Integrated Mathematics 3 NC (9-12)

Curriculum for the high school grades (9-12) for the state of North Carolina (US) : Curriculum for the high school grades (9-12) for the state of North Carolina (US)

Objectives : Objectives This module explains the curriculum for the high school grades (9-12) for the state of North Carolina (US) Country: United States State: North Carolina Time Zone: Eastern Standard Time (GMT-5) till 11th March 2007 Subject: Mathematics Topic: Integrated Mathematics 3 Language: English Grade: Grade 9-12

Introduction : Students will be expected to Describe and translate among graphic, algebraic, numeric, tabular, and verbal representations of relationships. Use those representations to solve problems. Appropriate technology, from manipulatives to calculators and application Software, should be used regularly for instruction and assessment. Introduction

Contents : Integrated Mathematics 3 includes: Continuation of study of topics from algebra, geometry, and statistics in a problem-centered, connected approach. Functions and the deductive methods of proof with geometric concepts are the principle topics of study. Contents

Pre-requisites : What your student must already know? Use the trigonometric ratios to model and solve problems. Apply geometric properties and relationships to solve problems. Use systems of linear equations or inequalities to model and solve problems. Define and use linear and exponential functions to model and solve problems. Pre-requisites

Competency Goal 1 : Competency Goal 1 1.01 Write equivalent forms of algebraic expressions to solve problems. Sample Problem: Solution: There were 43 snowy days last winter. This was 15 less than the number of snowy days during the winter the year before. Write an equation that models this situation. Words The number of snowy days last winter was 15 less than the number of snowy days during the winter the year before. Variable Let s represent the number of snowy days the year before. Equation 43 = s – 15

Competency Goal 1 : Competency Goal 1 1.02 Use algebraic expressions, including iterative and recursive forms, to model and solve problems. Sample Problem: Solution: Find the first three iterates x1, x2, x3 of the function f(x) = x 2 – 1 for an initial value of x0 = –2. To find the first iterate x1, find the value of the function for x0 = –2. x1 = f(x0) Iterate the function. = f(–2) x0 = –2 = (–2)2 – 1 or 3 Simplify. To find the second iterate x2, substitute x1 for x. x2 = f(x1) Iterate the function. = f(3) x1 = 3 = (3)2 – 1 or 8 Simplify.

Competency Goal 1 : Substitute x2 for x to find the third iterate. x3 = f(x2) Iterate the function. = f(8) x2 = 8 = (8)2 – 1 or 63 Simplify. Therefore, –2, 3, 8, 63 is an example of a sequence generated using iteration. Competency Goal 1 Solution: (contd.)

Competency Goal 1 : Competency Goal 1 1.03 Simplify and perform operations with rational exponents and logarithms to solve problems. Sample Problem: Solution: If 4 = logx 256, find the value of x. According to the logarithm law, 4 = logx 256 can be expressed as, x4 = 256 or, x4 = 44 or, x = 4

Competency Goal 1 : 1.04 Model and solve problems using direct, inverse, combined and joint variation. Sample Problem: Competency Goal 1 Solution: If A varies as B & C jointly, and if A = 2, when B = 3/5, C = 10/27; find C when A = 54 and B = 3.

Competency Goal 2 : Competency Goal 2 Sample Problem: 2.01 Use logic and deductive reasoning to draw conclusions and solve problems. Use the Law of Syllogism to determine a conclusion that follows from statements (1) and (2). (1) If it is raining, then I will not go on a picnic. (2) If I do not go on a picnic, then I will finish my homework. Let p, q, and r represent the parts of the statements. p: it is raining q: I will not go on a picnic r: I will finish my homework Use the Law of Syllogism to conclude p  r. Therefore, if it is raining, then I will finish my homework. Solution:

Competency Goal 2 : Competency Goal 2 Sample Problem 2.02 Apply the properties, definitions, and theorems of angles and lines to solve problems and write proofs. If EN bisects DEF and mDEF = 102, find m3 and m4. Solution: Since EN bisects DEF, m3 = m4. m3 + m4 = mDEF Postulate 3-3 m3 + m4 = 102 Replace mDEF with 102. m3 + m3 = 102 Replace m4 with m3. 2(m3) = 102 Combine like terms. m3 = 51 Divide each side by 2. Since m3 = m4, m4 = 51.

