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  • https://math.libretexts.org/Bookshelves/Algebra/Advanced_Algebra
    Intermediate Algebra is the second part of a two-part course in Algebra that builds on the basics learned in Elementary Algebra and introduces the more advanced topics required for further study of ap...Intermediate Algebra is the second part of a two-part course in Algebra that builds on the basics learned in Elementary Algebra and introduces the more advanced topics required for further study of applications found in most disciplines. Used as a standalone textbook, it offers plenty of review as well as something new to engage the student in each chapter. This textbook introduces functions early and stresses the geometry behind the algebra.
  • https://math.libretexts.org/Bookshelves/Algebra/Advanced_Algebra/03%3A_Solving_Linear_Systems/3.0E%3A_3.E%3A_Solving_Linear_Systems
    \(\left\{ \begin{array} { l } { 4 x - y + 2 z = - 1 } \\ { x - 4 y + 3 z = 11 } \\ { 3 x + 5 y - 4 z = 1 } \end{array} \right.\) \(\left\{ \begin{array} { l } { x - 4 y + 6 z = 5 } \\ { 2 x + 5 y - z ...\(\left\{ \begin{array} { l } { 4 x - y + 2 z = - 1 } \\ { x - 4 y + 3 z = 11 } \\ { 3 x + 5 y - 4 z = 1 } \end{array} \right.\) \(\left\{ \begin{array} { l } { x - 4 y + 6 z = 5 } \\ { 2 x + 5 y - z = - 3 } \\ { 3 x - 4 y + z = 25 } \end{array} \right.\) Determine whether or not \((−3, 2, −5)\) is a solution to \(\left\{ \begin{array} { l } { x - y + 2 z = - 15 } \\ { 2 x - 3 y + z = - 17 } \\ { 3 x + 5 y - 2 z = 10 } \end{array} \right.\).
  • https://math.libretexts.org/Courses/Cosumnes_River_College/STAT_300%3A_Introduction_to_Probability_and_Statistics_(Nam_Lam)/06%3A_The_Normal_Distribution/6.04%3A_The_Sample_Proportion
    Often sampling is done in order to estimate the proportion of a population that has a specific characteristic.
  • https://math.libretexts.org/Bookshelves/Algebra/Advanced_Algebra/03%3A_Solving_Linear_Systems/3.02%3A_Solving_Linear_Systems_with_Two_Variables
    In this section, we review a completely algebraic technique for solving systems, the substitution method11. The idea is to solve one equation for one of the variables and substitute the result into th...In this section, we review a completely algebraic technique for solving systems, the substitution method11. The idea is to solve one equation for one of the variables and substitute the result into the other equation. After performing this substitution step, we are left with a single equation with one variable, which can be solved using algebra.
  • https://math.libretexts.org/Bookshelves/Algebra/Elementary_Algebra_(LibreTexts)/06%3A_Factoring_and_Solving_by_Factoring/6.07%3A_Applications_Involving_Quadratic_Equations
    Recall that a right triangle is a triangle where one of the angles measures \(90\)°. The side opposite of the right angle is the longest side of the triangle and is called the hypotenuse. The Pythagor...Recall that a right triangle is a triangle where one of the angles measures \(90\)°. The side opposite of the right angle is the longest side of the triangle and is called the hypotenuse. The Pythagorean theorem gives us a relationship between the legs and hypotenuse of any right triangle, where \(a\) and \(b\) are the lengths of the legs and \(c\) is the length of the hypotenuse:
  • https://math.libretexts.org/Bookshelves/Algebra/Advanced_Algebra/07%3A_Exponential_and_Logarithmic_Functions/00%3A_Front_Matter/02%3A_InfoPage
    The LibreTexts libraries are Powered by NICE CXOne and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the Californi...The LibreTexts libraries are Powered by NICE CXOne and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot.
  • https://math.libretexts.org/Bookshelves/Algebra/Advanced_Algebra/06%3A_Solving_Equations_and_Inequalities
  • https://math.libretexts.org/Bookshelves/Algebra/Advanced_Algebra/01%3A_Algebra_Fundamentals/1.06%3A_Polynomials_and_Their_Operations
    A polynomial is a special algebraic expression with terms that consist of real number coefficients and variable factors with whole number exponents.
  • https://math.libretexts.org/Bookshelves/Algebra/Advanced_Algebra/03%3A_Solving_Linear_Systems/3.05%3A_Matrices_and_Gaussian_Elimination
    A linear system in upper triangular form can easily be solved using back substitution. The augmented coefficient matrix and Gaussian elimination can be used to streamline the process of solving linear...A linear system in upper triangular form can easily be solved using back substitution. The augmented coefficient matrix and Gaussian elimination can be used to streamline the process of solving linear systems.
  • https://math.libretexts.org/Bookshelves/Algebra/Elementary_Algebra_(LibreTexts)/06%3A_Factoring_and_Solving_by_Factoring/6.01%3A_Introduction_to_Factoring
    Step 3: The GCF of the monomials is the product of the common variable factors and the GCF of the coefficients. The GCF of two or more monomials is the product of the GCF of the coefficients and the c...Step 3: The GCF of the monomials is the product of the common variable factors and the GCF of the coefficients. The GCF of two or more monomials is the product of the GCF of the coefficients and the common variable factors with the smallest power. The surface area of a cylinder is given by the formula \(SA=2πr^{2}+2πrh\), where \(r\) represents the radius of the base and \(h\) is the height of the cylinder.
  • https://math.libretexts.org/Bookshelves/Algebra/Elementary_Algebra_(LibreTexts)/06%3A_Factoring_and_Solving_by_Factoring/6.04%3A_Factoring_Special_Binomials
    \(\begin{aligned} (a+b)(a^{2}-ab+b^{2})&=a^{3}-a^{2}b+ab^{2}-ab^{2}+b^{3} \\ &=a^{3}+b^{3}\quad\color{Cerulean}{\checkmark} \\ (a-b)(a^{2}+ab+b^{2})&=a^{3}+a^{2}b+ab^{2}=a^{2}b-ab^{2}-b^{3} \\ &=a^{3}...\(\begin{aligned} (a+b)(a^{2}-ab+b^{2})&=a^{3}-a^{2}b+ab^{2}-ab^{2}+b^{3} \\ &=a^{3}+b^{3}\quad\color{Cerulean}{\checkmark} \\ (a-b)(a^{2}+ab+b^{2})&=a^{3}+a^{2}b+ab^{2}=a^{2}b-ab^{2}-b^{3} \\ &=a^{3}-b^{3}\quad\color{Cerulean}{\checkmark} \end{aligned}\)

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