0904 ch 9 day 4
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The document discusses solving a system of linear equations in two variables. It provides an example problem involving finding the amounts of 5% and 20% solutions that make up a total 1 L solution with an overall concentration of 14%. The system is set up and solved, yielding amounts of 400 mL for the 5% solution and 600 mL for the 20% solution. Students are then instructed to work on homework in groups, with the teacher available to help.
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Σύγκριση κλασμάτων
Σύγκριση κλασμάτων , ΜΑΘΗΜΑΤΙΚΑ Α΄ΓΥΜΝΑΣΙΟΥ, Μέρος B' , Κεφάλαιο 2ο, ΚΛΑΣΜΑΤΑ, teagher, Εκπαιδευτικό βίντεο από τo ΚΑΛΙΤΣΑ : https://mcjmcjmcj.wordpress.com/ ,
Άρθρο του μαθήματος:https://wp.me/PJcUD-8yc/
#teagher, #ΕΚΠΑΙΔΕΥΣΗ, #ΜΑΘΗΜΑΤΑ, #ΓΥΜΝΑΣΙΟ #ΠΡΩΤΗΓΥΜΝΑΣΙΟΥ, #ΛΥΣΕΙΣ, #ΑΠΑΝΤΗΣΕΙΣ, #ΚΑΛΙΤΣΑ,
0905 ch 9 day 5
The document describes putting a system of linear equations into triangular form. It contains a 3x3 system of linear equations. The summary explains that through Gaussian elimination by eliminating variables from equations, the system can be transformed into an upper triangular matrix with equations arranged from top to bottom to easily solve for the variables through back substitution.
0903 ch 9 day 3
The document discusses systems of linear equations in two variables. It explains that two lines can intersect in 0, 1, or infinitely many points, corresponding to no solutions, exactly one solution, or infinitely many solutions (dependent system) for a system of two linear equations. Several examples are worked through, including inconsistent, dependent, and consistent systems. Matrices are introduced as another method for solving systems, but it is noted they have limitations.
0906 ch 9 day 6
The system of equations is dependent, meaning it has infinitely many solutions rather than a unique solution. To solve a dependent system, one can express one variable in terms of others and substitute into another equation to solve for a second variable in terms of the independent variable. Here, equation 2 is solved for y in terms of z, giving y = 3z + 2. Substituting this and z = k into equation 1 gives the solution x = -2k, where k can be any number.
0902 ch 9 day 2
The document contains examples of solving systems of equations and word problems involving geometry. It begins with an example of setting up and solving a system of two equations with two unknowns to determine how many 6-inch and 12-inch bricks were purchased given the total cost. It then works through finding the lengths of the sides of a right triangle given the area and hypotenuse. Finally, it solves problems involving finding dimensions of rectangles and cylinders given other geometric information like areas and volumes.
0901 ch 9 day 1
The document discusses various methods for solving systems of equations, including substitution, elimination, graphic, and matrix methods. It provides an example of using the substitution method to solve the system of equations 5x + y = 2 and 3x - 2y = 22. The solution obtained is x = 2 and y = -8.
0101 ch 1 day 1
The document summarizes key concepts from the first chapter of a Pre-Calculus textbook. It introduces interval notation and defines common inequality symbols like greater than, less than, greater than or equal to, and less than or equal to. It provides examples of writing inequalities using interval notation, such as x > 3 representing the interval (3, ∞).
1103 ch 11 day 3
This document provides information about sequences and series. It defines sequences as functions with positive integers as the domain. It distinguishes between infinite and finite sequences. Examples of sequences are provided and explicit formulas for finding terms are derived. Methods for finding the nth term of arithmetic and geometric sequences are described.
