Applications of Systems
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1) The document discusses solving systems of linear equations through examples involving finding unknown numbers, ticket sales, and mixture problems. 2) The examples show setting up the systems of equations, choosing variables to represent the unknowns, translating the word problems into algebraic equations, and solving the systems using substitution or addition methods. 3) The solutions are checked by substituting the values back into the original equations to verify the answers.
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The document provides information about functions and relations. It defines a function as a relation where each x-value is paired with exactly one y-value. To determine if a relation is a function, it describes using the vertical line test, where a relation is a function if a vertical line can only intersect the graph at one point. It gives examples of applying the vertical line test to graphs and determining the domain and range of relations.Relations and Functions
Relations and Functionstoni dimella
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The document discusses relations, functions, domains, and ranges. It defines a relation as a set of ordered pairs and a function as a relation where each x-value is mapped to only one y-value. It explains how to identify the domain and range of a relation, and use the vertical line test and mappings to determine if a relation is a function. Examples of evaluating functions are also provided.Multiplying Polynomials (no conjugates)
Multiplying Polynomials (no conjugates)toni dimella
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This document discusses techniques for multiplying polynomials including FOIL, the double distributive property, and squaring binomials. It provides examples of squaring binomial expressions like (x + 3)2 by multiplying (x + 3) by itself using FOIL. The document also lists more example problems but does not show the solutions.Fractions, Decimals, and Percents
Fractions, Decimals, and Percentstoni dimella
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This document defines decimals, fractions, and percents and provides steps for converting between them. Decimals are numbers with a decimal point, fractions show parts of a whole, and percents express amounts out of 100. To convert a fraction to a decimal, divide the numerator by the denominator. To convert a fraction to a percent, change it to a decimal then multiply by 100. Converting between other forms follows similar steps of changing the number to an equivalent decimal or percent value.C3 Study 際際滷s - MAT 151
C3 Study 際際滷s - MAT 151toni dimella
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This document contains lecture slides about statistics for describing, exploring, and comparing data. It discusses measures of center such as the mean, median, and mode. It also discusses variance and standard deviation as measures of spread. Additional topics covered include finding the mode, determining if a value is unusually high or low based on the mean and standard deviation, calculating percentiles, and comparing the detail provided by different graphic displays of data.C2 Study 際際滷s - MAT 151
C2 Study 際際滷s - MAT 151toni dimella
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The document contains multiple choice questions about summarizing and graphing data. It asks questions about outliers, frequency distributions, class boundaries, histograms, and types of charts including pie charts. One question asks the reader to use a pie chart showing housing types to find the number of people living in single family housing in a town of 12,200 people.C1 Study 際際滷s - MAT151
C1 Study 際際滷s - MAT151toni dimella
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The document contains lecture slides on introductory statistics topics including definitions of population, quantitative vs qualitative data, types of measurement scales, and examples of different sampling methods and study designs. Key points covered are the definition of a population as the complete collection of all elements, examples of quantitative data like weights vs qualitative nominal categories, ordinal scales involving ranking, and retrospective study designs using existing historical data.C3 test Doc
C3 test Doctoni dimella
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This document contains 16 logic and critical thinking problems involving statements, truth tables, logic symbols, validity of arguments, and Euler diagrams. The problems cover topics such as determining logical equivalences, constructing and analyzing truth tables, identifying inverse, converse and contrapositive statements, and determining the validity of arguments using logic rules or diagrams.C3 test
C3 testtoni dimella
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This document contains 16 logic and critical thinking problems involving statements, truth tables, logic symbols, validity of arguments, and Euler diagrams. The problems cover topics such as determining logical equivalence, constructing and analyzing truth tables, identifying inverse, converse and contrapositive statements, and determining the validity of arguments using logic rules or diagrams.Intro to Logs
Intro to Logstoni dimella
