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Texas Go Math Grade 6 Lesson 3.4 Answer Key Dividing Mixed Numbers

Refer to our Texas Go Math Grade 6 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 6 Lesson 3.4 Answer Key Dividing Mixed Numbers.

Question 1. Communicate Mathematical Ideas Which mathematical operation could you use to find the number of sushi rolls that Antoine can make? Explain. Answer: We can use dividing to find the number of sushi rolls that Antonie can make Actually, to find that number, we need to divide 2\(\frac{1}{2}\) by \(\frac{1}{4}\).

Question 2. Multiple Representations Write the division shown by the model. Answer: The division shown by the previous model is the following: 2\(\frac{1}{2}\) ÷ \(\frac{1}{4}\)

Question 3. What If ? Suppose Antoine instead uses \(\frac{1}{8}\) cup of rice for each sushi roll. How would his model change? How many rolls can he make? Explain. Answer: Instead dividing 2\(\frac{1}{2}\) by \(\frac{1}{8}\), he will divide 2\(\frac{1}{2}\) by \(\frac{1}{8}\) and in that case he will make more sushi rolls.

Go Math Lesson 3.4 Answer Key Dividing Mixed Numbers Question 4. Analyze Relationships Explain how you can check the answer. Answer: We can check the answer by dividing the result by the first of the fractions we were dividing or multiplying the result by the second of the fractions we were dividing.

Question 5. What If? Harold serves himself 1\(\frac{1}{2}\)-ounce servings of cereal each morning. How many servings does he get from a box of his favorite cereal? Show your work. Answer: We need to divide 1 by 1\(\frac{1}{2}\) in order to get how many servings Harold gets from a box. But first, we need to rewrite 1\(\frac{1}{2}\) as a fraction and find its reciprocal. So, we have the following: 1\(\frac{1}{2}\) = \(\frac{3}{2}\) Reciprocal of \(\frac{3}{2}\) is \(\frac{2}{3}\) Now, we have to multiply 1 by \(\frac{2}{3}\) and get: 1 × \(\frac{2}{3}=\frac{1 \times 2}{1 \times 3}=\frac{2}{3}\) So, Harold gets \(\frac{2}{3}\) servings of cereal from a box.

Question 7. Check for Reasonableness How can you determine if your answer is reasonable? Answer: If we have to divide two fractions, we can check our result on two ways: First is to multiply result by the second fraction and if product is the first fraction, then our original solution is correct. The second way is to divide the first fraction by result and if we get as a result of this dividing the second fraction, then our original solution is correct. Multiply result by the second fraction; Divide the fraction by the result.

Texas Go Math Grade 6 Lesson 3.4 Guided Practice Answer Key

Divide. Write each answer in simplest form.

Question 3. 4 ÷ 1\(\frac{1}{8}\) ____________ Answer: First we will rewrite the mixed numbers as fractions greater than 1, then again rewrite the problem as multiplication using the reciprocal ot the second fractions. We will simplify using the GCF in order to get the solution in simplest form and multiply numerators and denominators. At the end, we will rewrite the solution as a mixed number if it is greater than 1.

Question 4. 3\(\frac{1}{5}\) ÷ 1\(\frac{1}{7}\) ____________ Answer: First we will rewrite the mixed numbers as fractions greater than 1, then again rewrite the problem as multiplication using the reciprocal ot the second fractions. We will simplify using the GCF in order to get the solution in simplest form and multiply numerators and denominators. At the end, we will rewrite the solution as a mixed number if it is greater than 1.

Question 5. 8\(\frac{1}{3}\) ÷ 2\(\frac{1}{2}\) ____________ Answer: First we will rewrite the mixed numbers as fractions greater than 1, then again rewrite the problem as multiplication using the reciprocal ot the second fractions. We will simplify using the GCF in order to get the solution in simplest form and multiply numerators and denominators. At the end, we will rewrite the solution as a mixed number if it is greater than 1.

Question 6. 15\(\frac{1}{3}\) ÷ 3\(\frac{5}{6}\) ____________ Answer: Solution to this example is given below First we will rewrite the mixed numbers as fractions greater than 1, then again rewrite the problem as multiplication using the reciprocal ot the second fractions. We will simplify using the GCF in order to get the solution in simplest form and multiply numerators and denominators. At the end, we will rewrite the solution as a mixed number if it is greater than 1.

