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How to Use a Math Medic Answer Key

Written by Luke Wilcox published 3 years ago

Answer key might be the wrong term here. Sure, the Math Medic answer keys do provide the correct answers to the questions for a lesson, but they have been carefully designed to do much more than this. They are meant to be the official guide to teaching the lesson, providing specific instructions for what to do and say to make a successful learning experience for your students.

Before we look at the details of the answer key, let's make sure we understand the instructional model first.

Experience First, Formalize Later (EFFL)

A typical Math Medic lesson always has the same four parts: Activity, Debrief Activity, QuickNotes, and Check Your Understanding. Here are the cliff notes:

Activity: Students are in groups of 2 - 4 working collaboratively through the questions in the Activity. The teacher is checking in with groups and using questions, prompts, and cues to get students to refine their communication and understanding. As groups finish the activity, the teacher asks students to go to the whiteboard to write up their answers to the questions.

Debrief Activity: In the whole group setting, the teacher leads a discussion about the student responses to the questions in the activity, often asking students to explain their thinking and reasoning about their answers. The teacher then formalizes the learning by highlighting key concepts and introducing new vocabulary, notation, and formulas.

QuickNotes: The teacher uses direct instruction to summarize the learning from the activity in the QuickNotes box - making direct connections to the learning targets for the lesson.

Check Your Understanding: Students are then asked to apply their learning from the lesson to a new context in the Check Your Understanding (CYU) problem. This can be done individually or in small groups. The CYU is very flexible in it's use, as it can be used as an exit ticket, a homework problem, or a quick review the next day.

How Do I See EFFL in the Answer Key?

You will see EFFL in the answer key like this:

Anything written in blue is something we expect our students to produce. This might not be quite what we expect by the end of the lesson, but provides us with a starting point when we move to formalization.

Anything written in red is an idea added by the teacher - the formalization of the learning that happened during the Activity. Students are expected to add these "notes" to their Activity using a red pen or marker.

What Do Students Write Down For Notes?

By the end of the lesson, students will have written down everything you see on the Math Medic Answer Keys. The most important transition is when students finish the Activity and we move to Debrief Activity. "Students, now is the time for you to put down your pencils and get out your your red Paper Mate flair pens" We give each student a Paper Mate flair pen at the beginning of the school year and tell them they must cherish and protect it with their life. They all think we should be sponsored by Paper Mate (anyone have any leads on this?)

The lessons you see on Math Medic are all of the notes we use with our students. We do not have some secret collection of guided notes.

Do Students Have Access to Answer Keys?

Yes! Any student can create a free Math Medic account to get access to the answer keys. We often send students to the website when they are absent from a lesson or when we don't quite finish the lesson in class. We are comfortable with students having access to these answer keys because we do not think Math Medic lessons should be used as a summative assessment or be used for a grade (unless it's for completion). Our lessons are meant to be the first steps in the formative process of learning new concepts.

Math Medic Help

Home > CCG > Chapter 3 > Lesson 3.1.4

Lesson 3.1.1, lesson 3.1.2, lesson 3.1.3, lesson 3.1.4, lesson 3.2.1, lesson 3.2.2, lesson 3.2.3, lesson 3.2.4, lesson 3.2.5, lesson 3.2.6.

© 2022 CPM Educational Program. All rights reserved.

1.1 Real Numbers: Algebra Essentials

• ⓐ 11 1 11 1
• ⓒ − 4 1 − 4 1
• ⓐ 4 (or 4.0), terminating;
• ⓑ 0. 615384 ¯ , 0. 615384 ¯ , repeating;
• ⓒ –0.85, terminating
• ⓐ rational and repeating;
• ⓑ rational and terminating;
• ⓒ irrational;
• ⓓ rational and terminating;
• ⓔ irrational
• ⓐ positive, irrational; right
• ⓑ negative, rational; left
• ⓒ positive, rational; right
• ⓓ negative, irrational; left
• ⓔ positive, rational; right

• ⓐ 11, commutative property of multiplication, associative property of multiplication, inverse property of multiplication, identity property of multiplication;
• ⓑ 33, distributive property;
• ⓒ 26, distributive property;
• ⓓ 4 9 , 4 9 , commutative property of addition, associative property of addition, inverse property of addition, identity property of addition;
• ⓔ 0, distributive property, inverse property of addition, identity property of addition

