3e. Pedagogy of Mathematics (Part II) - Algebra (Ex 3.5)
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Pedagogy of Mathematics (Part II) - Algebra, Algebra, Maths, IX std Maths, Samacheerkalvi maths, II year B.Ed., Pedagogy, Mathematics, Polynomials, Constants, variables, algebraic expression
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Lattice points on the homogeneous cone 5(x2+y2) 9xy=23z2
Seven different method s of the non-zero non-negative solutions of homogeneous Diophantine equation 5(x2 + y2) – 9xy = 23z2 are obtained. Introducing the linear transformation x =u + v, y= u – v, u v0 in 5(x2+y2) -9xy = 23z2, it reduces to u2 + 19v2 =23z2. We are solved the above equation through various choices and are obtained seven different methods of solutions which are satisfied it. Some interesting relations among the special numbers and the solutions are exposed.
3d. Pedagogy of Mathematics (Part II) - Algebra (Ex 3.4)
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Final exam g8 correction
1) The student received a final exam in mathematics with 20 questions. Their total score was 60% or 12 out of 20 questions correct.
2) The exam covered topics like algebra, geometry, trigonometry, and statistics. Questions involved finding GCDs, simplifying expressions, solving inequalities, and calculating areas of shapes.
3) One statistics question asked them to calculate percentages based on a sample of 30 students where 60% liked a certain activity. They correctly found that 18 students would like the activity.
Taller final matematicas ii
1. The document evaluates determinants of 4x4 matrices using Sarrus' rule for 3x3 determinants. It finds the determinants to be 72, -81, and 26445.
2. It uses Cramer's rule to solve three systems of equations, finding the solutions to be (-1, 3, 7), (b+c/2, c+a/2, b+a/2), and (0, 3, 4).
3. It calculates the volumes of two geometric shapes with points given, finding the volumes to be 5 cubic units and 5/3 cubic units.
O Level A Maths Sec 3 Syallbus 2011 Simultaneous Equation, Indices, Surds, Ex...
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Prepared by Musicisintheworld.
Copyright © 2011. All Rights Reserved. No part of this publication may be reproduced without permission from the author, Musicisintheworld.
Simultaneous triple series equations
This academic article discusses solving simultaneous triple series equations associated with Laguerre polynomials with matrix arguments. It presents the following simultaneous triple series equations that arise in crack problems in fracture mechanics. It then provides the solution obtained for the unknown function Cnj by using a multiplying factor technique and applying integral properties and orthogonality relations for Laguerre polynomials with matrix arguments. The solution is determined for parameters where the equations are well defined.
Completing the square if a
The document provides 3 examples of solving quadratic equations by setting them equal to zero and using the quadratic formula. Each example shows the step-by-step work of isolating the constant term, factoring the equation, taking the square root of both sides to solve for the roots, and checking the solutions. The examples demonstrate how to solve quadratic equations from setting them equal to zero through finding the solution set.
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Lattice points on the homogeneous cone 5(x2+y2) 9xy=23z2
Seven different method s of the non-zero non-negative solutions of homogeneous Diophantine equation 5(x2 + y2) – 9xy = 23z2 are obtained. Introducing the linear transformation x =u + v, y= u – v, u v0 in 5(x2+y2) -9xy = 23z2, it reduces to u2 + 19v2 =23z2. We are solved the above equation through various choices and are obtained seven different methods of solutions which are satisfied it. Some interesting relations among the special numbers and the solutions are exposed.
3d. Pedagogy of Mathematics (Part II) - Algebra (Ex 3.4)
Pedagogy of Mathematics (Part II) - Algebra, Algebra, Maths, IX std Maths, Samacheerkalvi maths, II year B.Ed., Pedagogy, Mathematics, Polynomials, Constants, variables, algebraic expression
Final exam g8 correction
1) The student received a final exam in mathematics with 20 questions. Their total score was 60% or 12 out of 20 questions correct.
2) The exam covered topics like algebra, geometry, trigonometry, and statistics. Questions involved finding GCDs, simplifying expressions, solving inequalities, and calculating areas of shapes.
3) One statistics question asked them to calculate percentages based on a sample of 30 students where 60% liked a certain activity. They correctly found that 18 students would like the activity.