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Σύγκριση κλασμάτων
Σύγκριση κλασμάτων , ΜΑΘΗΜΑΤΙΚΑ Α΄ΓΥΜΝΑΣΙΟΥ, Μέρος B' , Κεφάλαιο 2ο, ΚΛΑΣΜΑΤΑ, teagher, Εκπαιδευτικό βίντεο από τo ΚΑΛΙΤΣΑ : https://mcjmcjmcj.wordpress.com/ ,
Άρθρο του μαθήματος:https://wp.me/PJcUD-8yc/
#teagher, #ΕΚΠΑΙΔΕΥΣΗ, #ΜΑΘΗΜΑΤΑ, #ΓΥΜΝΑΣΙΟ #ΠΡΩΤΗΓΥΜΝΑΣΙΟΥ, #ΛΥΣΕΙΣ, #ΑΠΑΝΤΗΣΕΙΣ, #ΚΑΛΙΤΣΑ,
0905 ch 9 day 5
The document describes putting a system of linear equations into triangular form. It contains a 3x3 system of linear equations. The summary explains that through Gaussian elimination by eliminating variables from equations, the system can be transformed into an upper triangular matrix with equations arranged from top to bottom to easily solve for the variables through back substitution.
0903 ch 9 day 3
The document discusses systems of linear equations in two variables. It explains that two lines can intersect in 0, 1, or infinitely many points, corresponding to no solutions, exactly one solution, or infinitely many solutions (dependent system) for a system of two linear equations. Several examples are worked through, including inconsistent, dependent, and consistent systems. Matrices are introduced as another method for solving systems, but it is noted they have limitations.
0906 ch 9 day 6
The system of equations is dependent, meaning it has infinitely many solutions rather than a unique solution. To solve a dependent system, one can express one variable in terms of others and substitute into another equation to solve for a second variable in terms of the independent variable. Here, equation 2 is solved for y in terms of z, giving y = 3z + 2. Substituting this and z = k into equation 1 gives the solution x = -2k, where k can be any number.
0902 ch 9 day 2
The document contains examples of solving systems of equations and word problems involving geometry. It begins with an example of setting up and solving a system of two equations with two unknowns to determine how many 6-inch and 12-inch bricks were purchased given the total cost. It then works through finding the lengths of the sides of a right triangle given the area and hypotenuse. Finally, it solves problems involving finding dimensions of rectangles and cylinders given other geometric information like areas and volumes.
0901 ch 9 day 1
The document discusses various methods for solving systems of equations, including substitution, elimination, graphic, and matrix methods. It provides an example of using the substitution method to solve the system of equations 5x + y = 2 and 3x - 2y = 22. The solution obtained is x = 2 and y = -8.
0101 ch 1 day 1
The document summarizes key concepts from the first chapter of a Pre-Calculus textbook. It introduces interval notation and defines common inequality symbols like greater than, less than, greater than or equal to, and less than or equal to. It provides examples of writing inequalities using interval notation, such as x > 3 representing the interval (3, ∞).
1103 ch 11 day 3
This document provides information about sequences and series. It defines sequences as functions with positive integers as the domain. It distinguishes between infinite and finite sequences. Examples of sequences are provided and explicit formulas for finding terms are derived. Methods for finding the nth term of arithmetic and geometric sequences are described.
1204 ch 12 day 4
The document discusses tangent lines to functions. It provides examples of finding the equation of a tangent line with a given slope to specific functions. It also discusses finding the average and instantaneous velocity of an object given its position function.
1203 ch 12 day 3
Here are the problems from the slides with their solutions:
1. Find the slope of the line tangent to the graph of the function f(x) = x^2 - 5x + 8 at the point P(1,4).
Slope = -3
2. Find the equation of the tangent line to the curve f(x) = 2x^2 - 3 at the point P(1,-1) using point-slope form.
y - (-1) = 4(x - 1)
3. Find the equation of the tangent line to the curve f(x) = x + 6 at the point P(3,3) using point-slope form.
y
1201 ch 12 day 1
This document contains examples and explanations of limits involving various functions. Some key points covered include:
- Substitution can be used to evaluate limits, such as substituting 2 into -2x^3.
- Left and right hand limits must agree for the overall limit to exist.
- The limit of a piecewise function exists if the left and right limits are the same.
- Graphs can help verify limit calculations and show discontinuities.
- Special limits involving trigonometric and greatest integer functions are evaluated.