Write each situation as a division problem. Then solve.

Essential Question Check-In

Question 9. How does dividing mixed numbers compare with dividing fractions? Answer: We have to rewrite mixed numbers as a fractions greater than 1 and then the steps of dividing are the same as dividing two fractions because we actually get two fractions. It is equivalent with dividing fractions, but first rewrite mixed numbers as mixed numbers.

H.O.T. Focus On Higher Order Thinking

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Answer Key 3.4

For questions 1 to 10, sketch the linear equation using the slope intercept method.

For questions 11 to 20, sketch the linear equation using the [latex]x[/latex] and [latex]y[/latex] intercepts.

For questions 21 to 28, sketch the linear equation using any method.

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Eureka Math Grade 6 Module 4 Lesson 6 Answer Key

Engage ny eureka math 6th grade module 4 lesson 6 answer key, eureka math grade 6 module 4 lesson 6 example answer key.

Example 1. Expressions with Only Addition, Subtraction, Multiplication, and Division

What operations are evaluated first? Answer: Multiplication and division are evaluated first, from left to right.

What operations are always evaluated last? Answer: Addition and subtraction are always evaluated last, from left to right.

Example 2. Expressions with Four Operations and Exponents 4 + 9 2 ÷ 3 × 2 – 2

What operation is evaluated first? Answer: Exponents (9 2 = 9 × 9 = 81)

What operations are evaluated next? Answer: Multiplication and division, from left to right (81 ÷ 3 = 27; 27 × 2 = 54)

What operations are always evaluated last? Answer: Addition and subtraction, from left to right (4 + 54 = 58; 58 – 2 = 56)

What is the final answer? Answer: 56

Example 3. Expressions with Parentheses Consider a family of 4 that goes to a soccer game. Tickets are $5.00 each. The mom also buys a soft drink for $2.00. How would you write this expression? Answer: 4 × 5 + 2

How much will this outing cost? Answer: $22

Consider a different scenario: The same family goes to the game as before, but each of the family members wants a drink. How would you write this expression? Answer: 4 × (5 + 2)

Why would you add the 5 and 2 first? Answer: We need to determine how much each person spends. Each person spends $7; then, we multiply by 4 people to figure out the total cost.

How much will this outing cost? Answer: $28

How many groups are there? Answer: 4

What does each group comprise? Answer: $5 + $2, or $7

Example 4. Expressions with Parentheses and Exponents 2 × (3 + 4) 2 Which value will we evaluate first within the parentheses? Evaluate. Answer: First, evaluate 4 2 which is 16; then, add 3.The value of the parentheses is 19. 2 × (3 + 4 2 ) 2 × (3 + 16) 2 × 19

Evaluate the rest of the expression. Answer: 2 × 19 = 38

What do you think will happen when the exponent in this expression is outside of the parentheses? 2 × (3 + 4) 2

Will the answer be the same? Answer: Answers will vary.

Which should we evaluate first? Evaluate. Answer: Parentheses 2 × (3 + 4) 2 2 × (7) 2

What happened differently here than in our last example? Answer: The 4 was not raised to the second power because it did not have an exponent. We simply added the values inside the parentheses.

What should our next step be?

We need to evaluate the exponent next. Answer: 7 2 = 7 × 7 = 49

Evaluate to find the final answer. Answer: 2 × 49 98

What do you notice about the two answers? Answer: The final answers were not the same.

What was different between the two expressions? Answer: Answers may vary. In the first problem, a value inside the parentheses had an exponent, and that value was evaluated first because it was inside of the parentheses. In the second problem, the exponent was outside of the parentheses, which made us evaluate what was in the parentheses first; then, we raised that value to the power of the exponent.

What conclusions can you draw about evaluating expressions with parentheses and exponents? Answer: Answers may vary. Regardless of the location of the exponent in the expression, evaluate the parentheses first. Sometimes there will be values with exponents inside the parentheses. If the exponent is outside the parentheses, evaluate the parentheses first, and then evaluate to the power of the exponent.