Taller final matematicas ii
1. The document evaluates determinants of 4x4 matrices using Sarrus' rule for 3x3 determinants. It finds the determinants to be 72, -81, and 26445.
2. It uses Cramer's rule to solve three systems of equations, finding the solutions to be (-1, 3, 7), (b+c/2, c+a/2, b+a/2), and (0, 3, 4).
3. It calculates the volumes of two geometric shapes with points given, finding the volumes to be 5 cubic units and 5/3 cubic units.
O Level A Maths Sec 3 Syallbus 2011 Simultaneous Equation, Indices, Surds, Ex...
O Level A Maths Sec 3 Syallbus 2011
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Prepared by Musicisintheworld.
Copyright © 2011. All Rights Reserved. No part of this publication may be reproduced without permission from the author, Musicisintheworld.
Simultaneous triple series equations
This academic article discusses solving simultaneous triple series equations associated with Laguerre polynomials with matrix arguments. It presents the following simultaneous triple series equations that arise in crack problems in fracture mechanics. It then provides the solution obtained for the unknown function Cnj by using a multiplying factor technique and applying integral properties and orthogonality relations for Laguerre polynomials with matrix arguments. The solution is determined for parameters where the equations are well defined.
Completing the square if a
The document provides 3 examples of solving quadratic equations by setting them equal to zero and using the quadratic formula. Each example shows the step-by-step work of isolating the constant term, factoring the equation, taking the square root of both sides to solve for the roots, and checking the solutions. The examples demonstrate how to solve quadratic equations from setting them equal to zero through finding the solution set.
MTK BAB 3 MATRIKS
The document discusses matrices and their operations including addition, subtraction, multiplication, transpose, determinant, and inverse. It provides examples of calculating the sum, difference, product, and inverse of various matrices. It also covers solving systems of linear equations using matrices and determinants.
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The document contains examples of matrix operations including addition, subtraction, multiplication, transpose, and inverse of matrices. It also contains examples of using determinants and inverse matrices to solve systems of linear equations. Some key examples are:
1) Finding the sum, difference, product, and transpose of various 2x2 matrices
2) Computing the determinants of several 2x2 and 3x3 matrices
3) Using determinants and inverse matrices to solve 3 systems of 2 linear equations
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1) The document contains notes from a math lesson on properties of addition, subtraction, multiplication, and division. It includes examples of applying various properties and solving equations using the order of operations. It also lists an assignment that is due the next day involving solving odd-numbered problems from Set 3.
MATH- FACTORIZE AND EXPAND EASY TO UNDERSTAND
Some example with formulas and solution
QUESTION # 1
If (a + b) 2 = 73 and ab = 65, find the value of a2 + b2.
Solution ; (a + b)2 = a2 + 2ab + b2
73 = a2 + 2(65) + b2
73 = a2 + b2 + 130
73 – 130 = a2 + b2
QUESTION # 2) FACTORISE
(I) (3g - h) 2 – g2
Solution ;
a2 - b2 = (a + b) (a - b)
[ ( 3g – h ) + g ] [ ( 3g – h ) – g ]
(II) f 4 + 8f + 16
Solution ;
(a + b) = (a)2 + 2(a)(b) + (b)2 (f 2)2 + 2(f2)(4) + (4)2
(III) 3x2 – 752
Solution ;
a2 – b2 = (a + b)(a - b)
3(x2 – 25y2)
3[ (x)2 – (5y)2 ]
(IV) 4x2 – y2 – 4 + 4y
Solution ;
(2x)2 - (y)2 + (-1 + y)
QUESTION # 3) EXPAND
(I) (p4 - 2)2
Solution ;
(a - b)2 = a2 – 2ab =b2
(p2)2 - 2(p2)(2) + (2)2
(II) ( x y )
( y x )
Solution ;
(a + b)2 = a2 – 2ab = b2
(x/y)2 + 2(x/y) (y/x) + (y/x)2