• ⓒ 121 3 π 121 3 π ;
• ⓐ −2 y −2 z or  −2 ( y + z ) ; −2 y −2 z or  −2 ( y + z ) ;
• ⓑ 2 t −1 ; 2 t −1 ;
• ⓒ 3 p q −4 p + q ; 3 p q −4 p + q ;
• ⓓ 7 r −2 s + 6 7 r −2 s + 6

A = P ( 1 + r t ) A = P ( 1 + r t )

1.2 Exponents and Scientific Notation

• ⓐ k 15 k 15
• ⓑ ( 2 y ) 5 ( 2 y ) 5
• ⓒ t 14 t 14
• ⓑ ( −3 ) 5 ( −3 ) 5
• ⓒ ( e f 2 ) 2 ( e f 2 ) 2
• ⓐ ( 3 y ) 24 ( 3 y ) 24
• ⓑ t 35 t 35
• ⓒ ( − g ) 16 ( − g ) 16
• ⓐ 1 ( −3 t ) 6 1 ( −3 t ) 6
• ⓑ 1 f 3 1 f 3
• ⓒ 2 5 k 3 2 5 k 3
• ⓐ t −5 = 1 t 5 t −5 = 1 t 5
• ⓑ 1 25 1 25
• ⓐ g 10 h 15 g 10 h 15
• ⓑ 125 t 3 125 t 3
• ⓒ −27 y 15 −27 y 15
• ⓓ 1 a 18 b 21 1 a 18 b 21
• ⓔ r 12 s 8 r 12 s 8
• ⓐ b 15 c 3 b 15 c 3
• ⓑ 625 u 32 625 u 32
• ⓒ −1 w 105 −1 w 105
• ⓓ q 24 p 32 q 24 p 32
• ⓔ 1 c 20 d 12 1 c 20 d 12
• ⓐ v 6 8 u 3 v 6 8 u 3
• ⓑ 1 x 3 1 x 3
• ⓒ e 4 f 4 e 4 f 4
• ⓓ 27 r s 27 r s
• ⓕ 16 h 10 49 16 h 10 49
• ⓐ $ 1.52 × 10 5 $ 1.52 × 10 5
• ⓑ 7.158 × 10 9 7.158 × 10 9
• ⓒ $ 8.55 × 10 13 $ 8.55 × 10 13
• ⓓ 3.34 × 10 −9 3.34 × 10 −9
• ⓔ 7.15 × 10 −8 7.15 × 10 −8
• ⓐ 703 , 000 703 , 000
• ⓑ −816 , 000 , 000 , 000 −816 , 000 , 000 , 000
• ⓒ −0.000 000 000 000 39 −0.000 000 000 000 39
• ⓓ 0.000008 0.000008
• ⓐ − 8.475 × 10 6 − 8.475 × 10 6
• ⓑ 8 × 10 − 8 8 × 10 − 8
• ⓒ 2.976 × 10 13 2.976 × 10 13
• ⓓ − 4.3 × 10 6 − 4.3 × 10 6
• ⓔ ≈ 1.24 × 10 15 ≈ 1.24 × 10 15

Number of cells: 3 × 10 13 ; 3 × 10 13 ; length of a cell: 8 × 10 −6 8 × 10 −6 m; total length: 2.4 × 10 8 2.4 × 10 8 m or 240 , 000 , 000 240 , 000 , 000 m.

1.3 Radicals and Rational Exponents

5 | x | | y | 2 y z . 5 | x | | y | 2 y z . Notice the absolute value signs around x and y ? That’s because their value must be positive!

10 | x | 10 | x |

x 2 3 y 2 . x 2 3 y 2 . We do not need the absolute value signs for y 2 y 2 because that term will always be nonnegative.

b 4 3 a b b 4 3 a b

14 −7 3 14 −7 3

• ⓒ 88 9 3 88 9 3

( 9 ) 5 = 3 5 = 243 ( 9 ) 5 = 3 5 = 243

x ( 5 y ) 9 2 x ( 5 y ) 9 2

28 x 23 15 28 x 23 15

1.4 Polynomials

The degree is 6, the leading term is − x 6 , − x 6 , and the leading coefficient is −1. −1.