Eureka Math Grade 6 Module 4 Lesson 6 Exercise Answer Key

Exercise 1. 4 + 2 × 7 Answer: 4 + 14 18

Exercise 2. 36 ÷ 3 × 4 Answer: 12 × 4 48

Exercise 3. 20 − 5 × 2 Answer: 20 − 10 10

Exercise 4. 90 − 5 2 × 3 Answer: 90 − 25 × 3 90 − 75 15

Exercise 5. 4 3 + 2 × 8 Answer: 64 + 2 × 8 64 + 16 80

Exercise 6. 2 + (9 2 ) Answer: 2 + (81 – 4) 2 + 77 79

Exercise 7. 2 . (13 + 5 – 14 ÷ (3 + 4) Answer: 2 . (13 + 5 – 14 ÷ 7) 2 . (13 + 5 – 2) 2 . 16 32

Exercise 8. 7 + (12 – 3 2 ) Answer: 7 + (12 – 9) 7 + 3 10

Exercise 9. 7 + (12 – 3) 2 Answer: 7 + 9 2 7 + 81 88

Eureka Math Grade 6 Module 4 Lesson 6 Problem Set Answer Key

Evaluate each expression.

Question 1. 3 × 5 + 2 × 8 + 2 Answer: 15 + 16 + 2 33

Question 2. ($1.75 + 2 × $0.25 + 5 × $0.05) × 24 Answer: ($1.75 + $0.50 + $0.25) × 24 $2.50 × 24 $60.00

Question 3. (2 × 6) + (8 × 4) + 1 Answer: 12 + 32 + 1 45

Question 4. ((8 × 1.95) + (3 × 2.95) + 10.95) × 1.06 Answer: (15.6 + 8.85 + 10.99) × 1.06 35.4 × 1.06 37.54

Question 5. ((12 ÷ 3) 2 – (18 ÷ 3 2 )) × (4 ÷ 2) Answer: (4 2 − (18 ÷ 9)) × (4 ÷ 2) (16 – 2) × 2 14 × 2 28

Eureka Math Grade 6 Module 4 Lesson 6 Exit Ticket Answer Key

Question 1. Evaluate this expression: 39 ÷ (2 + 1) – 2 × (4 + 1). Answer: 39 ÷ 3 − 2 × 5 13 − 10 3

Question 2. Evaluate this expression: 12 × (3 + 2 2 ÷ 2 − 10 Answer: 12 × (3 + 4) ÷2 − 10 12 × 7 ÷ 2 − 10 84 ÷ 2 − 10 42 − 10 32

Question 3. Evaluate this expression: 12 × (3 + 2) 2 ÷ 2 – 10. Answer: 12 × 5 2 ÷ 2 – 10 12 × 25 ÷ 2 − 10 300 ÷ 2 − 10 150 − 10 140

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4.1 Linear Functions

m = 4 − 3 0 − 2 = 1 − 2 = − 1 2 ; m = 4 − 3 0 − 2 = 1 − 2 = − 1 2 ; decreasing because m < 0. m < 0.

y = − 7 x + 3 y = − 7 x + 3

H ( x ) = 0.5 x + 12.5 H ( x ) = 0.5 x + 12.5

Possible answers include ( − 3 , 7 ) , ( − 3 , 7 ) , ( − 6 , 9 ) , ( − 6 , 9 ) , or ( − 9 , 11 ) . ( − 9 , 11 ) .

( 16 , 0 ) ( 16 , 0 )

• ⓐ f ( x ) = 2 x ; f ( x ) = 2 x ;
• ⓑ g ( x ) = − 1 2 x g ( x ) = − 1 2 x

y = – 1 3 x + 6 y = – 1 3 x + 6

4.2 Modeling with Linear Functions

ⓐ C ( x ) = 0.25 x + 25 , 000 C ( x ) = 0.25 x + 25 , 000 ⓑ The y -intercept is ( 0 , 25 , 000 ) ( 0 , 25 , 000 ) . If the company does not produce a single doughnut, they still incur a cost of $25,000.