1202 ch 12 day 2
The document provides an introduction to evaluating limits, including:
1. The limit of a constant function is the constant.
2. Common limit laws can be used to evaluate limits of sums, differences, products, and quotients if the individual limits exist.
3. Special techniques may be needed to evaluate limits that involve indeterminate forms, such as 0/0, infinity/infinity, or limits approaching infinity. These include factoring, graphing, and rationalizing.
1104 ch 11 day 4
The document discusses recursive rules for defining sequences. It explains that a recursive rule defines subsequent terms of a sequence using previous terms, with one or more initial terms provided. Examples are worked through, such as finding the first five terms of the sequence where a1 = 3 and an = 2an-1 - 1, which are 3, 5, 9, 17, 33. Other sequences discussed include the Fibonacci sequence and examples of finding recursive rules to define other given sequences.
1114 ch 11 day 14
The document discusses two methods for expanding binomial expressions: Pascal's triangle and the binomial theorem. Pascal's triangle uses a recursive method to provide the coefficients for expanding binomials, but is only practical for smaller values of n. The binomial theorem provides an explicit formula for expanding binomials of the form (a + b)n using factorials and combinations. It works better than Pascal's triangle when n is large. Examples are provided to demonstrate expanding binomials like (3 - xy)4 and (x - 2)6 using both methods.
1113 ch 11 day 13
The document discusses using mathematical induction to prove the formula:
3 + 5 + 7 +...+ (2k + 1) = k(k + 2)
It provides the base case of p(1) and shows that it is true. It then assumes p(k) is true, and shows that p(k+1) follows by algebraic manipulations. This completes the induction proof.
1112 ch 11 day 12
The document discusses mathematical induction. It provides examples of deductive and inductive reasoning. It then explains the principle of mathematical induction, which involves proving that a statement is true for a base case, and assuming the statement is true for some value k to prove it is also true for k+1. The document provides a full example of using mathematical induction to prove that the sum of the first k odd positive integers is equal to k^2. It demonstrates proving the base case of 1 and the induction step clearly.
1110 ch 11 day 10
The document discusses geometric sequences and series. It examines partial sums of geometric sequences, which involve adding a finite number of terms. It also explores whether infinite series, or adding an infinite number of terms, can converge to a limiting value. It provides an example of someone getting closer to a wall on successive trips, with the total distance traveled converging even as the number of trips approaches infinity. It analyzes the behavior of geometric series based on whether the common ratio r is less than, greater than, or equal to 1.
1109 ch 11 day 9
The document discusses geometric sequences. A geometric sequence is a sequence where each term after the first is found by multiplying the previous term by a fixed non-zero number called the common ratio. The common ratio is found by taking the quotient of any two consecutive terms. Explicit formulas are provided to calculate specific terms based on knowing the first term and common ratio. Examples are worked through, including finding a specific term for given sequences.
1108 ch 11 day 8
Here are the key steps:
- Find the formula for the nth term (an) of an arithmetic sequence
- Plug the values given into the formula to find a and d
- Use the formula for the sum of the first n terms (Sn) of an arithmetic sequence
- Set the formula equal to the total sum given and solve for n
The goal is to set up and solve the equation systematically rather than guessing and checking numbers. Documenting the work shows the logical steps and thought process. Keep exploring new approaches to solving problems more efficiently!
1107 ch 11 day 7
The document defines arithmetic sequences as sequences where the difference between consecutive terms is constant. It provides the formula for an arithmetic sequence as an = an-1 + d, where d is the common difference. It then gives several examples of arithmetic sequences and exercises identifying sequences as arithmetic and finding their common differences. It also explains how given any two terms of a sequence, the entire sequence is determined by finding the common difference d and using the formula an = a1 + (n-1)d.
1106 ch 11 day 6
This document discusses sequences and summation notation on day four. It references a bible verse about love and laying down one's life for others. It also contains instructions to be sure homework questions are addressed and for groups to begin the next homework assignment while working together. A quote by Henry Ford is included about dividing difficult tasks into smaller jobs.