(III)( a + b)2
( 3)
Solution ;
(a + b)2 = a2 + 2ab + b2
a2 + 2(a)(b/3) + (b/3)2
(IV) (ef - 2)2
Solution ;
(a - b)2 = a2 - 2ab + b2
(ef)2 – 2(ef)(2) + (2)2
QUESTION # 4) EVALUATE
(I) 412 + 738 + 81
Solution ;
[(40 + 1)2] + 738 + 81
(a + b)2 = a2 + 2(a)(b) + (b)2
[ (40)2 + 2(40)(1) + (1)2] + 738 +81
1600 + 80 + 1 + 738 + 81
1681 + 819
(II) 922 – 368 + 4
Solution ;
(a + b)2 = a2 + 2(a)(b) + (b)2
(90 + 2)2 – 368 + 4
[ (90)2 + 2(90)(2) + (2)2 ]– 368 + 4
8100 + 360 + 4 – 368 + 4
8460 – 360
(III) 2012 – 992
Solution ;
a2 – b2 = (a - b)2 (a + b)2
(200 + 1)2 - (100 - 1)2
2002 + 2(200)(1) + (1)2 – (100)2 – 2(100)(1) + (1)2
40000 + 400 + 1 – 10000 – 200 + 1
40401 - 9801
QUESTION # 5) FACTORISE
(I) 3a2 – 36a + 108
Solution ;
A2 – 2ab + b2 = (a-b)2
3(a2 – 362)
(II) a2 – 196
Solution ;
a2 – b2 = (a + b)(a - b)
a2 – 142
(III) p2 – 4q2 + 3(p – 2q)
Solution ;
a2 + b2 = (a + b)(a - b)
(p)2 – (2q)2 + 3(p – 2q)
(p – 2q)(p + 2q) + 3(p – 2q)
(IV) 2a4 – 32b2 c2
Solution ;
2(a4 – 16b2 c2)
2(a2)2 (4bc)2
2(a2 – 4bc) (a2 – 4bc)
2bc(a2 - 4) (a2 + 4bc)
2bc(a)2 – (2)2 (a2 +4bc)
2bc (a - 2) (a + 2)(a2 + 4bc)
(V) p4 – 81q4
Solution ;
(p2)2 – (9q2)2
a2 + b2 = (a + b)(a - b)
(p2 – 9q2) (p2 + 9q2)
(p)2 – (3q)2 (p2 – 9q2)
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Five different methods of the non-zero non-negative solutions of non- homogeneous cubic Diophantine equation x4 – y4 = 17( z2 –
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Keywords: The Method of Factorization, Integer Solutions, Linear Transformation, Relations and Special Numbers
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3) Simplifying (17 - 15)3 using the identity a3 - b3 = (a - b)(a2 + ab + b2)
4) Simplifying a3 - b2/(a-b) when a = 2 and b = -2
5) Simplifying a - b + b2/(1-a+b) when a = -1/2 and b = 3
Product of a binomial and a trinomial involving
The document discusses how to find the sum or difference of two cubes by multiplying a binomial with a trinomial of a certain form. It provides examples of multiplying binomial expressions of the form (a - b) with trinomials of the form (a^2 + ab + b^2) and using the rule (a - b)(a^2 + ab + b^2) = (a^3 - b^3). It also gives an example that does not follow this rule.
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sharda university noida and mte and their explanation about java
and there is marks of all pages and importance of this project
sharda university is located in noida and one of the best college for students
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Vectors intro
Vectors can represent journeys with distance and direction. They have components that work like coordinates and magnitude that is calculated using Pythagoras. Vectors can be added by placing them nose to tail and adding corresponding components, subtracted by adding the negative vector, and multiplied by a scalar by multiplying each component.
Latihan 3.3
The document contains examples of matrix operations including addition, subtraction, multiplication, transpose, and inverse of matrices. It also contains examples of using determinants and inverse matrices to solve systems of linear equations. Some key examples are:
1) Finding the sum, difference, product, and transpose of various 2x2 matrices
2) Computing the determinants of several 2x2 and 3x3 matrices
3) Using determinants and inverse matrices to solve 3 systems of 2 linear equations
7th math -l2
1) The document contains notes from a math lesson on properties of addition, subtraction, multiplication, and division. It includes examples of applying various properties and solving equations using the order of operations. It also lists an assignment that is due the next day involving solving odd-numbered problems from Set 3.
MATH- FACTORIZE AND EXPAND EASY TO UNDERSTAND
Some example with formulas and solution
QUESTION # 1
If (a + b) 2 = 73 and ab = 65, find the value of a2 + b2.
