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Integrated Mathematics 2 NC (9-12)

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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)

Slide2 :

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 2 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 2 continues: The study of topics from algebra, geometry, and statistics in a problem-centered, connected approach. Functions, matrix operations, and algebraic representations of geometric concepts are the principle topics of study. Contents

Pre-requisites : What your student must already know? Use linear expressions to model and solve problems. Collect, organize, analyze, and display data to solve problems. Apply geometric properties and relationships to solve problems. Apply the Pythagorean Theorem to solve problems. Pre-requisites

Competency Goal 1 : Competency Goal 1 1.01 Write equivalent forms of algebraic expressions to solve problems. Sample Problems: Solution: The opossum sleeps an average of 19 hours then the it sleeps for 19  7 or 19(7) hours in 7 days. Write a numerical expression to represent the number of hours an opossum sleeps in 7 days. An opossum sleeps an average of 19 hours per day.

Competency Goal 1 : Competency Goal 1 Sample Problems: Solution: Mary is three times as old as Sam. Five years later, Mary will be two and a half times as old as Sam. How old are Mary and Sam now? age will be y + 5 and then x + 5 = ( y + 5) Let Mary is x years old and Sam is years old. According to the question, x = 3y or, x – 3y = 0 --------(1) After five years Mary’s age will be x + 5 and Sam’s or, x+5 = (y + 5) or, 2x + 10 = 5y + 25 or, 2x -5y = 15 ------------------------------------(2) Multiplying equation no. (1) by 2 and subtracting from equation no. (2) we get, y = 15 putting y = 15 in equation number (1) we get, x = 3.(15) = 45 Therefore we get, Mary is 45 years old and Sam is 15 years old.

Competency Goal 1 : 1.02 Use algebraic expressions, including iterative and recursive forms, to model and solve problems. Sample Problem: Competency Goal 1 Solution: Find the first five terms of the sequence in which a1 = –5 and an + 1 = –2an – n + 5, n 1 an+ 1 = –2an – n + 5 Recursive formula a1+ 1 = –2a1 – (1) + 5 n = 1 a2 = –2(–5) – (1) + 5 or 14 a1 = –5 a2+ 1 = –2a2 – (2) + 5 n = 2 a3 = –2(14) – (2) + 5 or –25 a2 = 14 a3+ 1 = –2a3 – (3) + 5 n = 3 a4 = –2(–25) – (3) + 5 or 52 a3 = –25 a4+ 1 = –2a4 – (4) + 5 n = 4 a5 = –2(52) – (4) + 5 or –103 a4 = 52 The first five terms of the sequence are –5, 14, -25, 52, and –103.

Competency Goal 1 : Sample Problem: Competency Goal 1 Seven dozen oranges cost $56. Find the cost of 10 dozen oranges. Solution: The given information can be written as follows- X 56 Amount spent in dollars 10 7 Number of oranges in dozen. Clearly this is a case of direct variation. As the number of oranges increases, the amount spent also increases in the same ratio. Therefore, or, 7x = 560 or, x = or, x = 80 i.e cost of 10 dozen oranges is $ 80. 1.03 Model and solve problems using direct variation.

Competency Goal 1 : Sample Problem: Competency Goal 1 With two given numbers, when seven times the second number is subtracted from five times the first number we get 2. When five times the second number is subtracted from seven times the first number, the result is 3. Using matrices find the numbers. Solution: 1.04 Operate with matrices to model and solve problems.

Competency Goal 2 : Competency Goal 2 Sample Problem: 2.01 Find the lengths and midpoints of segments to solve problems. AB is diameter of a circle whose centre is (2, 5). If the coordinates of A are (- 4, 6) find the coordinates of B. Solution: Let the coordinate of B be (x, y) Hence C is the mid-point of line segment AB, where A = (4, -6) and C = (2, 5) Therefore, 2 = & 5 = or, 4 = -4 +x & 10 = 6 + y or, x = 0 & y = 4 Therefore, the coordinate of B is (0, 4)

Competency Goal 2 : Competency Goal 2 Sample Problem 2.02 Use the parallelism or perpendicularity of lines and segments to solve problems. In the figure, PS and QR are parallel lines , and q is a transversal. If m4 = 108, find m3, m5, and m6. 4 and 5 are consecutive interior angles, so by Theorem 4-2 they are supplementary. m4 + m5 = 180 or, 108 + m5 = 180 [as m4 = 108.] 108 - 108 + m5 = 180 – 108 Subtract 108 from each side. m5 = 72 4 and 6 are alternate interior angles, so by Theorem 4-1 they are congruent. Therefore, m6 = 108. 3 and 5 are alternate interior angles, so by Theorem 4-1 they are congruent. Therefore, m3 = 72. Solution:

Competency Goal 2 : Competency Goal 2 Sample Problem 2.03 Use the trigonometric ratios to model and solve problems. Yvette is visiting Toronto, Canada. She sights the top of the Canadian National Tower as shown in the diagram. Find the height of the tower. Round to the nearest foot. Solution:

Competency Goal 2 : Competency Goal 2 Sample Problem 2.04 Describe the transformation (translation, reflection, rotation, dilation) of polygons in the coordinate plane in simple algebraic terms. Miranda needs to enlarge a design for a project she is doing. The design consists of two triangles. Triangle ABC has vertices A(1, 2), B(2, 3), and C(2, 1). Triangle DEF has vertices D(2, 2), E(3, 3), and F(3, 1) She wants each side of the two joined triangles to be 3 times the length of the original sides. Find the coordinates of the dilation image of each vertex of the triangles, and graph the dilation image of the design. Solution: Since k > 1, this is an enlargement. To find the dilation image, multiply each coordinate in the ordered pairs by 3.

Competency Goal 2 : Competency Goal 2 Solution: (contd.)

Competency Goal 3 : 3.01 Describe data to solve problems. a) Apply and compare methods of data collection. b) Apply statistical principles and methods in sample surveys. c) Determine measures of central tendency and spread. d) Recognize, define, and use the normal distribution curve. e) Interpret graphical displays of data. f) Compare distributions of data. Competency Goal 3 Sample Problem: Find the measure of central tendency (mean, median & mode) for the given data: 22, 18, 24, 32, 24, 18. Solution: 6 138

Competency Goal 3 : 3.02 Create and use, for sets of data, calculator-generated models of linear, exponential, and quadratic functions to solve problems. a) Interpret the constants, coefficients, and bases in the context of the data. b) Check the model for goodness-of-fit and use the model, where appropriate, to draw conclusions or make predictions. Sample Problem: Competency Goal 3 The air temperatures at various heights on a mountain are given. Make a scatter plot using the data. Then draw a line that seems to best represent the data. 50 55 59 64 67 70 75 Temperature (F) 6,000 5,000 4,000 3,000 2,000 1,000 0 Mountain Height (feet)

Competency Goal 3 : Competency Goal 3 50 55 59 64 67 70 75 Temperature (F) 6,000 5,000 4,000 3,000 2,000 1,000 0 Mountain Height (feet) Solution: Each of the data points are graphed. A line that best fits the data is drawn.

Competency Goal 4 : Competency Goal 4 4.01 Use systems of linear equations or inequalities in two variables to model and solve problems. Solve using tables, graphs, and algebraic properties; justify steps used. Sample Problems: 37 books and 53 copies cost $320, while 53 books and 37 copies cost $ 400. Find the cost of a book and a copy. Let the cost of a book and a copy be x and y respectively. Then, 37x + 53y = 320 -------------(1) 53x + 37y = 400 -------------(2) Adding equations (1) & (2) we get, 90x + 90y = 720 or, x + y = 8 ---------(3) Subtracting (1) from (2) we get, 16x – 16y = 80 or, x – y = 5 ----------(4) Adding (3) & (4) we get, 2x = 13 or, x = 6.5 Substituting x = 6.5 in equation (3) we get, y = (8 – 6.5) = 1.5 Therefore, cost of a book is $6.5 and cost of a copy is $1.5 Solution:

Competency Goal 4 : Competency Goal 4 4.02 Use quadratic functions 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 Problems: The function h = 40 – 4.9t 2 represents the height (in meters) of a fireworks rocket after t seconds. Graph this function. Then use your graph to estimate the height of the rocket after 2 seconds. The equation h = 40 – 4.9t 2 is quadratic, since the variable t has an exponent of 2. Time cannot be negative, so use only positive values of t. Solution:

Competency Goal 4 : Competency Goal 4 Solution: (contd.) (2.5, 9.4) 40 – 4.9(2.5)2 = 9.4 2.5 (2, 20.4) 40 – 4.9(2)2 = 20.4 2 (1.5, 29.0) 40 – 4.9(1.5)2 = 29.0 1.5 (1, 35.1) 40 – 4.9(1)2 = 35.1 1 (0.5, 38.8) 40 – 4.9(0.5)2 = 38.8 0.5 (0, 40) 40 – 4.9(0)2 = 40 0 (t, h) h = 40 – 4.9t 2 t At a time of 2 seconds, the fireworks rocket would be 20.4 meters.

Competency Goal 4 : Competency Goal 4 4.03 Use power models to solve problems. a) Solve using tables, graphs, and algebraic properties. b) Interpret the constants, coefficients, and bases in the context of the problem. Sample Problems: Solve n for the equation: 5 5n + 1 = 125 n – 2 Solution: 5 5n + 1 = 125 n – 2 Original equation 5 5n + 1 = (53 ) n - 2 Rewrite 125 as 53 so each side has the same base. 5 5n + 1 = 5 3n – 6 Power of a Power 5n + 1 = 3n – 6 Property of Equality for Exponential Functions 2n + 1 = – 6 Subtract 3n from each side. 2n = –7 Subtract 1 from each side. n = –3.5 Divide each side by 2. The solution is –3.5.

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