1105 ch 11 day 5
The document discusses summation notation and properties of sums. It provides examples of writing sums using sigma notation, such as expressing the sum 2 + 5 + 8 + 11 + 14 + 17 + 20 + 23 + 26 as the summation of 3k - 1 from k = 1 to 9. It also covers properties of sums, such as the property that the sum of a sum of a terms and b terms is equal to the sum of a terms plus the sum of b terms. The document provides guidance on calculating sums using sigma notation on a calculator.
1115 ch 11 day 15
The document provides an explanation of the binomial theorem formula for finding a specific term in the expansion of a binomial expression. It gives the formula as:
⎛ n ⎞ n−r r
⎜ r ⎟ x y
⎝ ⎠
Where n is the total number of terms, r is 1 less than the term number being found, x and y are the terms being added or subtracted. It provides an example of finding the 5th term of (a + b)6. It also provides an example of finding the 5th term of (3x - 5y)
1007 ch 10 day 7
This document contains two problems about hyperbolas:
[1] It gives the vertices and foci of a hyperbola and asks to find the standard form equation. The vertices are (±2, 0) and the foci are (±3, 0). The standard form equation is calculated to be x^2/4 - y^2/5 = 1.
[2] It gives the vertices and asymptotes of another hyperbola and asks to find the equation and foci. The vertices are (0, ±4) and the asymptotes are y = ±4x. The standard form equation is calculated to be y^2/16 - x^2 = 1, and the
1006 ch 10 day 6
The document defines and explains hyperbolas through the following key points:
1. A hyperbola is the set of points where the absolute difference between the distance to two fixed points (foci) is a constant.
2. Key parts of a hyperbola include vertices, foci, transverse axis, and conjugate axis.
3. The standard equation of a hyperbola is (x2/a2) - (y2/b2) = 1
4. Examples are worked through to graph specific hyperbolas using their equations.
1005 ch 10 day 5
The document discusses homework assignments and working in groups. It reminds students to ensure all homework questions have been addressed and directs groups to start working together on homework number 5. It also includes a quote about the importance of direction over current position.
1004 ch 10 day 4
The document contains information about ellipses:
1) It defines the eccentricity of an ellipse and provides the formula for calculating it. Eccentricity represents how circular or stretched out an ellipse is.
2) It works through examples of finding the equation of an ellipse given properties like the vertices and foci.
3) It also includes an example of finding the foci, eccentricity, lengths of the major and minor axes, and sketching the graph of an ellipse given its equation.
1003 ch 10 day 3
The document defines an ellipse and provides its standard equation. It does this by:
1) Defining an ellipse as the set of points where the sum of the distances from two fixed points (foci) is a constant.
2) Deriving the standard equation of an ellipse centered at the origin using geometry and algebra.
3) Explaining how to sketch an ellipse given its standard equation, including identifying the lengths of the major and minor axes and plotting the foci.
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1204 ch 12 day 4
The document discusses tangent lines to functions. It provides examples of finding the equation of a tangent line with a given slope to specific functions. It also discusses finding the average and instantaneous velocity of an object given its position function.
1203 ch 12 day 3
Here are the problems from the slides with their solutions:
1. Find the slope of the line tangent to the graph of the function f(x) = x^2 - 5x + 8 at the point P(1,4).
Slope = -3
2. Find the equation of the tangent line to the curve f(x) = 2x^2 - 3 at the point P(1,-1) using point-slope form.
y - (-1) = 4(x - 1)
3. Find the equation of the tangent line to the curve f(x) = x + 6 at the point P(3,3) using point-slope form.
y
1201 ch 12 day 1
This document contains examples and explanations of limits involving various functions. Some key points covered include:
- Substitution can be used to evaluate limits, such as substituting 2 into -2x^3.
- Left and right hand limits must agree for the overall limit to exist.
- The limit of a piecewise function exists if the left and right limits are the same.
- Graphs can help verify limit calculations and show discontinuities.
- Special limits involving trigonometric and greatest integer functions are evaluated.
1202 ch 12 day 2
The document provides an introduction to evaluating limits, including:
1. The limit of a constant function is the constant.
2. Common limit laws can be used to evaluate limits of sums, differences, products, and quotients if the individual limits exist.