Solution ; (a + b)2 = a2 + 2ab + b2
73 = a2 + 2(65) + b2
73 = a2 + b2 + 130
73 – 130 = a2 + b2
QUESTION # 2) FACTORISE
(I) (3g - h) 2 – g2
Solution ;
a2 - b2 = (a + b) (a - b)
[ ( 3g – h ) + g ] [ ( 3g – h ) – g ]
(II) f 4 + 8f + 16
Solution ;
(a + b) = (a)2 + 2(a)(b) + (b)2 (f 2)2 + 2(f2)(4) + (4)2
(III) 3x2 – 752
Solution ;
a2 – b2 = (a + b)(a - b)
3(x2 – 25y2)
3[ (x)2 – (5y)2 ]
(IV) 4x2 – y2 – 4 + 4y
Solution ;
(2x)2 - (y)2 + (-1 + y)
QUESTION # 3) EXPAND
(I) (p4 - 2)2
Solution ;
(a - b)2 = a2 – 2ab =b2
(p2)2 - 2(p2)(2) + (2)2
(II) ( x y )
( y x )
Solution ;
(a + b)2 = a2 – 2ab = b2
(x/y)2 + 2(x/y) (y/x) + (y/x)2
(III)( a + b)2
( 3)
Solution ;
(a + b)2 = a2 + 2ab + b2
a2 + 2(a)(b/3) + (b/3)2
(IV) (ef - 2)2
Solution ;
(a - b)2 = a2 - 2ab + b2
(ef)2 – 2(ef)(2) + (2)2
QUESTION # 4) EVALUATE
(I) 412 + 738 + 81
Solution ;
[(40 + 1)2] + 738 + 81
(a + b)2 = a2 + 2(a)(b) + (b)2
[ (40)2 + 2(40)(1) + (1)2] + 738 +81
1600 + 80 + 1 + 738 + 81
1681 + 819
(II) 922 – 368 + 4
Solution ;
(a + b)2 = a2 + 2(a)(b) + (b)2
(90 + 2)2 – 368 + 4
[ (90)2 + 2(90)(2) + (2)2 ]– 368 + 4
8100 + 360 + 4 – 368 + 4
8460 – 360
(III) 2012 – 992
Solution ;
a2 – b2 = (a - b)2 (a + b)2
(200 + 1)2 - (100 - 1)2
2002 + 2(200)(1) + (1)2 – (100)2 – 2(100)(1) + (1)2
40000 + 400 + 1 – 10000 – 200 + 1
40401 - 9801
QUESTION # 5) FACTORISE
(I) 3a2 – 36a + 108
Solution ;
A2 – 2ab + b2 = (a-b)2
3(a2 – 362)
(II) a2 – 196
Solution ;
a2 – b2 = (a + b)(a - b)
a2 – 142
(III) p2 – 4q2 + 3(p – 2q)
Solution ;
a2 + b2 = (a + b)(a - b)
(p)2 – (2q)2 + 3(p – 2q)
(p – 2q)(p + 2q) + 3(p – 2q)
(IV) 2a4 – 32b2 c2
Solution ;
2(a4 – 16b2 c2)
2(a2)2 (4bc)2
2(a2 – 4bc) (a2 – 4bc)
2bc(a2 - 4) (a2 + 4bc)
2bc(a)2 – (2)2 (a2 +4bc)
2bc (a - 2) (a + 2)(a2 + 4bc)
(V) p4 – 81q4
Solution ;
(p2)2 – (9q2)2
a2 + b2 = (a + b)(a - b)
(p2 – 9q2) (p2 + 9q2)
(p)2 – (3q)2 (p2 – 9q2)
On homogeneous biquadratic diophantineequation x4 y4=17(z2-w2)r2
Abstract
Five different methods of the non-zero non-negative solutions of non- homogeneous cubic Diophantine equation x4 – y4 = 17( z2 –
w2) R2 are obtained. Some interesting relations among the special numbers and the solutions are exposed.
Keywords: The Method of Factorization, Integer Solutions, Linear Transformation, Relations and Special Numbers
Chapter 4 simultaneous equations
This document provides an overview of solving simultaneous equations between linear and quadratic equations with two unknowns. It includes 4 examples of solving simultaneous equations involving a linear equation equal to a quadratic equation. The examples show the steps of substituting one equation into the other and solving. The document also includes a chapter review with 6 practice problems involving simultaneous equations.
