my homework lesson 7 multiply by 0 and 1

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My Math 3 Volume 1 Common Core, Grade: 3 Publisher: McGraw-Hill

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Curriculum  /  Math  /  3rd Grade  /  Unit 2: Multiplication and Division, Part 1  /  Lesson 1

Multiplication and Division, Part 1

Lesson 1 of 21

Criteria for Success

Tips for teachers, anchor tasks.

Problem Set

Target Task

Additional practice.

Identify and create situations involving equal groups and describe these situations using the language and notation of multiplication.

Common Core Standards

Core standards.

The core standards covered in this lesson

Operations and Algebraic Thinking

3.OA.A.1 — Interpret products of whole numbers, e.g., interpret 5 × 7 as the total number of objects in 5 groups of 7 objects each. For example, describe a context in which a total number of objects can be expressed as 5 × 7.

Foundational Standards

The foundational standards covered in this lesson

Number and Operations in Base Ten

2.NBT.A.2 — Count within 1000; skip-count by 5s, 10s, and 100s.

2.OA.C.3 — Determine whether a group of objects (up to 20) has an odd or even number of members, e.g., by pairing objects or counting them by 2s; write an equation to express an even number as a sum of two equal addends.

2.OA.C.4 — Use addition to find the total number of objects arranged in rectangular arrays with up to 5 rows and up to 5 columns; write an equation to express the total as a sum of equal addends.

The essential concepts students need to demonstrate or understand to achieve the lesson objective

• Understand that equal groups  are groups made up of equal numbers of objects in each group. 
• Understand that multiplication is the operation that tells you the total number of objects when you have a certain number of equal groups and use the multiplication symbol to represent equal groups (MP.2).
• Understand that multiplication cannot be used to represent situations with unequal groups.
• Understand that the way we write a multiplication equation is a convention where the first factor represents the number of groups and the second factor represents the size of the groups (MP.6). 
• Identify the number of groups, the size of each group, and the total number of objects in a context or picture of equal groups. 
• Understand that a factor is the number of groups or size of each group in a multiplication equation and the product is the total number of objects in a multiplication equation. 

Suggestions for teachers to help them teach this lesson

• Students will learn to write factors in the conventional order, namely with the number of groups written as the first factor and the size of the groups as the second factor. But, since "some students bring th[e] interpretation of multiplication equations [where the meaning of the factors is switched] into the classroom, …it is useful to discuss the different interpretations and allow students to use whichever is used in their home" (OA Progression, p. 25), if it comes up. Thus, use the conventional way of writing factors when addressing the whole class, but allow individual students to write them in whatever order they like so long as they are able to explain the meaning of each factor. In Lesson 7, students will learn the commutative property, after which all students will have even more flexibility in the order in which they write their factors.
• In this lesson, finding the product is introduced in the context of writing equations, but efficient methods for finding the product are not discussed until Topics B and C. For now, because images are included with every problem, students can count all (Level 1 strategy) to find the total, or they may see that skip-counting or repeated addition (Level 2 strategies) can be used to find the total. Focus on the meaning of multiplication here, since they will have lots of time to discuss how to compute in later lessons. 
• Choral Counting: Choral counting is simply skip-counting out loud as a whole class. It’s a good routine to use when learning new skip-counting sequences or when you notice many students struggling with a particular skip-counting sequence. You could use a visual to help students with their counting, especially in the beginning, such as a hundreds chart. This also helps students notice patterns in the count sequence. You can involve movement, as well, by having students count on their fingers, do jumping jacks with every count, etc. It might be difficult to catch incorrect counts from students, which could reinforce incorrect counts, so you might transition to the routine below once students seem familiar enough with the count-by to do so.
• If we count by twos around the circle starting with Student A, what number do you think Student Z will say? If you didn’t count to figure that out, how did you solve? 
• If we count around the circle by fives and we go around twice, what will Student X say?
• Why did you choose ____ as an estimate?
• Why didn’t anyone choose ____ as an estimate?
• How did you know what comes next?
• (After a child gets stuck but figures it out): What did you do to figure it out? 
• (If you’ve written the numbers on the board as students counted): What do you notice? What do you wonder?
• As a supplement to the Problem Set, you can teach students the game Circles and Stars  from YouCubed.com. (Since students will only work with factors 2–5 and 10 in this unit, you could either block out the 1 on the dice, tell them to roll again if they encounter that factor, or tell them to treat 1 as 10. Similarly, you’ll need to tell students what to do if they roll double sixes (although 6 with any factor besides itself and 1 is fine)).

Lesson Materials

• Counters (20 per student or small group) — Students could use a common classroom material, such as paperclips, instead.

Unlock features to optimize your prep time, plan engaging lessons, and monitor student progress.

Tasks designed to teach criteria for success of the lesson, and guidance to help draw out student understanding

a.   Would you rather have Group A or Group B of Reese’s Pieces? Why?

b.   Would you rather have Group A or Group B of Milk Duds? Why?

Guiding Questions

a.   The $$2$$  groups of $$8$$  Milk Duds in Anchor Task 1 can be represented with the expression " $$2\times8$$ ," which we can read out loud as "two groups of eight," "two times eight," or "two multiplied by eight." We call this a multiplication expression whose factors are $$2$$  and $$8$$ . 

Using your counters, make $$6$$  groups of $$3$$ . Then write a multiplication expression that matches your model. 

b.   The total number of objects in equal groups is called a product . So the product of  $$2\times8$$  is  $$16$$  since that's how many total Milk Duds there are. We can write a multiplication equation to represent this,  $$2\times8 = 16$$ .

What is the product in your model? Then write a multiplication equation that matches your model, including the product of it.

Write a multiplication equation to represent the following picture.

Unlock the answer keys for this lesson's problem set and extra practice problems to save time and support student learning.

Discussion of Problem Set

• Identify the factors and their meanings from each image in #1. 
• What did you notice about the answers to Parts (a) and (b) and Parts (c) and (d) in #1? 
• Which models were correct in #5? How do you know? Which models were incorrect? Why?

A task that represents the peak thinking of the lesson - mastery will indicate whether or not objective was achieved

For each image below, 

• Determine whether it has equal groups. Explain how you know.
• If it does have equal groups, write a multiplication equation to represent it.

Student Response

The Extra Practice Problems can be used as additional practice for homework, during an intervention block, etc. Daily Word Problems and Fluency Activities are aligned to the content of the unit but not necessarily to the lesson objective, therefore feel free to use them anytime during your school day.

Extra Practice Problems

Answer keys for Problem Sets and Extra Practice Problems are available with a Fishtank Plus subscription.

Word Problems and Fluency Activities

Help students strengthen their application and fluency skills with daily word problem practice and content-aligned fluency activities.

Topic A: The Meaning of Multiplication and Division

Identify and create situations involving arrays and describe these situations using the language and notation of multiplication.

Identify and create situations involving unknown group size and find group size in situations.

3.OA.A.1 3.OA.A.2 3.OA.A.3

Identify and create situations involving an unknown number of groups and find the number of groups in situations.

Relate multiplication and division and understand that division can represent situations of unknown group size or an unknown number of groups.

3.OA.A.1 3.OA.A.2 3.OA.A.3 3.OA.B.6

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Topic B: Multiplication and Division by 2, 5, and 10

Build fluency with multiplication facts using units of 2, 5, and 10.

3.OA.A.1 3.OA.C.7

Demonstrate the commutativity of multiplication.

Build fluency with division facts using units of 2, 5, and 10.

3.OA.A.2 3.OA.B.6 3.OA.C.7

Solve one-step word problems involving multiplication and division using units of 2, 5, and 10.

Topic C: Multiplication and Division by 3 and 4

Build fluency with multiplication and division facts using units of 3.

3.OA.A.1 3.OA.A.2 3.OA.B.5 3.OA.B.6 3.OA.C.7

Build fluency with multiplication and division facts using units of 4.

Solve one-step word problems involving multiplication and division using units of 3 and 4.

Topic D: More Complex Multiplication and Division Problems

Determine the unknown whole number in a multiplication or division equation relating three whole numbers, including equations with a letter standing for the unknown quantity.

3.OA.A.4 3.OA.C.7 3.OA.D.8

Solve one-step word problems involving multiplication and division and write problem contexts to match expressions and equations.

Solve two-step word problems involving multiplication and division.

Solve two-step word problems involving all four operations.

Topic E: Scaled Picture and Bar Graphs

Create scaled picture graphs where the scale is provided.

Create scaled picture graphs where the scale must be determined.

Create scaled bar graphs where the scale is provided.

Create scaled bar graphs where the scale must be determined.

Solve one- and two-step word problems using information presented in scaled picture and bar graphs.

3.MD.B.3 3.OA.D.8

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ALGEBRA • MULTIPLY WITH 1 and 0 Chapter 3 Lesson 7 - 3rd Grade Go Math

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Do you find yourself searching for that perfect resource to accompany your 3rd grade Go Math curriculum? These Algebra Multiply With 1 and 0 activities for chapter 3 lesson 7 are filled with necessary practice to reinforce and enhance your lesson. No need to use your allotted paper supply on practice materials! All google slides are interactive and provide intriguing methods that will help students master math skills presented in the lesson.

