EPEI - RUSD Mathrusdmath.weebly.com/uploads/1/1/1/5/11156667/g6_u4...6.EE.2.c 13. f3 = 2 • 2 • 2 = 8 The volume of the cube is 8 cubic inches. Challenge Problem 6.EE.2.c 14. Expression

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ANSWERS FOR EXERCISES

MATH GRADE 6 UNIT 4

EXPRESSIONS


Grade 6 Unit 4: Expressions

ANSWERSLESSON 1: NUMERICAL EXPRESSIONS

ANSWERS

6.EE.1 2. I will test the trick with both odd and even original numbers between 1 and 100.

Original number = 1 1 • 2 = 2 2 + 12 = 14 14 ÷ 2 = 7 7 – 1 = 6

Original number = 4 4 • 2 = 8 8 + 12 = 20 20 ÷ 2 = 10 10 – 4 = 6

Original number = 5 5 • 2 = 10 10 + 12 = 22 22 ÷ 2 = 11 11 – 5 = 6

Original number = 31 31 • 2 = 62 62 + 12 = 74 74 ÷ 2 = 37 37 – 31 = 6

Original number = 88 88 • 2 = 176 176 + 12 = 188 188 ÷ 2 = 94 94 – 88 = 6

Original number = 100 100 • 2 = 200 200 + 12 = 212 212 ÷ 2 = 106 106 – 100 = 6

Yes, this trick will always give the number 6.




ANSWERS

6.EE.1 1. C 6

6.EE.1 2. 7

6.EE.1 3. 35

6.EE.1 4. 2

6.EE.1 5. 176

6.EE.1 6. 125

6.EE.1 7. B Subtract 5 from 12.

6.EE.1 8. D 99 – (50 ÷ 5) + 5 = 94

E 99 – [50 ÷ (5 + 5)] = 94

6.EE.1 9. (22 + 3) • (9 – 5) = 100

6.EE.1 10. 16 • (3 + 3) • 5 = 480

6.EE.1 11. 82 – 16 ÷ 8 • (5 + 25) = 22 OR 82 – (16 ÷ 8) • (5 + 25) = 22




Challenge Problem

6.EE.1 12. The order of operations that you follow is important because it affects the value of an expression. Two people could get very different values for the same expression if they followed a different order in performing the operations.

An example of a real-life situation that uses the order of operations is converting a temperature in Fahrenheit to a temperature in Celsius. For example, to convert 50°F to Celsius, you use this equation:

(50 – 32) • 5 ÷ 9 = (18) • 5 ÷ 9 = 90 ÷ 9 = 10º C

If you didn’t follow the proper order of operations—if you didn’t complete the operations inside the parentheses first—you might say that 50°F converts to about 32.2°C.

50 32 5 9 50 160 9

50 17 7

32 2

−( ) ÷ = − ÷

= −

≈

•

.

. �C



ANSWERSLESSON 3: ALGEBRAIC EXPRESSIONS

ANSWERS

6.EE.2.a 1.A

x3

16+

6.EE.2.a 2. n + 12 or 12 + n

6.EE.2 3. Algebraic Expression

Numerical Expression

Algebraic Equation

Numerical Equation

7 1n +( )

n n− +( )3 105 2

18

3

•( ) ÷

6 36n = 4 2 4

14

2÷ ( ) =•

6.EE.2.a 4. D 12(3 + a)

6.EE.2.a 5. C 9a + 3b

6.EE.2.a 6. 253

2513

−−( )n

n or •

6.EE.2.a 7. 2(n + 8) or (n + 8) • 2

6.EE.2.a 8. 52

425

4n

n+ ÷

+ or

6.EE.2.a 9. n2 – 7 or n • n – 7

6.EE.2.a 10. nn

++( ) ÷

93

9 3 or

6.EE.2.a 11. 72 + 3n

6.EE.2.a 12. Multiply any number n by 3, and then add 12.

6.EE.2.a 13. Add 15 to any number b, and then divide by 3.

or

Add 15 to any number b, and then multiply by 13



ANSWERSLESSON 3: ALGEBRAIC EXPRESSIONS

Challenge Problem

6.EE.2.a 14. Here is one example.