2 x 3 + 7 x 2 −4 x −3 2 x 3 + 7 x 2 −4 x −3

−11 x 3 − x 2 + 7 x −9 −11 x 3 − x 2 + 7 x −9

3 x 4 −10 x 3 −8 x 2 + 21 x + 14 3 x 4 −10 x 3 −8 x 2 + 21 x + 14

3 x 2 + 16 x −35 3 x 2 + 16 x −35

16 x 2 −8 x + 1 16 x 2 −8 x + 1

4 x 2 −49 4 x 2 −49

6 x 2 + 21 x y −29 x −7 y + 9 6 x 2 + 21 x y −29 x −7 y + 9

1.5 Factoring Polynomials

( b 2 − a ) ( x + 6 ) ( b 2 − a ) ( x + 6 )

( x −6 ) ( x −1 ) ( x −6 ) ( x −1 )

• ⓐ ( 2 x + 3 ) ( x + 3 ) ( 2 x + 3 ) ( x + 3 )
• ⓑ ( 3 x −1 ) ( 2 x + 1 ) ( 3 x −1 ) ( 2 x + 1 )

( 7 x −1 ) 2 ( 7 x −1 ) 2

( 9 y + 10 ) ( 9 y − 10 ) ( 9 y + 10 ) ( 9 y − 10 )

( 6 a + b ) ( 36 a 2 −6 a b + b 2 ) ( 6 a + b ) ( 36 a 2 −6 a b + b 2 )

( 10 x − 1 ) ( 100 x 2 + 10 x + 1 ) ( 10 x − 1 ) ( 100 x 2 + 10 x + 1 )

( 5 a −1 ) − 1 4 ( 17 a −2 ) ( 5 a −1 ) − 1 4 ( 17 a −2 )

1.6 Rational Expressions

1 x + 6 1 x + 6

( x + 5 ) ( x + 6 ) ( x + 2 ) ( x + 4 ) ( x + 5 ) ( x + 6 ) ( x + 2 ) ( x + 4 )

2 ( x −7 ) ( x + 5 ) ( x −3 ) 2 ( x −7 ) ( x + 5 ) ( x −3 )

x 2 − y 2 x y 2 x 2 − y 2 x y 2

1.1 Section Exercises

irrational number. The square root of two does not terminate, and it does not repeat a pattern. It cannot be written as a quotient of two integers, so it is irrational.

The Associative Properties state that the sum or product of multiple numbers can be grouped differently without affecting the result. This is because the same operation is performed (either addition or subtraction), so the terms can be re-ordered.

−14 y − 11 −14 y − 11

−4 b + 1 −4 b + 1

43 z − 3 43 z − 3

9 y + 45 9 y + 45

−6 b + 6 −6 b + 6

16 x 3 16 x 3

1 2 ( 40 − 10 ) + 5 1 2 ( 40 − 10 ) + 5

irrational number

g + 400 − 2 ( 600 ) = 1200 g + 400 − 2 ( 600 ) = 1200

inverse property of addition

1.2 Section Exercises

No, the two expressions are not the same. An exponent tells how many times you multiply the base. So 2 3 2 3 is the same as 2 × 2 × 2 , 2 × 2 × 2 , which is 8. 3 2 3 2 is the same as 3 × 3 , 3 × 3 , which is 9.

It is a method of writing very small and very large numbers.

12 40 12 40

1 7 9 1 7 9

3.14 × 10 − 5 3.14 × 10 − 5

16,000,000,000

b 6 c 8 b 6 c 8

a b 2 d 3 a b 2 d 3

q 5 p 6 q 5 p 6

y 21 x 14 y 21 x 14

72 a 2 72 a 2

c 3 b 9 c 3 b 9

y 81 z 6 y 81 z 6

1.0995 × 10 12 1.0995 × 10 12

0.00000000003397 in.

12,230,590,464 m 66 m 66

a 14 1296 a 14 1296

n a 9 c n a 9 c

1 a 6 b 6 c 6 1 a 6 b 6 c 6

0.000000000000000000000000000000000662606957

1.3 Section Exercises

When there is no index, it is assumed to be 2 or the square root. The expression would only be equal to the radicand if the index were 1.

The principal square root is the nonnegative root of the number.