3. Special techniques may be needed to evaluate limits that involve indeterminate forms, such as 0/0, infinity/infinity, or limits approaching infinity. These include factoring, graphing, and rationalizing.
1104 ch 11 day 4
The document discusses recursive rules for defining sequences. It explains that a recursive rule defines subsequent terms of a sequence using previous terms, with one or more initial terms provided. Examples are worked through, such as finding the first five terms of the sequence where a1 = 3 and an = 2an-1 - 1, which are 3, 5, 9, 17, 33. Other sequences discussed include the Fibonacci sequence and examples of finding recursive rules to define other given sequences.
1114 ch 11 day 14
The document discusses two methods for expanding binomial expressions: Pascal's triangle and the binomial theorem. Pascal's triangle uses a recursive method to provide the coefficients for expanding binomials, but is only practical for smaller values of n. The binomial theorem provides an explicit formula for expanding binomials of the form (a + b)n using factorials and combinations. It works better than Pascal's triangle when n is large. Examples are provided to demonstrate expanding binomials like (3 - xy)4 and (x - 2)6 using both methods.
1113 ch 11 day 13
The document discusses using mathematical induction to prove the formula:
3 + 5 + 7 +...+ (2k + 1) = k(k + 2)
It provides the base case of p(1) and shows that it is true. It then assumes p(k) is true, and shows that p(k+1) follows by algebraic manipulations. This completes the induction proof.
1112 ch 11 day 12
The document discusses mathematical induction. It provides examples of deductive and inductive reasoning. It then explains the principle of mathematical induction, which involves proving that a statement is true for a base case, and assuming the statement is true for some value k to prove it is also true for k+1. The document provides a full example of using mathematical induction to prove that the sum of the first k odd positive integers is equal to k^2. It demonstrates proving the base case of 1 and the induction step clearly.
1110 ch 11 day 10
The document discusses geometric sequences and series. It examines partial sums of geometric sequences, which involve adding a finite number of terms. It also explores whether infinite series, or adding an infinite number of terms, can converge to a limiting value. It provides an example of someone getting closer to a wall on successive trips, with the total distance traveled converging even as the number of trips approaches infinity. It analyzes the behavior of geometric series based on whether the common ratio r is less than, greater than, or equal to 1.
1109 ch 11 day 9
The document discusses geometric sequences. A geometric sequence is a sequence where each term after the first is found by multiplying the previous term by a fixed non-zero number called the common ratio. The common ratio is found by taking the quotient of any two consecutive terms. Explicit formulas are provided to calculate specific terms based on knowing the first term and common ratio. Examples are worked through, including finding a specific term for given sequences.
1108 ch 11 day 8
Here are the key steps:
- Find the formula for the nth term (an) of an arithmetic sequence
- Plug the values given into the formula to find a and d
- Use the formula for the sum of the first n terms (Sn) of an arithmetic sequence
- Set the formula equal to the total sum given and solve for n
The goal is to set up and solve the equation systematically rather than guessing and checking numbers. Documenting the work shows the logical steps and thought process. Keep exploring new approaches to solving problems more efficiently!
1107 ch 11 day 7
The document defines arithmetic sequences as sequences where the difference between consecutive terms is constant. It provides the formula for an arithmetic sequence as an = an-1 + d, where d is the common difference. It then gives several examples of arithmetic sequences and exercises identifying sequences as arithmetic and finding their common differences. It also explains how given any two terms of a sequence, the entire sequence is determined by finding the common difference d and using the formula an = a1 + (n-1)d.
1106 ch 11 day 6
This document discusses sequences and summation notation on day four. It references a bible verse about love and laying down one's life for others. It also contains instructions to be sure homework questions are addressed and for groups to begin the next homework assignment while working together. A quote by Henry Ford is included about dividing difficult tasks into smaller jobs.
1105 ch 11 day 5
The document discusses summation notation and properties of sums. It provides examples of writing sums using sigma notation, such as expressing the sum 2 + 5 + 8 + 11 + 14 + 17 + 20 + 23 + 26 as the summation of 3k - 1 from k = 1 to 9. It also covers properties of sums, such as the property that the sum of a sum of a terms and b terms is equal to the sum of a terms plus the sum of b terms. The document provides guidance on calculating sums using sigma notation on a calculator.
