COMMON CORE Standards Plus Mathematics GRADE 8 Teacher Packet Copyright 01 Learning Plus Associates All Rights Reserved; International Copyright Secured. Permission is hereby granted to teachers to reprint or photocopy in classroom quantities the pages in this set that carry a Learning Plus Associates copyright notice. These pages are designed to be reproduced by a teacher for use in his/her class with accompanying Learning Plus Associates materials. Such copies may not be sold and further distribution is expressly prohibited. Except as authorized above, prior written permission must be obtained from Learning Plus Associates to reproduce or transmit this work or portions thereof in any form or by any electronic or mechanical means, including any information storage or retrieval system, unless expressly permitted by federal copyright law.
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INTENSIVE TEST PREP Scope and Sequence COMMON CORE Standards Plus Grade 8 Mathematics Week Monday Tuesday Wednesday Thursday Friday 1 Lesson 1 (8.EE.) Lesson (8.EE.) Lesson 5 (8.EE.1) Lesson 7 (8.EE.1) Lesson 9 (8.EE.) Lesson 1 (8.EE.4) Lesson 17 (8.EE.5) Lesson 19 (8.EE.5) Lesson 1 (8.EE.6) Lesson (8.EE.6) Performance Lesson 4: What Is Slope? (8.EE.5, 8.EE.6) Lesson 5 (8.EE.7a) Lesson 7 (8.EE.7a-b) 4 Lesson 0 (8.EE.7b) Lesson 1 (8.EE.7b) Lesson 7 (8.EE.8a) Lesson 9 (8.EE.8a) Lesson 41 (8.EE.8b) 5 Lesson 47 (8.EE.8c) Performance Lesson 5: Systems of Equations (8.EE.7a, 8.EE.7b, 8.EE.8a, 8.EE.8b, 8.EE.8c) Functions Lesson 1 (8.F.1) 6 Functions Lesson (8.F.1) Functions Lesson (8.F.1) Functions Lesson 5 (8.F.) Functions Lesson 8 (8.F., 8.F.) Functions Lesson 9 (8.F., 8.F.) 7 Functions Lesson 10 (8.F.4) Functions Lesson 1 (8.F.) Functions Lesson 14 (8.F.) Performance Lesson 6: Functions Linear Functions & Relationships (8.F.1, 8.F., 8.F., 8.F.4) 8 Geometry Lesson 1 (8.G.1) Geometry Lesson (8.G.) Geometry Lesson 5 (8.G.) Geometry Lesson 6 (8.G.) Geometry Lesson 9 (8.G.4) 9 Geometry Lesson 1 (8.G.5) Geometry Lesson 14 (8.G.5) Geometry Lesson 1 (8.G.6) Geometry Lesson (8.G.6) Geometry Lesson 4 (8.G.6) 10 Geometry Lesson 5 (8.G.7) Geometry Lesson (8.G.8) Geometry Lesson 4 (8.G.8) Performance Lesson 11: Geometry Pythagorean Theorem (8.G.6, 8.G.7, 8.G.8) This 10-week set of lessons was selected from the year-long COMMON CORE Standards Plus materials. 1.877.505.915 www.intensivetestprep.org 01 Learning Plus Associates
Teacher Lesson Plan Domain: Expressions and Equations Focus: Square Numbers and Roots Lesson: #1 Standard: 8.EE.: Use square root and cube root symbols to represent solutions to equations of the form x = p and x = p, where p is a positive rational number. Evaluate square roots of small perfect squares and cube roots of small perfect cubes. Know that is irrational. Lesson Objective: Students will recognize perfect squares, evaluate square roots of perfect squares, and simplify irrational numbers with radical signs. Introduction: Today we will review perfect squares and their square roots. You will simplify irrational numbers with radical signs. Instruction: Project a multiplication table. A perfect square number is the product of an integer multiplied by itself. For example, 5 is a square number since it can be written as 5 5 or 5. Look at the multiplication table that I am projecting. Notice that the square numbers are the numbers in the diagonal. Perfect squares or square numbers are non-negative. But the factors can be negative. For example -5-5 = (-5) = 5. Remember a negative number times a negative number is a positive number. For perfect squares, we can easily evaluate the square root. For example the square root of 5 is 5. We will only use the positive square root when evaluating square roots. This also extends to rational numbers made up of perfect squares. For example, the square root of 4 9 is. In the last domain we reviewed irrational numbers. To evaluate the square root of a number that is not a perfect square, the value is irrational. For example, is not a perfect square, so the square root of is irrational. We can write down the decimal approximation to the square root of, or we can leave the number unchanged as square root of. But if the irrational number can be simplified by finding if one of the factors of the radicand (the number under the radical sign) is a perfect square, then we will simplify the number leaving the answer in exact form (using the radical sign, not a decimal approximation). Let s complete the first example together. Notice you are given the first 1 perfect squares in the table including zero. You may refer to the table throughout the lesson and the rest of the week. It is interesting to note that square numbers end in 0, 1, 4, 6, 9, or 5. In example 1, I need to evaluate the given number. First I decide if the number is a square number. I look at the table and see that 169 is a square number, so I write 1. Guided Practice: Now complete problem. I will give you a few seconds. What did you write down? 9 since 81 is a perfect square. Let s look at problem. Notice it is a rational number but the numerator and denominator are both perfect squares. Can you find these numbers in your table? Look for the numbers and write down the values. I will give you a few seconds. What did you write? ( 11 4 ) Now let s look at the last number. Is 8 a perfect square? Do you see it on your table? No. Therefore the square root of 8 is irrational, but we can simplify it. To do this we have to determine if any of the factors pairs of 8 is a perfect square. We know that 8 = 4 and 4 is a perfect square. We can rewrite 8 as 4 4. Notice a few things we did here. We can separate out the factors 4 and into separate radical signs. We can remove the multiplication sign at the last step and place the number outside the radical sign right in front of the number under the radical sign. We can do this by convention. So 8. This is the simplified exact form. Now take a moment and write down in your own words how we simplified 8. Independent Practice: Apply the same process to complete the practice problems. Review: When the students are finished, go over the problems. Closure: Today you simplified square roots that were not perfect squares. Turn to your elbow partner and explain how to find the square root of. Answers: 1. 6.. 17 4. 10 5. 5 6. Problem is irrational. Determine if there is a factor pair that has a perfect square as one of the factors. The factor pair of 1 that does have a perfect square is 4 and. The square root of 4 is. 1 4 4. 4 www.intensivetestprep.org 1.877.505.915 01 Learning Plus Associates