Algebraic expression: 7(5n + 2)

Written expression: Add 2 to 5 times any number n, and then multiply by 7.



ANSWERSLESSON 4: WRITING EXPRESSIONS

ANSWERS

6.EE.3 1. B 4x + 17

6.EE.2.a 2. B 7a

6.EE.2.a 3. A 3a

6.EE.2.a 4. D 2a + 3b

6.EE.2.a 5. D a + b + 2c

6.EE.2.a 6. a + 2(b + c + d) and a + 2b + 2c + 2d

6.EE.2.a 7. Any two of these equations are acceptable. 2(a + b + c) + 2b + c + d + e 2a + 4b + 3c + d + e 2(a + 2b) + 3c + d + e

6.EE.2.a 8. a + 3(c + d) + e and a + 3c + 3d + e

6.EE.2.a 9. a + 4(b + c) + b and a + 5b + 4c

6.EE.3 10.

Addition

Distributive property of multiplication over addition

Commutative property of addition

Associative property of additiona c b c d c a c c c b d

a c c c b

+ + + + + = + + + + += + + +( ) +

2 2 2 2 2 2 2 2

2 2 2 ++= + + += + + +( )

2

5 2 2

5 2

d

a c b d

a c b d

Challenge Problem

6.EE.2.a 6.EE.3

11. Here is one example.

Total length of train: 2a + 2b + 2c + 2d + a + 2b + 2c + 2d + a + 2b + 2c + 2d + a + 2b + 2c + 2d + e

Combine like terms: 2a + 2b + 2c + 2d + a + 2b + 2c + 2d + a + 2b + 2c + 2d + a + 2b + 2c + 2d + e (2a + a + a + a) + (2b + 2b + 2b + 2b) + (2c + 2c + 2c + 2c) + (2d + 2d + 2d + 2d) + e 5a + 8b + 8c + 8d + e



ANSWERSLESSON 5: EVALUATING EXPRESSIONS

ANSWERS

6.EE.2.c 1. D 20

6.EE.2.c 2. 54

6.EE.2.c 3. 102.8

6.EE.2.c 4. 4

6.EE.2.c 5. 4

6.EE.2.c 6. 123.6

6.EE.2.c 7. 16.2

6.EE.2.c 8. 24

6.EE.2.c 9. 20.25

6.EE.2.c 10. 12f

6.EE.2.c 11. 12f = 12 • 2 = 24

The total length of the edges is 24 inches.

6.EE.2.c 12. f3 is the volume of the cube.

6.EE.2.c 13. f3 = 2 • 2 • 2 = 8

The volume of the cube is 8 cubic inches.

Challenge Problem

6.EE.2.c 14. Expression in words: Subtract 2 from the number of sides.

Algebraic expression: n – 2

n – 2 = 100 – 2 = 98 98 triangles can be formed if n = 100.



ANSWERSLESSON 6: VOCABULARY OF EXPRESSIONS

ANSWERS

6.EE.2.b 1. A 2 terms

6.EE.2.b 2. A 43

D 7(t + 4b2)

6.EE.2.b 3. The coefficients are 4, 5, 1, and 9.

6.EE.2.b 4. The exponents are 3, 1, 2, and 1.

6.EE.2.c 5. 125

6.EE.2.b 6. There is only one constant: 11.

6.EE.2.b 7.Expression Variables Coefficients Exponents Number of

Terms Constants

s, t 4, 6 1, 1 2 25

s, t 8, 12 2, 1 3 50

w, z –3, 4 1, 1 2 none

noneb, c, d 2, 3, 4 1, 1, 1 1

+ +t s2(4 6 ) 25

+ +t s8 12 502

−w z4 3

+ +d b c6(4 3 2 )

6.EE.2.a 6.EE.2.b

8. Here are three examples.

7a + b + 5

2(x + 42) + 6y + 19

8(m + 4) + 4(3y – 5) + 3

6.EE.2.a 6.EE.2.b


3(x + 9)