1115 ch 11 day 15
The document provides an explanation of the binomial theorem formula for finding a specific term in the expansion of a binomial expression. It gives the formula as:
⎛ n ⎞ n−r r
⎜ r ⎟ x y
⎝ ⎠
Where n is the total number of terms, r is 1 less than the term number being found, x and y are the terms being added or subtracted. It provides an example of finding the 5th term of (a + b)6. It also provides an example of finding the 5th term of (3x - 5y)
1007 ch 10 day 7
This document contains two problems about hyperbolas:
[1] It gives the vertices and foci of a hyperbola and asks to find the standard form equation. The vertices are (±2, 0) and the foci are (±3, 0). The standard form equation is calculated to be x^2/4 - y^2/5 = 1.
[2] It gives the vertices and asymptotes of another hyperbola and asks to find the equation and foci. The vertices are (0, ±4) and the asymptotes are y = ±4x. The standard form equation is calculated to be y^2/16 - x^2 = 1, and the
1006 ch 10 day 6
The document defines and explains hyperbolas through the following key points:
1. A hyperbola is the set of points where the absolute difference between the distance to two fixed points (foci) is a constant.
2. Key parts of a hyperbola include vertices, foci, transverse axis, and conjugate axis.
3. The standard equation of a hyperbola is (x2/a2) - (y2/b2) = 1
4. Examples are worked through to graph specific hyperbolas using their equations.
1005 ch 10 day 5
The document discusses homework assignments and working in groups. It reminds students to ensure all homework questions have been addressed and directs groups to start working together on homework number 5. It also includes a quote about the importance of direction over current position.
1004 ch 10 day 4
The document contains information about ellipses:
1) It defines the eccentricity of an ellipse and provides the formula for calculating it. Eccentricity represents how circular or stretched out an ellipse is.
2) It works through examples of finding the equation of an ellipse given properties like the vertices and foci.
3) It also includes an example of finding the foci, eccentricity, lengths of the major and minor axes, and sketching the graph of an ellipse given its equation.
1003 ch 10 day 3
The document defines an ellipse and provides its standard equation. It does this by:
1) Defining an ellipse as the set of points where the sum of the distances from two fixed points (foci) is a constant.
2) Deriving the standard equation of an ellipse centered at the origin using geometry and algebra.
3) Explaining how to sketch an ellipse given its standard equation, including identifying the lengths of the major and minor axes and plotting the foci.
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Jemison, MacLaughlin, and Majumder "Broadening Pathways for Editors and Authors"
Jemison, MacLaughlin, and Majumder "Broadening Pathways for Editors and Authors"
Leveraging Generative AI to Drive Nonprofit Innovation
Leveraging Generative AI to Drive Nonprofit Innovation
Gender and Mental Health - Counselling and Family Therapy Applications and In...
Gender and Mental Health - Counselling and Family Therapy Applications and In...
Juneteenth Freedom Day 2024 David Douglas School District
Juneteenth Freedom Day 2024 David Douglas School District
0904 ch 9 day 4
- 1. 9.2 Systems of Linear Equations in Two Variables Day 2 Galatians 2:20 "I have been crucified with Christ. It is no longer I who live, but Christ who lives in me. And the life I now live in the flesh I live by faith in the Son of God, who loved me and gave himself for me."
- 2. Solution to HW 3 #50 Let A = amount of 5% solution Let B = amount of 20% solution ⎧ A + B = 1 ⎨ ⎧ − 5A − 5B = −5 ⎩ .05A + .2B = .14 (1) ⎨ ⎩ 5A + 20B = 14 ⎧ A + B = 1 ⎨ 15B = 9 ⎩ 5A + 20B = 14 3 2 B= A= 5 5 400 mL of 5% solution & 600 mL of 20% solution
- 3. You have the rest of class to work on your HW. Please work together in your groups. I’ll be walking around to give help if you need it. HW #4 Procrastination is opportunity’s assassin. Victor Kiam
Editor's Notes
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