Student Page Domain: Expressions and Equations Focus: Square Numbers and Roots Lesson: #1 Standard: 8.EE.: Use square root and cube root symbols to represent solutions to equations of the form x = p and x = p, where p is a positive rational number. Evaluate square roots of small perfect squares and cube roots of small perfect cubes. Know that is irrational. Example: Tables of exponentials and square numbers. Square Numbers 0 0 1 1 4 9 4 16 5 5 6 6 7 49 8 64 9 81 10 100 Exponentials Square Numbers 11 11 1 144 1 169 14 196 15 5 16 56 17 89 18 4 19 61 0 400 Exponentials Square numbers end in 0, 1, 4, 6, 9, or 5. Evaluate the following. If the number is irrational, simplify and keep answer in exact form. 1. 169. 81 11. 4. 8 16 Explain how you simplified 8. Directions: Evaluate the following. If the number is irrational, simplify and keep answer in exact form. 1. 6. 1 4. 4. 100 89 5. 5 9 6. Explain how you simplified the irrational number. www.intensivetestprep.org 1.877.505.915 01 Learning Plus Associates 5
Teacher Lesson Plan Domain: Expressions and Equations Focus: Evaluate Cube Roots Lesson: # Standard: 8.EE.: Use square root and cube root symbols to represent solutions to equations of the form x = p and x = p, where p is a positive rational number. Evaluate square roots of small perfect squares and cube roots of small perfect cubes. Know that is irrational. Lesson Objective: Students will recognize perfect cubes, evaluate the cube roots of perfect cubes, and simplify irrational numbers with radical signs. Introduction: Today we will review perfect cubes and their cube roots. You will also simplify irrational numbers with radical signs. Instruction: A perfect cube number is the product of an integer multiplied by itself three times. For example, 8 is a cube number since it can be written as. The cube root of 8 is. We have to distinguish between square roots and cube roots. The way we do this is by inserting a small in the check mark of the radical sign. We call this small number the index. An index can be any number. Today we will only work with cube roots. We write this as 8. Guided Practice: Let s complete the example together. Notice you are given the first 11 perfect cubes in the table including zero. You may refer to the table throughout the lesson and the rest of the week. We first must find the cube root of 15. Look to see if 15 is a perfect cube on the table. Do you see 15? Yes, it is on the table. So the cube root is 5. Next we are given a fraction. Look to see if the numerator and denominator are on the table. I will give you a few seconds to see and write down the cube root. This is similar to what we did with square roots. You should have written 9. Let s complete the last example together. We need to find the cube root of 16, but 16 is not a perfect cube. Since we can t evaluate the cube root of 16, we can simplify it. We need to find a factor pair that has a factor that is a perfect cube. Let s try 8 16. Eight is a perfect cube. Below the cube root write 8 8. Now take a moment to write how to simplify the cube root of 16 in your own words. Students should write down something about finding a factor pair of the radicand that includes a perfect cube. Then they should take the cube root of the perfect cube factor and leave the other factor under the radical. The final form of the answer should have both numbers written side by side by convention. Independent Practice: Apply the same process we did together to complete the practice problems. Review: When the students are finished, go over the problems. Closure: Today you evaluated cube roots and simplified irrational numbers using radical signs (keeping numbers in exact form). Turn to your elbow partner and explain the difference between a square root and a cube root. Call on students to share their answer. Answers: 1.. 1. 0.4 4. 5 4. 5. Since 5 5 5 5 and 5 5, then 5 5 5 4 5 4 5 4. 6 www.intensivetestprep.org 1.877.505.915 01 Learning Plus Associates
0.0 Domain: Expressions and Equations Focus: Evaluate Cube Roots Lesson: # Standard: 8.EE.: Use square root and cube root symbols to represent solutions to equations of the form x = p and x = p, where p is a positive rational number. Evaluate square roots of small perfect squares and cube roots of small perfect cubes. Know that is irrational. Example: Table of exponentials and cube numbers. Exponentials Cube Numbers 0 0 1 1 8 7 4 64 5 15 6 16 7 4 8 51 9 79 10 1000 Student Page n Cube root of n The number inside the check mark of the radical sign is called the index. The index for a cube root is. Evaluate the following. If the number is irrational, simplify and keep answer in exact form..79 1. 15 8. 16 Explain how you simplified 16. Directions: Evaluate the following. If the number is irrational, simplify and keep answer in exact form. 1. 7.1.64 4. 5 5 5 4 5. Explain how you simplified problem 4. www.intensivetestprep.org 1.877.505.915 01 Learning Plus Associates 7
Teacher Lesson Plan Domain: Expressions and Equations Focus: Properties of Exponents Lesson: #5 Standard: 8.EE.1: Know and apply the properties of integer exponents to generate equivalent numerical expressions. Lesson Objective: Students will apply the properties of integer exponents to generate equivalent expressions. Introduction: Today we will apply the properties of integer exponents to generate equivalent expressions. Instruction: When we work with power numbers or numbers raised to exponents, we have certain definitions to understand and rules we must follow to perform operations and to simplify expressions. We call these rules the Properties of Exponents. Today we will review some of the Properties of Exponents, and later this week we will review the rest of the properties. First, let s define exponential form. There are two numbers that make up a power. For example in the power number 4, is called the base, and 4 is called the exponent. The exponent indicates how many times a base multiplies itself. 