The document discusses expressions and equations. It provides examples of using expressions to calculate total costs given individual costs, and using equations to solve for unknown variables. Specifically, it gives an example of calculating the total cost of x pizzas using the expression "8x + 10" and solving the equation "8x + 10 = 810" to determine that x = 100 pizzas were ordered. It then discusses how to solve rational equations by clearing denominators using the lowest common denominator. An example problem demonstrates solving the rational equation (x - 2)/(x + 1) = 2/4 + 1 through multiplying both sides by the LCD (x - 2)(x + 1).
The document discusses mathematical expressions and polynomials. It provides examples of algebraic expressions involving variables and operations. Polynomial expressions are algebraic expressions that can be written in the form anxn + an-1xn-1 + ... + a1x + a0, where the ai coefficients are numbers. The document gives examples of factoring polynomials using formulas like a3b3 = (ab)(a2ab + b2). Factoring polynomials makes it easier to calculate outputs and simplify expressions for operations like addition and subtraction.
The document discusses expressions and polynomials. It provides examples of algebraic expressions and operations that can be performed on polynomials, such as factoring. Factoring polynomials is useful for easier evaluation, simplifying rational expressions, and solving equations. One example factors the polynomial 64x3 + 125 into (4x + 5)(16x2 - 20x + 25). Factoring the polynomial 2x3 - 5x2 + 2x is recommended before evaluating it for specific values of x.
The document discusses different methods for solving equations, including:
- Solving 1st and 2nd degree polynomial equations by setting them equal to 0 and using factoring or the quadratic formula.
- Solving rational equations by clearing all denominators using the lowest common denominator.
- Solving equations may require transforming them into polynomial equations first through methods like factoring or factoring by grouping.
47 operations of 2nd degree expressions and formulasalg1testreview
The document discusses operations involving binomials and trinomials. It defines a binomial as a two-term polynomial of the form ax + b and a trinomial as a three-term polynomial of the form ax2 + bx + c. It states that the product of two binomials is a trinomial that can be found using the FOIL method: multiplying the first, outer, inner, and last terms. The FOIL method is demonstrated through examples multiplying binomial expressions. Expanding products involving negative binomials requires distributing the negative sign before using FOIL.
1.0 factoring trinomials the ac method and making lists-xmath260
The document discusses factoring trinomials and making lists of numbers to help determine which trinomials are factorable. It states that trinomials are either factorable, where they can be written as the product of two binomials, or prime/unfactorable. Making lists of numbers that satisfy certain criteria, like having a product of the top number in a table, can help identify factorable trinomials and determine the factors.
1.2 review on algebra 2-sign charts and inequalitiesmath265
The document discusses sign charts and inequalities. It explains that sign charts can be used to determine if expressions are positive or negative by factoring them and evaluating at given values of x. Examples are provided to demonstrate how to construct a sign chart by: 1) solving for where the expression equals 0, 2) marking these values on a number line, and 3) evaluating the expression at sample points in each segment to determine the signs in between values where the expression equals 0. The sign chart then indicates the ranges where the expression is positive, negative or zero.
This document discusses first degree functions and linear equations. It explains that most real-world mathematical functions can be composed of algebraic, trigonometric, or exponential/log formulas. Linear equations of the form Ax + By = C represent straight lines that can be graphed by finding the x- and y-intercepts. If an equation contains only one variable, it represents a vertical or horizontal line. The slope-intercept form y = mx + b is introduced, where m is the slope and b is the y-intercept. Slope is defined as the ratio of the rise over the run between two points on a line.
The document discusses exponential and logarithmic functions. Exponential functions of the form f(x) = b^x are called exponential functions in base b. Logarithmic functions log_b(y) represent the exponent x needed to raise the base b to a power to get the output y. The exponential form b^x = y and logarithmic form x = log_b(y) describe the same relationship between the base b, exponent x, and output y. Questions can be translated between these forms by rewriting the exponential expression as a logarithm or vice versa. Examples demonstrate rewriting expressions and graphing logarithmic functions.
The document discusses mathematical expressions and polynomials. It provides examples of algebraic expressions involving variables and operations. Polynomial expressions are algebraic expressions that can be written in the form anxn + an-1xn-1 + ... + a1x + a0, where the ai coefficients are numbers. The document gives examples of factoring polynomials using formulas like a3b3 = (ab)(a2ab + b2). Factoring polynomials makes it easier to calculate outputs and simplify expressions for operations like addition and subtraction.
The document discusses expressions and polynomials. It provides examples of algebraic expressions and operations that can be performed on polynomials, such as factoring. Factoring polynomials is useful for easier evaluation, simplifying rational expressions, and solving equations. One example factors the polynomial 64x3 + 125 into (4x + 5)(16x2 - 20x + 25). Factoring the polynomial 2x3 - 5x2 + 2x is recommended before evaluating it for specific values of x.
The document discusses different methods for solving equations, including:
- Solving 1st and 2nd degree polynomial equations by setting them equal to 0 and using factoring or the quadratic formula.
- Solving rational equations by clearing all denominators using the lowest common denominator.
- Solving equations may require transforming them into polynomial equations first through methods like factoring or factoring by grouping.
47 operations of 2nd degree expressions and formulasalg1testreview
The document discusses operations involving binomials and trinomials. It defines a binomial as a two-term polynomial of the form ax + b and a trinomial as a three-term polynomial of the form ax2 + bx + c. It states that the product of two binomials is a trinomial that can be found using the FOIL method: multiplying the first, outer, inner, and last terms. The FOIL method is demonstrated through examples multiplying binomial expressions. Expanding products involving negative binomials requires distributing the negative sign before using FOIL.
1.0 factoring trinomials the ac method and making lists-xmath260
The document discusses factoring trinomials and making lists of numbers to help determine which trinomials are factorable. It states that trinomials are either factorable, where they can be written as the product of two binomials, or prime/unfactorable. Making lists of numbers that satisfy certain criteria, like having a product of the top number in a table, can help identify factorable trinomials and determine the factors.
1.2 review on algebra 2-sign charts and inequalitiesmath265
The document discusses sign charts and inequalities. It explains that sign charts can be used to determine if expressions are positive or negative by factoring them and evaluating at given values of x. Examples are provided to demonstrate how to construct a sign chart by: 1) solving for where the expression equals 0, 2) marking these values on a number line, and 3) evaluating the expression at sample points in each segment to determine the signs in between values where the expression equals 0. The sign chart then indicates the ranges where the expression is positive, negative or zero.
This document discusses first degree functions and linear equations. It explains that most real-world mathematical functions can be composed of algebraic, trigonometric, or exponential/log formulas. Linear equations of the form Ax + By = C represent straight lines that can be graphed by finding the x- and y-intercepts. If an equation contains only one variable, it represents a vertical or horizontal line. The slope-intercept form y = mx + b is introduced, where m is the slope and b is the y-intercept. Slope is defined as the ratio of the rise over the run between two points on a line.
The document discusses exponential and logarithmic functions. Exponential functions of the form f(x) = b^x are called exponential functions in base b. Logarithmic functions log_b(y) represent the exponent x needed to raise the base b to a power to get the output y. The exponential form b^x = y and logarithmic form x = log_b(y) describe the same relationship between the base b, exponent x, and output y. Questions can be translated between these forms by rewriting the exponential expression as a logarithm or vice versa. Examples demonstrate rewriting expressions and graphing logarithmic functions.
This document discusses algebraic expressions and polynomial expressions. It provides examples of algebraic expressions and defines them as formulas constructed with variables and numbers using basic arithmetic operations. Polynomials are defined as expressions of the form anxn + an-1xn-1 + ... + a1x + a0, where the ai's are numbers. The document gives examples of factoring polynomials and evaluating polynomial expressions at given values. It also discusses using factoring to find the roots of polynomial equations.
The document discusses solving literal equations by isolating the variable of interest on one side of the equation. It provides examples of solving equations for various variables by adding, subtracting, multiplying, or dividing both sides of the equation by the same quantity. The goal is to isolate the variable being solved for so it stands alone on one side of the equal sign. Steps include clearing fractions, moving all other terms to the other side of the equation, and then dividing both sides by the coefficient of the variable being solved for.
The document discusses the difference quotient formula for calculating the slope between two points (x1,y1) and (x2,y2) on a function y=f(x). It shows that the slope m is equal to (f(x+h)-f(x))/h, where h is the difference between x1 and x2. This "difference quotient" formula allows slopes to be calculated from the values of a function at two nearby points. Examples are given of simplifying the difference quotient for quadratic and rational functions.
The document discusses using sign charts to solve polynomial and rational inequalities. It provides examples of solving inequalities by setting one side equal to zero, factoring the expression, drawing the sign chart, and determining the solutions from the regions with the appropriate signs. Specifically, it works through examples of solving x^2 - 3x > 4, 2x^2 - x^3/(x^2 - 2x + 1) < 0, and (x - 2)/(2/(x - 1)) < 3.
The document discusses sign charts and how to determine the signs of outputs for polynomials and rational expressions. It provides examples of factoring polynomials to determine if the output is positive or negative for given values of x. The key steps to create a sign chart are: 1) solve for f=0 and any undefined values, 2) mark these values on a number line, 3) sample points in each segment to determine the sign in that region. Sign charts indicate the regions where a function is positive, negative or zero.
The document discusses polynomial expressions. A polynomial is the sum of monomial terms, where a monomial is a number multiplied by one or more variables raised to a non-negative integer power. Examples show evaluating polynomials by substituting values for variables and calculating each monomial term separately before combining them. A term refers to each monomial within a polynomial. Terms are identified by their variable part, such as the x2-term, x-term, or constant term.
The document discusses sign charts for factorable formulas. It provides examples of determining the sign (positive or negative) of expressions when evaluated at given values of x by factoring the expressions into their factored forms. The key steps to create a sign chart are: 1) solve for values where the expression is equal to 0, 2) mark these values on a number line, 3) select points in each segment to test the sign, 4) indicate the sign (positive or negative) in each segment based on the testing. Sign charts show the regions where an expression is positive, negative or equal to 0.
The document discusses applications of factoring expressions. The main purposes of factoring an expression E into a product E=AB is to utilize properties of multiplication. The most important application of factoring is to solve polynomial equations by setting each factor equal to 0 based on the zero-product property. Examples are provided to demonstrate solving polynomial equations by factoring, setting each factor equal to 0, and extracting the solutions.
The document discusses higher order derivatives. It defines the nth derivative of a function f(x) as f(n)(x). The first example finds the first five derivatives of f(x)=2x^4 - x^3 - 2. The second example finds the first three derivatives of f(x)=-x^2/3. The third example finds the first four derivatives of f(x)=ln(x) and discusses how derivatives of rational functions become more complicated with higher orders. It also provides examples of finding derivatives of other functions like sin(x).
The document discusses terms, factors, and cancellation in mathematics expressions. It defines a term as one or more quantities that are added or subtracted, and a factor as a quantity that is multiplied to other quantities. Cancellation can be used to simplify fractions by canceling common factors in the numerator and denominator. However, terms cannot be canceled as they represent distinct quantities being added or subtracted. Several examples demonstrate identifying terms and factors and applying cancellation when possible.
The document discusses exponents and the rules for working with them. Exponents describe the number of times a base is multiplied by itself. The main rules covered are:
- The multiply-add rule: When the same base has two exponents, you add the exponents and multiply the results, written as ANAK = AN+K
- The divide-subtract rule: When dividing exponents with the same base, you subtract the exponents and divide the results, written as AN/AK = AN-K
Examples are provided to demonstrate calculating exponents using these rules.
The document provides homework assignments and practice problems involving order of operations and evaluating expressions with exponents. It includes:
1) Assigning homework problems from the textbook pages 182 and 188 evaluating expressions and their divisibility.
2) Examples of evaluating expressions with exponents such as -x4 and (-x)4.
3) Practice problems simplifying expressions and evaluating expressions for given values using order of operations.
The document discusses polynomial division algorithms. It introduces long division and synthetic division as methods for dividing polynomials. Long division is analogous to dividing numbers, while synthetic division is simpler but only applies when dividing a polynomial by a monomial. The key points are:
- Long division allows dividing any polynomial P(x) by any polynomial D(x) to obtain a quotient Q(x) and remainder R(x) such that P(x) = Q(x)D(x) + R(x) and the degree of R(x) is less than the degree of D(x).
- Synthetic division is more efficient than long division when dividing a polynomial by a monomial of the form (
The document discusses notation and algebra of functions. It explains that functions assign unique outputs to inputs and are often written as formulas like f(x)=x^2-2x+3. The input box (x) holds the input value to be evaluated in the defining formula. New functions can be formed using addition, subtraction, multiplication, and division of existing functions.
The document discusses formulas for multiplying binomial expressions. It states that the conjugate of expressions like (A + B) is (A - B). The difference of squares formula is given as (A + B)(A - B) = A^2 - B^2. Examples of expanding expressions using this formula and the square formulas (A + B)^2 = A^2 + 2AB + B^2 and (A - B)^2 = A^2 - 2AB + B^2 are provided.