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The document introduces logarithms, defining them as the exponent that a fixed number (the base) must be raised to to equal the value. It provides examples of converting between logarithmic and exponential forms, and covers the key properties of logarithms including product, quotient, power, expanding, condensing, and the change of base formula.Logs
Logstoni dimella
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The document introduces logarithms, defining them as the exponent that a base number must be raised to to equal the value. It provides examples of converting between logarithmic and exponential forms, and discusses properties of logarithms such as the product, quotient, and power properties. It also covers expanding, condensing, and changing the base of logarithmic expressions.Intro to Polynomials
Intro to Polynomialstoni dimella
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This document discusses polynomial functions and how to graph them. It defines a polynomial as a sum of terms with non-negative integer exponents. Polynomial graphs are smooth curves that may be lines, parabolas, or higher-order curves. To graph a polynomial, one determines the end behavior from the leading term, finds the x-intercepts by setting the polynomial equal to 0, and uses intercepts and test points to plot the graph over intervals. Multiplicity of roots affects whether the graph crosses or is tangent to the x-axis at those points.Completing the Square
Completing the Squaretoni dimella
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The document explains how to solve quadratic equations by completing the square. It defines a perfect square trinomial as having the form x^2 + bx + c, where c is the square of half of b. It provides steps for completing the square, which involves adding a constant term to both sides of the equation such that the left side becomes a perfect square trinomial that can be factorized. This process results in the solution(s) to the quadratic equation. Two examples demonstrating this process are included.Graph Quadratics
Graph Quadraticstoni dimella
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The document discusses graphing quadratic functions in standard form (y=ax^2 + bx + c). It explains that the graph is a parabola that can open up or down depending on whether a is positive or negative. The line of symmetry for the parabola passes through the vertex and is given by the equation x=-b/2a. The steps to graph are: 1) find the line of symmetry, 2) plug the x-value into the original equation to find the vertex, 3) find two other points and reflect them across the line of symmetry.Composite functions
Composite functionstoni dimella
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Composite functions are formed by taking the output of one function and using it as the input of another function. This is shown notationally as f(g(x)), where the result of g(x) is used as the input for f. Changing the order of the functions changes the result, as the output of the inner function determines the input to the outer function. Examples show evaluating composite functions by substituting the output of the inner function into the outer function and simplifying.Parallel and Perpendicular lines
Parallel and Perpendicular linestoni dimella
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Parallel lines have the same slope and never intersect. Two lines are parallel if their slopes are equal. The equation of a line parallel to a given line can be found using the point-slope formula with the same slope.
Perpendicular lines intersect at a right angle. Two lines are perpendicular if the product of their slopes is -1. The equation of a line perpendicular to a given line can be found using the point-slope formula with the slope of the perpendicular line being the negative reciprocal of the given line's slope. Examples are provided of finding if lines are parallel or perpendicular and constructing parallel and perpendicular lines through a given point.Ad
Applications of Systems
- 1. Systems of Linear Equations and Problem Solving
- 2. Read and reread the problem. Suppose that the second number is 5. Then the first number, which is 4 more than twice the second number, would have to be 14 (4 + 25). Is their total 25? No: 14 + 5 = 19. Our proposed solution is incorrect, but we now have a better understanding of the problem.Since we are looking for two numbers, we letx = first numbery = second numberFinding an Unknown NumberExample:One number is 4 more than twice the second number. Their total is 25. Find the numbers.1.) UnderstandContinued
- 3. One number is 4 more than twice the second number.Their total is 25.Finding an Unknown NumberExample continued:2.) Translatex = 4 + 2yx + y = 25Continued
- 4. Finding an Unknown NumberExample continued:3.) SolveWe are solving the system x = 4 + 2y and x + y = 25Using the substitution method, we substitute the solution for x from the first equation into the second equation.x + y = 25 (4 + 2y) + y = 25Replace x with result from first equation.4 + 3y = 25Simplify left side.3y = 21Subtract 4 from both sides and simplify.y = 7Divide both sides by 3.Now we substitute the value for y into the first equation.Continuedx = 4 + 2y = 4 + 2(7) = 4 + 14 = 18
- 5. Finding an Unknown NumberExample continued:4.) InterpretCheck: Substitute x = 18 and y = 7 into both of the equations. First equation,x = 4 + 2y 18 = 4 + 2(7) true Second equation,x + y = 2518 + 7 = 25 trueState: The two numbers are 18 and 7.