ⓐ 41,100 ⓑ 2020

21.57 miles

4.3 Fitting Linear Models to Data

54 ° F 54 ° F

150.871 billion gallons; extrapolation

4.1 Section Exercises

Terry starts at an elevation of 3000 feet and descends 70 feet per second.

d ( t ) = 100 − 10 t d ( t ) = 100 − 10 t

The point of intersection is ( a , a ) . ( a , a ) . This is because for the horizontal line, all of the y y coordinates are a a and for the vertical line, all of the x x coordinates are a . a . The point of intersection is on both lines and therefore will have these two characteristics.

y = 3 5 x − 1 y = 3 5 x − 1

y = 3 x − 2 y = 3 x − 2

y = − 1 3 x + 11 3 y = − 1 3 x + 11 3

y = − 1.5 x − 3 y = − 1.5 x − 3

perpendicular

f ( 0 ) = − ( 0 ) + 2 f ( 0 ) = 2 y − int : ( 0 , 2 ) 0 = − x + 2 x − int : ( 2 , 0 ) f ( 0 ) = − ( 0 ) + 2 f ( 0 ) = 2 y − int : ( 0 , 2 ) 0 = − x + 2 x − int : ( 2 , 0 )

h ( 0 ) = 3 ( 0 ) − 5 h ( 0 ) = − 5 y − int : ( 0 , − 5 ) 0 = 3 x − 5 x − int : ( 5 3 , 0 ) h ( 0 ) = 3 ( 0 ) − 5 h ( 0 ) = − 5 y − int : ( 0 , − 5 ) 0 = 3 x − 5 x − int : ( 5 3 , 0 )

− 2 x + 5 y = 20 − 2 ( 0 ) + 5 y = 20 5 y = 20 y = 4 y − int : ( 0 , 4 ) − 2 x + 5 ( 0 ) = 20 x = − 10 x − int : ( − 10 , 0 ) − 2 x + 5 y = 20 − 2 ( 0 ) + 5 y = 20 5 y = 20 y = 4 y − int : ( 0 , 4 ) − 2 x + 5 ( 0 ) = 20 x = − 10 x − int : ( − 10 , 0 )

Line 1: m = –10 Line 2: m = –10 Parallel

Line 1: m = –2 Line 2: m = 1 Neither

Line 1 :   m = – 2       Line 2 :   m = – 2       Parallel Line 1 :   m = – 2       Line 2 :   m = – 2       Parallel

y = 3 x − 3 y = 3 x − 3

y = − 1 3 t + 2 y = − 1 3 t + 2

y = − 5 4 x + 5 y = − 5 4 x + 5

y = 3 x − 1 y = 3 x − 1

y = − 2.5 y = − 2.5

y = 3 y = 3

x = − 3 x = − 3

Linear, g ( x ) = − 3 x + 5 g ( x ) = − 3 x + 5

Linear, f ( x ) = 5 x − 5 f ( x ) = 5 x − 5

Linear, g ( x ) = − 25 2 x + 6 g ( x ) = − 25 2 x + 6

Linear, f ( x ) = 10 x − 24 f ( x ) = 10 x − 24

f ( x ) = − 58 x + 17.3 f ( x ) = − 58 x + 17.3

• ⓐ a = 11,900 , b = 1000.1 a = 11,900 , b = 1000.1
• ⓑ q ( p ) = 1000 p – 100 q ( p ) = 1000 p – 100

y = − 16 3 y = − 16 3

x = a x = a

y = d c – a x – a d c – a y = d c – a x – a d c – a

y = 100 x – 98 y = 100 x – 98

x < 1999 201 , x > 1999 201 x < 1999 201 , x > 1999 201

$45 per training session.

The rate of change is 0.1. For every additional minute talked, the monthly charge increases by $0.1 or 10 cents. The initial value is 24. When there are no minutes talked, initially the charge is $24.

The slope is –400. this means for every year between 1960 and 1989, the population dropped by 400 per year in the city.

4.2 Section Exercises

Determine the independent variable. This is the variable upon which the output depends.

To determine the initial value, find the output when the input is equal to zero.