The document discusses using sign charts to determine the sign (positive, negative, or zero) of polynomials and rational expressions for different values of x. It provides examples of drawing sign charts for various expressions and using them to solve inequality statements. Key steps include factoring expressions, identifying zeros and undefined values, and testing sample points in each interval to determine the sign over that interval. Sign charts can then be used to easily solve inequality statements by identifying the intervals where the expression is positive or negative.
The document discusses sign charts and inequalities. It provides examples of determining the sign (positive or negative) of expressions for given values of x. Specifically, it explains how to:
1. Factor polynomials or rational expressions to determine sign.
2. Construct a sign chart by solving for f=0, marking those values on a number line, and testing points in each segment.
3. Use a sign chart to indicate where an expression like f=x^2 - 3x - 4 is equal to 0, positive, or negative based on the value of x.
The document discusses solving numerical equations involving logarithmic and exponential functions in base 10 or base e. It provides examples of solving log and exponential equations by isolating the part containing the unknown, then rewriting the equation in the opposite form (log to exponential or exponential to log). The key steps outlined are: 1) isolate the exponential/log part containing the unknown, 2) rewrite the equation by "bringing down" exponents as logarithms or vice versa. Several examples are worked through demonstrating these steps.
The document discusses complex numbers. It begins by explaining that the equation x^2 = -1 has no real solutions, so an imaginary number i is defined such that i^2 = -1. A complex number is then defined as a number of the form a + bi, where a is the real part and bi is the imaginary part. Rules for adding, subtracting and multiplying complex numbers by treating i as a variable and setting i^2 to -1 are provided. Examples of solving equations and performing operations with complex numbers are given.
This document discusses solving numerical equations involving logarithmic and exponential functions. It provides examples of solving both log equations and exponential equations. To solve log equations, the problems are rewritten in exponential form by removing the log. To solve exponential equations, the exponents are brought down by rewriting the problem in logarithmic form. Steps include isolating the exponential or logarithmic term containing the unknown, rewriting the equation accordingly, and then solving for the unknown variable. Practice problems with solutions are provided to illustrate the process.
The document discusses solving equations by adding, subtracting, multiplying, and dividing. It provides examples of using the subtraction and addition properties of equality to isolate the variable by subtracting or adding the same quantity to both sides of an equation. Word problems are also presented where equations are set up and solved to find unknown values. Students are instructed to practice these solution techniques for homework.
Solving addition and subtraction equations power point copykatiewilkerosn
This document discusses how to solve addition and subtraction equations by using inverse operations. It explains that an equation contains an equal sign connecting two expressions, which may include variables representing unknown numbers. To solve an equation, the same operation must be applied to both sides so that the expressions remain equal. Examples demonstrate finding the value of variables by subtracting or adding to both sides of an equation until the variable is isolated. The key steps are identifying the operation, applying the inverse operation to both sides, writing the value of the variable, and checking the solution.
This document discusses algebraic expressions and polynomial expressions. It provides examples of algebraic expressions and defines them as formulas constructed with variables and numbers using basic arithmetic operations. Polynomials are defined as expressions of the form anxn + an-1xn-1 + ... + a1x + a0, where the ai's are numbers. The document gives examples of factoring polynomials and evaluating polynomial expressions at given values. It also discusses using factoring to find the roots of polynomial equations.
The document discusses solving literal equations by isolating the variable of interest on one side of the equation. It provides examples of solving equations for various variables by adding, subtracting, multiplying, or dividing both sides of the equation by the same quantity. The goal is to isolate the variable being solved for so it stands alone on one side of the equal sign. Steps include clearing fractions, moving all other terms to the other side of the equation, and then dividing both sides by the coefficient of the variable being solved for.
The document discusses the difference quotient formula for calculating the slope between two points (x1,y1) and (x2,y2) on a function y=f(x). It shows that the slope m is equal to (f(x+h)-f(x))/h, where h is the difference between x1 and x2. This "difference quotient" formula allows slopes to be calculated from the values of a function at two nearby points. Examples are given of simplifying the difference quotient for quadratic and rational functions.
The document discusses using sign charts to solve polynomial and rational inequalities. It provides examples of solving inequalities by setting one side equal to zero, factoring the expression, drawing the sign chart, and determining the solutions from the regions with the appropriate signs. Specifically, it works through examples of solving x^2 - 3x > 4, 2x^2 - x^3/(x^2 - 2x + 1) < 0, and (x - 2)/(2/(x - 1)) < 3.
The document discusses sign charts and how to determine the signs of outputs for polynomials and rational expressions. It provides examples of factoring polynomials to determine if the output is positive or negative for given values of x. The key steps to create a sign chart are: 1) solve for f=0 and any undefined values, 2) mark these values on a number line, 3) sample points in each segment to determine the sign in that region. Sign charts indicate the regions where a function is positive, negative or zero.
The document discusses polynomial expressions. A polynomial is the sum of monomial terms, where a monomial is a number multiplied by one or more variables raised to a non-negative integer power. Examples show evaluating polynomials by substituting values for variables and calculating each monomial term separately before combining them. A term refers to each monomial within a polynomial. Terms are identified by their variable part, such as the x2-term, x-term, or constant term.
The document discusses sign charts for factorable formulas. It provides examples of determining the sign (positive or negative) of expressions when evaluated at given values of x by factoring the expressions into their factored forms. The key steps to create a sign chart are: 1) solve for values where the expression is equal to 0, 2) mark these values on a number line, 3) select points in each segment to test the sign, 4) indicate the sign (positive or negative) in each segment based on the testing. Sign charts show the regions where an expression is positive, negative or equal to 0.
The document discusses applications of factoring expressions. The main purposes of factoring an expression E into a product E=AB is to utilize properties of multiplication. The most important application of factoring is to solve polynomial equations by setting each factor equal to 0 based on the zero-product property. Examples are provided to demonstrate solving polynomial equations by factoring, setting each factor equal to 0, and extracting the solutions.
The document discusses higher order derivatives. It defines the nth derivative of a function f(x) as f(n)(x). The first example finds the first five derivatives of f(x)=2x^4 - x^3 - 2. The second example finds the first three derivatives of f(x)=-x^2/3. The third example finds the first four derivatives of f(x)=ln(x) and discusses how derivatives of rational functions become more complicated with higher orders. It also provides examples of finding derivatives of other functions like sin(x).
The document discusses terms, factors, and cancellation in mathematics expressions. It defines a term as one or more quantities that are added or subtracted, and a factor as a quantity that is multiplied to other quantities. Cancellation can be used to simplify fractions by canceling common factors in the numerator and denominator. However, terms cannot be canceled as they represent distinct quantities being added or subtracted. Several examples demonstrate identifying terms and factors and applying cancellation when possible.
The document discusses exponents and the rules for working with them. Exponents describe the number of times a base is multiplied by itself. The main rules covered are:
- The multiply-add rule: When the same base has two exponents, you add the exponents and multiply the results, written as ANAK = AN+K
- The divide-subtract rule: When dividing exponents with the same base, you subtract the exponents and divide the results, written as AN/AK = AN-K
Examples are provided to demonstrate calculating exponents using these rules.
The document provides homework assignments and practice problems involving order of operations and evaluating expressions with exponents. It includes:
1) Assigning homework problems from the textbook pages 182 and 188 evaluating expressions and their divisibility.
2) Examples of evaluating expressions with exponents such as -x4 and (-x)4.
3) Practice problems simplifying expressions and evaluating expressions for given values using order of operations.
The document discusses polynomial division algorithms. It introduces long division and synthetic division as methods for dividing polynomials. Long division is analogous to dividing numbers, while synthetic division is simpler but only applies when dividing a polynomial by a monomial. The key points are:
- Long division allows dividing any polynomial P(x) by any polynomial D(x) to obtain a quotient Q(x) and remainder R(x) such that P(x) = Q(x)D(x) + R(x) and the degree of R(x) is less than the degree of D(x).
- Synthetic division is more efficient than long division when dividing a polynomial by a monomial of the form (
The document discusses notation and algebra of functions. It explains that functions assign unique outputs to inputs and are often written as formulas like f(x)=x^2-2x+3. The input box (x) holds the input value to be evaluated in the defining formula. New functions can be formed using addition, subtraction, multiplication, and division of existing functions.
The document discusses formulas for multiplying binomial expressions. It states that the conjugate of expressions like (A + B) is (A - B). The difference of squares formula is given as (A + B)(A - B) = A^2 - B^2. Examples of expanding expressions using this formula and the square formulas (A + B)^2 = A^2 + 2AB + B^2 and (A - B)^2 = A^2 - 2AB + B^2 are provided.
The document discusses using sign charts to determine the sign (positive, negative, or zero) of polynomials and rational expressions for different values of x. It provides examples of drawing sign charts for various expressions and using them to solve inequality statements. Key steps include factoring expressions, identifying zeros and undefined values, and testing sample points in each interval to determine the sign over that interval. Sign charts can then be used to easily solve inequality statements by identifying the intervals where the expression is positive or negative.
The document discusses sign charts and inequalities. It provides examples of determining the sign (positive or negative) of expressions for given values of x. Specifically, it explains how to:
1. Factor polynomials or rational expressions to determine sign.
2. Construct a sign chart by solving for f=0, marking those values on a number line, and testing points in each segment.
3. Use a sign chart to indicate where an expression like f=x^2 - 3x - 4 is equal to 0, positive, or negative based on the value of x.
The document discusses solving numerical equations involving logarithmic and exponential functions in base 10 or base e. It provides examples of solving log and exponential equations by isolating the part containing the unknown, then rewriting the equation in the opposite form (log to exponential or exponential to log). The key steps outlined are: 1) isolate the exponential/log part containing the unknown, 2) rewrite the equation by "bringing down" exponents as logarithms or vice versa. Several examples are worked through demonstrating these steps.
The document discusses complex numbers. It begins by explaining that the equation x^2 = -1 has no real solutions, so an imaginary number i is defined such that i^2 = -1. A complex number is then defined as a number of the form a + bi, where a is the real part and bi is the imaginary part. Rules for adding, subtracting and multiplying complex numbers by treating i as a variable and setting i^2 to -1 are provided. Examples of solving equations and performing operations with complex numbers are given.
This document discusses solving numerical equations involving logarithmic and exponential functions. It provides examples of solving both log equations and exponential equations. To solve log equations, the problems are rewritten in exponential form by removing the log. To solve exponential equations, the exponents are brought down by rewriting the problem in logarithmic form. Steps include isolating the exponential or logarithmic term containing the unknown, rewriting the equation accordingly, and then solving for the unknown variable. Practice problems with solutions are provided to illustrate the process.
The document discusses solving equations by adding, subtracting, multiplying, and dividing. It provides examples of using the subtraction and addition properties of equality to isolate the variable by subtracting or adding the same quantity to both sides of an equation. Word problems are also presented where equations are set up and solved to find unknown values. Students are instructed to practice these solution techniques for homework.
Solving addition and subtraction equations power point copykatiewilkerosn
This document discusses how to solve addition and subtraction equations by using inverse operations. It explains that an equation contains an equal sign connecting two expressions, which may include variables representing unknown numbers. To solve an equation, the same operation must be applied to both sides so that the expressions remain equal. Examples demonstrate finding the value of variables by subtracting or adding to both sides of an equation until the variable is isolated. The key steps are identifying the operation, applying the inverse operation to both sides, writing the value of the variable, and checking the solution.
Solving equations using addition and subtractionmcarlinjr0303
The document provides instructions for solving one-step equations with one variable by using addition or subtraction. It defines key terms like equation and solution. It explains that to find the solution, the variable must be isolated on one side of the equation by using the inverse of the operation on the other side of the equation. Several examples are provided and checked to demonstrate solving equations by adding or subtracting the same quantity to both sides.
This document provides instruction on solving one-step equations by adding or subtracting. It defines key terms like equation and variable. It presents examples of solving equations by adding or subtracting the opposite side of the equation. The examples are accompanied by step-by-step workings and checks of the solutions. Vocabulary and objectives are also outlined.
The document discusses solving equations. It defines key terms like open sentence and equation. It explains that an open sentence with variables is neither true nor false until the variables are replaced with numbers, with each valid replacement called a solution. It outlines properties of equality like reflexive, symmetric, and transitive properties that can be used to solve equations, such as adding or subtracting the same number to both sides.
This document provides instructions for solving two-step equations. It explains that two-step equations are solved in two steps: the first step uses addition or subtraction, and the second step uses multiplication or division. Examples are provided of solving equations by subtracting in the first step and dividing in the second step. Readers are prompted to practice solving sample two-step equations on their own.