9 5 5 9 5 5

6 10 19 6 10 19

− 1 + 17 2 − 1 + 17 2

7 2 3 7 2 3

20 x 2 20 x 2

17 m 2 m 17 m 2 m

2 b a 2 b a

15 x 7 15 x 7

5 y 4 2 5 y 4 2

4 7 d 7 d 4 7 d 7 d

2 2 + 2 6 x 1 −3 x 2 2 + 2 6 x 1 −3 x

− w 2 w − w 2 w

3 x − 3 x 2 3 x − 3 x 2

5 n 5 5 5 n 5 5

9 m 19 m 9 m 19 m

2 3 d 2 3 d

3 2 x 2 4 2 3 2 x 2 4 2

6 z 2 3 6 z 2 3

−5 2 −6 7 −5 2 −6 7

m n c a 9 c m n m n c a 9 c m n

2 2 x + 2 4 2 2 x + 2 4

1.4 Section Exercises

The statement is true. In standard form, the polynomial with the highest value exponent is placed first and is the leading term. The degree of a polynomial is the value of the highest exponent, which in standard form is also the exponent of the leading term.

Use the distributive property, multiply, combine like terms, and simplify.

4 x 2 + 3 x + 19 4 x 2 + 3 x + 19

3 w 2 + 30 w + 21 3 w 2 + 30 w + 21

11 b 4 −9 b 3 + 12 b 2 −7 b + 8 11 b 4 −9 b 3 + 12 b 2 −7 b + 8

24 x 2 −4 x −8 24 x 2 −4 x −8

24 b 4 −48 b 2 + 24 24 b 4 −48 b 2 + 24

99 v 2 −202 v + 99 99 v 2 −202 v + 99

8 n 3 −4 n 2 + 72 n −36 8 n 3 −4 n 2 + 72 n −36

9 y 2 −42 y + 49 9 y 2 −42 y + 49

16 p 2 + 72 p + 81 16 p 2 + 72 p + 81

9 y 2 −36 y + 36 9 y 2 −36 y + 36

16 c 2 −1 16 c 2 −1

225 n 2 −36 225 n 2 −36

−16 m 2 + 16 −16 m 2 + 16

121 q 2 −100 121 q 2 −100

16 t 4 + 4 t 3 −32 t 2 − t + 7 16 t 4 + 4 t 3 −32 t 2 − t + 7

y 3 −6 y 2 − y + 18 y 3 −6 y 2 − y + 18

3 p 3 − p 2 −12 p + 10 3 p 3 − p 2 −12 p + 10

a 2 − b 2 a 2 − b 2

16 t 2 −40 t u + 25 u 2 16 t 2 −40 t u + 25 u 2

4 t 2 + x 2 + 4 t −5 t x − x 4 t 2 + x 2 + 4 t −5 t x − x

24 r 2 + 22 r d −7 d 2 24 r 2 + 22 r d −7 d 2

32 x 2 −4 x −3 32 x 2 −4 x −3 m 2

32 t 3 − 100 t 2 + 40 t + 38 32 t 3 − 100 t 2 + 40 t + 38

a 4 + 4 a 3 c −16 a c 3 −16 c 4 a 4 + 4 a 3 c −16 a c 3 −16 c 4

1.5 Section Exercises

The terms of a polynomial do not have to have a common factor for the entire polynomial to be factorable. For example, 4 x 2 4 x 2 and −9 y 2 −9 y 2 don’t have a common factor, but the whole polynomial is still factorable: 4 x 2 −9 y 2 = ( 2 x + 3 y ) ( 2 x −3 y ) . 4 x 2 −9 y 2 = ( 2 x + 3 y ) ( 2 x −3 y ) .

Divide the x x term into the sum of two terms, factor each portion of the expression separately, and then factor out the GCF of the entire expression.

10 m 3 10 m 3

( 2 a −3 ) ( a + 6 ) ( 2 a −3 ) ( a + 6 )

( 3 n −11 ) ( 2 n + 1 ) ( 3 n −11 ) ( 2 n + 1 )

( p + 1 ) ( 2 p −7 ) ( p + 1 ) ( 2 p −7 )

( 5 h + 3 ) ( 2 h −3 ) ( 5 h + 3 ) ( 2 h −3 )

( 9 d −1 ) ( d −8 ) ( 9 d −1 ) ( d −8 )

( 12 t + 13 ) ( t −1 ) ( 12 t + 13 ) ( t −1 )

( 4 x + 10 ) ( 4 x − 10 ) ( 4 x + 10 ) ( 4 x − 10 )

( 11 p + 13 ) ( 11 p − 13 ) ( 11 p + 13 ) ( 11 p − 13 )

( 19 d + 9 ) ( 19 d − 9 ) ( 19 d + 9 ) ( 19 d − 9 )

