Eureka Math Module 4 Topic C Replacing Letters and Numbers
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1 Eureka Math Module 4 Topic C Replacing Letters and Numbers 6.EE.A.2c: Write, read, and evaluate expressions in which letters stand for numbers. 6.EE.A.4: Identify when two expressions are equivalent.
2 Copy all slides into your composition notebook.
3 Lesson 7 Replacing Letters with Numbers Objective: I can evaluate an expression by replacing a variable with a number. Guiding Question: (Write your answer on the lines below.) How do you evaluate an expression that contains variables?
4 Lesson 7 Replacing Letters with Numbers Expression a math expression that includes variables, numbers, and symbols Variable a letter that represents one number Unit a term used as the measurement of a number Formula for Area of a Square A = L W Formula for Perimeter of a Square P = L + W + L + W
5 Lesson 7 Replacing Letters with Numbers Notes: Expressions can also include variables which are letters used to represent numbers. We evaluate expressions by: 1. Replacing the variables with numbers (each variable will represent one number) 2. Using the Order of Operations to solve When we evaluate the area of a shape, we include the units. If the units are unknown, we keep the term units. For area we say units squared or square units. EX: A square as a length of 12 and width of 4. The area for the square can be expressed as The area for the square is 48. Since we do not know the measurement, the area is 48 units squared.
6 Lesson 7 Replacing Letters with Numbers Examples: Remember! A variable represents one number. An expression can be evaluated to one number. The length of this shape is 3 units. The width of this shape is also 3 units. The area for a square is l x w. Therefore, we can express the area for this square as 3 x 3. We evaluate this expression to 9 units squared or 9 square units.
7 Lesson 7 Replacing Letters with Numbers Examples: Since 8 cm and b cm are opposites in the first rectangle, they will have the same value. Therefore we know that the value of b = 8. We can use this same strategy to find the length of one side of the second rectangle. Each section is 4 cm long. Therefore the length can be expressed as 4 + (2 x 4) + 4. Using order of operations, we evaluate that to = = 16. The length of one side of the second rectangle is 16 cm 2.
8 Lesson 7 Replacing Letters with Numbers Examples: Volume = l x w x h We use the formula above to find the volume of geometric shapes such as prisms. The volume for the shape below can be expressed as 8 x 2 x 6 = 8 x 12 = 96 cm 3
9 Lesson 7 Replacing Letters with Numbers
10 Lesson 8 Replacing Numbers with Letters Objective: I can justify the rules of math properties by replacing numbers and letters. Guiding Question: (Write your answer on the lines below.) Give your own example of each of the math properties discussed in this lesson.
11 Lesson 8 Replacing Numbers with Letters Math Properties: Commutative Property of Addition Changing the order of the numbers does not change the value of the expression when adding = a + b = b + a Commutative Property of Multiplication Changing the order of the numbers does not change the value of the expression when multiplying. 2 x 4 = 4 x 2 a x b = b x a Additive Identity Property of Zero The value of a number does not change when zero is added to it. Any number added to zero equals the original number = 1 n + 0 = n Multiplicative Identity Property of One The value of a number does not change when multiplied by one. Any number multiplied by one equals the original number. 5 x 1 = 5 n x 1 = n
12 Lesson 8 Replacing Numbers with Letters
13 Lesson 8 Replacing Numbers with Letters
14 Lesson 8 Replacing Numbers with Letters
15 Lesson 8 Replacing Numbers with Letters Lesson Summary: We can replace any number with a variable to prove that each of the properties discussed in this lesson are true. Examples: Commutative Property of Addition = a + b = b + a Commutative Property of Multiplication 2 x 4 = 4 x 2 a x b = b x a Additive Identity Property of Zero = 1 n + 0 = n Multiplicative Identity Property of One 5 x 1 = 5 n x 1 = n
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