0.125(a + 9)

14

9m +( )

6.EE.2.a 6.EE.2.b

10. x 5



ANSWERSLESSON 6: VOCABULARY OF EXPRESSIONS

6.EE.2.a 6.EE.2.b


)

k

x

a

85

15

Challenge Problem

6.EE.2.a 6.EE.2.b

12. a. 2 4( )x + (number of terms: 1; coefficients: 1; exponents: 1) or 2 16x + (number of terms: 2; coefficients: 2; exponents: 1)

b. 2 4( )x + (number of terms: 1; coefficients: 1; exponents: 1) or 2 8x + (number of terms: 2; coefficients: 2; exponents: 1)

c. x x( )4 (number of terms: 1; coefficients: 1, 4; exponents: 1, 1) or 4 2x (number of terms: 1; coefficients: 4; exponents: 2)

d. 12

4x +( ) (number of terms: 1; coefficients: 1; exponents: 1)

or 12

2x + (number of terms: 2; coefficients: 12

exponents: 1)



ANSWERSLESSON 7: EQUIVALENT EXPRESSIONS

ANSWERS

6.EE.3 1. B 3a + 3b

6.EE.3 2. 7 28

6.EE.3 3. dx + 4d

6.EE.3 4.5

18

458

20x x−

= −

6.EE.3 5. Here are four examples.

4(x + 6 + x + 2.25) 4x + 24 + 4x + 9 8x + 33 4(x + 6) + 4(x + 2.25)

6.EE.2.c 6. 47

6.EE.3 7. D 8 a b+( )

6.EE.3 8. nx + ny = n(x + y)

6.EE.3 9. 12a + 4b = 4 • 3a + 4 • b = 4(3a + b)

6.EE.3 10. 19

23

19

69

19

6c d c d c d− = − = −( )



ANSWERSLESSON 7: EQUIVALENT EXPRESSIONS

Challenge Problem

6.EE.2.a 11. Baseball cards = x

Basketball cards = 13

x

Hockey cards = 713

73

• x x=

Football cards = 73

273

12

76

x x x÷ = =•

Total number of cards in Martin’s collection: x x x x+ + +13

73

76

Simplified expression: x x x x x x x+ + + = + + +

= + + +

=

13

73

76

113

73

76

66

26

146

76

296

The number of cards in Martin’s collection is 296

x



ANSWERSLESSON 8: GREATEST COMMON FACTOR

ANSWERS

6.NS.4 1. B 4

6.NS.4 2. D 15

6.NS.4 3. A 2

6.NS.4 4. C 15

6.NS.4 5. C 4

6.NS.4 6. Factors of 4 Only

Factors of 6 Only

Factors of Both 4 and 6

Not Factors of Either 4 or 6

43

6

1

25

6.NS.4 7. Factors of 12 Only

Factors of 15 Only

Factors of Both 12 and 15

Not Factors of Either 12 or 15

2 4

6 12

5

15

1

3

7 8 910 11 13 14




6.NS.4 8. Here is one example. I made a rectangle 16 units wide and 30 units long.

I tried different size squares to figure out which could fill the rectangle without any overlapping. I found the greatest common factor is 2. 1 is also a common factor for 16 and 30.

6.NS.4 9. Here is one example.

First, I found all of the factors of 12 : 1, 2, 3, 4, 6, 12 Then, starting with the greatest factor, I divided 16 by the factors of 12 to see which divided evenly without a remainder. When I got to a number that divided without a remainder, I knew that the number was a common factor. When I finished dividing 16 by all the factors of 12, I looked at the common factors and chose the greatest number as the greatest common factor.

16 ÷ 12 = 1 R4 16 ÷ 6 = 2 R2 16 ÷ 4 = 4 16 ÷ 3 = 5 R1 16 ÷ 2 = 8 16 ÷ 1 = 16

The greatest common factor is 4. 1 and 2 are also common factors of 12 and 16.

6.NS.4 10. Here is one example. I find the factors of each number.