( 12 b + 5 c ) ( 12 b − 5 c ) ( 12 b + 5 c ) ( 12 b − 5 c )

( 7 n + 12 ) 2 ( 7 n + 12 ) 2

( 15 y + 4 ) 2 ( 15 y + 4 ) 2

( 5 p − 12 ) 2 ( 5 p − 12 ) 2

( x + 6 ) ( x 2 − 6 x + 36 ) ( x + 6 ) ( x 2 − 6 x + 36 )

( 5 a + 7 ) ( 25 a 2 − 35 a + 49 ) ( 5 a + 7 ) ( 25 a 2 − 35 a + 49 )

( 4 x − 5 ) ( 16 x 2 + 20 x + 25 ) ( 4 x − 5 ) ( 16 x 2 + 20 x + 25 )

( 5 r + 12 s ) ( 25 r 2 − 60 r s + 144 s 2 ) ( 5 r + 12 s ) ( 25 r 2 − 60 r s + 144 s 2 )

( 2 c + 3 ) − 1 4 ( −7 c − 15 ) ( 2 c + 3 ) − 1 4 ( −7 c − 15 )

( x + 2 ) − 2 5 ( 19 x + 10 ) ( x + 2 ) − 2 5 ( 19 x + 10 )

( 2 z − 9 ) − 3 2 ( 27 z − 99 ) ( 2 z − 9 ) − 3 2 ( 27 z − 99 )

( 14 x −3 ) ( 7 x + 9 ) ( 14 x −3 ) ( 7 x + 9 )

( 3 x + 5 ) ( 3 x −5 ) ( 3 x + 5 ) ( 3 x −5 )

( 2 x + 5 ) 2 ( 2 x − 5 ) 2 ( 2 x + 5 ) 2 ( 2 x − 5 ) 2

( 4 z 2 + 49 a 2 ) ( 2 z + 7 a ) ( 2 z − 7 a ) ( 4 z 2 + 49 a 2 ) ( 2 z + 7 a ) ( 2 z − 7 a )

1 ( 4 x + 9 ) ( 4 x −9 ) ( 2 x + 3 ) 1 ( 4 x + 9 ) ( 4 x −9 ) ( 2 x + 3 )

1.6 Section Exercises

You can factor the numerator and denominator to see if any of the terms can cancel one another out.

True. Multiplication and division do not require finding the LCD because the denominators can be combined through those operations, whereas addition and subtraction require like terms.

y + 5 y + 6 y + 5 y + 6

3 b + 3 3 b + 3

x + 4 2 x + 2 x + 4 2 x + 2

a + 3 a − 3 a + 3 a − 3

3 n − 8 7 n − 3 3 n − 8 7 n − 3

c − 6 c + 6 c − 6 c + 6

d 2 − 25 25 d 2 − 1 d 2 − 25 25 d 2 − 1

t + 5 t + 3 t + 5 t + 3

6 x − 5 6 x + 5 6 x − 5 6 x + 5

p + 6 4 p + 3 p + 6 4 p + 3

2 d + 9 d + 11 2 d + 9 d + 11

12 b + 5 3 b −1 12 b + 5 3 b −1

4 y −1 y + 4 4 y −1 y + 4

10 x + 4 y x y 10 x + 4 y x y

9 a − 7 a 2 − 2 a − 3 9 a − 7 a 2 − 2 a − 3

2 y 2 − y + 9 y 2 − y − 2 2 y 2 − y + 9 y 2 − y − 2

5 z 2 + z + 5 z 2 − z − 2 5 z 2 + z + 5 z 2 − z − 2

x + 2 x y + y x + x y + y + 1 x + 2 x y + y x + x y + y + 1

2 b + 7 a a b 2 2 b + 7 a a b 2

18 + a b 4 b 18 + a b 4 b

a − b a − b

3 c 2 + 3 c − 2 2 c 2 + 5 c + 2 3 c 2 + 3 c − 2 2 c 2 + 5 c + 2

15 x + 7 x −1 15 x + 7 x −1

x + 9 x −9 x + 9 x −9

1 y + 2 1 y + 2

Review Exercises

y = 24 y = 24

3 a 6 3 a 6

x 3 32 y 3 x 3 32 y 3

1.634 × 10 7 1.634 × 10 7

4 2 5 4 2 5

7 2 50 7 2 50

3 x 3 + 4 x 2 + 6 3 x 3 + 4 x 2 + 6

5 x 2 − x + 3 5 x 2 − x + 3

k 2 − 3 k − 18 k 2 − 3 k − 18

x 3 + x 2 + x + 1 x 3 + x 2 + x + 1

3 a 2 + 5 a b − 2 b 2 3 a 2 + 5 a b − 2 b 2

4 a 2 4 a 2

( 4 a − 3 ) ( 2 a + 9 ) ( 4 a − 3 ) ( 2 a + 9 )