4 =. Now let s review some of the properties. The First Power Property says that anything raised to the first power or exponent of one is itself. For example, 18 1 = 18, and 18 = 18 1. Any number or variable without an exponent has a power or exponent of one. The Zero Power Property states that any non-zero base raised to a zero exponent equals 1. For example, 57 0 = 1, 0 0 1, x 1, etc. The Product Property states that when multiplying two or more powers with the 4 4 6 same base, add the exponents. For example. The Quotient Property states that when dividing two powers with the same base, subtract the exponents. For example, 8 8 6. Notice the proofs of the Product and Quotient Properties are provided on your page as well. The Negative Exponent Property states that any non-zero based raised to a negative exponent is equal to the reciprocal of the base raised to the positive exponent. For example, 1 1 and 1 9. 9 1 Guided Practice: Let s simplify the expressions with exponents. A power is in its simplest form when the properties and definitions of exponents can t be applied further to simplify it. If the exponent is very large then often times we leave the number written with the positive exponent. The first problem is 6 0. What does that simplify to? (1) Any non-zero number raised to the zero power is equal to one according to the Zero Property. The second problem is -6. What property do we use first? The Product Property. We must add the exponents. Negative six plus two is equal to 4 negative 4. We write down. Can we continue to simplify this expression? Yes, because we can 4 1 1 still apply the Negative Exponent Property.. Since we can easily simplify 4 we do. For 4 5 the last problem, we must simplify 4. What law do we use? The Quotient Property. We subtract the 5 exponents. Order matters. We subtract the denominator s exponent from the numerator s exponent. Four minus three is one. We now have 5 1. Using the First Power Property, we know this equals 5. Independent Practice: It s your turn to apply the laws to simplify each expression. Review: When the students are finished, go over the problems. Closure: Today you used some of the Properties of Exponents to simplify numeric expressions. With your elbow partner, explain the properties we used today. Call on students to share. 7 Answers: 1. 710 1 1 ; (Quotient & Negative Exponent Properties, Exponential Form 10 7 Definition). 0 1 9 9; (Zero & Negative Exponent Properties, Exponential Form Definition) 1. 4 4 1 1 1 4 4 4 4 ; (Product, Negative Exponent, and First Power Properties) 4 1 4 16 8 www.intensivetestprep.org 1.877.505.915 01 Learning Plus Associates
Domain: Expressions and Equations Focus: Properties of Exponents Lesson: #5 Standard: 8.EE.1: Know and apply the properties of integer exponents to generate equivalent numerical expressions. Definition and Properties of Exponents Definition of Exponents: 4 ; is called the base and 4 is called the exponent. First Power Property: Any base raised to an exponent of one is equal to itself. 1 Rule: a a Example: 18 1 = 18 Zero Power Property: Any non-zero base raised to a zero exponent equals 1. 0 Rule: a 1 Examples: 57 0 =1, 0 1 Product Property: When multiplying two or more powers with the same base, add the exponents. m n m n Rule: a a a 4 4 6 Example: 4 6 Proof: Quotient Property: When dividing two powers with the same base, subtract the exponents. m a mn Rule: a n a 6 6 4 Example: 6 Proof: 4 Negative Exponent Property: Any non-zero base raised to a negative exponent is equal to the reciprocal of the base raised to the positive exponent. m m 1 1 a m Rules: a and a m m a a 1 1 1 1 Examples: and 9 9 1 Example: Simplify the following expressions. Write the property of integer exponents used at each step. Student Page 1. 6 0 6.. 5 5 4 Directions: Simplify the following expressions. Write the property of integer exponents used at each step. 1. 7 10 90. - 4. 4 4 www.intensivetestprep.org 1.877.505.915 01 Learning Plus Associates 9
Teacher Lesson Plan Page 1 of Domain: Expressions and Equations Focus: Properties of Exponents Lesson: #7 Standard: 8.EE.1: Know and apply the properties of integer exponents to generate equivalent numerical expressions. Lesson Objective: Students will work with the properties of integer exponents to generate equivalent expressions. Introduction: Today we will work with the properties of integer exponents to generate equivalent expressions. 10 Instruction: Today we will review a few more properties of exponents. We have reviewed the Zero Power Property. This property only applies to non-zero bases. When zero is raised 0 to the power of zero, 0, the expression is undefined. The reason why 0 0 is undefined is too complex to explain here but you may revisit this expression in advanced mathematics. For now you only need to know that it is undefined. The Power to a Power Property states that when raising any power to an exponent, you multiply the exponents. For example, 9 ( ). The Product to a Power Property states that when the product of a base is powered by the same exponent, then both the factors are powered by the same exponent. For example, ( ) 4 9 6. Do you see another way to simplify this expression? Yes, you could multiply the factors inside the parentheses first and apply the definition of exponents so you get 6 6. But if the factors are not both numbers, but letters instead, then you must apply the property. The Quotient to a Power Property states when the quotient of a base is powered by the same exponent, then both the numerator and denominator are powered by the same exponent. For example, 4. Do you 9 see another way to simplify this expression? Yes, you can apply the definition of exponents first and get. Again if letters are in the expression instead of only numbers, it is best to apply the Quotient to a Power Property. Notice that the rules, examples, and proofs are on your student page for you to reference as you complete the problems. Guided Practice: Let s complete the examples together. We must simplify the expression using two different approaches. That means there is more than one property we can apply first. We have to use two different sequences of properties to arrive at the simplified expression. The first problem is (7 ). Let s apply the Power to Power Property first. We 6 multiply the exponents. 6. We now have 7. Now we apply the Negative Power 1 Property and we have 6. For our second approach, let s simplify the expression inside the 7 parentheses first and then apply the Negative Power Property first. We get apply the Quotient to Power Property and we get 1 1 6. 1 7 The second problem is 7 7. Now let s. Let s simplify what is inside parentheses first and then apply the Negative Power 1 1 Property followed by the definition of exponents. We get 6. ( 6) 6 For our second approach to this problem, let s apply the Product to a Power Property first. ( ). Next we apply the Negative Power Property and we get We get 1 1 1 1 1. The third problem is ( ) 9 4 6 5 www.intensivetestprep.org 1.877.505.915 01 Learning Plus Associates. For our first approach, let s apply