The document discusses expressions and equations. It provides an example of calculating the total cost of ordering x pizzas from Pizza Grande using the expression "8x + 10". It then shows how to solve the equation "8x + 10 = 810" to determine that x = 100 pizzas were ordered. The document explains that equations set two expressions equal and solving an equation means finding the value of the variable that makes the equation true. It distinguishes between linear and quadratic equations.
This document provides examples and explanations for solving various types of equations beyond linear and quadratic equations. These include polynomial equations, equations with fractional expressions, equations involving radicals, and equations of quadratic type. Step-by-step solutions are shown for sample equations of each type. Extraneous solutions are discussed. Applications involving dividing a lottery jackpot and calculating bird flight energy expenditure are presented.
This document discusses integrating rational functions. It begins by defining rational functions as functions of the form P(x)/Q(x) where P and Q are polynomials. It states that integrating rational functions involves decomposing them into simpler fractions using the Rational Decomposition Theorem. Examples are then provided to demonstrate how to decompose rational functions and integrate each term using substitution.
2.hyperbolic functions Further Mathematics Zimbabwe Zimsec Cambridgealproelearning
This chapter introduces hyperbolic functions including definitions, properties, graphs and calculus applications. The hyperbolic cosine and sine functions are defined analogously to trigonometric cosine and sine using exponential functions. Hyperbolic functions have similar properties and relationships as trigonometric functions which can be represented using Osborn's rule. Their graphs have distinct shapes important for applications in calculus, including integration and solving equations. Inverse hyperbolic functions are also introduced along with their relationship to corresponding hyperbolic functions.
This document provides an overview of quadratic equations and inequalities. It defines quadratic equations as equations of the form ax2 + bx + c = 0, where a, b, and c are real number constants and a ≠ 0. Examples of quadratic equations are provided. Methods for solving quadratic equations are discussed, including factoring, completing the square, and the quadratic formula. Properties of inequalities are outlined. The chapter also covers solving polynomial and rational inequalities, as well as equations and inequalities involving absolute value. Practice problems are included at the end.
The document discusses different methods for factoring polynomials:
1) Factoring the greatest common factor (GCF) involves finding a number or variable that is common to all terms and dividing each term by the GCF.
2) Factoring the difference of squares uses the formula a^2 - b^2 = (a+b)(a-b) and works for binomials where one term is a perfect square and the other is the negative of a perfect square.
3) Factoring trinomials of the forms x^2 + bx + c and ax^2 + bx + c involves finding two numbers whose product and sum meet certain criteria related to the coefficients in the trinomial.
Final Exam Name___________________________________Si.docxcharlottej5
Final Exam Name___________________________________
Silva Math 96 Spring 2020
YOU MUST SHOW ALL WORK AND BOX YOUR ANSWERS FOR CREDIT. WORK ALONE.
Solve the absolute value inequality. Write your answer
in interval notation.
1) |2x - 12 |> 2
Solve the compound inequality. Graph the solution set.
Write your answer in interval notation.
2) -4x > -8 and x + 4 > 3
Solve the three-part inequality. Write your answer in
interval notation.
3) -1 < 3x + 2 < 14
Solve the absolute value equation.
4) 4x + 9 = 2x + 7
Solve the compound inequality.
5) 3( x + 4 ) ≥ 0 or 4 ( x + 4 ) ≤ 4
Solve the inequality. Graph the solution set and write
your answer in interval notation.
6) |5k + 8| > -6
Solve the inequality graphically. Write your answer in
interval notation .
7) x + 3 ≥ 1
x-8 -6 -4 -2 2
y
8
6
4
2
x-8 -6 -4 -2 2
y
8
6
4
2
1
Graph the system of inequalities.
8) 2x + 8y ≥ -4
y < - 3
2
x + 6
x-10 -8 -6 -4 -2 2 4 6 8 10
y
10
8
6
4
2
-2
-4
-6
-8
-10
x-10 -8 -6 -4 -2 2 4 6 8 10
y
10
8
6
4
2
-2
-4
-6
-8
-10
Find the determinant of the given matrix.
9) 10 5
0 -4
Use Cramer's rule to solve the system of linear
equations.
10) 6x + 5y = -12
2x - 2y = -4
Write a system that models the situation. Then solve the
system using any method. Must show work for credit.
11)A vendor sells hot dogs, bags of potato chips,
and soft drinks. A customer buys 3 hot dogs,
4 bags of potato chips, and 5 soft drinks for
$14.00. The price of a hot dog is $0.25 more
than the price of a bag of potato chips. The
cost of a soft drink is $1.25 less than the price
of two hot dogs. Find the cost of each item.
Use row reduced echelon form to solve the system.
12) x + y + z = 3
x - y + 4z = 11
5x + y + z = -9
2
Find the domain of f. Write your answer in interval
notation.
13) f(x) = 13 - 9x
If possible, simplify the expression. If any variables
exist, assume that they are positive.
14) 2x + 6 32x + 6 8x
Match to the equivalent expression.
15) 100-1/2
A) 1
1000
B) 1
10
C) 1
100
D) 1
10
Write the expression in standard form.
16) (5 + 8i) - (-3 + i)
Simplify the expression. Assume that all variables are
positive.
17) 5 t
5
z10
Solve the equation.
18) 3x + 1 = 3 + x - 4
Write the expression in standard form.
19) 3 + 3i
5 + 3i
3
Write the equation in vertex form.
20) y = x2 + 5x + 2
The graph of ax2 + bx + c is given. Use this graph to solve
ax2 + bx + c = 0, if possible.
21)
x-5 5 10
y
50
40
30
20
10
-10
-20
-30
-40
-50
x-5 5 10
y
50
40
30
20
10
-10
-20
-30
-40
-50
Solve the equation. Write complex solutions in standard
form.
22) 4x2 + 5x + 5 = 0
Graph the quadratic function by its properties.
23) f(x) = 1
3
x2 - 2x + 3
x
y
x
y
Solve the equation. Find all real solutions.
24) 2(x - 1)2 + 11(x - 1) + 12 = 0
Solve the problem.
25) The length of a table is 12 inches more than its
width. If the area of the table is 2668 square
inches, what is its length?
4
Solve the equation..
The document provides examples and explanations for solving different types of equations, including:
1) Polynomial equations through factoring or the quadratic formula.
2) Rational equations by clearing denominators.
3) Radical equations by squaring both sides to remove radicals.
4) Absolute value equations by recognizing that |x-c|=r implies x=c±r.
The document also discusses solving power equations, finding zeros and domains of functions, and using properties of absolute values.
This document contains shortlisted problems for the Junior Balkan Mathematics Olympiad in 2016 in Romania. It lists the contributing countries that proposed problems, the problem selection committee members, and 4 algebra problems, 1 combinatorics problem, and 1 geometry problem. The combinatorics problem asks the student to find the least positive integer k such that k! multiplied by the sum of reciprocals of digits in numbers up to 2016 is an integer. It provides the solution by calculating this sum incrementally up to 999, then 1999, then 2016.
This document provides an overview of solving quadratic equations by factoring. It discusses identifying quadratic equations, rewriting them in standard form, factoring trinomials in the form x^2 + bx + c, and determining roots. Several examples of factoring trinomials and solving quadratic equations are shown. Activities include identifying quadratic equations, rewriting equations in standard form, factoring trinomials, and solving equations by factoring. The document provides resources for further learning about quadratic equations and factoring.
QUADRATIC EQUATIONS WITH MATHS PROPER VERIFYssuser2e348b
1) The document discusses theorems and proofs related to quadratic equations. It provides a necessary and sufficient condition for two quadratic equations to have a common root.
2) Several examples of solving equations that can be reduced to quadratic equations are presented. Substitutions are made to transform the equations into standard quadratic forms that can then be solved.
3) The last problem finds an expression for the sum of the reciprocals of the terms containing the roots of a quadratic equation.
This module introduces exponential functions and covers:
- Finding the roots of exponential equations using the property of equality for exponential equations.
- Simplifying expressions using laws of exponents.
- Determining the zeros of exponential functions by setting the function equal to 0 and solving for x.
The document provides examples and practice problems for students to learn skills in solving exponential equations and finding zeros of exponential functions.
Name _________________________ Score ______ ______1..docxlea6nklmattu
The document contains a series of math word problems and questions. It asks the reader to:
1) Solve various math equations and systems of equations, showing the work.
2) Write mathematical statements and prove they are true for different values of n.
3) Graph functions and find limits.
4) Solve optimization problems to maximize profits or minimize costs given certain constraints.
Introduction to Operations Research/ Management Science um1222
Here are the steps to solve this problem:
Let x = number of inches of orange beads
Let y = number of inches of black beads
Constraints:
x >= 0
y >= 0
x + y <= 24 (total length must be <= 24 inches)
y >= 2x (black beads must be >= 2x the length of orange beads)
y >= 5 (black beads must be >= 5 inches)
Objective: Maximize x + y (total length of necklace)
To sketch the problem:
Plot the lines y = 2x, x + y = 24, y = 5 on a xy-plane.
The shaded region satisfying all constraints is the feasible
This document discusses solving quadratic equations by factoring. It provides examples of quadratic equations in standard form and using the zero factor property to solve for the roots. The steps for solving a quadratic equation by factoring are outlined. Additional examples are provided solving quadratic equations that arise in applied problems using factoring techniques.
This document discusses methods for solving quadratic equations by factoring. It begins with a yes-no quiz on quadratic topics. Then, it explains the zero product property and outlines five steps for solving a quadratic equation by factoring: 1) write it in standard form, 2) factor the quadratic expression, 3) set each factor equal to 0 using the zero product property, 4) solve the resulting equations, and 5) check the solutions. Two examples demonstrating this process are shown. Finally, directions are given to solve five additional quadratic equations by factoring.
The document introduces linear programming through an example of a candy manufacturer that must determine how many cases of two candy products to make to maximize profit given constraints on available ingredients. It defines variables and constraints, graphs the feasible region, determines the optimal solution that maximizes profit, and discusses how linear programming can be applied to problems with multiple variables and constraints.
Trigonometric substitutions can help evaluate integrals involving trigonometric functions. The document provides examples of using trigonometric substitutions to evaluate integrals of powers of sine and cosine functions. Specifically:
1) Integrals of powers of sine and cosine can be evaluated by rewriting the functions using trigonometric identities and then substituting variables. For example, sin5x can be rewritten as sinx(sin2x)2 and evaluated using the substitution u=cosx.
2) More complex integrals may require multiple steps and identities. For example, evaluating sin6x requires rewriting sin2x using an identity and then integrating four resulting terms.
3) Trigonometric substitutions allow rewriting integrals involving
rational equation transformable to quadratic equation.pptxRizaCatli2
1. The document provides examples for solving quadratic equations that are not in standard form by transforming them into standard form ax2 + bx + c = 0 and then using methods like factoring or the quadratic formula.
2. It also gives examples for solving rational algebraic equations by multiplying both sides by the least common denominator to obtain a quadratic equation, transforming it into standard form, and then solving.
3. The examples cover topics like solving for the solution set, checking solutions, and using the quadratic formula to solve transformed equations.
This document provides information on mathematical concepts and formulas relevant to economics, including:
- Exponential functions such as y=ex and their graphs showing exponential growth and decay
- Quadratic functions of the form y=ax2+bx+c and total cost functions
- Differentiation rules for common functions like exponentials, logarithms, and the product, quotient and chain rules
- Integration basics and formulas for integrating common functions
- Concepts like inverse functions, the mean, variance and standard deviation in statistics
- Information is also provided on fractions, ratios, percentages, and algebraic rules involving exponents, logarithms and sigma notation.
The document introduces matrices and matrix operations. Matrices are rectangular tables of numbers that are used for applications beyond solving systems of equations. Matrix notation defines a matrix with R rows and C columns as an R x C matrix. The entry in the ith row and jth column is denoted as aij. Matrices can be added or subtracted if they are the same size by adding or subtracting the corresponding entries. There are two types of matrix multiplication: scalar multiplication multiplies a matrix by a constant, and matrix multiplication involves multiplying corresponding rows and columns where the number of columns of the left matrix equals the rows of the right matrix.
35 Special Cases System of Linear Equations-x.pptxmath260
The document discusses special cases of systems of linear equations, including inconsistent/contradictory systems where the equations are impossible to satisfy simultaneously, and dependent systems where there are infinitely many solutions. An inconsistent system is shown with equations x + y = 2 and x + y = 3, which has no solution since they cannot both be true. A dependent system is shown with equations x + y = 2 and 2x + 2y = 4, which has infinitely many solutions like (2,0) and (1,1). The row-reduced echelon form (rref) of a matrix is also discussed, which puts a system of equations in a standard form to help determine if it is consistent, dependent, or has
The document discusses conic sections and ellipses. Conic sections are graphs of quadratic equations of the form Ax2 + By2 + Cx + Dy = E, where A and B are not both 0. Their graphs include circles, ellipses, parabolas and hyperbolas. Ellipses are defined as the set of all points where the sum of the distances to two fixed foci is a constant. Ellipses have a center, two axes called the semi-major and semi-minor axes, and radii along the x and y axes called the x-radius and y-radius. The standard form of an ellipse equation is presented.
