are examples of decimals. 1. 42.6 The 6 is in the tenths position. $\text{42}\text{.}6=\text{42}\frac{6}{\text{10}}$ 2. 9.8014 The 8 is in the tenths position. The 0 is in the hundredths position. The 1 is in the thousandths position. The 4 is in the ten thousandths position. $9\text{.}\text{8014}=9\frac{\text{8014}}{\text{10},\text{000}}$ 3. 0.93 The 9 is in the tenths position. The 3 is in the hundredths position. $0\text{.}\text{93}=\frac{\text{93}}{\text{100}}$ Quite often a zero is inserted in front of a decimal point (in the units position) of a decimal fraction that has a value less than one. This zero helps keep us from overlooking the decimal point. 4. 0.7 The 7 is in the tenths position. $0\text{.}7=\frac{7}{\text{10}}$ We can insert zeros to the right of the right-most digit in a decimal fraction without changing the value of the number. $\frac{7}{\text{10}}=0\text{.}7=0\text{.}\text{70}=\frac{\text{70}}{\text{100}}=\frac{7}{\text{10}}$ 1. Read the whole number part as usual. (If the whole number is less than 1, omit steps 1 and 2.) 2. Read the decimal point as the word "and." 3. Read the number to the right of the decimal point as if it were a whole number. 4. Say the name of the position of the last digit. how can chip be made from sand are nano particles real yeah Joseph Hello, if I study Physics teacher in bachelor, can I study Nanotechnology in master? no can't Lohitha where we get a research paper on Nano chemistry....? nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review Ali what are the products of Nano chemistry? There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others.. learn Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level learn da no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts Bhagvanji hey Giriraj Preparation and Applications of Nanomaterial for Drug Delivery revolt da Application of nanotechnology in medicine has a lot of application modern world Kamaluddeen yes narayan what is variations in raman spectra for nanomaterials ya I also want to know the raman spectra Bhagvanji I only see partial conversation and what's the question here! what about nanotechnology for water purification please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment. Damian yes that's correct Professor I think Professor Nasa has use it in the 60's, copper as water purification in the moon travel. Alexandre nanocopper obvius Alexandre what is the stm is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.? Rafiq industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong Damian How we are making nano material? what is a peer What is meant by 'nano scale'? What is STMs full form? LITNING scanning tunneling microscope Sahil how nano science is used for hydrophobicity Santosh Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq Rafiq what is differents between GO and RGO? Mahi what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq Rafiq if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION Anam analytical skills graphene is prepared to kill any type viruses . Anam Any one who tell me about Preparation and application of Nanomaterial for drug Delivery Hafiz what is Nano technology ? write examples of Nano molecule? Bob The nanotechnology is as new science, to scale nanometric brayan nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale Damian how did you get the value of 2000N.What calculations are needed to arrive at it Privacy Information Security Software Version 1.1a Good In the number 779,844,205 how many ten millions are there? From 1973 to 1979, in the United States, there was an increase of 166.6% of Ph.D. social scientists to 52,000. How many were there in 1973? 7hours 36 min - 4hours 50 min
# Algorithms on Graphs: What is a graph? To simply put it, a graph is a collection of vertices and a collection of edges such that each edge connects a pair of vertices together. A visual representation would be: However, graphs can come with various properties and varying sizes. Here are a few examples: At this moment, we’re not going to delve into all these examples right away. With ease, we’re going to learn much about the varying types of graphs and the algorithms that are used for them. While it’s nice to see how the graphs are represented in a visual sense, let’s talk about how they are often represented in code. # Graph Representation Let’s take this graph as an example and see how we would represent it in code. One way it could be represented would be by an adjacency matrix that looks like this. Notice how this is a n x n matrix that’s symmetric along the diagonal. The symmetry is really due to the nature of our example. Since our graph is bidirectional, it makes sense that our matrix is symmetric since both vertices are neighbors with each other. Now, we know that