Domain: Expressions and Equations Focus: Properties of Exponents Lesson: #7 Standard: 8.EE.1: Know and apply the properties of integer exponents to generate equivalent numerical expressions. the Negative Exponent Property first. We get 5 15. 5. Next, let s apply the Quotient to a Power Property and we get 8 For our second approach, let s apply the Quotient to a Power Property first. We get. Next, let s apply the Negative Exponent Property. We get 5 1 1 5 1 1 1 15 15. Which way to you prefer? Why? Call on students to 8 15 8 1 8 explain why they prefer the method they chose. Independent Practice: It s your turn to apply a sequence of the properties to simplify the given expressions. You must also show two different approaches as we did in the example problems. Review: When the students are finished, go over the problems. Teacher Lesson Plan Page of Closure: Today we reviewed the remaining Properties of Exponents and you showed two different approaches to simplifying expressions. Answers: Sequences may vary slightly. 1. 4 9 4 (Apply Negative Power Property first) ( 4) 16 1 ( 4) 1 4 ( 4) 1 1 1 9 9 (Apply Quotient to 16 9 16 1 16 a Power Property first) 1 (10 ) 10 10 (Apply Power to a Power Property first) 9 10 9. 1 1 1 9 (10 ) (Apply Negative Power Property first) 10 10 10 1 1 1 1 1. ( 4) ( 4) (Apply Power to ( 4) 4 16 64 a Power Property first) 1 1 ( 4) ( 8) (Simplify inside parentheses first, then ( 8) 64 apply Negative Power Property) 4. Answer will vary. Possible answers include: 8 1 4 6 ; 6 ; 6 6; 6 6 ; 10 6 6 www.intensivetestprep.org 1.877.505.915 01 Learning Plus Associates 11
Student Page Domain: Expressions and Equations Focus: Properties of Exponents Lesson: #7 Standard: 8.EE.1: Know and apply the properties of integer exponents to generate equivalent numerical expressions. 0 Example: 0 is undefined. Power to a Power Property: When raising any power to an exponent, multiply the exponents. Rule: ( a m ) n a mn 9 Example: ( ) 9 Proof: ( ) ( ) ( ) ( ) Product to a Power Property: When the product of a base is powered by the same exponent, then both the factors are powered by the same exponent. Rule: ( ab) n a n b n Example: ( ) 4 9 6 Proof: ( ) ( ) ( ) 6 6 6 Quotient to a Power Property: When the quotient of a base is powered by the same exponent, then both the numerator and denominator are powered by the same exponent. n n Rule: a a n b b Example: 4 9 4 Proof: 9 Directions: Simplify the expressions. Show two different approaches. 1. (7 ).. 5 Directions: Simplify the expressions. Show two different approaches. 1. 4 (10 ). ( 4). 4. Using the Properties of Exponents, write 4 unique expressions that simplify to 1. 6 1 www.intensivetestprep.org 1.877.505.915 01 Learning Plus Associates
Student Page Lessons begin on the next page. www.intensivetestprep.org 1.877.505.915 01 Learning Plus Associates 1
Teacher Lesson Plan Domain: Expressions and Equations Focus: Scientific Notation Lesson: #9 Standard: 8.EE.: Use numbers expressed in the form of a single digit times an integer power of 10 to estimate very large or very small quantities, and to express how many times as much one is than the other. Lesson Objective: Students will understand the use of scientific notation and fluently translate back and forth between standard and scientific notation. Introduction: Today we will review the reason we use scientific notation and you will translate back and forth between standard and scientific notation. Instruction: When zero is used as a placeholder, it can make a number much greater or much lesser. Consider the number of atoms in an average nickel, which is about 6,0 with 0 zeros behind it. This is how we write the number in standard notation. That is a lot of atoms. Or consider the diameter of an average atom, which is 0.0000000 cm. That is very small. Writing all these zeros can be awkward. We see very large or very small numbers like these used in science. This is why scientists and mathematicians use a short cut notation called scientific notation. Scientific notation uses powers of 10 to keep the numbers simple and easy to write. Positive exponents are used for very large numbers such as in astronomy; negative exponents are used for very small numbers such as in chemistry. The number of atoms in an average nickel written in scientific notation is 6.0 10. We can think of scientific notation as the product of two numbers, the 6.0 is the digit term, and 10 is the exponential term. In scientific notation, the digit term indicates the number of significant figures in the number. The digit term must be a number equal to or greater than 1 and less than 10. The exponential term only places the decimal point. The exponent of indicates how many places you move the decimal point to the right. The diameter of an average atom is written as 10-8. The negative 8 in the exponent indicates how many places to move the decimal point to the left. You will need to add in 7 zeros to the left of, and then place a decimal point. The is the digit term, and 10-8 is the exponential term. Guided Practice: To help understand why we multiply the digit term by powers of 10, a table is provided to show the meaning of various powers of 10. You can see the connection between the exponent, the number of zeros, and therefore the number of places we move the decimal point over in the digit term when translating to standard notation. Let s complete the example problems together. Translate.5 10 5 into standard notation. The 5 exponent does not mean we add 5 zeros. Since the exponent is positive, it means we move the decimal point 5 places to the right. We will need to add in 4 place-holding zeros and the number in standard notation is 50,000. For the second example, translate 0.0000085 into scientific notation. Since the number is less than one, the exponent must be negative. The digit term must be equal to or greater than 1 and less than 10 and must include all the significant digits. The digit term is 8.5. The exponential term has the exponent of negative six since the decimal point in the digit term must move to the left 6 places to be translated back to the equivalent standard term. The term in scientific notation is 8.5 10 6. Independent Practice: For the first three problems, write your answers. For the remaining problems, apply the rules of translating to either standard or scientific notation. Review: When the students are finished, go over the problems. Closure: Today we reviewed the value of scientific notation and how to translate between standard and scientific notation. Answers: 1. We use scientific notation to make very large or very small numbers easier to write and work with.. Scientific notation is the product of two numbers. The first number, 4.7 is the digit term. It represents the number of significant digits in the number. The digit term must be between 1 and 10. The second number is the exponential term. It is a power of 10. The exponent represents the number of places the decimal point moves. For 10-5, the decimal point moves left 5 places from where it is in the digit term.. 1.4 10-4 is not written correctly in scientific notation since the digit term is not a number between 1 & 10: 1.4 10-9 4. 