- 6. Solving a Problem about PricesExample:Hilton University Drama club sold 311 tickets for a play. Student tickets cost 50 cents each; non-student tickets cost $1.50. If total receipts were $385.50, find how many tickets of each type were sold.1.) UnderstandRead and reread the problem. Suppose the number of students tickets was 200. Since the total number of tickets sold was 311, the number of non-student tickets would have to be 111 (311 200).Continued
- 7. Solving a Problem about PricesExample continued:1.) Understand (continued)Are the total receipts $385.50? Admission for the 200 students will be 200($0.50), or $100. Admission for the 111 non-students will be 111($1.50) = $166.50. This gives total receipts of $100 + $166.50 = $266.50. Our proposed solution is incorrect, but we now have a better understanding of the problem.Since we are looking for two numbers, we lets = the number of student ticketsn = the number of non-student ticketsContinued
- 8. Hilton University Drama club sold 311 tickets for a play.total receipts were $385.50Admission for studentsTotal receiptsAdmission for non students+=385.501.50nSolving a Problem about PricesExample continued:2.) Translates + n = 3110.50sContinued
- 9. s + n = 311s + n = 311s 3n = 7712(0.50s + 1.50n) = 2(385.50)Solving a Problem about PricesExample continued:3.) SolveWe are solving the system s + n = 311 and 0.50s + 1.50n = 385.50Since the equations are written in standard form (and we might like to get rid of the decimals anyway), well solve by the addition method. Multiply the second equation by 2.simplifies to2n = 460n = 230Now we substitute the value for n into the first equation.s + n = 311s + 230 = 311s = 81Continued
- 10. Solving a Problem about PricesExample continued:4.) InterpretCheck: Substitute s = 81 and n = 230 into both of the equations. First equation,s + n = 31181 + 230 = 311 true Second equation, 0.50s + 1.50n = 385.50 0.50(81) + 1.50(230) = 385.50 40.50 + 345 = 385.50 trueState: There were 81 student tickets and 230 non-student tickets sold.
- 11. price per unitnumber of units=price of all unitsSolving a Mixture ProblemExample:A Candy Barrel shop manager mixes M&Ms worth $2.00 per pound with trail mix worth $1.50 per pound. Find how many pounds of each she should use to get 50 pounds of a party mix worth $1.80 per pound.1.) UnderstandRead and reread the problem. We are going to propose a solution, but first we need to understand the formulas we will be using. To find out the cost of any quantity of items we use the formulaContinued
- 12. Solving a Mixture ProblemExample continued:1.) Understand (continued)Suppose the manage decides to mix 20 pounds of M&Ms. Since the total mixture will be 50 pounds, we need 50 20 = 30 pounds of the trail mix. Substituting each portion of the mix into the formula,M&Ms $2.00 per lb 20 lbs = $40.00trail mix $1.50 per lb 30 lbs = $45.00Mixture $1.80 per lb 50 lbs = $90.00Continued
- 13. Solving a Mixture ProblemExample continued:1.) Understand (continued)Since $40.00 + $45.00 $90.00, our proposed solution is incorrect (hey, we were pretty close again), but we now have a better understanding of the problem. Since we are looking for two quantities, we letx = the amount of M&Msy = the amount of trail mixContinued
- 14. Fifty pounds of party mixprice per unitprice of all unitsUsingnumber of units=Price of M&MsPrice of trail mixPrice of mixture+=2xSolving a Mixture ProblemExamplecontinued:2.) Translatex + y = 501.5y1.8(50) = 90Continued
- 15. 3(x + y) = 3(50) 3x + 3y = 150 4x 3y = 1802(2x + 1.50y) = 2(90)Solving a Mixture ProblemExample continued:3.) SolveWe are solving the system x + y = 50 and 2x + 1.50y = 90Since the equations are written in standard form (and we might like to get rid of the decimals anyway), well solve by the addition method. Multiply the first equation by 3 and the second equation by 2.simplifies to x = 30x = 30Now we substitute the value for x into the first equation.Continuedx + y = 5030 + y = 50y = 20
- 16. Solving a Mixture ProblemExample continued:4.) InterpretCheck: Substitute x = 30 and y = 20 into both of the equations. First equation,x + y = 5030 + 20 = 50 true Second equation, 2x + 1.50y = 90 2(30) + 1.50(20) = 90 60 + 30 = 90 trueState: The store manager needs to mix 30 pounds of M&Ms and 20 pounds of trail mix to get the mixture at $1.80 a pound.