( x + 5 ) 2 ( x + 5 ) 2

( 2 h − 3 k ) 2 ( 2 h − 3 k ) 2

( p + 6 ) ( p 2 − 6 p + 36 ) ( p + 6 ) ( p 2 − 6 p + 36 )

( 4 q − 3 p ) ( 16 q 2 + 12 p q + 9 p 2 ) ( 4 q − 3 p ) ( 16 q 2 + 12 p q + 9 p 2 )

( p + 3 ) 1 3 ( −5 p − 24 ) ( p + 3 ) 1 3 ( −5 p − 24 )

x + 3 x − 4 x + 3 x − 4

m + 2 m − 3 m + 2 m − 3

6 x + 10 y x y 6 x + 10 y x y

Practice Test

x = –2 x = –2

3 x 4 3 x 4

13 q 3 − 4 q 2 − 5 q 13 q 3 − 4 q 2 − 5 q

n 3 − 6 n 2 + 12 n − 8 n 3 − 6 n 2 + 12 n − 8

( 4 x + 9 ) ( 4 x − 9 ) ( 4 x + 9 ) ( 4 x − 9 )

( 3 c − 11 ) ( 9 c 2 + 33 c + 121 ) ( 3 c − 11 ) ( 9 c 2 + 33 c + 121 )

4 z − 3 2 z − 1 4 z − 3 2 z − 1

3 a + 2 b 3 b 3 a + 2 b 3 b

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Eureka Math Grade 3 Module 3 Lesson 3 Answer Key

Teachers and students can find this Eureka Answer Key for Grade 3 more helpful in raising students’ scores and supporting teachers to educate the students. Eureka Math Answer Key for Grade 3 aids teachers to differentiate instruction, building and reinforcing foundational mathematics skills that alter from the classroom to real life. With the help of Eureka’s primary school Grade 3 Answer Key, you can think deeply regarding what you are learning, and you will really learn math easily just like that.

Engage NY Eureka Math 3rd Grade Module 3 Lesson 3 Answer Key

Eureka math grade 3 module 3 lesson 3 problem set answer key.

Answer: e = 20, i = 1/ 6, s = 4, n = 10, k = 6, a = 2.6, l = 7, c = 3, t = 0.7, b = 2, h = 5.

Answer: The tables not have to learn is 10, 20, and 70.

Explanation: In the above-given question, given that, The tables are 9, 6, 3, 5, 70, 20, 10, 24, 2, 7, and 4. out of all the tables 10 and 20 are very easy. so we need not to learn.

Question 2. Lonna buys 3 t-shirts for $8 each. a. What is the total amount Lonna spends on 3 t-shirts? Use the letter m to represent the total amount of money Lonna spends, and then solve the problem.

Answer: The total amount of money Lonna spends = 24$.

Explanation: In the above-given question, given that, Lonna buys 3 t-shirts for $8 each. m x 3 = $8. m = 8 x 3. m = $24. so the total amount of money Lonna spends = $24.

b. If Lonna hands the cashier 3 ten dollar bills, how much change will she receive? Use the letter c in an equation to represent the change, and then find the value of c.

Answer: The change she received = $6.

Explanation: In the above-given question, given that, Lonna hands the cashier 3 ten dollar bills. c x 30 = 24. c = 30 – 24. c = $6. so the change Lonna received = $6.

Question 3. Miss Potts used a total of 28 cups of flour to bake some bread. She used 4 cups of flour for each loaf of bread. How many loaves of bread did she bake? Represent the problem using multiplication and division sentences and a letter for the unknown. Then, solve the problem. __7___ × ___4__ = __28____ __28___ ÷ __4___ = ___7___

Answer: The number of loaves of bread did she bake = 7.

Explanation: In the above-given question, given that, Miss potts used a total of 28 cups of flour to bake some bread. she used 4 cups of flour for each loaf of bread. 7 x 4 = 28. 28 / 4 = 7. so the number of loaves of bread did she bake = 7.

Answer: The other game takes to complete = 10 minutes.