often when we’re dealing with n x n matrices, it’s most often represented as a 2D array like this: Now this is great if we’re trying to find a value at (i,j) because an index lookup on an array is O(1) (constant time). However, having an array representation does come with some cons. O(1) access to nodes is nice if we already know which nodes to look for. However, when we’re traversing through a node’s adjacent neighbors, we have to look at each node to determine if it’s a neighbor which will take O(V) time, where V is the number of nodes in the graph. Not only that, but notice how we have a bunch of 0’s in the array to indicate that a node is not neighbors with another node. These 0’s can unnecessarily take up a lot of space and make our memory O(V²). Let’s look at another representation that we can use. Another way we can condense space and search time of neighbors is through an adjacency list: The adjacency list allows us to keep track of only neighboring nodes. This allows us to condense our memory usage from O(V²) to O(V + E), where E is the number of edges in the graph. Lookup time for a node’s neighbors are also shortened to O(E) because E represents the number of edges connected to the node (or another way to see it is that E represents the number of neighbors that the node has). Throughout much of these posts, I’ll be using the adjacency list over the adjacency matrix in order to optimize our algorithms. # Summary • Graphs come in varying size and shape. We’ll discuss how they influence our algorithms in future posts. • A graph can be represented either as an adjacency matrix or an adjacency list. • An adjacency matrix allows us O(1) access if we know the node but take up O(V) times for searching neighbors and O(V²) space. • An adjacency list cuts our search time to O(E) and our memory to O(V + E). ## More from Try Khov Software Engineer ## What is “Git”? Get the Medium app
# 3.4 - Difference in Proportions 3.4 - Difference in Proportions When the null hypothesis of independence is rejected, the nature of dependence---its direction and magnitude---needs to be measured. A statistic that measures the direction and magnitude of the relationship between two variables is called an effect size or a measure of association. One of the most intuitive measures of association is the difference in proportions, which compares the relative frequency of important characteristics between two groups. For example in the Vitamin C study, we want to know if the probability of a member of the placebo group contracting cold is the same as a probability of a member of the ascorbic group contracting cold. Regarding $$Z$$ as a response and $$Y$$ as explanatory variable, the difference in proportions for a $$2 \times 2$$ table is $$\delta =P(Z=1\|Y=1)-P(Z=1\|Y=2)= \dfrac{\pi_{11}}{\pi_{1+}}-\dfrac{\pi_{21}}{\pi_{2+}} = \pi_{1\|1}-\pi_{1\|2}$$ where $$\pi_{1\|1}$$ is the probability of "success" (e.g., "cold"), given row 1, and $$\pi_{1\|2}$$ is the probability of "success", given row 2. Recall that these are the conditional probabilities we already described for the Vitamin C example. Thus, the probability of "failure" (e.g., "no cold"), given row 1 is $$\pi_{2\|1}$$ and $$\pi_{1\|1}+ \pi_{2\|1}=1$$. Similarly, we can find conditional probabilities, given row 2. In social sciences and epidemiology, these are sometimes referred to as "risk" values. That is, we may refer to the probability that a person gets a cold, given that he/she took a placebo pill, as the risk of such an event. Furthermore, for diagnostic tests, the conditional probability that the diagnostic test is positive, given that the subject has a disease, is called sensitivity. The conditional probability that the diagnostic test is negative, given that the subject does NOT have a disease, is called specificity. Finally, because $$\delta$$ is a function only of the parameters of $$P(Z \| Y )$$, likelihood-based inferences will be the same, regardless if calculations assume Poisson sampling or multinomial sampling. ## Point Estimation for $$\delta$$ The natural estimate of $$\delta$$ is the difference in the conditional sample proportions: $$\hat{\delta}=d=\dfrac{n_{11}}{n_{1+}}-\dfrac{n_{21}}{n_{2+}}=p_{1\|1}-p_{1\|2}$$ #### Properties • It takes values between $$-1$$ and $$+1$$, • If variables are independent, the difference in the proportions equals 0. This is the maximum-likelihood estimate (MLE) because under product-multinomial sampling, the numerators are independent binomials: $$n_{11} \approx Bin( n_{1+}, \dfrac{\pi_{11}}{\pi_{1+}} )$$ $$n_{21} \approx Bin( n_{2+},\dfrac{\pi_{21}}{\pi_{2+}} )$$ For the Vitamin C example, the estimated or sample difference of proportions of getting a cold is $$d=17/139 - 31/140 = 0.12 - 0.22 = -0.10$$. Cold No Cold Totals Placebo 0.22 0.78 