Three Google Files included:

Google Slides

• Interactive practice slides allow students to type and click and drag their answers.
• Designed to be interesting and easy for students to manipulate.
• Use in Google Classroom.
• Great for centers, homework, seat work, morning work.
• Engaging practice of skills taught in the lesson.

Google Forms

• Great way to get an extra daily grade!
• Review Quiz measures how well students grasp skills taught. Self-graded in real-time, allowing both you and your students to view answers immediately.
• Individual results give each student’s response that allows you to know which skill needs to be retaught to which student.
• The questions are in Multiple Choice and Type Written format.

Google Slides Teacher Answer Key

• Provides answers to interactive activities

SKILLS ADDRESSED:

• Identity Property of Multiplication
• Zero Property of Multiplication
• Finding Products
• Multiplying With 1 and 0
• Multiplication Facts Practice (2's - 5's) - BONUS

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• 3rd Grade Math
• Title: My Math
• Author: McGraw Hill
• Edition: Volume 1

Lesson 1: Coming soon

Lesson 2: Coming soon

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7.1 Multiply and Divide Rational Expressions

Learning objectives.

By the end of this section, you will be able to:

• Determine the values for which a rational expression is undefined
• Simplify rational expressions
• Multiply rational expressions
• Divide rational expressions
• Multiply and divide rational functions

Be Prepared 7.1

Before you get started, take this readiness quiz.

Simplify: 90 y 15 y 2 . 90 y 15 y 2 . If you missed this problem, review Example 5.13 .

Be Prepared 7.2

Multiply: 14 15 · 6 35 . 14 15 · 6 35 . If you missed this problem, review Example 1.25 .

Be Prepared 7.3

Divide: 12 10 ÷ 8 25 . 12 10 ÷ 8 25 . If you missed this problem, review Example 1.26 .

We previously reviewed the properties of fractions and their operations. We introduced rational numbers, which are just fractions where the numerators and denominators are integers. In this chapter, we will work with fractions whose numerators and denominators are polynomials. We call this kind of expression a rational expression .

Rational Expression

A rational expression is an expression of the form p q , p q , where p and q are polynomials and q ≠ 0 . q ≠ 0 .

Here are some examples of rational expressions:

Notice that the first rational expression listed above, − 24 56 − 24 56 , is just a fraction. Since a constant is a polynomial with degree zero, the ratio of two constants is a rational expression, provided the denominator is not zero.

We will do the same operations with rational expressions that we did with fractions. We will simplify, add, subtract, multiply, divide and use them in applications.

Determine the Values for Which a Rational Expression is Undefined

If the denominator is zero, the rational expression is undefined. The numerator of a rational expression may be 0—but not the denominator.

When we work with a numerical fraction, it is easy to avoid dividing by zero because we can see the number in the denominator. In order to avoid dividing by zero in a rational expression, we must not allow values of the variable that will make the denominator be zero.

So before we begin any operation with a rational expression, we examine it first to find the values that would make the denominator zero. That way, when we solve a rational equation for example, we will know whether the algebraic solutions we find are allowed or not.

Determine the values for which a rational expression is undefined.

• Step 1. Set the denominator equal to zero.
• Step 2. Solve the equation.

Example 7.1

Determine the value for which each rational expression is undefined:

ⓐ 8 a 2 b 3 c 8 a 2 b 3 c ⓑ 4 b − 3 2 b + 5 4 b − 3 2 b + 5 ⓒ x + 4 x 2 + 5 x + 6 . x + 4 x 2 + 5 x + 6 .

The expression will be undefined when the denominator is zero.

Determine the value for which each rational expression is undefined.

ⓐ 3 y 2 8 x 3 y 2 8 x ⓑ 8 n − 5 3 n + 1 8 n − 5 3 n + 1 ⓒ a + 10 a 2 + 4 a + 3 a + 10 a 2 + 4 a + 3

ⓐ 4 p 5 q 4 p 5 q ⓑ y − 1 3 y + 2 y − 1 3 y + 2 ⓒ m − 5 m 2 + m − 6 m − 5 m 2 + m − 6

Simplify Rational Expressions

A fraction is considered simplified if there are no common factors, other than 1, in its numerator and denominator. Similarly, a simplified rational expression has no common factors, other than 1, in its numerator and denominator.

Simplified Rational Expression

A rational expression is considered simplified if there are no common factors in its numerator and denominator.

For example,

We use the Equivalent Fractions Property to simplify numerical fractions. We restate it here as we will also use it to simplify rational expressions.

Equivalent Fractions Property

If a , b , and c are numbers where b ≠ 0 , c ≠ 0 , b ≠ 0 , c ≠ 0 ,

Notice that in the Equivalent Fractions Property, the values that would make the denominators zero are specifically disallowed. We see b ≠ 0 , c ≠ 0 b ≠ 0 , c ≠ 0 clearly stated.

To simplify rational expressions, we first write the numerator and denominator in factored form. Then we remove the common factors using the Equivalent Fractions Property.

Be very careful as you remove common factors. Factors are multiplied to make a product. You can remove a factor from a product. You cannot remove a term from a sum.

Removing the x ’s from x + 5 x x + 5 x would be like cancelling the 2’s in the fraction 2 + 5 2 ! 2 + 5 2 !

Example 7.2

How to simplify a rational expression.

Simplify: x 2 + 5 x + 6 x 2 + 8 x + 12 x 2 + 5 x + 6 x 2 + 8 x + 12 .

Simplify: x 2 − x − 2 x 2 − 3 x + 2 . x 2 − x − 2 x 2 − 3 x + 2 .

Simplify: x 2 − 3 x − 10 x 2 + x − 2 . x 2 − 3 x − 10 x 2 + x − 2 .

We now summarize the steps you should follow to simplify rational expressions.

Simplify a rational expression.

• Step 1. Factor the numerator and denominator completely.
• Step 2. Simplify by dividing out common factors.

Usually, we leave the simplified rational expression in factored form. This way, it is easy to check that we have removed all the common factors.

We’ll use the methods we have learned to factor the polynomials in the numerators and denominators in the following examples.

Every time we write a rational expression, we should make a statement disallowing values that would make a denominator zero. However, to let us focus on the work at hand, we will omit writing it in the examples.

Example 7.3

Simplify: 3 a 2 − 12 a b + 12 b 2 6 a 2 − 24 b 2 3 a 2 − 12 a b + 12 b 2 6 a 2 − 24 b 2 .

Simplify: 2 x 2 − 12 x y + 18 y 2 3 x 2 − 27 y 2 2 x 2 − 12 x y + 18 y 2 3 x 2 − 27 y 2 .

Simplify: 5 x 2 − 30 x y + 25 y 2 2 x 2 − 50 y 2 5 x 2 − 30 x y + 25 y 2 2 x 2 − 50 y 2 .

Now we will see how to simplify a rational expression whose numerator and denominator have opposite factors. We previously introduced opposite notation: the opposite of a is − a − a and − a = −1 · a . − a = −1 · a .

The numerical fraction, say 7 −7 7 −7 simplifies to −1 −1 . We also recognize that the numerator and denominator are opposites.

The fraction a − a a − a , whose numerator and denominator are opposites also simplifies to −1 −1 .

This tells us that b − a b − a is the opposite of a − b . a − b .

In general, we could write the opposite of a − b a − b as b − a . b − a . So the rational expression a − b b − a a − b b − a simplifies to −1 . −1 .

Opposites in a Rational Expression

The opposite of a − b a − b is b − a . b − a .

An expression and its opposite divide to −1 . −1 .

We will use this property to simplify rational expressions that contain opposites in their numerators and denominators. Be careful not to treat a + b a + b and b + a b + a as opposites. Recall that in addition, order doesn’t matter so a + b = b + a a + b = b + a . So if a ≠ − b a ≠ − b , then a + b b + a = 1 . a + b b + a = 1 .

Example 7.4

Simplify: x 2 − 4 x − 32 64 − x 2 . x 2 − 4 x − 32 64 − x 2 .

Simplify: x 2 − 4 x − 5 25 − x 2 . x 2 − 4 x − 5 25 − x 2 .

Simplify: x 2 + x − 2 1 − x 2 . x 2 + x − 2 1 − x 2 .

Multiply Rational Expressions

To multiply rational expressions, we do just what we did with numerical fractions. We multiply the numerators and multiply the denominators. Then, if there are any common factors, we remove them to simplify the result.

Multiplication of Rational Expressions

If p , q , r , and s are polynomials where q ≠ 0 , s ≠ 0 , q ≠ 0 , s ≠ 0 , then

To multiply rational expressions, multiply the numerators and multiply the denominators.

Remember, throughout this chapter, we will assume that all numerical values that would make the denominator be zero are excluded. We will not write the restrictions for each rational expression, but keep in mind that the denominator can never be zero. So in this next example, x ≠ 0 , x ≠ 0 , x ≠ 3 , x ≠ 3 , and x ≠ 4 . x ≠ 4 .

Example 7.5

How to multiply rational expressions.

