A Circle Is the Set of All Points in a Plane That Are Equidistant from a Given Point Called the Center of the Circle
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Analytic Geometry
Guide Study Georgia End-Of-Course Tests Georgia ANALYTIC GEOMETRY TABLE OF CONTENTS INTRODUCTION ...........................................................................................................5 HOW TO USE THE STUDY GUIDE ................................................................................6 OVERVIEW OF THE EOCT .........................................................................................8 PREPARING FOR THE EOCT ......................................................................................9 Study Skills ........................................................................................................9 Time Management .....................................................................................10 Organization ...............................................................................................10 Active Participation ...................................................................................11 Test-Taking Strategies .....................................................................................11 Suggested Strategies to Prepare for the EOCT ..........................................12 Suggested Strategies the Day before the EOCT ........................................13 Suggested Strategies the Morning of the EOCT ........................................13 Top 10 Suggested Strategies during the EOCT .........................................14 TEST CONTENT ........................................................................................................15 -
Models of 2-Dimensional Hyperbolic Space and Relations Among Them; Hyperbolic Length, Lines, and Distances
Models of 2-dimensional hyperbolic space and relations among them; Hyperbolic length, lines, and distances Cheng Ka Long, Hui Kam Tong 1155109623, 1155109049 Course Teacher: Prof. Yi-Jen LEE Department of Mathematics, The Chinese University of Hong Kong MATH4900E Presentation 2, 5th October 2020 Outline Upper half-plane Model (Cheng) A Model for the Hyperbolic Plane The Riemann Sphere C Poincar´eDisc Model D (Hui) Basic properties of Poincar´eDisc Model Relation between D and other models Length and distance in the upper half-plane model (Cheng) Path integrals Distance in hyperbolic geometry Measurements in the Poincar´eDisc Model (Hui) M¨obiustransformations of D Hyperbolic length and distance in D Conclusion Boundary, Length, Orientation-preserving isometries, Geodesics and Angles Reference Upper half-plane model H Introduction to Upper half-plane model - continued Hyperbolic geometry Five Postulates of Hyperbolic geometry: 1. A straight line segment can be drawn joining any two points. 2. Any straight line segment can be extended indefinitely in a straight line. 3. A circle may be described with any given point as its center and any distance as its radius. 4. All right angles are congruent. 5. For any given line R and point P not on R, in the plane containing both line R and point P there are at least two distinct lines through P that do not intersect R. Some interesting facts about hyperbolic geometry 1. Rectangles don't exist in hyperbolic geometry. 2. In hyperbolic geometry, all triangles have angle sum < π 3. In hyperbolic geometry if two triangles are similar, they are congruent. -
Proofs with Perpendicular Lines
3.4 Proofs with Perpendicular Lines EEssentialssential QQuestionuestion What conjectures can you make about perpendicular lines? Writing Conjectures Work with a partner. Fold a piece of paper D in half twice. Label points on the two creases, as shown. a. Write a conjecture about AB— and CD — . Justify your conjecture. b. Write a conjecture about AO— and OB — . AOB Justify your conjecture. C Exploring a Segment Bisector Work with a partner. Fold and crease a piece A of paper, as shown. Label the ends of the crease as A and B. a. Fold the paper again so that point A coincides with point B. Crease the paper on that fold. b. Unfold the paper and examine the four angles formed by the two creases. What can you conclude about the four angles? B Writing a Conjecture CONSTRUCTING Work with a partner. VIABLE a. Draw AB — , as shown. A ARGUMENTS b. Draw an arc with center A on each To be prof cient in math, side of AB — . Using the same compass you need to make setting, draw an arc with center B conjectures and build a on each side of AB— . Label the C O D logical progression of intersections of the arcs C and D. statements to explore the c. Draw CD — . Label its intersection truth of your conjectures. — with AB as O. Write a conjecture B about the resulting diagram. Justify your conjecture. CCommunicateommunicate YourYour AnswerAnswer 4. What conjectures can you make about perpendicular lines? 5. In Exploration 3, f nd AO and OB when AB = 4 units. -