2f. Pedagogy of Mathematics - Part II (Numbers and Sequence - Ex 2.6)
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Square of trinomial
The document provides a formula and examples for finding the square of a trinomial. The formula is: F2 + M2 + L2 + 2FM + 2FL + 2ML, where F is the first term, M is the middle term, L is the last term, and the other terms represent twice the product of the terms. It then works through four examples applying the formula to find the square of trinomials such as (a + b + c)2, (x + y - 3)2, (x - 2y - 1)2, and (2a - b + b2)2.
On cubic diophantine equation x2+y2 xy=39 z3
Abstract
Four different methods of the non-zero non-negative solutions of non- homogeneous cubic Diophantine equation x2 + y2 – xy =
39z2 are obtained. Some interesting relations among the special numbers and the solutions are exposed.
Keywords: The Method of Factorization, Integer Solutions, Linear Transformation, Relations and Special Numbers
Ch02 13
This document contains solutions to 6 math problems:
1) Finding the value of x2 - x + 1 when x = 2
2) Evaluating (a + b - c)(a - b + c) when a = 1, b = 2, c = 3
3) Simplifying (17 - 15)3 using the identity a3 - b3 = (a - b)(a2 + ab + b2)
4) Simplifying a3 - b2/(a-b) when a = 2 and b = -2
5) Simplifying a - b + b2/(1-a+b) when a = -1/2 and b = 3
Product of a binomial and a trinomial involving
The document discusses how to find the sum or difference of two cubes by multiplying a binomial with a trinomial of a certain form. It provides examples of multiplying binomial expressions of the form (a - b) with trinomials of the form (a^2 + ab + b^2) and using the rule (a - b)(a^2 + ab + b^2) = (a^3 - b^3). It also gives an example that does not follow this rule.
Bvco u1 a2_jogf
This document contains an activity on complex numbers and functions of a complex variable. The activity includes two problems. The first problem involves calculating the modulus of complex expressions. Various steps are shown to solve the expressions, including using binomial theorem to expand powers of complex numbers. The second problem expresses vectors representing complex numbers in the form a + ib. Steps are outlined to calculate the components from the vector properties like length and angle. Graphical representations of the complex numbers are also provided.
2e. Pedagogy of Mathematics - Part II (Numbers and Sequence - Ex 2.5)
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Why is (a+b)2 = a2+b2+2ab
This document proves the formula (a+b)2 = a2 + b2 + 2ab using both geometric and algebraic approaches. Geometrically, it represents a+b as a line segment and draws the squares and rectangles that make up (a+b)2. Algebraically, it expands (a+b)2 using the distributive property and collects like terms. Both methods demonstrate that (a+b)2 equals the sum of the squares of a and b plus twice their product.
Colour in Mathematics
The document discusses how colour can be used in mathematics to add clarity. It provides examples of how colour highlights which terms are associated with signs in algebra examples. Colour can also emphasise the order of operations. Further examples show how colour aids understanding of topics like factorisation, composite functions, inequalities, coordinate geometry, binomial expansion, completing the square, circle geometry, integration, and decision mathematics algorithms. Worked solutions with colour coding are also suggested to help students understand steps at their own pace.
Mte (1)
sharda university noida and mte and their explanation about java
and there is marks of all pages and importance of this project
sharda university is located in noida and one of the best college for students
3i. Pedagogy of Mathematics (Part II) - Algebra (Ex 3.9)
This document contains solutions to examples finding the greatest common divisor (GCD) of various algebraic expressions. It provides step-by-step workings for expressions involving variables, exponents, and coefficients. Examples include finding the GCD of polynomials, expressions with common factors, and expressions where the GCD is 1. The document demonstrates multiple methods for determining the GCD algebraically and factorizing expressions.
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2d. Pedagogy of Mathematics - Part II (Numbers and Sequence - Ex 2.4)
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2f. Pedagogy of Mathematics - Part II (Numbers and Sequence - Ex 2.6)
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3i. Pedagogy of Mathematics (Part II) - Algebra (Ex 3.9)
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The property illustrated by the expressions 8 + [6 + (-7)] = (8+6) + (-7) and 12 + 6 = 18 is the associative property of addition.