The document discusses notation and algebra of functions. It defines a function as a procedure that assigns a unique output to each valid input. Most mathematical functions are represented by formulas like f(x) = x^2 - 2x + 3, where f(x) is the name of the function, x is the input variable, and the formula defines the relationship between input and output. New functions can be formed using basic operations like addition, subtraction, multiplication, and division of existing functions. Examples are provided to demonstrate evaluating functions at given inputs and combining functions algebraically.
The document discusses exponents and exponent rules. It defines exponents as the number of times a base is multiplied by 1. It presents rules for multiplying, dividing, and raising exponents. Examples are provided to demonstrate applying the rules, such as using the power-multiply rule to evaluate (22*34)3. Special exponent rules are also covered, such as the 0-power rule where A0 equals 1 when A is not 0. The document provides examples of calculating fractional exponents by first extracting the root and then raising it to the numerator power.
The document discusses functions and their basic language. It defines a function as a procedure that assigns each input exactly one output. It provides examples of functions, such as a license number to name function. It explains that a function must have a domain (set of inputs) and range (set of outputs). Functions can be represented graphically, through tables of inputs and outputs, or with mathematical formulas.
19 more parabolas a& hyperbolas (optional) xmath260
After dividing the general quadratic equation Ax2 + By2 + Cx + Dy = E by A, three types of conic sections can be obtained:
1) Parabolas occur when B = 0, resulting in equations of the form 1x2 + #x + #y = #.
2) Circles occur when A = B, resulting in the equation 1x2 + 1y2 = 1.
3) Hyperbolas occur when A and B have opposite signs, resulting in equations of the form 1x2 + ry2 + #x + #y = # with r < 0. Hyperbolas have two foci and asymptotes, and points on the hyperbola have
The document discusses conic sections, specifically circles and ellipses. It defines an ellipse as the set of points where the sum of the distances to two fixed foci is a constant. An ellipse has a center, two axes (semi-major and semi-minor), and can be represented by the standard form (x-h)2/a2 + (y-k)2/b2 = 1, where (h,k) is the center, a is the x-radius, and b is the y-radius. Examples are provided to demonstrate finding attributes of ellipses from their equations.
This document discusses conic sections and first degree equations. It begins by introducing conic sections as the shapes formed by slicing a cone at different angles. It then covers first degree equations, noting that their graphs are straight lines that can be written in the form of y=mx+b. Specific examples of first degree equations and their graphs are shown. The document ends by introducing the four types of conic sections - circles, ellipses, parabolas, and hyperbolas - and how graphs of second degree equations can represent these shapes.
The document discusses calculating the slope of a curve between two points (x, f(x)) and (x+h, f(x+h)) using the difference quotient formula. It defines the difference quotient as (f(x+h) - f(x))/h, where h is the difference between x and x+h. An example calculates the slope between the points (2, f(2)) and (2.2, f(2.2)) for the function f(x) = x^2 - 2x + 2, finding the slope to be 0.44.
The document discusses various transformations that can be performed on graphs of functions, including vertical translations, stretches, and compressions. Vertical translations move the entire graph up or down by adding or subtracting a constant to the function. Stretches elongate or compress the graph vertically by multiplying the function by a constant greater than or less than 1, respectively. These transformations can be represented by modifying the original function in a way that corresponds to the geometric transformation of its graph.
14 graphs of factorable rational functions xmath260
The document discusses graphs of rational functions. It defines rational functions as functions of the form R(x) = P(x)/Q(x) where P(x) and Q(x) are polynomials. It describes how vertical asymptotes occur where the denominator Q(x) is zero. The graph runs along either side of vertical asymptotes, going up or down depending on the sign chart. There are four cases for how the graph behaves at a vertical asymptote. The document uses examples to illustrate graphing rational functions and determining vertical asymptotes. It also mentions horizontal asymptotes will be discussed.
The document discusses factorable polynomials and graphing them. It defines a factorable polynomial P(x) as one that can be written as the product of linear factors P(x) = an(x - r1)(x - r2)...(x - rk), where r1, r2, etc. are the roots of P(x). It explains that for large values of |x|, the leading term of P(x) dominates so the graph resembles that of the leading term, while near the roots other terms contribute to the shape of the graph. Examples of graphs of polynomials like x^n are provided to illustrate the approach.
The document discusses quadratic functions and parabolas. It defines quadratic functions as functions of the form y = ax2 + bx + c, where a ≠ 0. It states that the graphs of quadratic equations are called parabolas. Parabolas are symmetric around a central line, with the vertex (highest/lowest point) located on this line. The vertex formula is given as x = -b/2a. Steps for graphing a parabola are outlined, including finding the vertex, another point, and reflections across the central line. An example graphs the parabola y = x2 - 4x - 12, finding the vertex as (2, -16) and x-intercepts as -
The document describes the rectangular coordinate system. Each point in a plane can be located using an ordered pair (x,y) where x represents the distance right or left from the origin and y represents the distance up or down. Changing the x-value moves the point right or left, and changing the y-value moves the point up or down. The plane is divided into four quadrants based on the sign of the x and y values. Reflecting a point across an axis results in another point with the same magnitude but opposite sign for the corresponding coordinate.
The document discusses first degree (linear) functions. It states that most real-world mathematical functions can be composed of formulas from three families: algebraic, trigonometric, and exponential-logarithmic. It focuses on linear functions of the form f(x)=mx+b, where m is the slope and b is the y-intercept. Examples are given of equations and how to determine the slope and y-intercept to write the equation in slope-intercept form as a linear function.
The document discusses the basic language of functions. It defines a function as a procedure that assigns each input exactly one output. Functions can be represented by formulas using typical variables like f(x) = x^2 - 2x + 3, where x is the input and f(x) is the output. Functions have a domain, which is the set of all possible inputs, and a range, which is the set of all possible outputs. Functions can be depicted graphically or via tables listing inputs and outputs.
The document discusses inverse functions. An inverse function reverses the input and output of a function. For a function f(x) to have an inverse function f^-1(y), it must be one-to-one, meaning that different inputs map to different outputs. The inverse of f(x) is obtained by solving the original function equation for x in terms of y. Examples show how to determine if a function has an inverse and how to calculate the inverse function. For non one-to-one functions like f(x)=x^2, the inverse procedure is not a well-defined function.
This document discusses two types of log and exponential equations: those that do not require calculators and numerical equations that do require calculators. Equations that do not require calculators can be solved by putting both sides into a common base, consolidating exponents, and dropping the base to solve the resulting equation. For log equations, logs are consolidated on each side first before dropping the log. Two examples demonstrating these solution methods are provided.
The document discusses exponential and logarithmic expressions. Exponential expressions like 43, 82, 26 all equal 64. Their corresponding logarithmic forms are log4(64), log8(64), log2(64) and equal 3, 2, 6 respectively. When working with exponential or logarithmic expressions, the base number must be identified first. Both numbers in the logarithmic expression logb(y) must be positive.
Salesforce Integration for Bonterra Impact Management (fka Social Solutions A...Jeffrey Haguewood
Sidekick Solutions uses Bonterra Impact Management (fka Social Solutions Apricot) and automation solutions to integrate data for business workflows.
We believe integration and automation are essential to user experience and the promise of efficient work through technology. Automation is the critical ingredient to realizing that full vision. We develop integration products and services for Bonterra Case Management software to support the deployment of automations for a variety of use cases.
This video focuses on integration of Salesforce with Bonterra Impact Management.
Interested in deploying an integration with Salesforce for Bonterra Impact Management? Contact us at sales@sidekicksolutionsllc.com to discuss next steps.
Unlock the Future of Search with MongoDB Atlas_ Vector Search Unleashed.pdfMalak Abu Hammad
Discover how MongoDB Atlas and vector search technology can revolutionize your application's search capabilities. This comprehensive presentation covers:
* What is Vector Search?
* Importance and benefits of vector search
* Practical use cases across various industries
* Step-by-step implementation guide
* Live demos with code snippets
* Enhancing LLM capabilities with vector search
* Best practices and optimization strategies
Perfect for developers, AI enthusiasts, and tech leaders. Learn how to leverage MongoDB Atlas to deliver highly relevant, context-aware search results, transforming your data retrieval process. Stay ahead in tech innovation and maximize the potential of your applications.
#MongoDB #VectorSearch #AI #SemanticSearch #TechInnovation #DataScience #LLM #MachineLearning #SearchTechnology
Main news related to the CCS TSI 2023 (2023/1695)Jakub Marek
An English 🇬🇧 translation of a presentation to the speech I gave about the main changes brought by CCS TSI 2023 at the biggest Czech conference on Communications and signalling systems on Railways, which was held in Clarion Hotel Olomouc from 7th to 9th November 2023 (konferenceszt.cz). Attended by around 500 participants and 200 on-line followers.
The original Czech 🇨🇿 version of the presentation can be found here: https://www.slideshare.net/slideshow/hlavni-novinky-souvisejici-s-ccs-tsi-2023-2023-1695/269688092 .
The videorecording (in Czech) from the presentation is available here: https://youtu.be/WzjJWm4IyPk?si=SImb06tuXGb30BEH .
Threats to mobile devices are more prevalent and increasing in scope and complexity. Users of mobile devices desire to take full advantage of the features
available on those devices, but many of the features provide convenience and capability but sacrifice security. This best practices guide outlines steps the users can take to better protect personal devices and information.
TrustArc Webinar - 2024 Global Privacy SurveyTrustArc
How does your privacy program stack up against your peers? What challenges are privacy teams tackling and prioritizing in 2024?
In the fifth annual Global Privacy Benchmarks Survey, we asked over 1,800 global privacy professionals and business executives to share their perspectives on the current state of privacy inside and outside of their organizations. This year’s report focused on emerging areas of importance for privacy and compliance professionals, including considerations and implications of Artificial Intelligence (AI) technologies, building brand trust, and different approaches for achieving higher privacy competence scores.
See how organizational priorities and strategic approaches to data security and privacy are evolving around the globe.
This webinar will review:
- The top 10 privacy insights from the fifth annual Global Privacy Benchmarks Survey
- The top challenges for privacy leaders, practitioners, and organizations in 2024
- Key themes to consider in developing and maintaining your privacy program
Introduction of Cybersecurity with OSS at Code Europe 2024Hiroshi SHIBATA
I develop the Ruby programming language, RubyGems, and Bundler, which are package managers for Ruby. Today, I will introduce how to enhance the security of your application using open-source software (OSS) examples from Ruby and RubyGems.
The first topic is CVE (Common Vulnerabilities and Exposures). I have published CVEs many times. But what exactly is a CVE? I'll provide a basic understanding of CVEs and explain how to detect and handle vulnerabilities in OSS.
Next, let's discuss package managers. Package managers play a critical role in the OSS ecosystem. I'll explain how to manage library dependencies in your application.
I'll share insights into how the Ruby and RubyGems core team works to keep our ecosystem safe. By the end of this talk, you'll have a better understanding of how to safeguard your code.
Programming Foundation Models with DSPy - Meetup SlidesZilliz
Prompting language models is hard, while programming language models is easy. In this talk, I will discuss the state-of-the-art framework DSPy for programming foundation models with its powerful optimizers and runtime constraint system.
GraphRAG for Life Science to increase LLM accuracyTomaz Bratanic
GraphRAG for life science domain, where you retriever information from biomedical knowledge graphs using LLMs to increase the accuracy and performance of generated answers
How to Interpret Trends in the Kalyan Rajdhani Mix Chart.pdfChart Kalyan
A Mix Chart displays historical data of numbers in a graphical or tabular form. The Kalyan Rajdhani Mix Chart specifically shows the results of a sequence of numbers over different periods.
5th LF Energy Power Grid Model Meet-up SlidesDanBrown980551
5th Power Grid Model Meet-up
It is with great pleasure that we extend to you an invitation to the 5th Power Grid Model Meet-up, scheduled for 6th June 2024. This event will adopt a hybrid format, allowing participants to join us either through an online Mircosoft Teams session or in person at TU/e located at Den Dolech 2, Eindhoven, Netherlands. The meet-up will be hosted by Eindhoven University of Technology (TU/e), a research university specializing in engineering science & technology.
Power Grid Model
The global energy transition is placing new and unprecedented demands on Distribution System Operators (DSOs). Alongside upgrades to grid capacity, processes such as digitization, capacity optimization, and congestion management are becoming vital for delivering reliable services.
Power Grid Model is an open source project from Linux Foundation Energy and provides a calculation engine that is increasingly essential for DSOs. It offers a standards-based foundation enabling real-time power systems analysis, simulations of electrical power grids, and sophisticated what-if analysis. In addition, it enables in-depth studies and analysis of the electrical power grid’s behavior and performance. This comprehensive model incorporates essential factors such as power generation capacity, electrical losses, voltage levels, power flows, and system stability.
