Example applications of theorems
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Given that m∠2 = 47°, the document finds the measures of the other angles. It states that ∠2 and ∠4 are vertical angles, so m∠4 = 47°. It also states that ∠2 and ∠3 are a linear pair, so m∠3 = 180° - 47° = 133°. By the same logic, m∠1 = 133°.
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This document discusses parallel lines and transversals. It defines key terms like corresponding angles, alternate interior angles, alternate exterior angles, and consecutive interior angles. It presents theorems about angle relationships that are created when parallel lines are cut by a transversal, such as corresponding angles being congruent and consecutive interior angles being supplementary. Examples are provided to demonstrate applying these concepts and theorems to find missing angle measures. Students are assigned practice problems to reinforce their understanding of using parallel lines and transversals to solve for unknown angle measures.
3.2 use parallel lines and transversals
This document discusses parallel lines and transversals. It provides examples of using theorems about alternate interior angles, corresponding angles, vertical angles, and consecutive interior angles to solve for unknown angle measures. Theorems demonstrated include the alternate interior angles theorem, vertical angles congruence theorem, and consecutive interior angles theorem. Students are guided through practice problems applying these theorems to find missing angle measures given information about parallel lines cut by a transversal.
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e
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A parallelogram is a quadrilateral where opposite sides are equal and opposite angles are equal, and a diagonal of a parallelogram divides it into two congruent triangles.
Unidad 2 paso 3
This document discusses trigonometric identities. It describes the basic types of identities such as basic identities, sum and difference identities, double angle identities, and half angle identities. It also discusses product-sum and sum-product identities. Examples of specific identities are given such as the Pythagorean identity, reciprocal identities, and even-odd identities. Methods for deriving identities from angle addition or subtraction are explained. Several examples of using identities to solve for trigonometric functions of specific angles are provided.
Variation revision card
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Regression analysis is a mathematical measure of the average relationship between two or more variables in terms of the original units of the data.
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transforming equation to slope intercept form.pptx
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5) Equations
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This document contains information from a mathematics teacher training on topics related to similar polygons and triangles. It includes:
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Digital text book
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Integers and matrices (slides)
This document discusses several key concepts regarding integers:
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2) It defines prime numbers as integers greater than 1 that are only divisible by 1 and themselves. It provides an algorithm for determining if a number is prime.
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POTENCIAS Y RAÍCES DE NÚMEROS COMPLEJOS-LAPTOP-3AN2F8N2.pptx
The document discusses complex numbers and their properties. It covers representing complex numbers in polar form, exponential form, products and powers of complex numbers, and finding square roots of complex numbers. Some key points include:
1. A complex number z can be written in polar form as z = r(cosθ + i sinθ) where r is the modulus and θ is the argument.
2. Complex numbers can also be written in exponential form as z = reiθ using Euler's formula.
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4. Taking powers
Expresiones algebraicas
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Neo_Phase_1_Vectors_HN_S10_PPT.pptx
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GCSE-Vectors.pptx
Vectors can represent both positions and movements. Positions use coordinates (x,y) while movements use vectors with the change in x and y written vertically. Vectors can be added by adding the x and y components. Vectors can also be scaled by multiplying each component by a scalar value. Exam questions often involve finding a vector in part (a) and then using it to find another vector in part (b).
Semana 13 ecuaciones polinomiales ii álgebra-uni ccesa007
The document discusses cubic equations and their applications. It provides examples of solving cubic equations by factorizing them into linear factors using the rational root theorem or Cardano's formula. The key steps are factorizing the equation, setting each factor equal to zero to find the roots, and determining the number of solutions. The document also presents theorems regarding the relationship between the number of roots and solutions, and the sums and products of the roots.