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1) The document discusses the process of completing the square to solve quadratic equations. It provides examples of factorizing quadratics and using completing the square to solve equations like x^2 - 8x - 7 = 0.
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1. The diagram shows angles MBC and ABC measuring 25 degrees each.
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- 1. PEDAGOGY OF MATHEMATICS – PART II BY Dr. I. UMA MAHESWARI Principal Peniel Rural College of Education,Vemparali, Dindigul District iuma_maheswari@yahoo.co.in
- 2. STD IX CHAPTER 3 – ALGEBRA Ex – 3.5
- 12. Solution: (i) 2a² + 4a²b + 8a²c = 2a²(1 + 2b + 4c) (ii) ab – ac – mb + mc = a(b – c) – m(b – c) = (b – c) (a – m)
- 13. Solution: (i) x2 + 4x + 4 = x2 + 2 × x × 2 + 22 [a2 + 2ab+ b2 = (a + b)2] = (x + 2)2 (ii) 3a2 – 24ab + 48b2 = 3[a2 – 8ab + 16b2] = 3[a2 – 2 × a × 4b + (4b)2] = 3(a- 4b)2 [a2 – 2ab + b2 = (a – b)2]
- 16. Solution: [a2 + b2 + c2 + 2ab + 2bc + 2ac = (a + b + c)2] = (2x)2 + (3y)2 + (5z)2 + 2(2x) (3y) + 2(3y) (5z) + 2(5z) (2x) = (2x + 3y + 5z)2 (ii) 25x2 + 4y2 + 9z2 – 20xy + 12yz – 30xz Solution: = (5x)2 + (2y)2 + (3z)2 + 2(5x) (-2y) + 2(-2y) (- 3z) + 2(-3z) (5x) 2
- 17. Solution: (i) 8x3 + 125y3 = (2x)3 + (5y)3 [a3 + b3 = (a + b)(a2 – ab + b2) = (2x + 5y) [(2x)2 – (2x) (5y) + (5y)2] = (2x + 5y) (4x2 – 10xy + 25y2) (ii) 27x3 – 8y3 = (3x)3 – (2y)3 [a3 – b3 = (a – b)(a2 + ab + b2)] = (3x – 2y) [(3x)2 + (3x) (2y) + (2y)2] = (3x – 2y) (9x2 + 6xy + Ay2)
- 18. (iii) a6 – 64 = a6 – 26 = (a2)3 – (22)3 [a2 – b3 = (a- b) + (a2 + ab + b2)] = (a2 – 22) [(a2)2 + (a2) (22) + (22)2] = (a + 2) (a – 2) (a4 + 4a2 + 16) = (a + 2) (a – 2) [(a2)2 + 42 + 8a2 – 4a2] = (a + 2)(a- 2) [(a2 + 4)2 – (2a)2] {a2 – b2 = (a + b) (a – b)} = (a + 2) (a – 2) [(a2 + 4 + 2a) (a2 + 4 – 2a) = (a + 2) (a – 2) (a2 + 2a + 4) (a2 – 2a + 4)
- 19. Using the formula [a3 + b3 + c3 – 3abc] = (a + b + c) (a2 + b2 + c2 – ab – bc – ac) Solution: (i) x3 + 8y2 + 6xy – 1 = -(-x3 – 8y3 – 6xy + 1) = – (-x3 – 8y3 + 1 – 6xy) = -[(-x)3 + (-2y)3 + 1 – 3(x) (2y) (1)] = -[-x – 2y + 1] [(-x)2 + (-2y)2 + 12 – (-x) (-2y) – (-2y) (1) – (1) (-x)] = (x + 2y – 1)(x2 + 4y2 + 1 – 2xy + 2y + x)
- 20. (ii) l3 – 8m3 – 27n3 – 18lmn = l3 + (-2m)3 + (-3n)2 – 3(l) (-2m) (-3n) = (l – 2m – 3n) [l2 + (-2m)2 + (-3n)2 -1 (-2m)] – (-2m)(-3n) – (-3n)(l) = (l – 2m – 3n) (l2 + 4m2 + 9n2 + 2lm – 6mn + 3ln)