Simplify: 2 x x 2 − 7 x + 12 · x 2 − 9 6 x 2 . 2 x x 2 − 7 x + 12 · x 2 − 9 6 x 2 .

Simplify: 5 x x 2 + 5 x + 6 · x 2 − 4 10 x . 5 x x 2 + 5 x + 6 · x 2 − 4 10 x .

Try It 7.10

Simplify: 9 x 2 x 2 + 11 x + 30 · x 2 − 36 3 x 2 . 9 x 2 x 2 + 11 x + 30 · x 2 − 36 3 x 2 .

Multiply rational expressions.

• Step 1. Factor each numerator and denominator completely.
• Step 2. Multiply the numerators and denominators.
• Step 3. Simplify by dividing out common factors.

Example 7.6

Multiply: 3 a 2 − 8 a − 3 a 2 − 25 · a 2 + 10 a + 25 3 a 2 − 14 a − 5 . 3 a 2 − 8 a − 3 a 2 − 25 · a 2 + 10 a + 25 3 a 2 − 14 a − 5 .

Try It 7.11

Simplify: 2 x 2 + 5 x − 12 x 2 − 16 · x 2 − 8 x + 16 2 x 2 − 13 x + 15 . 2 x 2 + 5 x − 12 x 2 − 16 · x 2 − 8 x + 16 2 x 2 − 13 x + 15 .

Try It 7.12

Simplify: 4 b 2 + 7 b − 2 1 − b 2 · b 2 − 2 b + 1 4 b 2 + 15 b − 4 . 4 b 2 + 7 b − 2 1 − b 2 · b 2 − 2 b + 1 4 b 2 + 15 b − 4 .

Divide Rational Expressions

Just like we did for numerical fractions, to divide rational expressions, we multiply the first fraction by the reciprocal of the second.

Division of Rational Expressions

If p , q , r, and s are polynomials where q ≠ 0 , r ≠ 0 , s ≠ 0 , q ≠ 0 , r ≠ 0 , s ≠ 0 , then

To divide rational expressions, multiply the first fraction by the reciprocal of the second.

Once we rewrite the division as multiplication of the first expression by the reciprocal of the second, we then factor everything and look for common factors.

Example 7.7

How to divide rational expressions.

Divide: p 3 + q 3 2 p 2 + 2 p q + 2 q 2 ÷ p 2 − q 2 6 . p 3 + q 3 2 p 2 + 2 p q + 2 q 2 ÷ p 2 − q 2 6 .

Try It 7.13

Simplify: x 3 + 8 3 x 2 − 6 x + 12 ÷ x 2 − 4 6 . x 3 + 8 3 x 2 − 6 x + 12 ÷ x 2 − 4 6 .

Try It 7.14

Simplify: 2 z 2 z 2 − 1 ÷ z 3 − z 2 + z z 3 + 1 . 2 z 2 z 2 − 1 ÷ z 3 − z 2 + z z 3 + 1 .

Divide rational expressions.

• Step 1. Rewrite the division as the product of the first rational expression and the reciprocal of the second.
• Step 2. Factor the numerators and denominators completely.
• Step 3. Multiply the numerators and denominators together.
• Step 4. Simplify by dividing out common factors.

Recall from Use the Language of Algebra that a complex fraction is a fraction that contains a fraction in the numerator, the denominator or both. Also, remember a fraction bar means division. A complex fraction is another way of writing division of two fractions.

Example 7.8

Divide: 6 x 2 − 7 x + 2 4 x − 8 2 x 2 − 7 x + 3 x 2 − 5 x + 6 . 6 x 2 − 7 x + 2 4 x − 8 2 x 2 − 7 x + 3 x 2 − 5 x + 6 .

Try It 7.15

Simplify: 3 x 2 + 7 x + 2 4 x + 24 3 x 2 − 14 x − 5 x 2 + x − 30 . 3 x 2 + 7 x + 2 4 x + 24 3 x 2 − 14 x − 5 x 2 + x − 30 .

Try It 7.16

Simplify: y 2 − 36 2 y 2 + 11 y − 6 2 y 2 − 2 y − 60 8 y − 4 . y 2 − 36 2 y 2 + 11 y − 6 2 y 2 − 2 y − 60 8 y − 4 .

If we have more than two rational expressions to work with, we still follow the same procedure. The first step will be to rewrite any division as multiplication by the reciprocal. Then, we factor and multiply.

Example 7.9

Perform the indicated operations: 3 x − 6 4 x − 4 · x 2 + 2 x − 3 x 2 − 3 x − 10 ÷ 2 x + 12 8 x + 16 . 3 x − 6 4 x − 4 · x 2 + 2 x − 3 x 2 − 3 x − 10 ÷ 2 x + 12 8 x + 16 .

Try It 7.17

Perform the indicated operations: 4 m + 4 3 m − 15 · m 2 − 3 m − 10 m 2 − 4 m − 32 ÷ 12 m − 36 6 m − 48 . 4 m + 4 3 m − 15 · m 2 − 3 m − 10 m 2 − 4 m − 32 ÷ 12 m − 36 6 m − 48 .

Try It 7.18

Perform the indicated operations: 2 n 2 + 10 n n − 1 ÷ n 2 + 10 n + 24 n 2 + 8 n − 9 · n + 4 8 n 2 + 12 n . 2 n 2 + 10 n n − 1 ÷ n 2 + 10 n + 24 n 2 + 8 n − 9 · n + 4 8 n 2 + 12 n .

Multiply and Divide Rational Functions

We started this section stating that a rational expression is an expression of the form p q , p q , where p and q are polynomials and q ≠ 0 . q ≠ 0 . Similarly, we define a rational function as a function of the form R ( x ) = p ( x ) q ( x ) R ( x ) = p ( x ) q ( x ) where p ( x ) p ( x ) and q ( x ) q ( x ) are polynomial functions and q ( x ) q ( x ) is not zero.

Rational Function

A rational function is a function of the form

where p ( x ) p ( x ) and q ( x ) q ( x ) are polynomial functions and q ( x ) q ( x ) is not zero.

The domain of a rational function is all real numbers except for those values that would cause division by zero. We must eliminate any values that make q ( x ) = 0 . q ( x ) = 0 .

Determine the domain of a rational function.

• Step 3. The domain is all real numbers excluding the values found in Step 2.

Example 7.10

Find the domain of R ( x ) = 2 x 2 − 14 x 4 x 2 − 16 x − 48 . R ( x ) = 2 x 2 − 14 x 4 x 2 − 16 x − 48 .

The domain will be all real numbers except those values that make the denominator zero. We will set the denominator equal to zero , solve that equation, and then exclude those values from the domain.

Try It 7.19

Find the domain of R ( x ) = 2 x 2 − 10 x 4 x 2 − 16 x − 20 . R ( x ) = 2 x 2 − 10 x 4 x 2 − 16 x − 20 .

Try It 7.20

Find the domain of R ( x ) = 4 x 2 − 16 x 8 x 2 − 16 x − 64 . R ( x ) = 4 x 2 − 16 x 8 x 2 − 16 x − 64 .

To multiply rational functions, we multiply the resulting rational expressions on the right side of the equation using the same techniques we used to multiply rational expressions.

Example 7.11

Find R ( x ) = f ( x ) · g ( x ) R ( x ) = f ( x ) · g ( x ) where f ( x ) = 2 x − 6 x 2 − 8 x + 15 f ( x ) = 2 x − 6 x 2 − 8 x + 15 and g ( x ) = x 2 − 25 2 x + 10 . g ( x ) = x 2 − 25 2 x + 10 .

Try It 7.21

Find R ( x ) = f ( x ) · g ( x ) R ( x ) = f ( x ) · g ( x ) where f ( x ) = 3 x − 21 x 2 − 9 x + 14 f ( x ) = 3 x − 21 x 2 − 9 x + 14 and g ( x ) = 2 x 2 − 8 3 x + 6 . g ( x ) = 2 x 2 − 8 3 x + 6 .

Try It 7.22

Find R ( x ) = f ( x ) · g ( x ) R ( x ) = f ( x ) · g ( x ) where f ( x ) = x 2 − x 3 x 2 + 27 x − 30 f ( x ) = x 2 − x 3 x 2 + 27 x − 30 and g ( x ) = x 2 − 100 x 2 − 10 x . g ( x ) = x 2 − 100 x 2 − 10 x .

To divide rational functions, we divide the resulting rational expressions on the right side of the equation using the same techniques we used to divide rational expressions.

Example 7.12

Find R ( x ) = f ( x ) g ( x ) R ( x ) = f ( x ) g ( x ) where f ( x ) = 3 x 2 x 2 − 4 x f ( x ) = 3 x 2 x 2 − 4 x and g ( x ) = 9 x 2 − 45 x x 2 − 7 x + 10 . g ( x ) = 9 x 2 − 45 x x 2 − 7 x + 10 .

Try It 7.23

Find R ( x ) = f ( x ) g ( x ) R ( x ) = f ( x ) g ( x ) where f ( x ) = 2 x 2 x 2 − 8 x f ( x ) = 2 x 2 x 2 − 8 x and g ( x ) = 8 x 2 + 24 x x 2 + x − 6 . g ( x ) = 8 x 2 + 24 x x 2 + x − 6 .