Lesson 3: Rectangles Inscribed in Circles
NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 3 M5 GEOMETRY Lesson 3: Rectangles Inscribed in Circles Student Outcomes . Inscribe a rectangle in a circle. Understand the symmetries of inscribed rectangles across a diameter. Lesson Notes Have students use a compass and straightedge to locate the center of the circle provided. If necessary, remind students of their work in Module 1 on constructing a perpendicular to a segment and of their work in Lesson 1 in this module on Thales’ theorem. Standards addressed with this lesson are G-C.A.2 and G-C.A.3. Students should be made aware that figures are not drawn to scale. Classwork Scaffolding: Opening Exercise (9 minutes) Display steps to construct a perpendicular line at a point. Students follow the steps provided and use a compass and straightedge to find the center of a circle. This exercise reminds students about constructions previously . Draw a segment through the studied that are needed in this lesson and later in this module. point, and, using a compass, mark a point equidistant on Opening Exercise each side of the point. Using only a compass and straightedge, find the location of the center of the circle below. Label the endpoints of the Follow the steps provided. segment 퐴 and 퐵. Draw chord 푨푩̅̅̅̅. Draw circle 퐴 with center 퐴 . Construct a chord perpendicular to 푨푩̅̅̅̅ at and radius ̅퐴퐵̅̅̅. endpoint 푩. Draw circle 퐵 with center 퐵 . Mark the point of intersection of the perpendicular chord and the circle as point and radius ̅퐵퐴̅̅̅. 푪. Label the points of intersection . -
Squaring the Circle a Case Study in the History of Mathematics the Problem
Squaring the Circle A Case Study in the History of Mathematics The Problem Using only a compass and straightedge, construct for any given circle, a square with the same area as the circle. The general problem of constructing a square with the same area as a given figure is known as the Quadrature of that figure. So, we seek a quadrature of the circle. The Answer It has been known since 1822 that the quadrature of a circle with straightedge and compass is impossible. Notes: First of all we are not saying that a square of equal area does not exist. If the circle has area A, then a square with side √A clearly has the same area. Secondly, we are not saying that a quadrature of a circle is impossible, since it is possible, but not under the restriction of using only a straightedge and compass. Precursors It has been written, in many places, that the quadrature problem appears in one of the earliest extant mathematical sources, the Rhind Papyrus (~ 1650 B.C.). This is not really an accurate statement. If one means by the “quadrature of the circle” simply a quadrature by any means, then one is just asking for the determination of the area of a circle. This problem does appear in the Rhind Papyrus, but I consider it as just a precursor to the construction problem we are examining. The Rhind Papyrus The papyrus was found in Thebes (Luxor) in the ruins of a small building near the Ramesseum.1 It was purchased in 1858 in Egypt by the Scottish Egyptologist A. -
Holley GM LS Race Single-Plane Intake Manifold Kits
Holley GM LS Race Single-Plane Intake Manifold Kits 300-255 / 300-255BK LS1/2/6 Port-EFI - w/ Fuel Rails 4150 Flange 300-256 / 300-256BK LS1/2/6 Carbureted/TB EFI 4150 Flange 300-290 / 300-290BK LS3/L92 Port-EFI - w/ Fuel Rails 4150 Flange 300-291 / 300-291BK LS3/L92 Carbureted/TB EFI 4150 Flange 300-294 / 300-294BK LS1/2/6 Port-EFI - w/ Fuel Rails 4500 Flange 300-295 / 300-295BK LS1/2/6 Carbureted/TB EFI 4500 Flange IMPORTANT: Before installation, please read these instructions completely. APPLICATIONS: The Holley LS Race single-plane intake manifolds are designed for GM LS Gen III and IV engines, utilized in numerous performance applications, and are intended for carbureted, throttle body EFI, or direct-port EFI setups. The LS Race single-plane intake manifolds are designed for hi-performance/racing engine applications, 5.3 to 6.2+ liter displacement, and maximum engine speeds of 6000-7000 rpm, depending on the engine combination. This single-plane design provides optimal performance across the RPM spectrum while providing maximum performance up to 7000 rpm. These intake manifolds are for use on non-emissions controlled applications only, and will not accept stock components and hardware. Port EFI versions may not be compatible with all throttle body linkages. When installing the throttle body, make certain there is a minimum of ¼” clearance between all linkage and the fuel rail. SPLIT DESIGN: The Holley LS Race manifold incorporates a split feature, which allows disassembly of the intake for direct access to internal plenum and port surfaces, making custom porting and matching a snap. -