6,1,000,000 = 6.1 10 5. 9.4 10 9,400 6. 15,000 1.5 10 4 7. 0.0011 1.1 10 6 8. 8.9 10 0.0000089 4 9. 6.75 10 0.000675 14 www.intensivetestprep.org 1.877.505.915 01 Learning Plus Associates
Student Page Domain: Expressions and Equations Focus: Scientific Notation Lesson: #9 Standard: 8.EE.: Use numbers expressed in the form of a single digit times an integer power of 10 to estimate very large or very small quantities, and to express how many times as much one is than the other. Example: Scientific Notation 1 10 Meaning 1 1 1 10 101010 1,000 Example 5. 10 5. 0.005 1 5. 5 1,000 1,000 10,000 1 10 1 10 1 (not usually used) 1 10 0 (not usually used) 1 10 1 (not usually used) 1 10 1 10 1 1 1 10 1010 100 1 5. 5 5. 10 5. 0.05 100 100 1,000 1 1 10 1 10 1 1 5. 5 10 10 100 0 10 1 1 10 10 0 1 10 10 10 100 5. 10 5. 0.5 5. 10 5. 1 5. 5. 10 5. 10 5 5. 10 5. 100 50 10 10 10 10 1,000 5. 10 5. 1,000 5,00 Translate into standard notation. Translate into scientific notation. 5 1..5 10. 0.0000085 Directions: Complete the following problems. 1. Explain why we use scientific notation. 5. Explain the meaning of scientific notation using the number 4.7 10. 4. Explain why 1.4 10 is not correctly written in scientific notation. Write it correctly in scientific notation. Directions: Translate to either standard or scientific notation. 4. 6,1,000,000 7. 0.0011 6 5. 9.4 10 8. 8.9 10 4 6. 15,000 9. 6.75 10 www.intensivetestprep.org 1.877.505.915 01 Learning Plus Associates 15
Teacher Lesson Plan Domain: Expressions and Equations Focus: Operations Using Scientific Notation Lesson: #1 Standard: 8.EE.4: Perform operations with numbers expressed in scientific notation, including problems where both decimal and scientific notation are used. Use scientific notation and choose units of appropriate size for measurements of very large or very small quantities (e.g., use millimeters per year for seafloor spreading). Interpret scientific notation that has been generated by technology. Lesson Objective: Students will perform operations with numbers in scientific notation. Introduction: Today you will add and subtract with numbers in scientific notation. Instruction: We have multiplied and divided with decimals and numbers in scientific notation. You applied the Properties of Exponents when you multiplied and divided. For multiplying, the digits are multiplied and the exponents are added. For dividing, the digits are divided and the exponents are subtracted. Today we will focus on adding and subtracting with numbers in scientific notation. Unlike multiply or dividing, when adding and subtracting, all the numbers are changed to the same power of 10 and the digits are added or subtracted. The sum or difference is then rewritten in scientific notation. Follow along as I complete the first example. You follow this same process for adding. Sometimes the last step is not needed if the result is already in scientific notation. You can choose either power of 10 of the two numbers you are adding or subtracting, but usually it is easier if you choose the power of 10 of the larger number. Guided Practice: Let s complete the second example together. We need to find the result 5 4 of 9.8 10 5. 10. First we convert to the same power of 10. We will rewrite the 5 second number to 0.5 10. Next we group the digit terms together by factoring out the 5 exponential term. We write (9.8 0.5) 10. Next we add the digit terms and we get 5 10. 10. Since this is not written correctly in scientific notation, we rewrite the number 6 into scientific notation as 1.0 10. Independent Practice: It s your turn to apply the same process to finding the sum or difference of decimals and numbers in scientific notation. Write your answers in scientific notation. Review: When the students are finished, go over the problems. Closure: Today you added and subtracted decimals and numbers in scientific notation. You first had to change the numbers to the same power of 10 then you added or subtracted the digit terms. You converted the result back into scientific notation if you needed to. Answers: 15 14 1. 7.7 10 9 10 15 (7.7 0.9) 10 15 8.6 10 8.. 10 9 9 10 8 (. 0.9) 10 8 1.4 10. 0.98 6.7 10 (98 6.7) 10 404.7 10 4.047 10 OR 0.98 6.7 10 1 (.98 0.067) 10 1 4.047 10 4. 5,510,000 4.5 10 5 (5.51 0.45) 10 6 5.96 10 6 4 5. 6.7 10 4 0.0005 (6.7.5) 10 4 10. 10 1.0 10 1 16 www.intensivetestprep.org 1.877.505.915 01 Learning Plus Associates
Domain: Expressions and Equations Focus: Operations Using Scientific Notation Lesson: #1 Standard: 8.EE.4: Perform operations with numbers expressed in scientific notation, including problems where both decimal and scientific notation are used. Use scientific notation and choose units of appropriate size for measurements of very large or very small quantities (e.g., use millimeters per year for seafloor spreading). Interpret scientific notation that has been generated by technology. Example: 1. Find the difference. Write the answer in scientific notation. - -4 1.7 10-9. 10 1.7 10 4 9. 10 given problem 1.7 10 0.9 10 converted to the same power of 10 (1.7 0.9) 10 grouped the digit terms together by factoring 0.78 10 subtracted the digit terms 4 7.8 10 rewrote into scientific notation Find the sum. Write the answer in scientific notation. 9.8 10 5. 10 5 4. Student Page Directions: Find the sum or difference. Write answers in scientific notation. 7.7 10 9 10 15 14 1. 8 9.. 10 9 10. 0.98 6.7 10 5 4. 5,510,000 4.5 10 4 5. 6.7 10 0.0005 www.intensivetestprep.org 1.877.505.915 01 Learning Plus Associates 17