Power Grid Model is currently being applied in a wide variety of use cases, including grid planning, expansion, reliability, and congestion studies. It can also help in analyzing the impact of renewable energy integration, assessing the effects of disturbances or faults, and developing strategies for grid control and optimization.
What to expect
For the upcoming meetup we are organizing, we have an exciting lineup of activities planned:
-Insightful presentations covering two practical applications of the Power Grid Model.
-An update on the latest advancements in Power Grid -Model technology during the first and second quarters of 2024.
-An interactive brainstorming session to discuss and propose new feature requests.
-An opportunity to connect with fellow Power Grid Model enthusiasts and users.
UiPath Test Automation using UiPath Test Suite series, part 6DianaGray10
Welcome to UiPath Test Automation using UiPath Test Suite series part 6. In this session, we will cover Test Automation with generative AI and Open AI.
UiPath Test Automation with generative AI and Open AI webinar offers an in-depth exploration of leveraging cutting-edge technologies for test automation within the UiPath platform. Attendees will delve into the integration of generative AI, a test automation solution, with Open AI advanced natural language processing capabilities.
Throughout the session, participants will discover how this synergy empowers testers to automate repetitive tasks, enhance testing accuracy, and expedite the software testing life cycle. Topics covered include the seamless integration process, practical use cases, and the benefits of harnessing AI-driven automation for UiPath testing initiatives. By attending this webinar, testers, and automation professionals can gain valuable insights into harnessing the power of AI to optimize their test automation workflows within the UiPath ecosystem, ultimately driving efficiency and quality in software development processes.
What will you get from this session?
1. Insights into integrating generative AI.
2. Understanding how this integration enhances test automation within the UiPath platform
3. Practical demonstrations
4. Exploration of real-world use cases illustrating the benefits of AI-driven test automation for UiPath
Topics covered:
What is generative AI
Test Automation with generative AI and Open AI.
UiPath integration with generative AI
Speaker:
Deepak Rai, Automation Practice Lead, Boundaryless Group and UiPath MVP
Monitoring and Managing Anomaly Detection on OpenShift.pdfTosin Akinosho
Monitoring and Managing Anomaly Detection on OpenShift
Overview
Dive into the world of anomaly detection on edge devices with our comprehensive hands-on tutorial. This SlideShare presentation will guide you through the entire process, from data collection and model training to edge deployment and real-time monitoring. Perfect for those looking to implement robust anomaly detection systems on resource-constrained IoT/edge devices.
Key Topics Covered
1. Introduction to Anomaly Detection
- Understand the fundamentals of anomaly detection and its importance in identifying unusual behavior or failures in systems.
2. Understanding Edge (IoT)
- Learn about edge computing and IoT, and how they enable real-time data processing and decision-making at the source.
3. What is ArgoCD?
- Discover ArgoCD, a declarative, GitOps continuous delivery tool for Kubernetes, and its role in deploying applications on edge devices.
4. Deployment Using ArgoCD for Edge Devices
- Step-by-step guide on deploying anomaly detection models on edge devices using ArgoCD.
5. Introduction to Apache Kafka and S3
- Explore Apache Kafka for real-time data streaming and Amazon S3 for scalable storage solutions.
6. Viewing Kafka Messages in the Data Lake
- Learn how to view and analyze Kafka messages stored in a data lake for better insights.
7. What is Prometheus?
- Get to know Prometheus, an open-source monitoring and alerting toolkit, and its application in monitoring edge devices.
8. Monitoring Application Metrics with Prometheus
- Detailed instructions on setting up Prometheus to monitor the performance and health of your anomaly detection system.
9. What is Camel K?
- Introduction to Camel K, a lightweight integration framework built on Apache Camel, designed for Kubernetes.
10. Configuring Camel K Integrations for Data Pipelines
- Learn how to configure Camel K for seamless data pipeline integrations in your anomaly detection workflow.
11. What is a Jupyter Notebook?
- Overview of Jupyter Notebooks, an open-source web application for creating and sharing documents with live code, equations, visualizations, and narrative text.
12. Jupyter Notebooks with Code Examples
- Hands-on examples and code snippets in Jupyter Notebooks to help you implement and test anomaly detection models.
2. Recall the prices for the pizzas from Pizza Grande.
Each pizza is $8 and there is a $10 delivery charge.
Expressions
3. Recall the prices for the pizzas from Pizza Grande.
Each pizza is $8 and there is a $10 delivery charge.
Hence if we want x pizzas delivered, then the total cost
is given by the expression “8x + 10”.
Expressions
4. Recall the prices for the pizzas from Pizza Grande.
Each pizza is $8 and there is a $10 delivery charge.
Hence if we want x pizzas delivered, then the total cost
is given by the expression “8x + 10”. If we order
x = 100 pizzas, then the cost is 8(100) + 10 = $810.
Expressions
5. Recall the prices for the pizzas from Pizza Grande.
Each pizza is $8 and there is a $10 delivery charge.
Hence if we want x pizzas delivered, then the total cost
is given by the expression “8x + 10”. If we order
x = 100 pizzas, then the cost is 8(100) + 10 = $810.
Expressions calculate future results.
Expressions
6. Recall the prices for the pizzas from Pizza Grande.
Each pizza is $8 and there is a $10 delivery charge.
Hence if we want x pizzas delivered, then the total cost
is given by the expression “8x + 10”. If we order
x = 100 pizzas, then the cost is 8(100) + 10 = $810.
Expressions calculate future results.
Now if we know the cost is $810 but forgot how many
pizzas were ordered, we may backtrack to find x:
Expressions
7. Recall the prices for the pizzas from Pizza Grande.
Each pizza is $8 and there is a $10 delivery charge.
Hence if we want x pizzas delivered, then the total cost
is given by the expression “8x + 10”. If we order
x = 100 pizzas, then the cost is 8(100) + 10 = $810.
Expressions calculate future results.
Now if we know the cost is $810 but forgot how many
pizzas were ordered, we may backtrack to find x:
8x + 10 = 810
Expressions
8. Recall the prices for the pizzas from Pizza Grande.
Each pizza is $8 and there is a $10 delivery charge.
Hence if we want x pizzas delivered, then the total cost
is given by the expression “8x + 10”. If we order
x = 100 pizzas, then the cost is 8(100) + 10 = $810.
Expressions calculate future results.
Now if we know the cost is $810 but forgot how many
pizzas were ordered, we may backtrack to find x:
8x + 10 = 810
8x = 800
–10 –10 subtract the delivery cost,
so the pizzas cost $800,
Expressions
9. Recall the prices for the pizzas from Pizza Grande.
Each pizza is $8 and there is a $10 delivery charge.
Hence if we want x pizzas delivered, then the total cost
is given by the expression “8x + 10”. If we order
x = 100 pizzas, then the cost is 8(100) + 10 = $810.
Expressions calculate future results.
Now if we know the cost is $810 but forgot how many
pizzas were ordered, we may backtrack to find x:
8x + 10 = 810
8x = 800
–10 –10 subtract the delivery cost,
so the pizzas cost $800,
divide this by $8/per pizza,
x = 100 (pizzas)
8 8
Expressions
10. Recall the prices for the pizzas from Pizza Grande.
Each pizza is $8 and there is a $10 delivery charge.
Hence if we want x pizzas delivered, then the total cost
is given by the expression “8x + 10”. If we order
x = 100 pizzas, then the cost is 8(100) + 10 = $810.
Expressions
Expressions calculate future results.
Now if we know the cost is $810 but forgot how many
pizzas were ordered, we may backtrack to find x:
8x + 10 = 810
8x = 800
–10 –10 subtract the delivery cost,
so the pizzas cost $800,
divide this by $8/per pizza,
x = 100
8 8
“8x + 10 = 810” is called an equation.
(pizzas)
11. Recall the prices for the pizzas from Pizza Grande.
Each pizza is $8 and there is a $10 delivery charge.
Hence if we want x pizzas delivered, then the total cost
is given by the expression “8x + 10”. If we order
x = 100 pizzas, then the cost is 8(100) + 10 = $810.
Expressions
Expressions calculate future results.
Now if we know the cost is $810 but forgot how many
pizzas were ordered, we may backtrack to find x:
8x + 10 = 810
8x = 800
–10 –10 subtract the delivery cost,
so the pizzas cost $800,
divide this by $8/per pizza,
x = 100
8 8
“8x + 10 = 810” is called an equation.
Equations are set up to backtrack to the original
input x or x’s, i.e. we want to solve equations.
(pizzas)
13. An equation is made by setting two expressions
equal to each other:
Solving Equations
expression 1 = expression 2
14. An equation is made by setting two expressions
equal to each other:
Solving Equations
expression 1 = expression 2
To solve an equation means to find value(s)
for the variables that makes the equation true.
15. An equation is made by setting two expressions
equal to each other:
Solving Equations
Example A. 3x2 – 2x = 8
We solve polynomial equations by factoring.
expression 1 = expression 2
To solve an equation means to find value(s)
for the variables that makes the equation true.
16. An equation is made by setting two expressions
equal to each other:
Solving Equations
Example A. 3x2 – 2x = 8
We solve polynomial equations by factoring.
Set one side to 0, 3x2 – 2x – 8 = 0
expression 1 = expression 2
To solve an equation means to find value(s)
for the variables that makes the equation true.
17. An equation is made by setting two expressions
equal to each other:
Solving Equations
Example A. 3x2 – 2x = 8
We solve polynomial equations by factoring.
Set one side to 0, 3x2 – 2x – 8 = 0 factor this
(3x + 4)(x – 2) = 0
expression 1 = expression 2
To solve an equation means to find value(s)
for the variables that makes the equation true.
18. An equation is made by setting two expressions
equal to each other:
Solving Equations
Example A. 3x2 – 2x = 8
We solve polynomial equations by factoring.
Set one side to 0, 3x2 – 2x – 8 = 0 factor this
(3x + 4)(x – 2) = 0 extract answers
so x = –4/3, 2
expression 1 = expression 2
To solve an equation means to find value(s)
for the variables that makes the equation true.
19. An equation is made by setting two expressions
equal to each other:
Solving Equations
Example A. 3x2 – 2x = 8
We solve polynomial equations by factoring.
Set one side to 0, 3x2 – 2x – 8 = 0 factor this
(3x + 4)(x – 2) = 0 extract answers
so x = –4/3, 2
expression 1 = expression 2
To solve an equation means to find value(s)
for the variables that makes the equation true.
To solve other equations such as rational equations,
or equations with radicals, we have to transform them
into polynomial equations.
22. Rational Equations
Solve rational equations by clearing all denominators
using the LCD. Check the answers afterwards.
Example B. Solve x – 2
2 = x + 1
4 + 1
23. Rational Equations
Solve rational equations by clearing all denominators
using the LCD. Check the answers afterwards.
Example B. Solve
LCD = (x – 2)(x + 1), multiply the LCD to both sides
of the equation:
x – 2
2 = x + 1
4 + 1
24. Rational Equations
Solve rational equations by clearing all denominators
using the LCD. Check the answers afterwards.
Example B. Solve
LCD = (x – 2)(x + 1), multiply the LCD to both sides
of the equation:
p
(x – 2)(x + 1) * [ ]x – 2
2 =
x + 1
4
+ 1
x – 2
2 = x + 1
4 + 1
25. Rational Equations
Solve rational equations by clearing all denominators
using the LCD. Check the answers afterwards.
Example B. Solve
LCD = (x – 2)(x + 1), multiply the LCD to both sides
of the equation:
(x – 2)(x + 1) * [ ]x – 2
2 =
x + 1
4
+ 1
(x + 1)
x – 2
2 = x + 1
4 + 1
26. Rational Equations
Solve rational equations by clearing all denominators
using the LCD. Check the answers afterwards.
Example B. Solve
LCD = (x – 2)(x + 1), multiply the LCD to both sides
of the equation:
(x – 2)(x + 1) * [ ]x – 2
2 =
x + 1
4
+ 1
(x + 1) (x – 2)
x – 2
2 = x + 1
4 + 1
27. Rational Equations
Solve rational equations by clearing all denominators
using the LCD. Check the answers afterwards.
Example B. Solve
LCD = (x – 2)(x + 1), multiply the LCD to both sides
of the equation:
(x – 2)(x + 1) * [ ]x – 2
2 =
x + 1
4
+ 1
(x + 1) (x – 2) (x + 1)(x – 2)
x – 2
2 = x + 1
4 + 1
28. Rational Equations
Solve rational equations by clearing all denominators
using the LCD. Check the answers afterwards.