Try It 7.24

Find R ( x ) = f ( x ) g ( x ) R ( x ) = f ( x ) g ( x ) where f ( x ) = 15 x 2 3 x 2 + 33 x f ( x ) = 15 x 2 3 x 2 + 33 x and g ( x ) = 5 x − 5 x 2 + 9 x − 22 . g ( x ) = 5 x − 5 x 2 + 9 x − 22 .

Section 7.1 Exercises

Practice makes perfect.

In the following exercises, determine the values for which the rational expression is undefined.

ⓐ 2 x 2 z 2 x 2 z , ⓑ 4 p − 1 6 p − 5 4 p − 1 6 p − 5 , ⓒ n − 3 n 2 + 2 n − 8 n − 3 n 2 + 2 n − 8

ⓐ 10 m 11 n 10 m 11 n , ⓑ 6 y + 13 4 y − 9 6 y + 13 4 y − 9 , ⓒ b − 8 b 2 − 36 b − 8 b 2 − 36

ⓐ 4 x 2 y 3 y 4 x 2 y 3 y , ⓑ 3 x − 2 2 x + 1 3 x − 2 2 x + 1 , ⓒ u − 1 u 2 − 3 u − 28 u − 1 u 2 − 3 u − 28

ⓐ 5 p q 2 9 q 5 p q 2 9 q , ⓑ 7 a − 4 3 a + 5 7 a − 4 3 a + 5 , ⓒ 1 x 2 − 4 1 x 2 − 4

In the following exercises, simplify each rational expression.

− 44 55 − 44 55

56 63 56 63

8 m 3 n 12 m n 2 8 m 3 n 12 m n 2

36 v 3 w 2 27 v w 3 36 v 3 w 2 27 v w 3

8 n − 96 3 n − 36 8 n − 96 3 n − 36

12 p − 240 5 p − 100 12 p − 240 5 p − 100

x 2 + 4 x − 5 x 2 − 2 x + 1 x 2 + 4 x − 5 x 2 − 2 x + 1

y 2 + 3 y − 4 y 2 − 6 y + 5 y 2 + 3 y − 4 y 2 − 6 y + 5

a 2 − 4 a 2 + 6 a − 16 a 2 − 4 a 2 + 6 a − 16

y 2 − 2 y − 3 y 2 − 9 y 2 − 2 y − 3 y 2 − 9

p 3 + 3 p 2 + 4 p + 12 p 2 + p − 6 p 3 + 3 p 2 + 4 p + 12 p 2 + p − 6

x 3 − 2 x 2 − 25 x + 50 x 2 − 25 x 3 − 2 x 2 − 25 x + 50 x 2 − 25

8 b 2 − 32 b 2 b 2 − 6 b − 80 8 b 2 − 32 b 2 b 2 − 6 b − 80

−5 c 2 − 10 c −10 c 2 + 30 c + 100 −5 c 2 − 10 c −10 c 2 + 30 c + 100

3 m 2 + 30 m n + 75 n 2 4 m 2 − 100 n 2 3 m 2 + 30 m n + 75 n 2 4 m 2 − 100 n 2

5 r 2 + 30 r s − 35 s 2 r 2 − 49 s 2 5 r 2 + 30 r s − 35 s 2 r 2 − 49 s 2

a − 5 5 − a a − 5 5 − a

5 − d d − 5 5 − d d − 5

20 − 5 y y 2 − 16 20 − 5 y y 2 − 16

4 v − 32 64 − v 2 4 v − 32 64 − v 2

w 3 + 216 w 2 − 36 w 3 + 216 w 2 − 36

v 3 + 125 v 2 − 25 v 3 + 125 v 2 − 25

z 2 − 9 z + 20 16 − z 2 z 2 − 9 z + 20 16 − z 2

a 2 − 5 a − 36 81 − a 2 a 2 − 5 a − 36 81 − a 2

In the following exercises, multiply the rational expressions.

12 16 · 4 10 12 16 · 4 10

32 5 · 16 24 32 5 · 16 24

5 x 2 y 4 12 x y 3 · 6 x 2 20 y 2 5 x 2 y 4 12 x y 3 · 6 x 2 20 y 2

12 a 3 b b 2 · 2 a b 2 9 b 3 12 a 3 b b 2 · 2 a b 2 9 b 3

5 p 2 p 2 − 5 p − 36 · p 2 − 16 10 p 5 p 2 p 2 − 5 p − 36 · p 2 − 16 10 p

3 q 2 q 2 + q − 6 · q 2 − 9 9 q 3 q 2 q 2 + q − 6 · q 2 − 9 9 q

2 y 2 − 10 y y 2 + 10 y + 25 · y + 5 6 y 2 y 2 − 10 y y 2 + 10 y + 25 · y + 5 6 y

z 2 + 3 z z 2 − 3 z − 4 · z − 4 z 2 z 2 + 3 z z 2 − 3 z − 4 · z − 4 z 2

28 − 4 b 3 b − 3 · b 2 + 8 b − 9 b 2 − 49 28 − 4 b 3 b − 3 · b 2 + 8 b − 9 b 2 − 49

72 m − 12 m 2 8 m + 32 · m 2 + 10 m + 24 m 2 − 36 72 m − 12 m 2 8 m + 32 · m 2 + 10 m + 24 m 2 − 36

3 c 2 − 16 c + 5 c 2 − 25 · c 2 + 10 c + 25 3 c 2 − 14 c − 5 3 c 2 − 16 c + 5 c 2 − 25 · c 2 + 10 c + 25 3 c 2 − 14 c − 5

2 d 2 + d − 3 d 2 − 16 · d 2 − 8 d + 16 2 d 2 − 9 d − 18 2 d 2 + d − 3 d 2 − 16 · d 2 − 8 d + 16 2 d 2 − 9 d − 18

6 m 2 − 13 m + 2 9 − m 2 · m 2 − 6 m + 9 6 m 2 + 23 m − 4 6 m 2 − 13 m + 2 9 − m 2 · m 2 − 6 m + 9 6 m 2 + 23 m − 4

2 n 2 − 3 n − 14 25 − n 2 · n 2 − 10 n + 25 2 n 2 − 13 n + 21 2 n 2 − 3 n − 14 25 − n 2 · n 2 − 10 n + 25 2 n 2 − 13 n + 21

In the following exercises, divide the rational expressions.

v − 5 11 − v ÷ v 2 − 25 v − 11 v − 5 11 − v ÷ v 2 − 25 v − 11

10 + w w − 8 ÷ 100 − w 2 8 − w 10 + w w − 8 ÷ 100 − w 2 8 − w

3 s 2 s 2 − 16 ÷ s 3 + 4 s 2 + 16 s s 3 − 64 3 s 2 s 2 − 16 ÷ s 3 + 4 s 2 + 16 s s 3 − 64

r 2 − 9 15 ÷ r 3 − 27 5 r 2 + 15 r + 45 r 2 − 9 15 ÷ r 3 − 27 5 r 2 + 15 r + 45

p 3 + q 3 3 p 2 + 3 p q + 3 q 2 ÷ p 2 − q 2 12 p 3 + q 3 3 p 2 + 3 p q + 3 q 2 ÷ p 2 − q 2 12

v 3 − 8 w 3 2 v 2 + 4 v w + 8 w 2 ÷ v 2 − 4 w 2 4 v 3 − 8 w 3 2 v 2 + 4 v w + 8 w 2 ÷ v 2 − 4 w 2 4

x 2 + 3 x − 10 4 x ÷ ( 2 x 2 + 20 x + 50 ) x 2 + 3 x − 10 4 x ÷ ( 2 x 2 + 20 x + 50 )

2 y 2 − 10 y z − 48 z 2 2 y − 1 ÷ ( 4 y 2 − 32 y z ) 2 y 2 − 10 y z − 48 z 2 2 y − 1 ÷ ( 4 y 2 − 32 y z )

2 a 2 − a − 21 5 a + 20 a 2 + 7 a + 12 a 2 + 8 a + 16 2 a 2 − a − 21 5 a + 20 a 2 + 7 a + 12 a 2 + 8 a + 16

3 b 2 + 2 b − 8 12 b + 18 3 b 2 + 2 b − 8 2 b 2 − 7 b − 15 3 b 2 + 2 b − 8 12 b + 18 3 b 2 + 2 b − 8 2 b 2 − 7 b − 15

12 c 2 − 12 2 c 2 − 3 c + 1 4 c + 4 6 c 2 − 13 c + 5 12 c 2 − 12 2 c 2 − 3 c + 1 4 c + 4 6 c 2 − 13 c + 5

4 d 2 + 7 d − 2 35 d + 10 d 2 − 4 7 d 2 − 12 d − 4 4 d 2 + 7 d − 2 35 d + 10 d 2 − 4 7 d 2 − 12 d − 4

For the following exercises, perform the indicated operations.