14 = 1, 2, 7, 14

70 = 1, 2, 5, 7, 10, 14, 35, 70

I circle all of the common factors and look for the greatest one. The greatest common factor is 14.




6.NS.4 11. Here is one example. I made a table of the factors for each number. Then I highlighted the common factors.

19 95

1 1

19 5

19

95

The greatest common factor is 19.

Challenge Problem

6.NS.4 12. Multiply the common factors together using the lowest exponent that appears with each factor. For example, the greatest common factor of 72 and 126 is 2 × 32 = 18. This method can be used to find the greatest common factor of any set of numbers.

One method for determining the greatest common factor of a set of numbers inovlves finding the prime factorization of those numbers. This is done by finding the list of prime numbers whose product gives the original integer. For example, consider the numbers 126 and 72.

The prime factorization of 126 is 2 × 3 × 3 × 7.

2

3

126

63

37

21

126 = 2 × 3 × 3 × 7

The prime factorization of 72 is 2 × 3 × 3 × 2 × 2.

2

3

72

36

3 12

224

72 = 2 × 3 × 3 × 2 × 2




Find the common factors and multiply all of the common factors together. For example, the common factors in the prime factorization of 72 and 126 are 2, 3, and 3.

126 = 2 × 3 × 3 × 772 = 2 × 3 × 3 × 2 × 2

Multiplying these together, the greatest common factor of 72 and 126 is 2 × 3 × 3 = 18. This method can be used to find the greatest common factor of any set of numbers.



ANSWERSLESSON 9: LEAST COMMON MULTIPLE

ANSWERS

6.NS.4 1. B 4

6.NS.4 2. C 10

6.NS.4 3. D 12

6.NS.4 4. D 21

6.NS.4 5. D 20

6.NS.4 6. Multiples of 4 Only

Multiples of 8 Only

Multiples of Both 4 and 8

Not Multiples of Either 4 or 8

4

12

20

8

16

24

6 10 14

18 22

6.NS.4 7. Multiples of 5: 5, 10, 15, 20, 25, 30, 35, 40 … Multiples of 7: 7, 14, 21, 28, 35, 42 … Least common multiple: 35

Multiples of 5 Only

Multiples of 7 Only

Multiples of Both 5 and 7

Not Multiples of Either 5 or 7

5 10 15

20 25

30 40

7 14 21

28 4235





0

8

16

24

32

12 24 36

On graph paper, I made squares using 8-unit-by-12-unit rectangles. The smallest square I was able to make was 24 units by 24 units. So, the least common multiple is 24.


Multiples of 10:

10 × 1 = 1010 × 2 = 2010 × 3 = 3010 × 4 = 4010 × 5 = 5010 × 6 = 6010 × 7 = 7010 × 8 = 8010 × 9 = 90

10 × 10 = 10010 × 11 = 11010 × 12 = 120

Multiples of 11:

11 × 1 = 1111 × 2 = 2211 × 3 = 3311 × 4 = 4411 × 5 = 5511 × 6 = 6611 × 7 = 7711 × 8 = 8811 × 9 = 99

11 × 10 = 110

The least common multiple is 110.




6.NS.4 10. Here is one example. I made a table and listed the first 10 multiples of each number.

Multiples of 6 6 12 18 24 30 36 42 48 54 60

Multiples of 10 10 20 30 40 50 60 70 80 90 100

I highlighted the multiples that the two numbers have in common and found the least one. So, the least common multiple is 30.


Multiples of 7 Multiples of 9

7 9

14 18

21 27

28 36

35 45

42 54

49 63

56

63

The least common multiple is 63.

Challenge Problem

6.NS.4 12. The least common multiple is the product of each prime number raised to the highest power that appears in each factorization.

Prime factorization of 12 and 21: 12: 22 × 3 21: 3 × 7

Prime numbers in the factorizations: 2, 3, 7 Highest power of 2: 22 Highest power of 3: 3 Highest power of 7: 7

The least common multiple of 12 and 21 is 22 × 3 × 7 = 84.