1 Ascorbic Acid 0.12 0.88 1 Question: Is this difference "big" or "small"? ## Confidence interval for $$\delta$$ If $$n_{1+}$$ and $$n_{2+}$$ are large, the estimate $$d$$ is approximately normal with variance $$\displaystyle V(d)=\left[\frac{ \frac{\pi_{11}}{\pi_{1+}} (1-\frac{\pi_{11}}{\pi_{1+}})} {n_{1+}} + \frac{\frac{\pi_{21}}{\pi_{2+}} (1-\frac{\pi_{21}}{\pi_{2+}})} {n_{2+}} \right]$$ This expression follows from the fact that if $$X_1$$ and $$X_2$$ are independent random variables, then $$Var(X_1 − X_2) = Var(X_1) + Var(X_2)$$. Both $$X_1$$ and $$X_2$$ here are Bernoulli random variables. Plugging in the estimates, we get the estimate of the variance $$\hat{V}(d)=\dfrac{n_{11}n_{12}}{(n_{1+})^3}+\dfrac{n_{21}n_{22}}{(n_{2+})^3}$$ which is used for computing the standard errors and confidence intervals. A large sample, $$(1- \alpha) 100\%$$ CI for the Vitamin C example is $$SE(d)=\sqrt{\hat{V}(d)}=\sqrt{\dfrac{0.12\times 0.88}{139}+\dfrac{0.22\times 0.78}{140}}=0.045$$ $$\Rightarrow -0.10\pm 1.96\times 0.045=(-0.19,-0.01)$$ #### Interpret We are 95% confident that the true difference in proportions of people getting cold given the placebo or a vitamin C is somewhere between 1% and 19%. Note that the value of 0 (corresponding to no difference) does not fall within that boundary. ## Hypothesis testing for $$\delta$$ Under the null hypothesis of no difference, $$H_0: \delta = 0$$, the rows of the table can be pooled to get an estimate of the common proportion, $$P(Z = 1 \| Y = 1) = P(Z = 1 \| Y = 2)$$. The pooled estimate is $$\hat{\pi}=\dfrac{n_{11}+n_{21}}{n_{1+}+n_{2+}}$$ Under $$H_0: \delta = 0$$, a more efficient estimate of variance $$V (d)$$ is $$\dfrac{\hat{\pi}(1-\hat{\pi})}{n_{1+}}+\dfrac{\hat{\pi}(1-\hat{\pi})}{n_{2+}}$$ and the test statistic $$z=\dfrac{d}{\sqrt{\frac{\hat{p}(1-\hat{p})}{n_{1+}}+\frac{\hat{p}(1-\hat{p})}{n_{2+}}}}$$ is approximately distributed as $$N(0,1)$$. Many elementary textbooks in statistics use this test to compare two proportions. For the Vitamin C example, $$z=\dfrac{31/140-17/139}{\sqrt{\frac{48}{279}\times \frac{231}{279}\times (\frac{1}{140}+\frac{1}{139})}}=2.19$$ This value is significant at the 0.05 level, so if data are taken at face value, we could conclude that the proportion of colds in vitamin C group is smaller than in the placebo group. The result is consistent with confidence interval inference. For computation in SAS, see VitaminC.sas, Vitamin C SAS Output. The analysis can be done with PROC FREQ, using options MEASURES or RISKDIFF. Compare the values of the above calculations to relevant SAS output under heading "Statistics for Table of Treatment and Response". Notice that $$d=-0.099$$; we just rounded the value to $$-0.01$$ in our calculations. For computation in R, see VitaminC.R. We can use, among others, either prop.test() or binom.test() to get these probabilities. ## Equivalence to statistical independence test It is useful to note that the null hypothesis $$H_0:\delta = 0$$ is equivalent to independence. Using conditional probabilities, this corresponds to $$\pi_{j\|1} = \pi_{j\|2}$$ Thus we can test $$\delta = 0$$ by the usual $$X^2$$ or $$G^2$$ test for independence in a $$2\times 2$$ table already discussed in the previous sections. In fact, we can show that $$(\text{z-statistic})^2$$ above is algebraically equal to $$X^2$$. So a two-sided test based on comparing the z-statistic to a $$N(0, 1)$$ distribution is identical to comparing $$X^2$$ from the test of independence to a chi-squared distribution with $$\chi^2_1$$. In the Vitamin C example, note that $$X^2 = 4.81 \approx (2.19)^2$$. Even though the difference of two proportions is very easy to interpret, one problem with using $$\delta$$ is that when $$Z = 1$$ is a rare event, the individual probabilities $$P(Z = 1 \| Y = 1)$$ and $$P(Z = 1 \| Y = 2)$$ are both small, i.e., close to zero. The absolute value of $$\delta$$ will be close to zero even when the effect is strong. In the following sections, we study two other common measures of association which compare the relative value of the proportions, rather than the absolute values. [1] Link ↥ Has Tooltip/Popover Toggleable Visibility
solve-system-of-linear-equations-by-substitution-method-solving-word-problems-by-substitution # Interactive video lesson plan for: Solve system of linear equations by substitution method \| Solving word problems by substitution #### Activity overview: Solve system of linear equations by substitution method \| Solving word problems by substitution 00:06 Algebraic Methods of Solving a Pair of Linear Equations 06:01 Solve Systems of equations word problems 11:36 Solving word problems based on the real life situation http://www.learncbse.in/ncert-class-10-math-solutions/ http://www.learncbse.in/pair-of-linear-equations-in-two-variables-cbse-extra-questions/ http://www.learncbse.in/rd-sharma-class-10-solutions-pair-of-linear-equations-in-two-variables/ Procedure for Solve pair of equations by substitution method: Step 1 : We pick either of the equations and write one variable in terms of the other.Say x=f(y). Step 2 : Substitute the value of x in other Equation.solve for y Step 3 : Substituting this value of y in Equation x=f(y).Solve for x. We have substituted the value of one variable by expressing it in terms of the other variable to solve the pair of linear equations. That is why the method is known as the substitution method. Objective of this video content • To solve the pair of linear equations by substitution method. • Learn Algebraic Methods of Solving a Pair of Linear Equations Description: In this video students will algebraically solve the pair of linear equations by substitution method. They will find the value of one variable in terms of other from the first equations and substitute its value in the second equation to get linear equation in one variable. They will solve the equation obtained and get the value of one variable. Further they will find the value of another variable using the value of first variable. Execution: Teacher will distribute the worksheets and students will solve the worksheets. Teacher will discuss the questions on Linear systems once the students complete their worksheets. Parameters for Assessment: . Able to write find the value of one variable in terms of other from first equation. . Able to substitute the above value in the second equation to get a linear equation in one variable. . Able to solve the equation thus obtained and get the value of one variable. . Able to find the value of second variable by using the value of first variable. . Able to write the solution of pair of linear equation. Solve word problems \| Application in word problems \| Procedure • Able to identify the variables from problems given statement • Able to frame pair of linear equations correctly • Able to solve the equations by any of the above methods • Able to verify the solution. Pinterest https://in.pinterest.com/LearnCBSE/ Wordpress https://cbselabs.wordpress.com/ learncbse.in CBSE solutions for class 10 maths Chapter 3 Pair of linear equations Exercise 3.4 CBSE class 10 maths NCERT Solutions Chapter 3 Pair of linear equations Ex 3.2 Solve the system of linear equations in two variables by substitution method Solutions for CBSE class 10 Maths Pair of linear equations Linear systems word problem with substitution NCERT solutions for CBSE class 10 maths Pair of linear equations graphical representation \| Graphical Method \| algebraical Method \| parallel coincident \| consistent \| inconsistent \| Algebraic Methods \| Solving a Pair of Linear Equations \| Substitution Method \| Elimination Method \| Cross - Multiplication Method \| Solving system of linear equations CBSE class 10 maths Solutions Pair of linear equations Exercise 3.5 Common core Math System of linear equations ICSE Class 10 Pair of linear equations Tagged under: ncert solutions,learncbse.,gyanpub,Solve system linear equations,workedout examples,Substitution Method,CCSS.Math.Content,Algebraic Methods,Algebra 1,Solve substitution Method,Solve word problems,Solve pair equations Clip makes it super easy to turn any public video into a formative assessment activity in your classroom. Add multiple choice quizzes, questions and browse hundreds of approved, video lesson ideas for Clip Make YouTube one of your teaching aids - Works perfectly with lesson micro-teaching plans Play this activity 1. Students enter a simple code 2. You play the video 3. The students comment 4. You review and reflect * Whiteboard required for teacher-paced activities ## Ready to see what elsecan do? With four apps, each designed around existing classroom activities, Spiral gives you the power to do formative assessment with anything you teach. Quickfire Carry out a quickfire formative assessment to see what the whole class is thinking Discuss Create interactive presentations to spark creativity in class Team Up Student teams can create and share collaborative presentations from linked devices Clip Turn any public video into a live chat with questions and quizzes ### Spiral Reviews by Teachers and Digital Learning Coaches @kklaster Tried out the canvas response option on @SpiralEducation & it's so awesome! Add text or drawings AND annotate an image! #R10tech Using @SpiralEducation in class for math review. Student approved! Thumbs up! Thanks. @ordmiss Absolutely amazing collaboration from year 10 today. 100% engagement and constant smiles from all #lovetsla #spiral @strykerstennis Students show better Interpersonal Writing skills than Speaking via @SpiralEducation Great #data #langchat folks!