10 m 2 + 80 m 3 m − 9 · m 2 + 4 m − 21 m 2 − 9 m + 20 ÷ 5 m 2 + 10 m 2 m − 10 10 m 2 + 80 m 3 m − 9 · m 2 + 4 m − 21 m 2 − 9 m + 20 ÷ 5 m 2 + 10 m 2 m − 10

4 n 2 + 32 n 3 n + 2 · 3 n 2 − n − 2 n 2 + n − 30 ÷ 108 n 2 − 24 n n + 6 4 n 2 + 32 n 3 n + 2 · 3 n 2 − n − 2 n 2 + n − 30 ÷ 108 n 2 − 24 n n + 6

12 p 2 + 3 p p + 3 ÷ p 2 + 2 p − 63 p 2 − p − 12 · p − 7 9 p 3 − 9 p 2 12 p 2 + 3 p p + 3 ÷ p 2 + 2 p − 63 p 2 − p − 12 · p − 7 9 p 3 − 9 p 2

6 q + 3 9 q 2 − 9 q ÷ q 2 + 14 q + 33 q 2 + 4 q − 5 · 4 q 2 + 12 q 12 q + 6 6 q + 3 9 q 2 − 9 q ÷ q 2 + 14 q + 33 q 2 + 4 q − 5 · 4 q 2 + 12 q 12 q + 6

In the following exercises, find the domain of each function.

R ( x ) = x 3 − 2 x 2 − 25 x + 50 x 2 − 25 R ( x ) = x 3 − 2 x 2 − 25 x + 50 x 2 − 25

R ( x ) = x 3 + 3 x 2 − 4 x − 12 x 2 − 4 R ( x ) = x 3 + 3 x 2 − 4 x − 12 x 2 − 4

R ( x ) = 3 x 2 + 15 x 6 x 2 + 6 x − 36 R ( x ) = 3 x 2 + 15 x 6 x 2 + 6 x − 36

R ( x ) = 8 x 2 − 32 x 2 x 2 − 6 x − 80 R ( x ) = 8 x 2 − 32 x 2 x 2 − 6 x − 80

For the following exercises, find R ( x ) = f ( x ) · g ( x ) R ( x ) = f ( x ) · g ( x ) where f ( x ) f ( x ) and g ( x ) g ( x ) are given.

f ( x ) = 6 x 2 − 12 x x 2 + 7 x − 18 f ( x ) = 6 x 2 − 12 x x 2 + 7 x − 18 g ( x ) = x 2 − 81 3 x 2 − 27 x g ( x ) = x 2 − 81 3 x 2 − 27 x

f ( x ) = x 2 − 2 x x 2 + 6 x − 16 f ( x ) = x 2 − 2 x x 2 + 6 x − 16 g ( x ) = x 2 − 64 x 2 − 8 x g ( x ) = x 2 − 64 x 2 − 8 x

f ( x ) = 4 x x 2 − 3 x − 10 f ( x ) = 4 x x 2 − 3 x − 10 g ( x ) = x 2 − 25 8 x 2 g ( x ) = x 2 − 25 8 x 2

f ( x ) = 2 x 2 + 8 x x 2 − 9 x + 20 f ( x ) = 2 x 2 + 8 x x 2 − 9 x + 20 g ( x ) = x − 5 x 2 g ( x ) = x − 5 x 2

For the following exercises, find R ( x ) = f ( x ) g ( x ) R ( x ) = f ( x ) g ( x ) where f ( x ) f ( x ) and g ( x ) g ( x ) are given.

f ( x ) = 27 x 2 3 x − 21 f ( x ) = 27 x 2 3 x − 21 g ( x ) = 3 x 2 + 18 x x 2 + 13 x + 42 g ( x ) = 3 x 2 + 18 x x 2 + 13 x + 42

f ( x ) = 24 x 2 2 x − 8 f ( x ) = 24 x 2 2 x − 8 g ( x ) = 4 x 3 + 28 x 2 x 2 + 11 x + 28 g ( x ) = 4 x 3 + 28 x 2 x 2 + 11 x + 28

f ( x ) = 16 x 2 4 x + 36 f ( x ) = 16 x 2 4 x + 36 g ( x ) = 4 x 2 − 24 x x 2 + 4 x − 45 g ( x ) = 4 x 2 − 24 x x 2 + 4 x − 45

f ( x ) = 24 x 2 2 x − 4 f ( x ) = 24 x 2 2 x − 4 g ( x ) = 12 x 2 + 36 x x 2 − 11 x + 18 g ( x ) = 12 x 2 + 36 x x 2 − 11 x + 18

Writing Exercises

Explain how you find the values of x for which the rational expression x 2 − x − 20 x 2 − 4 x 2 − x − 20 x 2 − 4 is undefined.

Explain all the steps you take to simplify the rational expression p 2 + 4 p − 21 9 − p 2 . p 2 + 4 p − 21 9 − p 2 .

ⓐ Multiply 7 4 · 9 10 7 4 · 9 10 and explain all your steps. ⓑ Multiply n n − 3 · 9 n + 3 n n − 3 · 9 n + 3 and explain all your steps. ⓒ Evaluate your answer to part ⓑ when n = 7 n = 7 . Did you get the same answer you got in part ⓐ ? Why or why not?

ⓐ Divide 24 5 ÷ 6 24 5 ÷ 6 and explain all your steps. ⓑ Divide x 2 − 1 x ÷ ( x + 1 ) x 2 − 1 x ÷ ( x + 1 ) and explain all your steps. ⓒ Evaluate your answer to part ⓑ when x = 5 . x = 5 . Did you get the same answer you got in part ⓐ ? Why or why not?

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ If most of your checks were:

…confidently. Congratulations! You have achieved your goals in this section! Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific!

…with some help. This must be addressed quickly as topics you do not master become potholes in your road to success. Math is sequential - every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Whom can you ask for help?Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?

…no - I don’t get it! This is critical and you must not ignore it. You need to get help immediately or you will quickly be overwhelmed. See your instructor as soon as possible to discuss your situation. Together you can come up with a plan to get you the help you need.

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• Section URL: https://openstax.org/books/intermediate-algebra-2e/pages/7-1-multiply-and-divide-rational-expressions

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Chapter 7, Lesson 1: Multiplying Monomials

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Math Expressions Grade 3 Unit 1 Lesson 7 Answer Key Multiply and Divide with 10

Solve the questions in Math Expressions Grade 3 Homework and Remembering Answer Key Unit 1 Lesson 7 Answer Key Multiply and Divide with 10 to attempt the exam with higher confidence. https://mathexpressionsanswerkey.com/math-expressions-grade-3-unit-1-lesson-7-answer-key/

Math Expressions Common Core Grade 3 Unit 1 Lesson 7 Answer Key Multiply and Divide with 10

Math Expressions Grade 3 Unit 1 Lesson 7 Homework

Use this chart to practice your 10s count-bys, multiplications, and divisions. Then have your Homework Helper test you.

Multiply or divide to find the unknown numbers. Then check your answers at the bottom of this page.

Math Expressions Lesson 7 Multiply with 10 Answer Key Question 1. 2 × 10 = ___ Answer: 20

Explanation: By multiplying 2 with 10 we get 20.

Multiply and Divide with 10 Grade 5 Answer Key Unit 1 Math Expressions  Question 2. 15 ÷ 5 = ___ Answer: 3

Explanation: By dividing 15 with 5 we get 3.

Math Expressions Unit 1 Multiply and Divide with 10 Grade 3 Answer Key Question 3. 4 * 2 = ___ Answer: 8

Explanation: By multiplying 4 with 2 we get 8.

Question 4. 80/10 = ___ Answer: 8 By dividing 80 with 10 we get 8.

Question 5. 5 • __ = 40 Answer: 8

Explanation: By multiplying 5 with 8 we get 40.

Question 6. \(\frac{60}{10}\) = ___ Answer: 6

Explanation: By dividing 60 with 10 we get 6.

Question 7. ___ × 5 = 30 Answer: 6

Explanation: By multiplying 6 with 5 we get 30.

Question 8. \(\frac{20}{2}\) = ___ Answer: 10

Explanation: By dividing 20 with 2 we get 10.

Question 9. 6 × 10 = __ Answer: 60

Explanation: By multiplying 6 with 10 we get 60.

Question 10. 25/5 = ___ Answer: 5

Explanation: By dividing 25 with 5 we get 5.

Question 11. 10 • 7 = ___ Answer: 70

Explanation: By multiplying 10 with 7 we get 70.

Question 12. 14 ÷ 2 = ___ Answer: 7

Explanation: By dividing 14 with 2 we get 7.

Question 13. 9 * 2 = ___ Answer: 18

Explanation: By multiplying 9 with 2 we get 18.

Question 14. \(\frac{45}{5}\) = ___ Answer: 9

Explanation: by dividing 45 with 5 we get 9.

Question 15. 10 • 4 = ___ Answer: 40

Explanation: by multiplying 10 with 4 we get 40.

Question 17. 70 ÷ 10 = ___ Answer: 7 By dividing 70 with 10 we get 7.

Question 18. 9 * __ = 18 Answer: 2

Explanation: By multiplying 9 with 2 we get 18

Question 19. __ × 5 = 35 Answer: 7 By multiplying 7 with 5 we get 35.