Example B. Solve
LCD = (x – 2)(x + 1), multiply the LCD to both sides
of the equation:
(x – 2)(x + 1) * [ ]
2(x + 1) = 4(x – 2) + 1*(x + 1)(x – 2)
x – 2
2 =
x + 1
4
+ 1
(x + 1) (x – 2) (x + 1)(x – 2)
x – 2
2 = x + 1
4 + 1
29. Rational Equations
Solve rational equations by clearing all denominators
using the LCD. Check the answers afterwards.
Example B. Solve
LCD = (x – 2)(x + 1), multiply the LCD to both sides
of the equation:
(x – 2)(x + 1) * [ ]
2(x + 1) = 4(x – 2) + 1*(x + 1)(x – 2)
2x + 2 = 4x – 8 + x2 – x – 2
x – 2
2 =
x + 1
4
+ 1
(x + 1) (x – 2) (x + 1)(x – 2)
x – 2
2 = x + 1
4 + 1
30. Rational Equations
Solve rational equations by clearing all denominators
using the LCD. Check the answers afterwards.
Example B. Solve
LCD = (x – 2)(x + 1), multiply the LCD to both sides
of the equation:
(x – 2)(x + 1) * [ ]
2(x + 1) = 4(x – 2) + 1*(x + 1)(x – 2)
2x + 2 = 4x – 8 + x2 – x – 2
2x + 2 = x2 + 3x – 10
x – 2
2 =
x + 1
4
+ 1
(x + 1) (x – 2) (x + 1)(x – 2)
x – 2
2 = x + 1
4 + 1
31. Rational Equations
Solve rational equations by clearing all denominators
using the LCD. Check the answers afterwards.
Example B. Solve
LCD = (x – 2)(x + 1), multiply the LCD to both sides
of the equation:
(x – 2)(x + 1) * [ ]
2(x + 1) = 4(x – 2) + 1*(x + 1)(x – 2)
2x + 2 = 4x – 8 + x2 – x – 2
2x + 2 = x2 + 3x – 10
0 = x2 + x – 12
x – 2
2 =
x + 1
4
+ 1
(x + 1) (x – 2) (x + 1)(x – 2)
x – 2
2 = x + 1
4 + 1
32. Rational Equations
Solve rational equations by clearing all denominators
using the LCD. Check the answers afterwards.
Example B. Solve
LCD = (x – 2)(x + 1), multiply the LCD to both sides
of the equation:
(x – 2)(x + 1) * [ ]
2(x + 1) = 4(x – 2) + 1*(x + 1)(x – 2)
2x + 2 = 4x – 8 + x2 – x – 2
2x + 2 = x2 + 3x – 10
0 = x2 + x – 12
0 = (x + 4)(x – 3) x = -4, 3
x – 2
2 =
x + 1
4
+ 1
(x + 1) (x – 2) (x + 1)(x – 2)
x – 2
2 = x + 1
4 + 1
33. Rational Equations
Solve rational equations by clearing all denominators
using the LCD. Check the answers afterwards.
Example B. Solve
LCD = (x – 2)(x + 1), multiply the LCD to both sides
of the equation:
(x – 2)(x + 1) * [ ]
2(x + 1) = 4(x – 2) + 1*(x + 1)(x – 2)
2x + 2 = 4x – 8 + x2 – x – 2
2x + 2 = x2 + 3x – 10
0 = x2 + x – 12
0 = (x + 4)(x – 3) x = -4, 3
Both are good.
x – 2
2 =
x + 1
4
+ 1
(x + 1) (x – 2) (x + 1)(x – 2)
x – 2
2 = x + 1
4 + 1
34. Factoring By Grouping
Some polynomials may be factored by pulling out
common factors twice. We call this factor by
grouping.
35. Example C. Solve 2x3 – 3x2 – 8x + 12 = 0
Some polynomials may be factored by pulling out
common factors twice. We call this factor by
grouping.
Factoring By Grouping
36. Example C. Solve 2x3 – 3x2 – 8x + 12 = 0
Some polynomials may be factored by pulling out
common factors twice. We call this factor by
grouping.
2x3 – 3x2 – 8x + 12 = 0
Factoring By Grouping
37. Example C. Solve 2x3 – 3x2 – 8x + 12 = 0
Some polynomials may be factored by pulling out
common factors twice. We call this factor by
grouping.
2x3 – 3x2 – 8x + 12 = 0
x2(2x – 3) – 4(2x – 3) = 0
Factoring By Grouping
38. Example C. Solve 2x3 – 3x2 – 8x + 12 = 0
Some polynomials may be factored by pulling out
common factors twice. We call this factor by
grouping.
2x3 – 3x2 – 8x + 12 = 0
x2(2x – 3) – 4(2x – 3) = 0
(2x – 3)(x2 – 4) = 0
Factoring By Grouping
39. Example C. Solve 2x3 – 3x2 – 8x + 12 = 0
Some polynomials may be factored by pulling out
common factors twice. We call this factor by
grouping.
2x3 – 3x2 – 8x + 12 = 0
x2(2x – 3) – 4(2x – 3) = 0
(2x – 3)(x2 – 4) = 0
(2x – 3)(x – 2)(x + 2) = 0
Factoring By Grouping
40. Example C. Solve 2x3 – 3x2 – 8x + 12 = 0
Some polynomials may be factored by pulling out
common factors twice. We call this factor by
grouping.
So x = 3/2, 2, –2
2x3 – 3x2 – 8x + 12 = 0
x2(2x – 3) – 4(2x – 3) = 0
(2x – 3)(x2 – 4) = 0
(2x – 3)(x – 2)(x + 2) = 0
Factoring By Grouping
41. Example C. Solve 2x3 – 3x2 – 8x + 12 = 0
Some polynomials may be factored by pulling out
common factors twice. We call this factor by
grouping.
So x = 3/2, 2, –2
2x3 – 3x2 – 8x + 12 = 0
x2(2x – 3) – 4(2x – 3) = 0
(2x – 3)(x2 – 4) = 0
(2x – 3)(x – 2)(x + 2) = 0
Except for some special cases, polynomial equations
with degree 3 or more are solved with computers.
Factoring By Grouping
42. Example C. Solve 2x3 – 3x2 – 8x + 12 = 0
Some polynomials may be factored by pulling out
common factors twice. We call this factor by
grouping.
So x = 3/2, 2, –2
2x3 – 3x2 – 8x + 12 = 0
x2(2x – 3) – 4(2x – 3) = 0
(2x – 3)(x2 – 4) = 0
(2x – 3)(x – 2)(x + 2) = 0
We may also use the quadratic formula to solve all
2nd degree polynomial equations.
Except for some special cases, polynomial equations
with degree 3 or more are solved with computers.
Factoring By Grouping
44. Quadratic Formula and Discriminant
Quadratic Formula (QF):
The roots for the equation ax2 + bx + c = 0 are
x =
–b ± b2 – 4ac
2a
A “root” is a solution for
the equation “# = 0”.
45. Quadratic Formula and Discriminant
Quadratic Formula (QF):
The roots for the equation ax2 + bx + c = 0 are
b2 – 4ac is the discriminant because its value
indicates the type of roots we have.
x =
–b ± b2 – 4ac
2a
A “root” is a solution for
the equation “# = 0”.
46. Quadratic Formula and Discriminant
Quadratic Formula (QF):
The roots for the equation ax2 + bx + c = 0 are
b2 – 4ac is the discriminant because its value
indicates the type of roots we have.
I. If b2 – 4ac > 0, we have real roots,
x =
–b ± b2 – 4ac
2a
A “root” is a solution for
the equation “# = 0”.
47. Quadratic Formula and Discriminant
Quadratic Formula (QF):
The roots for the equation ax2 + bx + c = 0 are
b2 – 4ac is the discriminant because its value
indicates the type of roots we have.
I. If b2 – 4ac > 0, we have real roots, furthermore if
b2 – 4ac is a perfect square, the roots are rational.
x =
–b ± b2 – 4ac
2a
A “root” is a solution for
the equation “# = 0”.
48. Quadratic Formula and Discriminant
Quadratic Formula (QF):
The roots for the equation ax2 + bx + c = 0 are
b2 – 4ac is the discriminant because its value
indicates the type of roots we have.
I. If b2 – 4ac > 0, we have real roots, furthermore if
b2 – 4ac is a perfect square, the roots are rational.
II. If b2 – 4ac < 0, the roots are not real.
x =
–b ± b2 – 4ac
2a
A “root” is a solution for
the equation “# = 0”.
49. Quadratic Formula and Discriminant
Quadratic Formula (QF):
The roots for the equation ax2 + bx + c = 0 are
b2 – 4ac is the discriminant because its value
indicates the type of roots we have.
I. If b2 – 4ac > 0, we have real roots, furthermore if
b2 – 4ac is a perfect square, the roots are rational.
II. If b2 – 4ac < 0, the roots are not real.
x =
–b ± b2 – 4ac
2a
Example D. Find the values of k where the
solutions are real for x2 + 2x + (2 – 3k) = 0
A “root” is a solution for
the equation “# = 0”.
50. Quadratic Formula and Discriminant
Quadratic Formula (QF):
The roots for the equation ax2 + bx + c = 0 are
b2 – 4ac is the discriminant because its value
indicates the type of roots we have.
I. If b2 – 4ac > 0, we have real roots, furthermore if
b2 – 4ac is a perfect square, the roots are rational.
II. If b2 – 4ac < 0, the roots are not real.
x =
–b ± b2 – 4ac
2a
Example D. Find the values of k where the
solutions are real for x2 + 2x + (2 – 3k) = 0
We need b2 – 4ac > 0,
A “root” is a solution for
the equation “# = 0”.
51. Quadratic Formula and Discriminant
Quadratic Formula (QF):
The roots for the equation ax2 + bx + c = 0 are
b2 – 4ac is the discriminant because its value
indicates the type of roots we have.
I. If b2 – 4ac > 0, we have real roots, furthermore if
b2 – 4ac is a perfect square, the roots are rational.
II. If b2 – 4ac < 0, the roots are not real.
x =
–b ± b2 – 4ac
2a
Example D. Find the values of k where the
solutions are real for x2 + 2x + (2 – 3k) = 0
We need b2 – 4ac > 0, i.e 4 – 4(2 – 3k) > 0.
A “root” is a solution for
the equation “# = 0”.
52. Quadratic Formula and Discriminant
Quadratic Formula (QF):
The roots for the equation ax2 + bx + c = 0 are
b2 – 4ac is the discriminant because its value
indicates the type of roots we have.
I. If b2 – 4ac > 0, we have real roots, furthermore if
b2 – 4ac is a perfect square, the roots are rational.
II. If b2 – 4ac < 0, the roots are not real.
x =
–b ± b2 – 4ac
2a
Example D. Find the values of k where the
solutions are real for x2 + 2x + (2 – 3k) = 0
We need b2 – 4ac > 0, i.e 4 – 4(2 – 3k) > 0.
– 4 + 12k > 0
or k > 1/3
A “root” is a solution for
the equation “# = 0”.
57. Power Equations
The solution to the equation
x 3 = –8 is
x = √–8 = –2.
3
Using fractional exponents, we write these steps as
if x3 = –8
Equations of the Form xp/q = c
58. Power Equations
The solution to the equation
x 3 = –8 is
x = √–8 = –2.
3
Using fractional exponents, we write these steps as
if x3 = –8 then
The reciprocal of the power 3
Equations of the Form xp/q = c
x = (–8)1/3 = –2
59. Power Equations
The solution to the equation
x 3 = –8 is
x = √–8 = –2.
3
Using fractional exponents, we write these steps as
if x3 = –8 then
The reciprocal of the power 3
Equations of the Form xp/q = c
x = (–8)1/3 = –2
Rational power equations are equations of the type
xR = c where R = p/q is a rational number.
60. Power Equations
The solution to the equation
x 3 = –8 is
x = √–8 = –2.
3
Using fractional exponents, we write these steps as
if x3 = –8 then
To solve a power equation, take the reciprocal power,
The reciprocal of the power 3
Equations of the Form xp/q = c
x = (–8)1/3 = –2
Rational power equations are equations of the type
xR = c where R = p/q is a rational number.
61. Power Equations
The solution to the equation
x 3 = –8 is
x = √–8 = –2.
3
Using fractional exponents, we write these steps as
if x3 = –8 then
To solve a power equation, take the reciprocal power,
so if xR = c,
The reciprocal of the power 3
xp/q = c
Equations of the Form xp/q = c
x = (–8)1/3 = –2
Rational power equations are equations of the type
xR = c where R = p/q is a rational number.
62. Power Equations
The solution to the equation
x 3 = –8 is
x = √–8 = –2.
3
Using fractional exponents, we write these steps as
if x3 = –8 then
To solve a power equation, take the reciprocal power,
so if xR = c,
then x = (±)c1/R
The reciprocal of the power 3
xp/q = c
Equations of the Form xp/q = c
x = (–8)1/3 = –2
Rational power equations are equations of the type
xR = c where R = p/q is a rational number.
Reciprocate the powers
63. Power Equations
The solution to the equation
x 3 = –8 is
x = √–8 = –2.