Explanation: In the above-given question, given that, At a table tennis tournament, two games went on for a total of 32 minutes. one game took 12 minutes longer than the other. 32 + t = 12. t = 32 – 12. t = 20. so the other game takes to complete = 10 minutes.

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

Find the value of the unknown in Problems 1 – 4.

Question 1. z = 5 × 9 z = ___45___

Answer: z = 45.

Explanation: In the above-given question, given that, the equation z = 5 x 9. z = 45.

Question 2. 30 ÷ 6 = v v = ___5___

Answer: v = 5.

Explanation: In the above-given question, given that, the equation 30/ 6 = v. v = 5.

Question 3. 8 × w = 24 w = ____3__

Answer: w = 3.

Explanation: In the above-given question, given that, the equation 8 x w = 24. w = 24/8. w = 3.

Question 4. y ÷ 4 = 7 y = __1.75____

Answer: y = 1.75.

Explanation: In the above-given question, given that, y / 4 = 7. y = 7 / 4. y = 1.75.

Question 5. Mr. Strand waters his rose bushes for a total of 15 minutes. He waters each rose bush for 3 minutes. How many rose bushes does Mr. Strand water? Represent the problem using multiplication and division sentences and a letter for the unknown. Then, solve the problem. __3___ × __5___ = __15___ __15___ ÷ __3___ = ___5__

Answer: The number of rose bushes Mr. strand water = 15.

Explanation: In the above-given question, given that, Mr. Strand waters his rose bushes for a total of 15 minutes. He waters each rose bush for 3 minutes. 3 x 5 = 15. 15 / 3 = 5. so the number of rose bushes Mr. Strand water = 15.

Eureka Math Grade 3 Module 3 Lesson 3 Homework Answer Key

Answer: The missing numbers in the pattern are 40, 50, 70, 80, and 100.

b. Find the value of the unknown.

10 × 2 = d   d = __20___ 3 × 10 = e   e = __30___ f = 4 × 10   f = ___40__ p = 5 × 10  p = __50___ 10 × 6 = w  w = ___60__ 10 × 7 = n  n = __70___ g = 8 × 10  g = __80___

Answer: d = 20. e = 30. f = 40. p = 50. w = 60. n = 70. g = 80.

Explanation: In the above-given question, given that, the equations 10 x 2 = d, d = 20. 3 x 10 = e, e = 30. f = 4 x 10, f = 40. p = 5 x 10, p = 50. 10 x 6 = w, w = 60. 10 x 7 = n, n = 70. g = 8 x 10, g = 80.

Question 2. Each equation contains a letter representing the unknown. Find the value of the unknown.

Answer: n = 4. a = 4. p = 5. c = 3. d = 6. h = 1.4. f = 18. y = 8.

Explanation: In the above-given question, given that, the equations 8 / 2 = n, n = 4. 3 x a = 12, a = 12 / 3, a = 4. p x 8 = 40, p = 40 / 8, p = 5. 18 / 6 = c, c = 3. d x 4 = 24, d = 24 / 4, d = 6. h / 7 = 5, h = 1.4. 6 x 3 = f, f = 18. 32 / y = 4. y = 8.

Question 3. Pedro buys 4 books at the fair for $7 each. a. What is the total amount Pedro spends on 4 books? Use the letter b to represent the total amount Pedro spends, and then solve the problem.

Answer: The total amount Pedro spends on 4 books = $28.

Explanation: In the above-given question, given that, Pedro buys 4 books at the fair for $7 each. b x 4 = 7. b = 7 x 4. b = $28. so the total amount Pedro spends on 4 books = $28.

b. Pedro hands the cashier 3 ten dollar bills. How much change will he receive? Write an equation to solve. Use the letter c to represent the unknown.

Answer: The change Pedro receives = $2.

Explanation: In the above-given question, given that, Pedro hands the cashier 3 ten dollar bills. c + 30 = 28. c = 30 – 28. c = $2.

Question 4. On field day, the first-grade dash is 25 meters long. The third-grade dash is twice the distance of the first-grade dash. How long is the third-grade dash? Use a letter to represent the unknown and solve.

Answer: The third-grade dash = 75 meters long.

Explanation: In the above-given question, given that, On field day, the first-grade dash is 25 meters long. The third-grade dash is twice the distance of the first-grade dash. g = 25 + 50. g = 75. so the third-grade dash = 75 meters long.

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