ANSWERSLESSON 10: USING EXPRESSIONS

ANSWERS

6.EE.1 6.EE.2.a

1.Number of Hours 0.5 1 1.5 2 t

Expression for Total Distance Jason Runs (5 miles per hour)

5 • 0.5 5 • 1 5 • 1.5 5 • 2 5t

6.NS.4 2. 8

6.EE.2.a 3. A 2 4n −

6.EE.2.a 4. The wheels on the bicycles is represented by 2n. The wheels on the tricycles is represented by 3m. The total number of wheels is represented by 2n + 3m.

6.EE.2.a 5. Martin walked 2w + 5.3 miles.

6.EE.3 6.EE.4

6. C x(3x + 2)

D 3x2 + 2x

F 2x(x + 1) + x2

6.EE.3 6.EE.4

7. A 2(3x + 2) + 2x

B 2(4x + 2)

C 6(x + 1) + 2(x – 1)

F 8x + 4

6.EE.3 6.EE.4

8. B 50 20x y+

D 5 10 4x y+( )

6.EE.2.c 9. 335 or 335.00

6.NS.4 10. 24



ANSWERSLESSON 10: USING EXPRESSIONS

Challenge Problem

6.EE.2 11. Here is one example for x – 3: Emma took out x books from the library. Mia took out 3 fewer books than Emma. How many books did Mia take out?

Here is one example for 5n + 10d: Denzel planted 5 rows of tomato plants and 10 rows of cucumbers. There are n tomato plants in each row and d cucumber plants in each row. Write an expression for the total number of plants (tomato and cucumber) Denzel planted.



ANSWERSLESSON 11: PUTTING IT TOGETHER

ANSWERS

6.EE.2.b 6.EE.4

3. Here are two examples.

Word or Phrase Definition Examples

expression consists of numbers, letters, or both connected by one or more of the arithmetic operations (addition, subtraction, multiplication, division, and raising to a power)

Algebraic expressions have one or more variables and numerical expressions have no variables.

Algebraic Expressions

x2

3(y + 2)

16

4 4 53m n g− + ( ) +

Numerical Expressions

5 – 2

412

10 ÷ •

2 4 6

3

2 −( )

Expression Vocabulary

1

64 4 3 5m n g− + ( ) +

coefficientvariable

exponent

term

constant

} } }Not Expressions (equations):

5 – 2 = 3 3(y + 2) = 24

numerical expression

consists of numbers linked by operation signs—addition, subtraction, multiplication, division, and raising to a power

has no variables

5 – 2

412

10 ÷ •

2 4 6

3

2 −( )

Not examples (these are numerical equations):

5 – 2 = 3

412

10 9 ÷ • =



ANSWERSLESSON 11: PUTTING IT TOGETHER

ANSWERS

algebraic expression

consists of letters (or letters and numbers) connected by the operation signs—addition, subtraction, multiplication, division, and raising to a power

has variables

x2

3(y + 2)

16

4 4 53m n g− + ( ) +

Not examples (these are algebraic equations):

x2 3=

3(y + 2) = 24coefficient a number that appears

before a variable and multiplies it

1

64 4 3 5m n g− + ( ) +

constant terms in an algebraic expression that contain only numbers

1

64 4 3 5m n g− + ( ) +

exponent the number of copies of a number or variable that are multiplied together

1

64 4 3 5m n g− + ( ) +

x2

term number, variable, or combination of numbers and variables separated by operation signs in an expression

x2 There is one term: x2

3(y + 2) There is one term: 3(y + 2)

16

4 4 53m n g− + ( ) +

There are four terms: 16

m and 4n and 4 3g( ) and 5

variable a symbol (usually a letter) that stands for an unknown value in an expression or equation

( )1

64 4 3 5m n g− + +

x2

3 2y +( )

Documents

EPEI - RUSD Mathrusdmath.weebly.com/uploads/1/1/1/5/11156667/g6_u4...6.EE.2.c 13. f3 = 2 • 2 • 2 = 8 The volume of the cube is 8 cubic inches. Challenge Problem 6.EE.2.c 14. Expression