3
Using fractional exponents, we write these steps as
if x3 = –8 then
To solve a power equation, take the reciprocal power,
so if xR = c,
then x = (±)c1/R
Reciprocate the powers
The reciprocal of the power 3
xp/q = c
x = (±)cq/por
Equations of the Form xp/q = c
x = (–8)1/3 = –2
Rational power equations are equations of the type
xR = c where R = p/q is a rational number.
64. Equations of the Form xp/q = c
Solve xp/q = c by raising both sides to the reciprocal
exponent q/p and check your answers afterwards.
65. Equations of the Form xp/q = c
Solve xp/q = c by raising both sides to the reciprocal
exponent q/p and check your answers afterwards.
Example E. Solve x-2/3 = 16
66. Equations of the Form xp/q = c
Solve xp/q = c by raising both sides to the reciprocal
exponent q/p and check your answers afterwards.
Example E. Solve x-2/3 = 16
Raise both sides to -3/2 power.
67. Equations of the Form xp/q = c
Solve xp/q = c by raising both sides to the reciprocal
exponent q/p and check your answers afterwards.
Example E. Solve x-2/3 = 16
x-2/3 = 16 (x-2/3)-3/2 = (16)-3/2
Raise both sides to -3/2 power.
68. Equations of the Form xp/q = c
Solve xp/q = c by raising both sides to the reciprocal
exponent q/p and check your answers afterwards.
Example E. Solve x-2/3 = 16
Raise both sides to -3/2 power.
x-2/3 = 16 (x-2/3)-3/2 = (16)-3/2 = 1/64
We check that –1/64 is also a solution.
69. Equations of the Form xp/q = c
Solve xp/q = c by raising both sides to the reciprocal
exponent q/p and check your answers afterwards.
Example E. Solve x-2/3 = 16
Example F. Solve (2x – 3)3/2 = -8
Raise both sides to -3/2 power.
x-2/3 = 16 (x-2/3)-3/2 = (16)-3/2 = 1/64
We check that –1/64 is also a solution.
70. Equations of the Form xp/q = c
Solve xp/q = c by raising both sides to the reciprocal
exponent q/p and check your answers afterwards.
Example E. Solve x-2/3 = 16
Example F. Solve (2x – 3)3/2 = -8
Raise both sides to -3/2 power.
Raise both sides to 2/3 power.
x-2/3 = 16 (x-2/3)-3/2 = (16)-3/2 = 1/64
We check that –1/64 is also a solution.
71. Equations of the Form xp/q = c
Solve xp/q = c by raising both sides to the reciprocal
exponent q/p and check your answers afterwards.
Example E. Solve x-2/3 = 16
Example F. Solve (2x – 3)3/2 = -8
Raise both sides to -3/2 power.
Raise both sides to 2/3 power.
(2x – 3)3/2 = -8 [(2x – 3)3/2]2/3 = (-8)2/3
x-2/3 = 16 (x-2/3)-3/2 = (16)-3/2 = 1/64
We check that –1/64 is also a solution.
72. Equations of the Form xp/q = c
Solve xp/q = c by raising both sides to the reciprocal
exponent q/p and check your answers afterwards.
Example E. Solve x-2/3 = 16
Example F. Solve (2x – 3)3/2 = -8
Raise both sides to -3/2 power.
Raise both sides to 2/3 power.
(2x – 3)3/2 = -8 [(2x – 3)3/2]2/3 = (-8)2/3
(2x – 3) = 4
x-2/3 = 16 (x-2/3)-3/2 = (16)-3/2 = 1/64
We check that –1/64 is also a solution.
73. Equations of the Form xp/q = c
Solve xp/q = c by raising both sides to the reciprocal
exponent q/p and check your answers afterwards.
Example E. Solve x-2/3 = 16
Example F. Solve (2x – 3)3/2 = -8
Raise both sides to -3/2 power.
Raise both sides to 2/3 power.
(2x – 3)3/2 = -8 [(2x – 3)3/2]2/3 = (-8)2/3
(2x – 3) = 4
2x = 7 x = 7/2
x-2/3 = 16 (x-2/3)-3/2 = (16)-3/2 = 1/64
We check that –1/64 is also a solution.
74. Equations of the Form xp/q = c
Solve xp/q = c by raising both sides to the reciprocal
exponent q/p and check your answers afterwards.
Example E. Solve x-2/3 = 16
x-2/3 = 16 (x-2/3)-3/2 = (16)-3/2 = 1/64
We check that –1/64 is also a solution.
Example F. Solve (2x – 3)3/2 = -8
Raise both sides to -3/2 power.
Raise both sides to 2/3 power.
(2x – 3)3/2 = -8 [(2x – 3)3/2]2/3 = (-8)2/3
(2x – 3) = 4
2x = 7 x = 7/2
Since x = 7/2 doesn't work because 43/2 = -8,
there is no solution.
77. Radical Equations
Solve radical equations by squaring both sides to
remove the square root. Do it again if necessary.
Reminder: (A ± B)2 = A2 ± 2AB + B2
78. Radical Equations
Solve radical equations by squaring both sides to
remove the square root. Do it again if necessary.
Reminder: (A ± B)2 = A2 ± 2AB + B2
Example G. Solve
x + 4 = 5x + 4
79. Radical Equations
Solve radical equations by squaring both sides to
remove the square root. Do it again if necessary.
Reminder: (A ± B)2 = A2 ± 2AB + B2
Example G. Solve
x + 4 = 5x + 4 square both sides;
(x + 4)2 = (5x + 4 )2
80. Radical Equations
Solve radical equations by squaring both sides to
remove the square root. Do it again if necessary.
Reminder: (A ± B)2 = A2 ± 2AB + B2
Example G. Solve
x + 4 = 5x + 4 square both sides;
(x + 4)2 = (5x + 4 )2
x + 2 * 4 x + 16 = 5x + 4
81. Radical Equations
Solve radical equations by squaring both sides to
remove the square root. Do it again if necessary.
Reminder: (A ± B)2 = A2 ± 2AB + B2
Example G. Solve
x + 4 = 5x + 4 square both sides;
(x + 4)2 = (5x + 4 )2
x + 2 * 4 x + 16 = 5x + 4 isolate the radical;
8x = 4x – 12
82. Radical Equations
Solve radical equations by squaring both sides to
remove the square root. Do it again if necessary.
Reminder: (A ± B)2 = A2 ± 2AB + B2
Example G. Solve
x + 4 = 5x + 4 square both sides;
(x + 4)2 = (5x + 4 )2
x + 2 * 4 x + 16 = 5x + 4 isolate the radical;
8x = 4x – 12 divide by 4;
2x = x – 3
83. Radical Equations
Solve radical equations by squaring both sides to
remove the square root. Do it again if necessary.
Reminder: (A ± B)2 = A2 ± 2AB + B2
Example G. Solve
x + 4 = 5x + 4 square both sides;
(x + 4)2 = (5x + 4 )2
x + 2 * 4 x + 16 = 5x + 4 isolate the radical;
8x = 4x – 12 divide by 4;
2x = x – 3 square again;
( 2x)2 = (x – 3)2
84. Radical Equations
Solve radical equations by squaring both sides to
remove the square root. Do it again if necessary.
Reminder: (A ± B)2 = A2 ± 2AB + B2
Example G. Solve
x + 4 = 5x + 4 square both sides;
(x + 4)2 = (5x + 4 )2
x + 2 * 4 x + 16 = 5x + 4 isolate the radical;
8x = 4x – 12 divide by 4;
2x = x – 3 square again;
( 2x)2 = (x – 3)2
4x = x2 – 6x + 9
85. Radical Equations
Solve radical equations by squaring both sides to
remove the square root. Do it again if necessary.
Reminder: (A ± B)2 = A2 ± 2AB + B2
Example G. Solve
x + 4 = 5x + 4 square both sides;
(x + 4)2 = (5x + 4 )2
x + 2 * 4 x + 16 = 5x + 4 isolate the radical;
8x = 4x – 12 divide by 4;
2x = x – 3 square again;
( 2x)2 = (x – 3)2
4x = x2 – 6x + 9
0 = x2 – 10x + 9
86. Radical Equations
Solve radical equations by squaring both sides to
remove the square root. Do it again if necessary.
Reminder: (A ± B)2 = A2 ± 2AB + B2
Example G. Solve
x + 4 = 5x + 4 square both sides;
(x + 4)2 = (5x + 4 )2
x + 2 * 4 x + 16 = 5x + 4 isolate the radical;
8x = 4x – 12 divide by 4;
2x = x – 3 square again;
( 2x)2 = (x – 3)2
4x = x2 – 6x + 9
0 = x2 – 10x + 9
0 = (x – 9)(x – 1)
x = 9, x = 1
87. Radical Equations
Solve radical equations by squaring both sides to
remove the square root. Do it again if necessary.
Reminder: (A ± B)2 = A2 ± 2AB + B2
Example G. Solve
x + 4 = 5x + 4 square both sides;
(x + 4)2 = (5x + 4 )2
x + 2 * 4 x + 16 = 5x + 4 isolate the radical;
8x = 4x – 12 divide by 4;
2x = x – 3 square again;
( 2x)2 = (x – 3)2
4x = x2 – 6x + 9
0 = x2 – 10x + 9
0 = (x – 9)(x – 1)
x = 9, x = 1 Only 9 is good.
88. The absolute value of x is the distance measured
from x to 0 and it is denoted as |x|.
Absolute Value Equations
89. The absolute value of x is the distance measured
from x to 0 and it is denoted as |x|. Because it is
distance, |x| is nonnegative.
Absolute Value Equations
90. The absolute value of x is the distance measured
from x to 0 and it is denoted as |x|. Because it is
distance, |x| is nonnegative.
Absolute Value Equations
Algebraic definition of absolute value:
|x|=
x if x is positive or 0.
–x (opposite of x) if x is negative.{
91. The absolute value of x is the distance measured
from x to 0 and it is denoted as |x|. Because it is
distance, |x| is nonnegative.
Absolute Value Equations
Algebraic definition of absolute value:
|x|=
x if x is positive or 0.
–x (opposite of x) if x is negative.{
Hence | -5 | = –(-5) = 5.
92. The absolute value of x is the distance measured
from x to 0 and it is denoted as |x|. Because it is
distance, |x| is nonnegative.
Absolute Value Equations
Algebraic definition of absolute value:
|x|=
x if x is positive or 0.
–x (opposite of x) if x is negative.{
Hence | -5 | = –(-5) = 5. Since the absolute value is
never negative, an equation such as
|x4 – 3x + 1 | = – 2 doesn't have any solution.
93. The absolute value of x is the distance measured
from x to 0 and it is denoted as |x|. Because it is
distance, |x| is nonnegative.
Absolute Value Equations
Algebraic definition of absolute value:
|x|=
x if x is positive or 0.
–x (opposite of x) if x is negative.{
Hence | -5 | = –(-5) = 5. Since the absolute value is
never negative, an equation such as
|x4 – 3x + 1 | = – 2 doesn't have any solution.
Fact I. |x*y| = |x|*|y|. For example, |-2*3 | = |-2|*|3| = 6.
94. The absolute value of x is the distance measured
from x to 0 and it is denoted as |x|. Because it is
distance, |x| is nonnegative.
Absolute Value Equations
Algebraic definition of absolute value:
|x|=
x if x is positive or 0.
–x (opposite of x) if x is negative.{
Hence | -5 | = –(-5) = 5. Since the absolute value is
never negative, an equation such as
|x4 – 3x + 1 | = – 2 doesn't have any solution.
Fact I. |x*y| = |x|*|y|. For example, |-2*3 | = |-2|*|3| = 6.
Warning: In general |x ± y| |x| ± |y|.
95. The absolute value of x is the distance measured
from x to 0 and it is denoted as |x|. Because it is
distance, |x| is nonnegative.
Absolute Value Equations
Algebraic definition of absolute value:
|x|=
x if x is positive or 0.
–x (opposite of x) if x is negative.{
Hence | -5 | = –(-5) = 5. Since the absolute value is
never negative, an equation such as
|x4 – 3x + 1 | = – 2 doesn't have any solution.
Fact I. |x*y| = |x|*|y|. For example, |-2*3 | = |-2|*|3| = 6.
Warning: In general |x ± y| |x| ± |y|.
For instance, | 2 – 3 | |2| – |3| |2| + |3|.
96. Because |x±y| |x|±|y|, we have to solve absolute
value equations by rewriting it into two equations
without absolute values.
Absolute Value Equations
97. Because |x±y| |x|±|y|, we have to solve absolute
value equations by rewriting it into two equations
without absolute values.
Absolute Value Equations
Fact II: If |x| = a where a is a positive number, then
x = a or x = –a.
98. Because |x±y| |x|±|y|, we have to solve absolute
value equations by rewriting it into two equations
without absolute values.