P. 1 Math 490 Notes 7 Zero Dimensional Spaces for (SΩ,Τo)
p. 1 Math 490 Notes 7 Zero Dimensional Spaces For (SΩ, τo), discussed in our last set of notes, we can describe a basis B for τo as follows: B = {[λ, λ] ¯ λ is a non-limit ordinal } ∪ {[µ + 1, λ] ¯ λ is a limit ordinal and µ < λ}. ¯ ¯ The sets in B are τo-open, since they form a basis for the order topology, but they are also closed by the previous Prop N7.1 from our last set of notes. Sets which are simultaneously open and closed relative to the same topology are called clopen sets. A topology with a basis of clopen sets is defined to be zero-dimensional. As we have just discussed, (SΩ, τ0) is zero-dimensional, as are the discrete and indiscrete topologies on any set. It can also be shown that the Sorgenfrey line (R, τs) is zero-dimensional. Recall that a basis for τs is B = {[a, b) ¯ a, b ∈R and a < b}. It is easy to show that each set [a, b) is clopen relative to τs: ¯ each [a, b) itself is τs-open by definition of τs, and the complement of [a, b)is(−∞,a)∪[b, ∞), which can be written as [ ¡[a − n, a) ∪ [b, b + n)¢, and is therefore open. n∈N Closures and Interiors of Sets As you may know from studying analysis, subsets are frequently neither open nor closed. However, for any subset A in a topological space, there is a certain closed set A and a certain open set Ao associated with A in a natural way: Clτ A = A = \{B ¯ B is closed and A ⊆ B} (Closure of A) ¯ o Iτ A = A = [{U ¯ U is open and U ⊆ A}. -
Construction Surveying Curves
Construction Surveying Curves Three(3) Continuing Education Hours Course #LS1003 Approved Continuing Education for Licensed Professional Engineers EZ-pdh.com Ezekiel Enterprises, LLC 301 Mission Dr. Unit 571 New Smyrna Beach, FL 32170 800-433-1487 [email protected] Construction Surveying Curves Ezekiel Enterprises, LLC Course Description: The Construction Surveying Curves course satisfies three (3) hours of professional development. The course is designed as a distance learning course focused on the process required for a surveyor to establish curves. Objectives: The primary objective of this course is enable the student to understand practical methods to locate points along curves using variety of methods. Grading: Students must achieve a minimum score of 70% on the online quiz to pass this course. The quiz may be taken as many times as necessary to successful pass and complete the course. Ezekiel Enterprises, LLC Section I. Simple Horizontal Curves CURVE POINTS Simple The simple curve is an arc of a circle. It is the most By studying this course the surveyor learns to locate commonly used. The radius of the circle determines points using angles and distances. In construction the “sharpness” or “flatness” of the curve. The larger surveying, the surveyor must often establish the line of the radius, the “flatter” the curve. a curve for road layout or some other construction. The surveyor can establish curves of short radius, Compound usually less than one tape length, by holding one end Surveyors often have to use a compound curve because of the tape at the center of the circle and swinging the of the terrain. -
Molecular Symmetry
Molecular Symmetry Symmetry helps us understand molecular structure, some chemical properties, and characteristics of physical properties (spectroscopy) – used with group theory to predict vibrational spectra for the identification of molecular shape, and as a tool for understanding electronic structure and bonding. Symmetrical : implies the species possesses a number of indistinguishable configurations. 1 Group Theory : mathematical treatment of symmetry. symmetry operation – an operation performed on an object which leaves it in a configuration that is indistinguishable from, and superimposable on, the original configuration. symmetry elements – the points, lines, or planes to which a symmetry operation is carried out. Element Operation Symbol Identity Identity E Symmetry plane Reflection in the plane σ Inversion center Inversion of a point x,y,z to -x,-y,-z i Proper axis Rotation by (360/n)° Cn 1. Rotation by (360/n)° Improper axis S 2. Reflection in plane perpendicular to rotation axis n Proper axes of rotation (C n) Rotation with respect to a line (axis of rotation). •Cn is a rotation of (360/n)°. •C2 = 180° rotation, C 3 = 120° rotation, C 4 = 90° rotation, C 5 = 72° rotation, C 6 = 60° rotation… •Each rotation brings you to an indistinguishable state from the original. However, rotation by 90° about the same axis does not give back the identical molecule. XeF 4 is square planar. Therefore H 2O does NOT possess It has four different C 2 axes. a C 4 symmetry axis. A C 4 axis out of the page is called the principle axis because it has the largest n . By convention, the principle axis is in the z-direction 2 3 Reflection through a planes of symmetry (mirror plane) If reflection of all parts of a molecule through a plane produced an indistinguishable configuration, the symmetry element is called a mirror plane or plane of symmetry . -