Lecture-4 Reduction of Quadratic Form.pdf
The document discusses reducing a quadratic form to canonical form using an orthogonal transformation. It begins by defining quadratic forms and representing them using matrices. It then provides examples of writing the matrix and quadratic form for functions of 2 and 3 variables. The document explains finding the eigenvectors and eigenvalues of the coefficient matrix to form an orthogonal transformation matrix B. Premultiplying the coefficient matrix A by the inverse of B results in the diagonal canonical form matrix D, where the diagonal elements are the eigenvalues of A. The quadratic form is then in canonical (sum of squares) form. An example problem demonstrates reducing a 3D quadratic form to canonical form using this process.
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transforming equation to slope intercept form.pptx
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The document shows the steps to solve the system of equations 5x - y = 5 and 4x - y = 3 by substitution. It involves solving the first equation for y in terms of x, substituting this expression for y into the second equation, solving the resulting equation for x, and then substituting x back into the original equation to solve for y. The solution is the ordered pair (2,5).
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The document shows the step-by-step workings of two base conversions:
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This document explains how to write numbers in expanded form. It shows working through writing 472 and 12,357 in expanded form. To write a number in expanded form, you start with the rightmost digit and multiply it by increasing powers of 10 as you move left. So 472 in expanded form is (7 × 101) + (4 × 102) + (2 × 100).
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This document provides step-by-step instructions for dividing fractions. It shows dividing 2/3 by 6/7, which involves inverting the second fraction and multiplying. The resulting fraction is then simplified by factoring out common factors to get 1/3. A second example divides -5/8 by 4/5, again inverting the second fraction and multiplying before giving the final simplified fraction of -25/32.
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The document provides steps for multiplying fractions. It shows:
1) Multiplying 2/3 by 7/16 by finding a common factor of 2 in the denominators and simplifying to get 7/24.
2) Multiplying 1 3/4 by 2 1/2 by first converting to improper fractions, then multiplying numerators and denominators to get 35/8.
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The document provides steps to find the prime factors of 198 using the division method. It is first divided by 2, then the resulting number 99 is divided by 3 to get 33, which is then divided by 3 to get 11. Since 11 is prime, the prime factors of 198 are 2, 3, and 11.
Branching Example
This document outlines the steps to use the branching method to find the prime factors of 315. It explains that 315 is divisible by 5, 3, and 9, so the first step is to divide 315 by 15 to get 21. Then 21 is divided by 3 and 7, the prime factors of 21. The prime factors of 315 are then expressed as the product 3 x 3 x 5 x 7.