# What does 500 grams equal in cups? 500 grams is equal to 2.1 cups. To convert grams to cups, divide the grams by the conversion ratio of 1 gram to 0.0042267528 cups. Since 500 grams divided by 0.0042267528 cups per gram equals 118.131 cups, rounding to the nearest tenth gives us 2.1 cups. When baking or cooking, you’ll often need to convert between different units of measurement for ingredients. Two common units used are grams and cups. Knowing how to convert between the two allows you to easily substitute ingredients and adjust recipe sizes. One question that often comes up is how many cups are in 500 grams? Let’s walk through the conversion step-by-step: ### Step 1: Identify the Units We are starting with 500 grams and want to convert to cups. ### Step 2: Find the Conversion Ratio To go between any two units, we need a conversion ratio. Luckily, conversion ratios are standardized, so you can easily look them up. The conversion ratio between grams and cups is: 1 gram = 0.0042267528 cups This lets us convert between the two units. ### Step 3: Set up the Unit Conversion Using the conversion ratio, we can now set up our unit conversion: 500 grams * (1 cup / 0.0042267528 grams) ### Step 4: Perform the Calculation Multiplying out the unit conversion gives us: 500 grams * (1 cup / 0.0042267528 grams) = 118.131 cups ### Step 5: Round the Answer For convenience, we’ll round 118.131 cups to 2.1 cups. Therefore, 500 grams is approximately equal to 2.1 cups. ## Conversion Table To make conversions easier, here is table showing some common gram to cup conversions: Grams Cups 100 grams 0.4 cups 200 grams 0.8 cups 250 grams 1 cup 500 grams 2.1 cups 750 grams 3.2 cups 1000 grams (1 kg) 4.2 cups As you can see, 500 grams falls between 250 grams (1 cup) and 750 grams (3.2 cups). So 2.1 cups is a reasonable conversion. ## When to Use Grams vs Cups Now that you know how to convert between them, when should you use grams versus cups in cooking and baking? Here are some general guidelines: ### Use Grams For: – Dry ingredients like flour, sugar, spices, etc. Grams provide the most accurate measurement. – Small quantities under 1 cup. Grams give you more precision. – When following scientific recipes that require exact measurements. ### Use Cups For: – Liquid ingredients like milk, water, oil, etc. Cups are easier to measure out. – Ingredients over 1 cup. Cup measurements make scaling recipe sizes easier. – When you don’t need as much precision, like for casual cooking. – Following conventional recipes where ingredients are given in cups, tablespoons, etc. So in summary: – Use grams for dry ingredients and precision. – Use cups for liquids and convenience. Choosing the right unit will ensure your recipes come out perfectly every time! ## Factors That Change the Gram to Cup Conversion It’s important to note that the grams to cups conversion can vary slightly depending on the ingredient you are measuring. Here are some factors that impact the conversion: ### 1. Density Denser ingredients take up less volume per gram than lighter ingredients. For example, 500g of flour fits into about 2 cups of volume. But 500g of powdered sugar takes up over 2.5 cups of volume due to air pockets between the fine granules. ### 2. Packing How tightly you pack an ingredient into a cup also affects its weight. Loosely packed brown sugar, for example, weighs less per cup than firmly packed brown sugar. ### 3. Humidity For hygroscopic ingredients that absorb moisture, humidity can change the weight per cup. Flour weighs more per cup on humid days. ### 4. Sifting Sifting ingredients like flour or cocoa powder aerates them and decreases their density. Unsifted versus sifted measurements will weigh differently. ### 5. Grind Size Coarsely ground ingredients like rolled oats weigh less per cup than finely ground ingredients like all-purpose flour. So while the standard gram to cup conversion is a good starting point, these factors mean you may