And Are Congruent Chords, So the Corresponding Arcs RS and ST Are Congruent
9-3 Arcs and Chords ALGEBRA Find the value of x. 3. SOLUTION: 1. In the same circle or in congruent circles, two minor SOLUTION: arcs are congruent if and only if their corresponding Arc ST is a minor arc, so m(arc ST) is equal to the chords are congruent. Since m(arc AB) = m(arc CD) measure of its related central angle or 93. = 127, arc AB arc CD and . and are congruent chords, so the corresponding arcs RS and ST are congruent. m(arc RS) = m(arc ST) and by substitution, x = 93. ANSWER: 93 ANSWER: 3 In , JK = 10 and . Find each measure. Round to the nearest hundredth. 2. SOLUTION: Since HG = 4 and FG = 4, and are 4. congruent chords and the corresponding arcs HG and FG are congruent. SOLUTION: m(arc HG) = m(arc FG) = x Radius is perpendicular to chord . So, by Arc HG, arc GF, and arc FH are adjacent arcs that Theorem 10.3, bisects arc JKL. Therefore, m(arc form the circle, so the sum of their measures is 360. JL) = m(arc LK). By substitution, m(arc JL) = or 67. ANSWER: 67 ANSWER: 70 eSolutions Manual - Powered by Cognero Page 1 9-3 Arcs and Chords 5. PQ ALGEBRA Find the value of x. SOLUTION: Draw radius and create right triangle PJQ. PM = 6 and since all radii of a circle are congruent, PJ = 6. Since the radius is perpendicular to , bisects by Theorem 10.3. So, JQ = (10) or 5. 7. Use the Pythagorean Theorem to find PQ. -
Circle Theorems
Circle theorems A LEVEL LINKS Scheme of work: 2b. Circles – equation of a circle, geometric problems on a grid Key points • A chord is a straight line joining two points on the circumference of a circle. So AB is a chord. • A tangent is a straight line that touches the circumference of a circle at only one point. The angle between a tangent and the radius is 90°. • Two tangents on a circle that meet at a point outside the circle are equal in length. So AC = BC. • The angle in a semicircle is a right angle. So angle ABC = 90°. • When two angles are subtended by the same arc, the angle at the centre of a circle is twice the angle at the circumference. So angle AOB = 2 × angle ACB. • Angles subtended by the same arc at the circumference are equal. This means that angles in the same segment are equal. So angle ACB = angle ADB and angle CAD = angle CBD. • A cyclic quadrilateral is a quadrilateral with all four vertices on the circumference of a circle. Opposite angles in a cyclic quadrilateral total 180°. So x + y = 180° and p + q = 180°. • The angle between a tangent and chord is equal to the angle in the alternate segment, this is known as the alternate segment theorem. So angle BAT = angle ACB. Examples Example 1 Work out the size of each angle marked with a letter. Give reasons for your answers. Angle a = 360° − 92° 1 The angles in a full turn total 360°. = 268° as the angles in a full turn total 360°. -
Descriptive Geometry Section 10.1 Basic Descriptive Geometry and Board Drafting Section 10.2 Solving Descriptive Geometry Problems with CAD
10 Descriptive Geometry Section 10.1 Basic Descriptive Geometry and Board Drafting Section 10.2 Solving Descriptive Geometry Problems with CAD Chapter Objectives • Locate points in three-dimensional (3D) space. • Identify and describe the three basic types of lines. • Identify and describe the three basic types of planes. • Solve descriptive geometry problems using board-drafting techniques. • Create points, lines, planes, and solids in 3D space using CAD. • Solve descriptive geometry problems using CAD. Plane Spoken Rutan’s unconventional 202 Boomerang aircraft has an asymmetrical design, with one engine on the fuselage and another mounted on a pod. What special allowances would need to be made for such a design? 328 Drafting Career Burt Rutan, Aeronautical Engineer Effi cient travel through space has become an ambi- tion of aeronautical engineer, Burt Rutan. “I want to go high,” he says, “because that’s where the view is.” His unconventional designs have included every- thing from crafts that can enter space twice within a two week period, to planes than can circle the Earth without stopping to refuel. Designed by Rutan and built at his company, Scaled Composites LLC, the 202 Boomerang aircraft is named for its forward-swept asymmetrical wing. The design allows the Boomerang to fl y faster and farther than conventional twin-engine aircraft, hav- ing corrected aerodynamic mistakes made previously in twin-engine design. It is hailed as one of the most beautiful aircraft ever built. Academic Skills and Abilities • Algebra, geometry, calculus • Biology, chemistry, physics • English • Social studies • Humanities • Computer use Career Pathways Engineers should be creative, inquisitive, ana- lytical, detail oriented, and able to work as part of a team and to communicate well.