Explanation: By dividing 30 with 3 we get 10.

Question 21. ___ • 2 = ___ Answer: 20

Explanation: by multiplying 10 with 2 we get 20.

Write an equation and solve the problem.

Question 1. Wendy has 100 cents. She wants to buy some marbles that cost 10 cents each. How many marbles can she buy? Answer: 10

Explanation: Given Wendy has 100 cents. She wants to buy some marbles that cost 10 cents each To find how many marbles can she buy we have to divide 100 with 10 we get 10 marbles.

Question 2. Natalie turned off 2 lights in each of the 6 rooms of her house. How many lights did she turn off? Answer: 12

Explanation: Given Natalie turned off 2 lights in each of the 6 rooms of her house. To find how many lights did she turn off We have to multiple 2 with 6 we get 12 lights.

Question 3. Luis has 18 single socks. How many pairs of socks does he have? Answer: 9

Explanation: Given Luis has 18 single socks one pair has 2 socks By dividing 18 with 2 we get 9 pairs

Question 4. Lana has 9 nickels. She wants to buy an apple that costs 40 cents. Does she have enough money? Answer:  yes

Explanation: Given Lana has 9 nickels 1 nickel equals to 5 cents so by multiplying 9 nickel with 5 cents 45 cents. yes she had enough money to buy an apple.

Question 5. Annabelle had 20 crayons. She gave 5 of them to each of her sisters. How many sisters does Annabelle have? Complete the table. Answer:

Question 6. Harvey wrote letters to 10 of his friends. Each letter was 3 pages long. How many pages did Harvey write? Answer:

Complete the table

Math Expressions Grade 3 Unit 1 Lesson 7 Remembering

Write a multiplication equation and a division equation for each problem. Then solve the problem. Show your work.

Question 1. Tara folds 25 sweaters. She puts the same number of sweaters in each pile. There are 5 piles. How many sweaters are in each pile? Answer:

Question 2. Mr. McBride orders 30 new pencils. There are 5 pencils in each box. How many boxes of pencils does Mr. McBride order? Answer:

Multiply or divide to find the unknown numbers.

Question 3. 5 • __ = 45 Answer: 9

Explanation: By multiplying 5 with 9 we 45

Question 5. 14 ÷ 2 = ___ Answer: 7

Question 6. __ * 5 = 20 Answer: 4

Explanation: By multiplying 4 with 5 we get 20.

Question 7. 2 × __ = 8 Answer: 4

Explanation: By multiplying 2 with 4 we get 8.

Question 8. \(\frac{16}{2}\) = ___ Answer: 8

Explanation: By dividing 16 with 2 we get 8.

Question 9. The books were put in 5 equal rows on display. There were 45 books. How many are in each row? Answer: 9

Given The books were put in 5 equal rows on display. There were 45 books By dividing 45 with 5 we get 9 books in each row.

Question 10. The class lined up in 2 rows with 8 students in each row. How many students are in the class? Answer: 16

Given The class lined up in 2 rows with 8 students in each row By multiplying 2 rows with 8 students we get 16 students.

Question 11. Stretch Your Thinking Explain how you know if a number can be divided by 10 evenly. Answer: Numbers that are divisible by 10 need to be even and divisible by 5

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Eureka Math Grade 4 Module 3 Lesson 7 Answer Key

Engage ny eureka math 4th grade module 3 lesson 7 answer key, eureka math grade 4 module 3 lesson 7 sprint answer key.

Question 1. 3 × 2 = Answer: 3 X 2 = 6,

Explanation: Given 3 X 2 = multiplying 3 with 2 we get 6, 3 X 2 = 6.

Question 2. 30 × 2 = Answer: 30 X 2 = 3 X 10 X 2 = 60,

Explanation: Given 30 X 2 = writing 30 as 3 X 10 and multiplying with 2 as  3 X 10 X 2 we get 60, So 30 X 2 = 60.

Question 3. 300 × 2 = Answer: 300 X 2 = 3 X 100 X 2 = 600,

Explanation: Given 300 X 2 = writing 300 as 3 X 100 and multiplying 2 we get 600, 3 X 100 X 2 = 600, so 300 X 2 = 600.

Question 4. 3,000 × 2 = Answer: 3,000 X 2 = 3 X 1,000 X 2 = 6,000,

Explanation: Given 3,000 X 2 = writing 3,000 as 3 X 1,000 and multiplying with 2 we get 6,000 as 3 X 1,000 X 2 = 6,000, So 3,000 X 2 = 6,000.

Question 5. 2 × 3,000 = Answer: 2 X 3,000 = 2 X 3 X 1,000 = 6,000,

Explanation: Given 2 X 3,000 = writing 3,000 as 3 X 1,000 and multiplying with 2 we get 6,000,2 X 3 X 1,000 = 6,000, So 2 X 3,000 = 6,000.

Question 6. 2 × 4 = Answer: 2 X 4 = 8,

Explanation: Given 2 X 4 = multiplying 2 with 4 we get 8, 2 X 4 = 8.

Question 7. 2 × 40 = Answer: 2 X 40 =

2 X 4 X 10 = 80,

Explanation: Given 2 X 40 = writing 40 as 4 X 10 and multiplying with 2 we get 80, as 2 X 4 x 10 = 80, So 2 X 40 = 80.

Question 8. 2 × 400 = Answer: 2 X 400 = 2 X 4 X 100 = 800,

Explanation: Given 2 X 400 = writing 400 as 4 X 100 and multiplying with 2 we get 800 as 2 X 4 X 100 = 800, So 2 X 400 = 800.

Question 9. 2 × 4,000 = Answer: 2 X 4,000 = 2 X 4 X 1,000 = 8,000,

Explanation: Given 2 X 4,000 = writing 4,000 as 4 X 1,000 and multiplying with 2, we get 8,000, as 2 X 4 X 1,000 = 8,000, So 2 X 4,000 = 8,000.

Question 10. 3 × 3 = Answer: 3 X 3 = 9,

Explanation: Given 3 X 3 = multiplying 3 with 3 we get 9, 3 X 3 = 9.

Question 11. 30 × 3 = Answer: 30 X 3 = 3 X 10 X 3 = 90,

Explanation: Given 30 X 3 = writing 30 as 3 X 10 and multiplying with 3 we get 90 as 3 X 10 X 3 = 90, So 30 X 3 = 90.

Question 12. 300 × 3 = Answer: 300 X 3 = 3 X 100 X 3 = 900,

Explanation: Given 300 X 3 = writing 300 as 3 X 100 and multiplying with 3 we get 900 as 3 X 100 X 3 = 900, So 300 X 3 = 900.

Question 13. 3,000 × 3 = Answer: 3,000 X 3 = 3 X 1,000 X 3 = 9,000,

Explanation: Given 3,000 X 3 = writing 3,000 as 3 X 1,000 and multiplying with 3 we get 9,000 as 3 X 1,000 X 3 = 9,000, So 3,000 X 3 = 9,000.

Question 14. 4,000 × 3 = Answer: 4,000 X 3 = 4 X 1,000 X 3 = 12,000,

Explanation: Given 4,000 X 3 = writing 4,000 as 4 X 1,000 and multiplying with 3 we get 12,000 so 4 X 1,000 X 3 = 12,000, So 4,000 X 3 = 12,000.

Question 15. 400 × 3 = Answer: 400 X 3 = 4 X 100 X 3 = 1,200,

Explanation: Given 400 X 3 = writing 400 as 4 X 100 and multiplying with 3 we get 1,200 as 4 X 100 X 3 = 1,200, So 400 X 3 = 1,200.

Question 16. 40 × 3 = Answer: 40 X 3 = 4 X 10 X 3 = 120,

Explanation: Given 40 X 3 = writing 40 as 4 X 10 multiplying with 3 we get 120 as 4 X 10 X 3 = 120, So 40 X 3 = 120.

Question 17. 5 × 3 = Answer: 5 X 3 = 15,

Explanation: Given 5 X 3 = multiplying 5 with 3 we get 15, 5 X 3 = 15.

Question 18. 500 × 3 = Answer: 500 X 3 = 5 X 100 X 3 = 1,500,

Explanation: Given 500 X 3 = writing 500 as 5 X 100 and multiplying with 3 we get 1,500 as 5 x 100 X 3 = 1,500, So 500 X 3 = 1,500.

Question 19. 7 × 2 = Answer: 7 X 2 = 14,

Explanation: Given 7 X 2 = multiplying 7 with 2 we get 14, 7 X 2 = 14.

Question 20. 70 × 2 = Answer: 70 X 2 = 7 X 10 X 2 = 140,

Explanation: Given 70 X 2 = writing 70 as 7 X 10 and multiplying with 2 we get 140 as 7 X 10 X 2 = 140, So 70 X 2 = 140.

Question 21. 4 × 4 = Answer: 4 X 4 = 16,

Explanation: Given 4 X 4 = multiplying 4 with 4 we get 16, 4 X 4 = 16.