Teacher Lesson Plan Page 1 of Domain: Expressions and Equations Focus: Graph Proportional Relationships & Determine Unit Rate Lesson: #17 Standard: 8.EE.5: Graph proportional relationships, interpreting the unit rate as the slope of the graph. Compare two different proportional relationships represented in different ways. Lesson Objective: The students will graph proportional relationships and determine the unit rates. The students will understand that the unit rate is the slope of the graphed line representing the proportional relationship. Introduction: Today you will graph proportional relationships and determine the unit rate. Instruction: Two quantities or variables are proportional if they have the same ratio. A proportional relationship between two quantities is one in which the two quantities vary directly with one another. Another way of saying this is the quantity y is proportional to x if y = kx, where k is the constant of proportionality. If x is equal to zero, then y is also equal to zero. When proportional relationships are grouped, they create a line that contains the origin. The variable x represents the independent variable and y represents the dependent variable. We graph the independent variable on the horizontal axis and the dependent variable on the vertical axis. Guided Practice: Let s complete the examples together. A helium balloon starts from the ground and ascends at a constant rate of 100 feet every 40 seconds. To ascend means to rise or go up. Since it starts from the ground, at 0 seconds the balloon is at 0 feet. Since it ascends at a constant rate, then this is a proportional relationship. We are going to graph this relationship on the grid. Let s title our graph Height of Ascending Balloon. Next we will define the dependent and independent variables. The height of the balloon depends on the time that has passed. So height is the dependent variable and is on the y-axis and time is the independent variable and is on the x-axis. Draw in your axes and label the x-axis time (seconds) and the y-axis height (feet). Let s label the origin since every proportional relationship contains the point (0, 0). We will use a scale of 5 for the x-axis and a scale of 10 for the y-axis. Create the axes, labels, and scale now. Next we will plot the points and draw a line through the points. (See the answer key on the teacher page for a completed graph.) Now that we have an accurate graph, let s answer the questions. Does our line contain the point (0, 0)? Yes, it does because all proportional relationships contain the point (0, 0). What is the unit rate of the helium balloon s ascent? In other words, what is the number of feet the balloon rises per second? To find the unit rate through calculation we divided 100 feet by 40 seconds and we get.5 feet per second. Where can we find this information on the graph? The unit rate, r, is always the y-value at the x-value of 1. To generalize this statement, in the point (1, r), r is the unit rate. In our case it is represented at the point (1,.5). In one second the balloon rises.5 feet. The unit rate is.5 feet per second or.5 ft/s. What is the slope of your line? The slope is the change in feet over the change in time. We can choose any two points on the line to calculate slope. Since our line contains (0, 0), we should choose that point. It will make our calculations simpler. We will choose (40, 100) as the other point. Now we find the slope by finding the change in feet over 100 0 100 the change in seconds. m.5. We see that the slope is the same as 40 0 40 the unit rate. For the last problem, a second helium balloon ascends at a constant rate of 00 feet every 80 seconds. Is the second balloon ascending at a slower or faster rate than the first balloon? What are some different methods that we can apply to compare? We could graph this relationship on the same graph and compare the slopes of the lines. What is another method? We could calculate the unit rate of the second balloon and compare unit rates. What is another method? We could reason through the problem by comparing the two balloons for the same amount of time. We could compare both heights at 40 or 80 seconds 18 www.intensivetestprep.org 1.877.505.915 01 Learning Plus Associates
Domain: Expressions and Equations Focus: Graph Proportional Relationships & Determine Unit Rate Lesson: #17 Standard: 8.EE.5: Graph proportional relationships, interpreting the unit rate as the slope of the graph. Compare two different proportional relationships represented in different ways. and see which is higher. Choose a method and find the answer then explain how you found your answer. I will call on you to share. Independent Practice: It s your turn to apply the same process and reasoning to complete the practice problems. Remember to include a title, to label the axes with units, and to use an appropriate scale on the axes. Review: When the students are finished, go over the problems. Teacher Lesson Plan Page of Closure: Today you graphed proportional relationships and determined the unit rate. You showed that the slope is equal to the unit rate for proportional relationships. Answers: 1. Completed graph Number of Cups in Gallons St. Ed. Pg. 5. Yes. 16 cups in one gallon. On the graph, you can find this information at the point (1, r) where r is the unit rate, in this case the number of cups per gallon. 4. The slope is 16. The slope is equal to the unit rate for proportional relationships. The unit rate in this case is the number of cups per gallon. You can determine the number of cups in one gallon through calculation by dividing 56 by.5 OR you can find the y-value when at x = 1 OR you can find the slope of the line which is the number of cups per gallon. www.intensivetestprep.org 1.877.505.915 01 Learning Plus Associates 19