Competency Goal 2 : Competency Goal 2 Sample Problem: 2.03 Apply the properties, definitions, and theorems of two-dimensional figures to solve problems and write proofs: a) Triangles. b) Quadrilaterals. c) Other polygons. d) Circles. Justify the steps for the proof of the conditional. If mACG = 90, mCGI = 90, and mECG = mCGK, then mACE = mIGK. Given: mACG = 90 mCGI = 90 mECG = mCGK Prove: mACE = mIGK

Competency Goal 2 : Competency Goal 2 5. ? 5. mACE = mIGK 4. ? 4. mACE + mECG = mIGK + mCGK 3. ? 3. mACG = mACE + mECG mCGI = mIGK + mCGK 2. ? 2. mACG = mCGI 1. ? 1. mACG = 90 mCGI = 90 mECG = mCGK Reasons Statements Reason 1: Given Reason 2: Substitution Property of Equality Reason 3: Angle Addition Postulate Reason 4: Substitution Property of Equality Reason 5: Subtraction Property of Equality Solution:

Competency Goal 2 : Competency Goal 2 Sample Problem 2.04 Use the law of sines and law of cosines to solve problems. Show that (sec A + tan A)(1 – sin A) = cos A Solution:

Competency Goal 3 : 3.01 Use systems of two or more equations or inequalities to model and solve problems; justify results. Solve using tables, graphs, matrix operations, and algebraic properties. Sample Problem: Competency Goal 3 Solution: Use elimination to solve the system of equations. 2x - 5y = 13 -4x + 3y = -12 Multiply the first equation by 2 so the x terms are additive inverses. 4x - 10y = 26 (+) - 4x + 3y = - 12 0 - 7y = 14 -7y = 14 y = - 2 Multiply by 2. 2x - 5y = 13 -4x + 3y = -12

Competency Goal 3 : Competency Goal 3 Solution: (contd.)

Competency Goal 3 : 3.02 Use quadratic functions and inequalities to model and solve problems; justify results. a) Solve using tables, graphs, and algebraic properties. b) Interpret the constants and coefficients in the context of the problem. Sample Problem: Competency Goal 3 The length  of a rectangle is 10 feet less than twice its width w. The area of the rectangle is 300 square inches. Find the measures of the sides. The formula for the area of a rectangle is A = w and  = 2w - 10. 300 = (2w – 10)w or, 300 = 2w2 - 10w 0 = 2w2 - 10w – 300 or, 0 = 2(w2 - 5w - 150) 0 = w2 - 5w – 150 or, 0 = (w - 15)(w + 10) w – 15 = 0 or, w + 10 = 0 w = 15 or, w = -10 Since width cannot be negative, the width must be equal to 15 feet. Substituting 15 for w, the length of the rectangle is 2(15) - 10 or 20 feet. Solution:

Competency Goal 3 : 3.03 Use rational equations to model and solve problems; justify results. a) Solve using tables, graphs, and algebraic properties. b) Interpret the constants and coefficients in the context of the problem. c) Identify the asymptotes and intercepts graphically and algebraically. Sample Problem: Competency Goal 3 Graph the reciprocal function, 1/x, and show the vertical asymptote of x = 0 and horizontal asymptote of y= 0. Solution:

Competency Goal 3 : 3.04 Use equations and inequalities with absolute value to model and solve problems; justify results. a) Solve using tables, graphs, and algebraic properties. b) Interpret the constants and coefficients in the context of the problem. Sample Problem: Competency Goal 3 Check if following in-equations have solution or not. 3y > -x + 6 2x + 6y < -18 Solution: The graphs of 3y = -x + 6 and 2x + 6y = -18 are parallel lines. Because the regions in the solution of 3y > -x + 6 (the blue region) and 2x + 6y < -18 (the yellow region) have no points in common, the system of inequalities has no solution.

Competency Goal 3 : 3.05 Transform functions in two dimensions; describe the results algebraically and geometrically. Sample Problem: Competency Goal 3 O(0,0) Relation between two quantities is given in the table, write the equation best suits with the given relations. 9 -9 18 18 7 -7 3.5 3.5 1 -1 1 1 y x Solution: As we find that for any positive value of x the value of y is same. And for negative value of x the value of y is numerically equal but the sign is positive. Therefore the equation of the given data can be written as y = I x I. The graph shows the nature of the line.