Question 22. 4,000 × 4 = Answer: 4,000 X 4 = 4 X 1,000 X 4 = 1,600,

Explanation: Given 4,000 X 4 = writing 4,000 as 4 X 1,000 and multiplying with 4 we get 1,600 as 4 X 1,000 X 4 = 1,600, So 4,000 X 4 = 1,600.

Question 23. 7 × 5 = Answer: 7 X 5 = 35,

Explanation: Given 7 X 5 = multiplying 7 with 5 we get 35, 7 X 5 = 35.

Question 24. 700 × 5 = Answer: 700 X 5 = 7 x 100 X 5 = 3,500,

Explanation: Given 700 X 5 = writing 700 as 7 X 100 and multiplying with 5 we get 3,500 as 7 X 100 X 5 = 700, So 700 X 5 = 3,500.

Question 25. 8 × 3 = Answer: 8 X 3 = 24,

Explanation: Given 8 X 3 = multiplying 8 with 3 we get 24, 8 X 3 = 24.

Question 26. 80 × 3 = Answer: 80 X 3 = 8 X 10 X 3 = 240,

Explanation: Given 80 X 3 = writing 80 as 8 X 10 and multiplying with 3 we get 240 as 8 X 10 X 3 = 240, So 80 X 3 = 240.

Question 27. 9 × 4 = Answer: 9 X 4 = 36,

Explanation: Given 9 X 4 = multiplying 9 with 4 we get 36, 9 X 4 = 36.

Question 28. 9,000 × 4 = Answer: 9,000 x 4 = 9 X 1,000 X 4 = 36,000,

Explanation: Given 9,000 X 4 = writing 9,000 as 9 X 1,000 and multiplying with 4 we get 36,000 as 9 X 1,000 X 4 = 36,000, So 9,000 X 4 = 36,000.

Question 29. 7 × 6 = Answer: 7 X 6 = 42,

Explanation: Given 7 X 6 = multiplying 7 with 6 we get 42, 7 X 6 = 42.

Question 30. 7 × 600 = Answer: 7 X 600 = 7 X 6 X 100 = 4,200,

Explanation: Given 7 X 600 = writing 600 as 6 X 100 and multiplying with 7 we get 4,200 as 7 X 6 X 100 = 4,200 or 7 X 600 = 4,200.

Question 31. 8 × 9 = Answer: 8 X 9 = 72,

Explanation: Given 8 X 9 = multiplying 8 with 9 we get 72, 8 X 9 = 72.

Question 32. 8 × 90 = Answer: 8 X 90 = 8 X 9 X 10 = 720,

Explanation: Given 8 X 90 = writing 90 as 9 X 10 and multiplying with 8 we get 720 as 8 x 9 X 10 = 720, So 8 X 90 = 720.

Question 33. 6 × 9 = Answer: 6 X 9 = 54,

Explanation: Given 6 X 9 = multiplying 6 with 9 we get 54, 6 X 9 = 54.

Question 34. 6 × 9,000 = Answer: 6 X 9,000 = 6 X 9 X 1,000 = 54,000,

Explanation: Given 6 X 9,000 = writing 9,000 as 9 X 1,000 and multiplying with 6 we get 54,000 as 6 x 9 X 1,000 = 54,000, So 6 X 9,000 = 54,000.

Question 35. 900 × 9 = Answer: 900 X 9 = 9 X 100 X 9 = 8,100,

Explanation: Given 900 X 9 = writing 900 as 9 X 100 and multiplying with 9 we get 8,100 as 9 X 100 X 9 = 8,100, So 900 X 9 = 8,100.

Question 36. 8,000 × 8 = Answer: 8,000 X 8 = 8 X 1,000 X 8 = 64,000,

Explanation: Given 8,000 X 8 = writing 8,000 as 8 X 1,000 and multiplying with 8 we get 64,000 as 8 X 1,000 X 8 = 64,000, So 8,000 X 8 = 64,000.

Question 37. 7 × 70 = Answer: 7 X 70 = 7 X 7 X 10 = 490,

Explanation: Given 7 X 70 = writing 70 as 7 X 10 and multiplying with 7 we get 490 as 7 X 7 X 10 = 490, So 7 X 70 = 490.

Question 38. 6 × 600 = Answer: 6 X 600 = 6 X 6 X 100 = 3,600,

Explanation: Given 6 X 600 = writing 600 as 6 X 100 and multiplying with 6 we get 3,600 as 6 x 6 X 100 = 3,600, So 6 X 600 = 3,600.

Question 39. 800 × 7 = Answer: 800 X 7 = 8 X 100 X 7 = 5,600,

Explanation: Given 800 X 7 = writing 800 as 8 x 100 and multiplying with 7 we get 5,600 as 8 X 100 X 7 = 5,600, So 800 X 7 = 5,600.

Question 40. 7 × 9,000 = Answer: 7 X 9,000 = 7 X 9 X 1,000 = 63,000,

Explanation: Given 7 X 9,000 = writing 9,000 as 9 X 1,000 and multiplying with 7 we get 63,000 as 7 X 9 X 1,000 = 63,000, So 7 X 9,000 = 63,000.

Question 41. 200 × 5 = Answer: 200 X 5 = 2 X 100 X 5 = 1,000,

Explanation: Given 200 X 5 = writing 200 as 2 X 100 and multiplying with 5 we get 1,000 as 2 X 100 X 5 = 1,000, So 200 X 5 = 1,000.

Question 42. 5 × 60 = Answer: 5 X 60 = 5 X 6 X 10 =300,

Explanation: Given 5 X 60 = writing 60 as 6 X 10 and multiplying with 5 we get 300 as 5 X 6 X 10 = 300, So 5 X 60 = 300.

Question 43. 4,000 × 5 = Answer: 4,000 X 5 = 4 X 1,000 X 5 = 20,000,

Explanation: Given 4,000 X 5 = writing 4,000 as 4 X 1,000 and multiplying with 5 we get 20,000 as 4 x 1,000 X 5 = 20,000, So 4,000 X 5 = 20,000.

Question 44. 800 × 5 = Answer: 800 X 5 = 4,000, 8 X 100 X 5 = 4,000,

Explanation: Given 800 X 5 = writing 800 as 8 X 100 and multiplying with 5 we get 4,000 as 8 X 100 X 5 = 4,000, So 800 X 5 = 4,000.

Question 1. 4 × 2 = Answer: 4 X 2 = 8,

Explanation: Given 4 X 2 = multiplying 4 with 2 we get 8, 4 X 2 = 8.

Question 2. 40 × 2 = Answer: 40 X 2 = 4 X 10 X 2 = 80,

Explanation: Given 40 X 2 = writing 40 as 4 X 10 and multiplying with 2 we get 80 as 4 X 10 X 2 = 80, So 40 X 2 = 80.

Question 3. 400 × 2 = Answer: 400 X 2 = 4 X 100 X 2 = 800,

Explanation: Given 400 X 2 = writing 400 as 4 X 100 and multiplying with 2 we get 800 as 4 X 100 X 2 = 800, So 400 X 2 = 800.

Question 4. 4,000 × 2 = Answer: 4,000 X 2 = 4 X 1,000 X 2 = 8,000,

Explanation: Given 4,000 X 2 = writing 4,000 as 4 X 1,000 and multiplying with 2, we get 8,000, as 4 X 1,000 X 2 = 8,000, So 4,000 X 2 = 8,000.

Question 5. 2 × 4,000 = Answer: 2 X 4,000 = 2 X 4 X 1,000 = 8,000,

Question 6. 3 × 3 = Answer: 3 X 3 = 9,

Question 7. 3 × 30 = Answer: 3 X 30 = 3 X 3 X 10 = 90,

Explanation: Given 3 X 30 = writing 30 as 3 X 10 and multiplying with 3 we get 90 as 3 X 3 X 10 = 90, So 3 X 30 = 90.

Question 8. 3 × 300 = Answer: 3 X 300 = 3 X 3 X 100 = 900,

Explanation: Given 3 X 300 = writing 300 as 3 X 100 and multiplying with 3 we get 900 as 3 X 3 X 100 = 900, So 3 X 300 = 900.

Question 9. 3 × 3,000 = Answer: 3 X 3,000 = 3 X 3 X 1,000 = 9,000,

Explanation: Given 3 X 3,000 = writing 3,000 as 3 X 1,000 and multiplying with 3 we get 9,000 as 3 X 3 X 1,000 = 9,000, So 3 X 3,000 = 9,000.

Question 10. 2 × 3 = Answer: 2 X 3 = 6,

Explanation: Given 2 X 3 = multiplying 2 with 3 we get 6, 2 X 3 = 6.

Question 11. 20 × 3 = Answer: 20 X 3 = 2 x 10 X 3 = 60,

Explanation: Given 20 X 3 = writing 20 as 2 X 10 and multiplying with 3 we get 60 as 2 X 10 X 3 = 60, So 20 X 3 = 60.

Question 12. 200 × 3 = Answer: 200 X 3 = 2 X 100 X 3 = 600,

Explanation: Given 200 X 3 = writing 200 as 2 X 100 and multiplying with 3 we get 600 as 2 X 100 X 3 = 600, So 200 X 3 = 600.