Teacher Lesson Plan Domain: Expressions and Equations Focus: Comparing Proportional Relationships Lesson: #19 Standard: 8.EE.5: Graph proportional relationships, interpreting the unit rate as the slope of the graph. Compare two different proportional relationships represented in different ways. Lesson Objective: Students will compare different proportional relationships. Introduction: Today you will compare proportional relationships that are represented graphically. Instruction: We have been learning that the graphs of proportional relationships contain the origin and are linear. We also know that the unit rate and slope are equal. We saw how defining the independent and dependent variables affect how we interpret the graph. Today you will be given the graph. The lines are already placed on the graph for you. You will need to define the independent and dependent variables. You will not need to scale the axes. Instead you will focus on the relative relationship among the proportional situations. The steeper the line, the greater the unit rate. The flatter the line, the lesser the unit rate. Guided Practice: Let s complete the example. We are given a chart of produce prices. Each line on the graph represents one type of produce. We need to label each line appropriately. We are only given unit prices for the peaches and the yellow squash. Copy the unit prices for those two produce over to the last column now. To be able to correctly label the lines, we must first determine the unit prices for the other three produce. To find unit price, we divide the dollar amount by the number of pounds. Find the unit prices and write the unit prices in the last column. You should have written bananas as $0.48/lb., strawberries as $1.78/lb., and zucchinis as $1.8/lb. Next we need to complete the graph. We first need to determine independent and dependent variables. Since unit price is dollars per pound, we know the price depends on weight. Therefore, price is the dependent variable and weight is the independent variable. Label the x-axis as Weight (pounds) and the y-axis as Price (dollars). We need to title the graph as well. We can title the graph using the same title of the chart. Write Produce Prices above the graph. We don t need to include a scale. Now that we have the graph labeled, we can label the lines with the produce names. Remember, the steeper the line, the greater the unit price. Label the lines on the graph. Let s look at problem. Avocados cost more than peaches but less than strawberries. We need to draw in a line on the graph that could represent the price of x pounds of avocados. Where should we place the line? Yes, we should place the line somewhere between the lines that represent peaches and strawberries. The line should contain the origin. Draw in a line now. Label the line Avocados. Read problem. By looking at the graph, how can you determine which produce is the least expensive? Yes, the flattest line represents the least expensive produce. Read problem 4. Where can we find the answer? In the last column we filled out for unit price. Bananas are the least expensive and they are $0.48/lb. Does our answer confirm our answer for problem? Yes, both answers are the same. Read problem 5. Our equation will be written in the form, y = mx. We know m, the slope, is 0.48, so our equation will be y = 0.48x; y represents the price, and x represents the weight. Independent Practice: It is now your turn to apply the same concepts on the practice problems. Review: When the students are finished, go over the problems. Closure: Today you compared proportional relationships that were represented graphically. 0 Answers: 1. Labeled graph. Distance Traveled. Plane C travels slower than Plane B. The slopes of the lines represent the unit rate in meters per second. Since the slope of the line representing Plane C is flatter, or less than the slope of the line representing Plane B, Plane C travels at a slower rate.. y 45x 4. Answers will vary. The coefficient > 45. Example: y 60 x. 5. The line representing the distance traveled for Plane D would be positioned slightly below the line representing Plane B. The unit rate for Plane D is 41 m/s which is slightly slower than Plane B. The slope is the unit rate so the line must be placed below the line for Plane B since the slope is less. www.intensivetestprep.org 1.877.505.915 01 Learning Plus Associates
Domain: Expressions and Equations Focus: Comparing Proportional Relationships Lesson: #19 Standard: 8.EE.5: Graph proportional relationships, interpreting the unit rate as the slope of the graph. Compare two different proportional relationships represented in different ways. Example: Use the chart below to complete the problems. Produce Prices Produce Type Price Unit Price Bananas 5 lbs for $.40 Peaches $1.48/lb Strawberries lbs for $.56 Yellow Squash $0.98/lb Zucchini lbs for $.84 StudenStudent Student Page 1 Page of 1. Each line on the graph below represents one of the produce types listed in the chart. Label each line. Also label the axes and title the graph. You do not need to add a scale.. Avocados cost more than peaches but less than strawberries. Draw in a line on the graph that could represent the price of x pounds of avocados. Label the line.. By looking at the graph, how can you determine which produce type is the least expensive? 4. What is the unit rate of the least expensive produce type? 5. Write an equation for bananas that represents the relationship between the price, y, and the weight, x. www.intensivetestprep.org 1.877.505.915 01 Learning Plus Associates 1
Student Page of Domain: Expressions and Equations Focus: Comparing Proportional Relationships Lesson: #19 Standard: 8.EE.5: Graph proportional relationships, interpreting the unit rate as the slope of the graph. Compare two different proportional relationships represented in different ways. Directions: The graph below shows the distance three planes have traveled over a period of time. Plane B travels at 45 meters per second (m/s). 1. Label the axes and title the graph. You do not need to include a scale.. Does Plane C travel slower or faster than Plane B? Explain.. Write an equation for Plane B that represents the relationship between the distance traveled, y, and the time, x. 4. Write an equation that could represent Plane A. 5. Plane D s data is shown in the table below. Time (seconds) Plane D Distance (meters) 5 1,05 15,615 5 8,45 If you graph Plane D s data on the same graph, where would you place the line relative to plane B? Explain. www.intensivetestprep.org 1.877.505.915 01 Learning Plus Associates
Student Page Lessons begin on the next page. www.intensivetestprep.org 1.877.505.915 01 Learning Plus Associates