need to adjust the conversion up or down slightly depending on the specific ingredient and how it’s measured. ## Common Baking Ingredient Conversions Here are some more specific gram to cup conversions for common baking ingredients: ### All-purpose flour – 120g all-purpose flour = 1 cup – 1 cup all-purpose flour = 4.8 oz = 136 g ### Granulated white sugar – 200g granulated sugar = 1 cup – 1 cup granulated sugar = 7 oz = 198 g ### Brown sugar – 210g lightly packed brown sugar = 1 cup – 1 cup firmly packed brown sugar = 7.4 oz = 210 g ### Butter – 230g butter = 1 cup – 1 cup butter = 8 oz = 227 g ### Powdered sugar – 120g powdered sugar = 1 cup sifted – 1 cup powdered sugar = 4 oz = 113 g ### Cocoa powder – 110g unsifted cocoa powder = 1 cup – 1 cup unsifted cocoa powder = 3.5 oz = 100 g Keep in mind these conversions are approximates. The exact weight will vary based on factors like humidity, packing, and brand of ingredient. When in doubt, weigh ingredients directly in grams for the most accuracy. ## Converting Other Common Food Weights In addition to ingredients, it’s also useful to know gram conversions for common food item weights: ### 1 ounce = 28 grams Handy for converting smaller food amounts like meat portions or cheese servings. ### 8 ounces = 227 grams Useful when working with packaged foods listed in ounces, like cans or boxes. ### 1 pound = 454 grams To convert pounds of meat, produce, etc to grams. ### 1000 grams = 1 kilogram = 2.2 pounds For converting kilograms from international recipes to pounds. ### 1 gram of water = 1 mL water = 0.03 oz water Helpful for liquid volume to weight conversions for water-based ingredients. Knowing these by heart makes converting all kinds of recipes and food package sizes a breeze! ## Tips for Accurate Measuring To get accurate, consistent results when measuring ingredients in grams and cups: – Use a kitchen scale for ingredients measured in grams. Digital scales provide the most precision. – When measuring cups, use standard dry measuring cups (not liquid ones) for dry ingredients. Scoop and level off the top. – For liquids, use clear liquid measuring cups. Read at eye level to get the correct amount. – When measuring small amounts under 1/4 cup, use measuring spoons to ensure accuracy. – Follow any packing instructions in the recipe, like “lightly packed” or “firmly packed”. – Unless specified, don’t sift flour or other ingredients before measuring. Scoop directly from the container. – When doubling recipes, double the gram measurements. Don’t double volume cups. By mastering proper measuring techniques, your recipes will turn out perfect every time. No more failed cakes or salty cookies from inaccurate measures! ## Common Baking Substitutions Using Gram Conversions Knowing gram conversions allows you to easily substitute ingredients in baking recipes: ### Flour – Don’t have all-purpose? Use 120g bread flour or 110g cake flour per 1 cup all-purpose flour. ### Sugar – Replace 1 cup (200g) granulated sugar with 1 cup (210g) packed brown sugar. – Replace 1 cup (200g) granulated sugar with 1 1/4 cup (160g) powdered sugar. ### Butter – Instead of 1 cup (230g) butter, use 3/4 cup (180g) vegetable oil or shortening. Adjust other wet ingredients. – For dairy-free, replace 1 cup (230g) butter with 1 cup (240g) unsweetened applesauce. ### Eggs – Swap out 1 large egg (50g) for 1/4 cup (60g) unsweetened applesauce or 1/2 a mashed banana. – For vegan recipes, replace 1 egg with 1 tablespoon (15g) ground flaxseed whisked with 3 tablespoons (45g) water. With the gram amounts, you can easily scale the substitutions up or down as needed. ## How Cup Measurements Can Go Wrong While very convenient, cup measurements can sometimes lead to baking fails due to their inherent inaccuracies. Here are some things that can go wrong: – Estimating fractions of cups inaccurately, like 1/2 cup versus 3/4 cup. Being even slightly off throws the whole recipe off. – Not leveling off