Absolute Value Equations
Fact II: If |x| = a where a is a positive number, then
x = a or x = –a.
Example H.
a. If | x | = 3
99. Because |x±y| |x|±|y|, we have to solve absolute
value equations by rewriting it into two equations
without absolute values.
Absolute Value Equations
Fact II: If |x| = a where a is a positive number, then
x = a or x = –a.
Example H.
a. If | x | = 3 then x = 3 or x= –3
100. Because |x±y| |x|±|y|, we have to solve absolute
value equations by rewriting it into two equations
without absolute values.
Absolute Value Equations
Fact II: If |x| = a where a is a positive number, then
x = a or x = –a.
Example H.
a. If | x | = 3 then x = 3 or x= –3
b. If | 2x – 3 | = 5 then
101. Because |x±y| |x|±|y|, we have to solve absolute
value equations by rewriting it into two equations
without absolute values.
Absolute Value Equations
Fact II: If |x| = a where a is a positive number, then
x = a or x = –a.
Example H.
a. If | x | = 3 then x = 3 or x= –3
b. If | 2x – 3 | = 5 then
2x – 3 = 5 or 2x – 3 = –5
102. Because |x±y| |x|±|y|, we have to solve absolute
value equations by rewriting it into two equations
without absolute values.
Absolute Value Equations
Fact II: If |x| = a where a is a positive number, then
x = a or x = –a. Also if |x| = |y| then x = y or x = –y.
Example H.
a. If | x | = 3 then x = 3 or x= –3
b. If | 2x – 3 | = 5 then
2x – 3 = 5 or 2x – 3 = –5
103. Because |x±y| |x|±|y|, we have to solve absolute
value equations by rewriting it into two equations
without absolute values.
Absolute Value Equations
Fact II: If |x| = a where a is a positive number, then
x = a or x = –a. Also if |x| = |y| then x = y or x = –y.
Example H.
a. If | x | = 3 then x = 3 or x= –3
b. If | 2x – 3 | = 5 then
2x – 3 = 5 or 2x – 3 = –5
c. Solve |2x – 3| = |3x + 1|
104. Because |x±y| |x|±|y|, we have to solve absolute
value equations by rewriting it into two equations
without absolute values.
Absolute Value Equations
Fact II: If |x| = a where a is a positive number, then
x = a or x = –a. Also if |x| = |y| then x = y or x = –y.
Example H.
a. If | x | = 3 then x = 3 or x= –3
b. If | 2x – 3 | = 5 then
2x – 3 = 5 or 2x – 3 = –5
c. Solve |2x – 3| = |3x + 1|
Rewrite: 2x – 3 = 3x + 1 or 2x – 3 = –(3x + 1)
105. Because |x±y| |x|±|y|, we have to solve absolute
value equations by rewriting it into two equations
without absolute values.
Absolute Value Equations
Fact II: If |x| = a where a is a positive number, then
x = a or x = –a. Also if |x| = |y| then x = y or x = –y.
Example H.
a. If | x | = 3 then x = 3 or x= –3
b. If | 2x – 3 | = 5 then
2x – 3 = 5 or 2x – 3 = –5
c. Solve |2x – 3| = |3x + 1|
Rewrite: 2x – 3 = 3x + 1 or 2x – 3 = –(3x + 1)
–4 = x
–4 = x
106. Because |x±y| |x|±|y|, we have to solve absolute
value equations by rewriting it into two equations
without absolute values.
Absolute Value Equations
Fact II: If |x| = a where a is a positive number, then
x = a or x = –a. Also if |x| = |y| then x = y or x = –y.
Example H.
a. If | x | = 3 then x = 3 or x= –3
b. If | 2x – 3 | = 5 then
2x – 3 = 5 or 2x – 3 = –5
c. Solve |2x – 3| = |3x + 1|
Rewrite: 2x – 3 = 3x + 1 or 2x – 3 = –(3x + 1)
–4 = x or 2x – 3 = –3x – 1
–4 = x or x = 2/5
107. The geometric meaning of “|x – y|” or “|y – x|” is
“the distance between x and y”.
Absolute Value Equations
108. The geometric meaning of “|x – y|” or “|y – x|” is
“the distance between x and y”.
So the absolute value equation “| x – c | = r” is claiming
that “the distance from x to c is r”.
Absolute Value Equations
109. The geometric meaning of “|x – y|” or “|y – x|” is
“the distance between x and y”.
So the absolute value equation “| x – c | = r” is claiming
that “the distance from x to c is r”.
Absolute Value Equations
c
110. The geometric meaning of “|x – y|” or “|y – x|” is
“the distance between x and y”.
So the absolute value equation “| x – c | = r” is claiming
that “the distance from x to c is r”.
Absolute Value Equations
c
r r
111. The geometric meaning of “|x – y|” or “|y – x|” is
“the distance between x and y”.
So the absolute value equation “| x – c | = r” is claiming
that “the distance from x to c is r”. Hence x = c ± r.
Absolute Value Equations
c
r
x = c + r
r
x = c – r
112. The geometric meaning of “|x – y|” or “|y – x|” is
“the distance between x and y”.
So the absolute value equation “| x – c | = r” is claiming
that “the distance from x to c is r”. Hence x = c ± r.
Absolute Value Equations
Example G. Draw and solve for x geometrically
if |x – 7| = 12.
c
r
x = c + r
r
x = c – r
113. The geometric meaning of “|x – y|” or “|y – x|” is
“the distance between x and y”.
So the absolute value equation “| x – c | = r” is claiming
that “the distance from x to c is r”. Hence x = c ± r.
Absolute Value Equations
Example G. Draw and solve for x geometrically
if |x – 7| = 12.
The equation asks for the locations of x's that are
12 units away from 7.
c
r
x = c + r
r
x = c – r
114. The geometric meaning of “|x – y|” or “|y – x|” is
“the distance between x and y”.
So the absolute value equation “| x – c | = r” is claiming
that “the distance from x to c is r”. Hence x = c ± r.
Absolute Value Equations
Example G. Draw and solve for x geometrically
if |x – 7| = 12.
The equation asks for the locations of x's that are
12 units away from 7.
c
r
x = c + r
7
1212
r
x = c – r
115. The geometric meaning of “|x – y|” or “|y – x|” is
“the distance between x and y”.
So the absolute value equation “| x – c | = r” is claiming
that “the distance from x to c is r”. Hence x = c ± r.
Absolute Value Equations
Example G. Draw and solve for x geometrically
if |x – 7| = 12.
The equation asks for the locations of x's that are
12 units away from 7.
c
r
x = c + r
So to the right x = 7 + 12 = 19.
and to the left x = 7 – 12 = – 5,
7
1212
x = 19x = – 5
r
x = c – r
116. Given a formula f, the domain of f are all the numbers
that may be computed by f.
Zeroes and Domains
117. Given a formula f, the domain of f are all the numbers
that may be computed by f. We may specify the
domain of a formula by saying what’s not in it.
Zeroes and Domains
118. Given a formula f, the domain of f are all the numbers
that may be computed by f. We may specify the
domain of a formula by saying what’s not in it.
For example, the domain of the formula are
“all the numbers except x = 0”.
Zeroes and Domains
1
x
119. Given a formula f, the domain of f are all the numbers
that may be computed by f. We may specify the
domain of a formula by saying what’s not in it.
For example, the domain of the formula are
“all the numbers except x = 0”.
Zeroes and Domains
1
x
For a reduced fractional formula f = ,
the zeroes of f are where N = 0
the domain of f are “all numbers except where D = 0”
N
D
120. Given a formula f, the domain of f are all the numbers
that may be computed by f. We may specify the
domain of a formula by saying what’s not in it.
For example, the domain of the formula are
“all the numbers except x = 0”.
Zeroes and Domains
Example I.
1
x
For a reduced fractional formula f = ,
the zeroes of f are where N = 0
the domain of f are “all numbers except where D = 0”
N
D
x2 – 4.Find the zeros and the domain f = x2 – 1
121. Given a formula f, the domain of f are all the numbers
that may be computed by f. We may specify the
domain of a formula by saying what’s not in it.
For example, the domain of the formula are
“all the numbers except x = 0”.
Zeroes and Domains
Example I.
1
x
For a reduced fractional formula f = ,
the zeroes of f are where N = 0
the domain of f are “all numbers except where D = 0”
N
D
x2 – 4.Find the zeros and the domain f = x2 – 1
The zeros of f are where x2 – 1 = 0 or x = ±1.
122. Given a formula f, the domain of f are all the numbers
that may be computed by f. We may specify the
domain of a formula by saying what’s not in it.
For example, the domain of the formula are
“all the numbers except x = 0”.
Zeroes and Domains
Example I.
1
x
For a reduced fractional formula f = ,
the zeroes of f are where N = 0
the domain of f are “all numbers except where D = 0”
N
D
x2 – 4.Find the zeros and the domain f = x2 – 1
The zeros of f are where x2 – 1 = 0 or x = ±1.
The domain of f are
“all numbers except x2 – 4 = 0, or except x = ±2”.
123. A. Solve the following equations by factoring.
5. x2 – 3x = 10
9. x3 – 2x2 = 0
6. x2 = 4
7. 2x(x – 3) + 4 = 2x – 4
10. 2x2(x – 3) = –4x
8. x(x – 3) + x + 6 = 2x2 + 3x
1. x2 – 3x – 4 = 0 2. x2 – 2x – 15 = 0 3. x2 + 7x + 12 = 0
4. –x2 – 2x + 8 = 0
11. 4x2 = x4
12. 7x2 = –4x3 – 3x 13. 5 = (x + 2)(2x + 1)
14. (x + 1)2 = x2 + (x – 1)2 15. (x + 3)2 – (x + 2)2 = (x + 1)2
B. Solve the following equations by the quadratic formula.
If the answers are not real numbers, just state so.
1. x2 – x + 1 = 0 2. x2 – x – 1 = 0
3. x2 – 3x – 2 = 0 4. x2 – 2x + 3 = 0
5. 2x2 – 3x – 1 = 0 6. 3x2 = 2x + 3
Equations
124. Rational Equations
7. 1
x
+
1
x – 1
=
5
6
8. 1
x
+
1
x + 2
=
3
4
9. 2
x
+ 1
x + 1
= 3
2
10. + 5
x + 2
= 22
x – 1
11. – 1
x + 1
= 3
2
12.6
x + 2
– 4
x + 1
= 11
x – 2
x
6 3
1 2
3
5
2
3–+ = x1. x
4 6
–3 1
8
–5
– 1– = x2.
x
4 5
3 2
10
7
4
3+– = x3. x
8 12
–5 7
16
–5
+ 1+ = x
(x – 20) = x – 3
100
30
100
205.
(x + 5) – 3 = (x – 5)
100
25
100
206.
C. Clear the denominators of the equations, then solve.
125. Radical Equations and Power Equations
D. Isolate one radical if needed, square both sides, do it again
if necessary, to solve for x. Check your answers.
1. x – 2 = x – 4 2. x + 3 = x + 1
3. 2x – 1 = x + 5 4. 4x + 1 – x + 2 = 1
5. x – 2 = x + 3 – 1 6. 3x + 4 = 3 – x – 1
7. 2x + 5 = x + 4 8. 5 – 4x – 3 – x = 1
E. Solve by raising both sides to an appropriate power.
No calculator.
1. x –2 = 1/4 2. x –1/2 = 1/4
3. x –3 = –8 4. x –1/3 = –8
5. x –2/3 = 4 6. x –3/2 = 8
7. x –2/3 = 1/4 8. x –3/2 = – 1/8
9. x 1.5 = 1/27 10. x 1.25 = 32
11. x –1.5 = 27 12. x –1. 25 = 1/32
126. F. Solve for x.
1. Is it always true that I+x| = x? Give reason for your answer.
2. Is it always true that |–x| = x? Give reason for your answer.
Absolute Value Equations
3. |4 – 5x| = 3 4. |3 + 2x| = 7 5. |–2x + 3| = 5
6. |4 – 5x| = –3 7. |2x + 1| – 1= 5 8. 3|2x + 1| – 1= 5
9. |4 – 5x| = |3 + 2x|
11. |4 – 5x| = |2x + 1| 12. |3x + 1| = |5 – x|
10. |–2x + 3|= |3 – 2x|
Solve geometrically for x. Draw the solution.
13. |x – 2| = 1 14. |3 + x| = 5 15. | –9 + x| = –7
x – 4
G. Find the zeros and the domain of the following rational
formulas. (See 2.1)
2x – 1
1. x2 – 1.
x2 – 43.
5x + 7
2.
3x + 5
x2 – x
x2 – x – 24.
x2 – 4x5.
x2 + x – 2
x2 + 2x6.
x2 + x + 2
2x2 – x – 17.
x3 + 2x
x4 – 4x8.
x3 – 8
16. |2 + x| = 1 17. |3 – x| = –5 18. | –9 – x| = 8