Question 13. 2,000 × 3 = Answer: 2,000 X 3 = 2 X 1,000 X 3 = 6,000,

Explanation: Given 2,000 X 3 = writing 2,000 as 2 X 1,000 and multiplying with 3 we get 6,000 as 2 X 1,000 X 3 = 6,000, So 2,000 X 3 = 6,000.

Question 14. 3,000 × 4 = Answer: 3,000 X 4 = 3 X 1,000 X 4 = 12,000,

Explanation: Given 3,000 X 4 = writing 3,000 as 3 X 1,000 and multiplying with 4 we get 12,000 as 3 X 1,000 X 4 = 12,000, So 3,000 X 4 = 12,000.

Question 15. 300 × 4 = Answer: 300 X 4 = 3 X 100 X 4 = 1,200,

Explanation: Given 300 X 4 = writing 300 as 3 X 100 and multiplying with 4 we get 1,200 as 3 X 100 X 4 = 1,200, So 300 X 4 = 1,200.

Question 16. 30 × 4 = Answer: 30 X 4 = 3 X 10 X 4 = 120,

Explanation: Given 30 X 4 = writing 30 as 3 X 10 and multiplying with 4 we get 120 as 3 X 10 X 4 = 120, So 30 X 4 = 120.

Question 17. 3 × 5 = Answer: 3 X 5 = 15,

Explanation: Given 3 X 5 = multiplying 3 with 5 we get 15 as 3 X 5 = 15, So 3 X 5 = 15.

Question 18. 30 × 5 = Answer: 30 X 5 = 3 X 10 X 5 = 150,

Explanation: Given 30 X 5 = writing 30 as 3 X 10 and multiplying with 5 we get 150 as 3 X 10 X 5 = 150, So 30 X 5 = 150.

Question 19. 6 × 2 = Answer: 6 X 2 = 12,

Explanation: Given 6 X 2 = multiplying 6 with 2 we get 12 as 6 X 2 = 12, So 6 X 2 = 12.

Question 20. 60 × 2 = Answer: 60 X 2 = 6 X 10 X 2 = 120,

Explanation: Given 60 X 2 = writing 60 as 6 X 10 and multiplying with 2 we get 120 as 6 X 10 X 2 = 120, So 60 X 2 = 120.

Explanation: Given 4 X 4 = multiplying 4 with 4 we get 16 as 4 X 4 = 16, So 4 X 4 = 16.

Question 22. 400 × 4 = Answer: 400 X 4 = 4 X 100 X 4 = 1,600,

Explanation: Given 400 X 4 = writing 400 as 4 X 100 and multiplying with 4 we get 1,600 as 4 X 100 X 4 = 1,600, So 400 X 4 = 1,600.

Question 23. 9 × 5 = Answer: 9 X 5 = 45,

Explanation: Given 9 X 5 = multiplying 9 with 5 we get 45 as 9 X 5 = 45.

Question 24. 900 × 5 = 9 X 100 X 5 = 4,500,

Explanation: Given 900 X 5 = writing 900 as 9 X 100 and multiplying with 5 we get 4,500 as 9 X 100 X 5 = 4,500, So 900 X 5 = 4,500.

Question 25. 8 × 4 = Answer: 8 X 4 = 32,

Explanation: Given 8 X 4 = multiplying 8 with 4 we get 32 as 8 X 4 = 32.

Question 26. 80 × 4 = Answer: 80 X 4 = 8 X 10 X 4 = 320,

Explanation: Given 80 X 4 = writing 80 as 8 X 10 and multiplying with 4 we get 320 as 8 X 10 X 4 = 320, So 80 X 4 = 320.

Question 27. 9 × 3 = Answer: 9 X 3 = 27,

Explanation: Given 9 X 3 = multiplying 9 with 3 we get 27 as 9 X 3 = 27.

Question 28. 9,000 × 3 = Answer: 9,000 X 3 = 9 X 1,000 X 3 = 27,000,

Explanation: Given 9,000 X 3 = writing 9,000 as 9 X 1,000 and multiplying with 3 we get 27,000 as 9 X 1,000 X 3 = 27,000, So 9,000 X 3 = 27,000.

Question 29. 6 × 7 = Answer: 6 X 7 = 42,

Explanation: Given 6 X 7 = multiplying 6 with 7 we get 42 as 6 X 7 = 42.

Question 30. 6 × 700 = Answer: 6 X 700 = 6 X 7 X 100 = 4,200,

Explanation: Given 6 X 700 = writing 700 as 7 X 100 and multiplying with 6 we get 4,200 as 6 X 7 X 100 = 4,200, So 6 X 700 = 4,200.

Question 31. 8 × 7 = Answer: 8 X 7 = 56,

Explanation: Given 8 X 7 = multiplying 8 with 7 we get 56 as 8 X 7 = 56.

Question 32. 8 × 70 = Answer: 8 X 70 = 8 X 7 X 10 = 560,

Explanation: Given 8 X 70 = writing 70 as 7 X 10 and multiplying with 8 we get 560 as 8 x 7 X 10 = 560, So 8 X 70 = 560.

Question 33. 9 × 6 = Answer: 9 X 6 = 54,

Explanation: Given 9 X 6 = multiplying 9 with 6 we get 54 as 9 X 6 = 54.

Question 34. 9 × 6,000 = Answer: 9 X 6,000 = 9 X 6 X 1,000 = 54,000,

Explanation: Given 9 X 6,000 = writing 6,000 as 6 X 1,000 and multiplying with 9, we get 54,000, as 9 X 6 X 1,000 = 54,000, So 9 X 6,000 = 54,000.

Question 35. 800 × 8 = Answer: 800 X 8 = 8 X 100 X 8 = 6,400,

Explanation: Given 800 X 8 = writing 800 as 8 X 100 and multiplying with 8 we get 6,400 as 8 X 100 X 8 = 6,400, So 800 X 8 = 6,400.

Question 36. 9,000 × 9 = Answer: 9,000 x 9 = 9 X 1,000 X 9 = 81,000,

Explanation: Given 9,000 X 9 = writing 9,000 as 9 X 1,000 and multiplying with 9 we get 81,000 as 9 X 1,000 X 9 = 81,000, So 9,000 X 9 = 81,000.

Question 37. 7 × 700 = Answer: 7 X 700 = 7 X 7 X 100 = 4,900,

Explanation: Given 7 X 700 = writing 700 as 7 X 100 and multiplying with 7 we get 4,900 as 7 X 7 X 100 = 4,900, So 7 X 700 = 4,900.

Question 38. 6 × 60 = Answer: 6 X 60 = 6 X 6 X 10 = 360,

Explanation: Given 6 X 60 = writing 60 as 6 X 10 and multiplying with 6 we get 360 as 6 x 6 X 10 = 360, So 6 X 60 = 360.

Question 39. 700 × 8 = Answer: 700 X 8 = 7 X 100 X 8 = 5,600,

Explanation: Given 700 X 8 = writing 700 as 7 X 100 and multiplying with 8 we get 5,600 as 7 X 100 X 8 = 5,600, So 700 X 8 = 5,600.

Question 40. 9 × 7,000 = Answer: 9 X 7000 = 9 X 7 X 1,000 = 63,000,

Explanation: Given 9 X 7,000 = writing 7,000 as 7 X 1,000 and multiplying with 9, we get 63,000, as 9 X 7 X 1,000 = 63,000, So 9 X 7,000 = 63,000.

Question 41. 20 × 5 = Answer: 20 X 5 = 2 X 10 X 5 = 100,

Explanation: Given 20 X 5 = writing 20 as 2 X 10 and multiplying with 5 we get 100 as 2 X 10 X 5 = 100, So 20 X 5 = 100.

Question 42. 5 × 600 = Answer: 5 X 600 = 5 X 6 X 100 = 3,000,

Explanation: Given 5 X 600 = writing 600 as 6 X 100 and multiplying with 5 we get 3,000 as 5 X 6 X 100 = 3,000, So 5 X 600 = 3,000.

Question 43. 400 × 5 = Answer: 400 X 5 = 4 X 100 X 5 = 2,000,

Explanation: Given 400 X 5 = writing 400 as 4 X 100 and multiplying with 5 we get 2,000 as 4 X 100 X 5 = 2,000, So 400 X 5 = 2,000.

Question 44. 8,000 × 5 = Answer: 8,000 x 5 = 8 X 1,000 X 5 = 40,000,

Explanation: Given 8,000 X 5 = writing 8,000 as 8 X 1,000 and multiplying with 5 we get 40,000 as 8 X 1,000 X 5 = 40,000, So 8,000 X 5 = 40,000.

Eureka Math Grade 4 Module 3 Lesson 7 Problem Set Answer Key

Explanation: Represented the following expression 1 X43 with disks, regrouped as necessary, wrote a matching expression, and recorded the partial products vertically as shown above 1 X 43 = 43.

Eureka Math Grade 4 Module 3 Lesson 7 Exit Ticket Answer Key

Represent the following expressions with disks, regrouping as necessary. To the right, record the partial products vertically.

Eureka Math Grade 4 Module 3 Lesson 7 Homework Answer Key

Eureka Math Grade 4 Module 3 Lesson 7 Template Answer Key

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