dry ingredients properly leading to too much or too little flour, sugar, etc. – Measuring cups that are worn down with the markings partially erased. – Not using proper measuring cups, like liquid cups versus dry cups. – Scooping baking soda or baking powder directly from the box. Compaction leads to using too much. – Assuming all 1 cup measures are equal. In reality, ingredients weigh differently per cup. – Not following packing instructions in a recipe, loosely packing instead of firmly packing brown sugar for example. – Making substitutions using the same cup measures instead of weights. 3/4 cup oil does not weigh the same as 3/4 cup applesauce. – Doubling volume amounts inaccurately. Two cups of flour is not exactly double one cup. Using a kitchen scale and converting recipes to grams provides protection against many of these issues and helps remove some of the human error element from baking. ## Conclusion So in summary, 500 grams equals approximately 2.1 cups, depending on the ingredient. To convert between grams and cups: – Use the conversion ratio of 1 gram = 0.0042267528 cups – For convenience, use a conversion table for common ingredients – Factor in density, packing, humidity, sifting, and grind size – Use proper measuring techniques for accuracy – Utilize gram weights for better precision in baking recipes Mastering converting between grams and cups gives you more flexibility in the kitchen. So don’t be afraid to mix and match units and substitute ingredients using the right conversions. Understanding gram and cup equivalents helps ensure your baked goods turn out perfectly every time!
# How do you add and subtract mentally fast? ## How do you add and subtract mentally fast? 1. To add 9 to another number, add 10 and then subtract 1: 36 + 9 = 36 + 10 – 1 = 45. 2. To add 18 to another number, add 20 and then subtract 2: 48 + 18 = 48 + 20 – 2 = 66. 3. To add 97 to another number, add 100 and then subtract 3: 439 + 97 = 439 + 100 – 3 = 536. ### How can I increase my addition speed? First add the hundreds place digit in second number with the first number: 953 + 800 = 1753. Then add the then place digit in second number with the result obtained in previous step: 1753 + 60 = 1813. Next add the units place digit in the second number with the result obtained in Step 2: 1813 + 7 = 1820. How can I learn addition quickly? How to Teach Addition \| 7 Simple Steps 1. Introduce the concept using countable manipulatives. Using countable manipulatives (physical objects) will make addition concrete and much easier to understand. 2. Transition to visuals. 3. Use a number line. 4. Counting Up. 5. Finding the ten. 6. Word problems. 7. Memorize the math facts. What is the mental math strategy? Mental math means possessing “tools” that can be used to solve an equation. The goal in our classrooms should not necessarily be to “teach” the strategies, but rather to foster an environment where students construct their own understanding. Instead, we want students to be able to think flexibly when solving problems. ## How are mental math tricks used in math? With mental math tricks you will be able to work out sums in your head more rapidly – a critical skill in math. Here are examples of some mental math strategies for addition. This strategy is used when regrouping is required . One of the addends is broken up into its expanded form and added in parts to the other addend.
# Difference between revisions of "1986 AIME Problems/Problem 2" ## Problem Evaluate the product $(\sqrt 5+\sqrt6+\sqrt7)(-\sqrt 5+\sqrt6+\sqrt7)(\sqrt 5-\sqrt6+\sqrt7)(\sqrt 5+\sqrt6-\sqrt7)$. ## Solution Simplify by repeated application of the difference of squares. $\left((\sqrt{6} + \sqrt{7})^2 - \sqrt{5}^2\right)\left(\sqrt{5}^2 - (\sqrt{6} - \sqrt{7})^2\right)$ $= (13 + 2\sqrt{42} - 5)(5 - (13 - 2\sqrt{42}))$ $= (2\sqrt{42} + 8)(2\sqrt{42} - 8)$ $= (2\sqrt{42})^2 - 8^2 =$ $\boxed{104}$

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