Section 3-1: Congruent Triangles
Two triangles are congruent if their corresponding six parts (three angles and three sides) are congruent. Congruence can also be described as one triangle coinciding perfectly with the other after rotation or flipping. In the figure, triangle ABC is congruent to triangle DEF as demonstrated by their corresponding angles and sides being congruent. Congruence of triangles is an equivalence relation that is reflexive, symmetric, and transitive.
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Example applications of theorems
- 1. Given: 𝑚∠2 = 47 𝑜 Find: 𝑚∠1, 𝑚∠3, and 𝑚∠4 12 3 4 m n
- 2. Given: 𝑚∠2 = 47 𝑜 Find: 𝑚∠1, 𝑚∠3, and 𝑚∠4 ∠2 and ∠4 are vertical angles and therefore congruent 𝑚∠4 = 𝑚∠2 = 47 𝑜 m n 12 3 4
- 3. Given: 𝑚∠2 = 47 𝑜 Find: 𝑚∠1, 𝑚∠3, and 𝑚∠4 ∠2 and ∠4 are vertical angles and therefore congruent 𝑚∠4 = 𝑚∠2 = 47 𝑜 ∠2 and ∠3 form a linear pair, so they are supplementary 𝑚∠3 = 180 𝑜 − 𝑚∠2 = 133 𝑜 m n 12 3 4
- 4. Given: 𝑚∠2 = 47 𝑜 Find: 𝑚∠1, 𝑚∠3, and 𝑚∠4 ∠2 and ∠4 are vertical angles and therefore congruent 𝑚∠4 = 𝑚∠2 = 47 𝑜 ∠2 and ∠3 form a linear pair, so they are supplementary 𝑚∠3 = 180 𝑜 − 𝑚∠2 = 133 𝑜 ∠3 and ∠1 are vertical angles and therefore congruent 𝑚∠1 = 𝑚∠3 = 133 𝑜 m n 12 3 4
- 5. Given: 𝑚∠2 = 𝑥 + 15 and 𝑚∠3 = 2𝑥 Find: 𝑥 and 𝑚∠2 12 3 4 m n
- 6. Given: 𝑚∠2 = 𝑥 + 15 and 𝑚∠3 = 2𝑥 Find: 𝑥 and 𝑚∠2 ∠2 and ∠3 form a linear pair, so they are supplementary 𝑚∠2 + 𝑚∠3 = 180 𝑜 12 3 4 m n
- 7. Given: 𝑚∠2 = 𝑥 + 15 and 𝑚∠3 = 2𝑥 Find: 𝑥 and 𝑚∠2 ∠2 and ∠3 form a linear pair, so they are supplementary 𝑚∠2 + 𝑚∠3 = 180 𝑜 Substituting with the given expressions: 𝑥 + 15 + 2𝑥 = 180 𝑜 12 3 4 m n
- 8. Given: 𝑚∠2 = 𝑥 + 15 and 𝑚∠3 = 2𝑥 Find: 𝑥 and 𝑚∠2 ∠2 and ∠3 form a linear pair, so they are supplementary 𝑚∠2 + 𝑚∠3 = 180 𝑜 Substituting with the given expressions: 𝑥 + 15 + 2𝑥 = 180 𝑜 Simplify: 3𝑥 + 15 = 180 𝑜 12 3 4 m n
- 9. Given: 𝑚∠2 = 𝑥 + 15 and 𝑚∠3 = 2𝑥 Find: 𝑥 and 𝑚∠2 ∠2 and ∠3 form a linear pair, so they are supplementary 𝑚∠2 + 𝑚∠3 = 180 𝑜 Substituting with the given expressions: 𝑥 + 15 + 2𝑥 = 180 𝑜 Simplify: 3𝑥 + 15 = 180 𝑜 Solve for x: 3𝑥 = 165 𝑜 12 3 4 m n
- 10. Given: 𝑚∠2 = 𝑥 + 15 and 𝑚∠3 = 2𝑥 Find: 𝑥 and 𝑚∠2 ∠2 and ∠3 form a linear pair, so they are supplementary 𝑚∠2 + 𝑚∠3 = 180 𝑜 Substituting with the given expressions: 𝑥 + 15 + 2𝑥 = 180 𝑜 Simplify: 3𝑥 + 15 = 180 𝑜 Solve for x: 3𝑥 = 165 𝑜 𝑥 = 55 𝑜 12 3 4 m n
- 11. Given: 𝑚∠2 = 𝑥 + 15 and 𝑚∠3 = 2𝑥 Find: 𝑥 and 𝑚∠2 ∠2 and ∠3 form a linear pair, so they are supplementary 𝑚∠2 + 𝑚∠3 = 180 𝑜 Substituting with the given expressions: 𝑥 + 15 + 2𝑥 = 180 𝑜 Simplify: 3𝑥 + 15 = 180 𝑜 Solve for x: 3𝑥 = 165 𝑜 𝑥 = 55 𝑜 Substitute to find 𝑚∠2: 𝑚∠2 = 55 + 15 12 3 4 m n
- 12. Given: 𝑚∠2 = 𝑥 + 15 and 𝑚∠3 = 2𝑥 Find: 𝑥 and 𝑚∠2 ∠2 and ∠3 form a linear pair, so they are supplementary 𝑚∠2 + 𝑚∠3 = 180 𝑜 Substituting with the given expressions: 𝑥 + 15 + 2𝑥 = 180 𝑜 Simplify: 3𝑥 + 15 = 180 𝑜 Solve for x: 3𝑥 = 165 𝑜 𝑥 = 55 𝑜 Substitute to find 𝑚∠2: 𝑚∠2 = 55 + 15 𝑚∠2 = 70 𝑜 12 3 4 m n