Teacher Lesson Plan Domain: Expressions and Equations Focus: Similar Triangles and Slope Lesson: #1 Standard: 8.EE.6: Use similar triangles to explain why the slope m is the same between any two distinct points on a non-vertical line in the coordinate plane; derive the equation y = mx for a line through the origin and the equation y = mx + b for a line intercepting the vertical axis at b. Lesson Objective: Students will apply the concept of similar triangles and proportional reasoning to understanding slope. Students will use proportions to find a missing coordinate of the vertex of a triangle on a coordinate plane. Introduction: Today you will understand the connection between similar triangles and slope. You will apply proportional reasoning to find missing coordinates of vertices of similar triangles on a coordinate plane. Instruction: The unit rate in a proportional relationship and the slope of its graph depends on a connection with the geometry of similar triangles. A line has a defined slope. The ratio between the rise and run for any two points on the line is always the same and that is why the slope is always the same. Look at Example 1. There are two right triangles: 1 and 456. Angle 1 corresponds to angle 4 and they are congruent both are right angles. Angle corresponds to angle 5 and they are congruent. Angle corresponds to angle 6 and they are congruent. Since corresponding angles are congruent, 1 and 456 are similar. Since the triangles are similar, all corresponding sides are in the same proportion. The ratio of rise to run is the same for each triangle. In our example, 4. All triangles drawn 6 8 where the hypotenuses of the triangles are on the same line are similar, and corresponding sides are proportional. Guided Practice: Applying the concept of similar triangles, let s complete Example together. In the coordinate plane, ABC is similar to ADE. What is the value of x? Since AB AD the triangles are similar, we can set up the proportion. When we substitute in the BC DE side lengths of the triangle we get 6. When we solve for the length of DE, we get 10 DE DE = 6. Since the x-value of point D is 0, then we only need to add 6 to 0 to determine the value of x. So x = 6. Independent Practice: Use the concept of similar triangles to complete the practice problem. Remember you are finding the length of the side of a triangle. Since you are working on a coordinate plane, pay close attention to the signs of the ordered pairs. You must translate the length of the side of the triangle to the actual ordered pair that defines the vertex of the triangle. Review: When the students are finished, go over the problem. Closure: Today we reviewed the connection between similar triangles and slope. You applied proportional reasoning to find missing coordinates of vertices of similar triangles on a coordinate plane. Answers: AD AB 5 AB ; ; AB 10; y 7 DE BC 4 4 www.intensivetestprep.org 1.877.505.915 01 Learning Plus Associates
Student Page Domain: Expressions and Equations Focus: Similar Triangles and Slope Lesson: #1 Standard: 8.EE.6: Use similar triangles to explain why the slope m is the same between any two distinct points on a non-vertical line in the coordinate plane; derive the equation y = mx for a line through the origin and the equation y = mx + b for a line intercepting the vertical axis at b. Example: 1. 1 4 5 6 Therefore, 1 is similar to 456 and corresponding sides are proportional. slope vertical change 4 horiztonal change 6 8. In the coordinate plane below, ABC is similar to ADE. What is the value of x? Directions: In the coordinate plane below, y? Show your work. ABC is similar to ADE. What is the value of www.intensivetestprep.org 1.877.505.915 01 Learning Plus Associates 5
Teacher Lesson Plan Domain: Expressions and Equations Focus: Derive the Equation y = mx Lesson: # Standard: 8.EE.6: Use similar triangles to explain why the slope m is the same between any two distinct points on a non-vertical line in the coordinate plane; derive the equation y = mx for a line through the origin and the equation y = mx + b for a line intercepting the vertical axis at b. Lesson Objective: Students will derive the equation y = mx and write the equation of a line through the origin. Introduction: Today you will use similar triangles and proportional reasoning to derive the equation y = mx. You will write the equations of lines through the origin given the graph. Instruction: For a line through the origin, the right triangle whose hypotenuse is the line segment from (0, 0) to a point (x,y) on the line is similar to the right triangle from (0,0) to the point (1,m) on the line. Look at the graph on your student page; m equals the slope or unit rate, so m is the y-coordinate when x = 1 on a line through the origin. Guided Practice: We need to set up a proportion to describe this situation and solve for y. You will be expected to be able to show how to derive the equation y mx on your weekly evaluation. Write this on your page. The m 0 y 0 m y proportion is. This simplifies to. Then cross multiply and 1 0 x 0 1 x we get y = mx. This is how you derive the equation y = mx. We use the concept of similar triangles and set up a proportion. Independent Practice: It s your turn to determine the slope of each line. The slope is the number you put in the place of m in the equation y = mx. Remember the slope is the vertical change over the horizontal change. The slope is negative if the line decreases from left to right. You can draw in a triangle to help you determine the vertical change over the horizontal change. Review: When the students are finished, go over the problems. Closure: Today you used similar triangles and proportional reasoning to derive the equation y = mx. You wrote the equations of lines through the origin given the graph in the form y = mx. Answers: 1. y = x 1. y x 6 www.intensivetestprep.org 1.877.505.915 01 Learning Plus Associates
Domain: Expressions and Equations Focus: Derive the Equation y = mx Lesson: # Standard: 8.EE.6: Use similar triangles to explain why the slope m is the same between any two distinct points on a non-vertical line in the coordinate plane; derive the equation y = mx for a line through the origin and the equation y = mx + b for a line intercepting the vertical axis at b. Example: Derive the equation y = mx for a line through the origin using similar triangles. For a line through the origin, the right triangle whose hypotenuse is the line segment from (0, 0) to a point (x,y) on the line is similar to the right triangle from (0,0) to the point (1,m) on the line. m = slope or unit rate m is the y-coordinate when x = 1 on a line through the origin Student Page Set up a proportion to describe this situation and solve for y. Directions: Write an equation in the form y = mx to describe the following lines. 1.. www.intensivetestprep.org 1.877.505.915 01 Learning Plus Associates 7
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