6 square units

20.01 square units

P ( t ) = 75 , 000 + 2500 t P ( t ) = 75 , 000 + 2500 t

(–30, 0) Thirty years before the start of this model, the town had no citizens. (0, 75,000) Initially, the town had a population of 75,000.

Ten years after the model began

W ( t ) = 0.5 t + 7.5 W ( t ) = 0.5 t + 7.5

( − 15 , 0 ) ( − 15 , 0 ) : The x -intercept is not a plausible set of data for this model because it means the baby weighed 0 pounds 15 months prior to birth. ( 0 , 7 . 5 ) ( 0 , 7 . 5 ) : The baby weighed 7.5 pounds at birth.

At age 5.8 months

C ( t ) = 12 , 025 − 205 t C ( t ) = 12 , 025 − 205 t

( 58 . 7 , 0 ) : ( 58 . 7 , 0 ) : In roughly 59 years, the number of people inflicted with the common cold would be 0. ( 0 , 12 , 0 25 ) ( 0 , 12 , 0 25 ) Initially there were 12,025 people afflicted by the common cold.

y = − 2 t +180 y = − 2 t +180

In 2070, the company’s profit will be zero.

y = 3 0 t − 3 00 y = 3 0 t − 3 00

(10, 0) In the year 1990, the company’s profits were zero

During the year 1933

• ⓐ 696 people
• ⓒ 174 people per year
• ⓓ 305 people
• ⓔ P(t) = 305 + 174t
• ⓕ 2,219 people
• ⓐ C(x) = 0.15x + 10
• ⓑ The flat monthly fee is $10 and there is a $0.15 fee for each additional minute used

P(t) = 190t + 4,360

• ⓐ R ( t ) = − 2 . 1 t   +   16 R ( t ) = − 2 . 1 t   +   16
• ⓑ 5.5 billion cubic feet
• ⓒ During the year 2017

More than 133 minutes

More than $42,857.14 worth of jewelry

More than $66,666.67 in sales

4.3 Section Exercises

When our model no longer applies, after some value in the domain, the model itself doesn’t hold.

We predict a value outside the domain and range of the data.

The closer the number is to 1, the less scattered the data, the closer the number is to 0, the more scattered the data.

61.966 years

Interpolation. About 60 ° F . 60 ° F .

This value of r indicates a strong negative correlation or slope, so C This value of r indicates a strong negative correlation or slope, so C

This value of r indicates a weak negative correlation, so B This value of r indicates a weak negative correlation, so B

Yes, trend appears linear because r = 0. 985 r = 0. 985 and will exceed 12,000 near midyear, 2016, 24.6 years since 1992.

y = 1 . 64 0 x + 13 . 8 00 , y = 1 . 64 0 x + 13 . 8 00 , r = 0. 987 r = 0. 987

y = − 0.962 x + 26.86 ,       r = − 0.965 y = − 0.962 x + 26.86 ,       r = − 0.965

y = − 1 . 981 x + 6 0. 197; y = − 1 . 981 x + 6 0. 197; r = − 0. 998 r = − 0. 998

y = 0. 121 x − 38.841 , r = 0.998 y = 0. 121 x − 38.841 , r = 0.998

( −2 , −6 ) , ( 1 , −12 ) , ( 5 , −20 ) , ( 6 , −22 ) , ( 9 , −28 ) ; ( −2 , −6 ) , ( 1 , −12 ) , ( 5 , −20 ) , ( 6 , −22 ) , ( 9 , −28 ) ; Yes, the function is a good fit.

( 189 .8 , 0 ) ( 189 .8 , 0 ) If 18,980 units are sold, the company will have a profit of zero dollars.

y = 0.00587 x + 1985 .4 1 y = 0.00587 x + 1985 .4 1

y = 2 0. 25 x − 671 . 5 y = 2 0. 25 x − 671 . 5

y = − 1 0. 75 x + 742 . 5 0 y = − 1 0. 75 x + 742 . 5 0

Review Exercises

y = − 3 x + 26 y = − 3 x + 26

y = 2 x − 2 y = 2 x − 2

Not linear.

( –9 , 0 ) ; ( 0 , –7 ) ( –9 , 0 ) ; ( 0 , –7 )

Line 1: m = − 2 ; m = − 2 ; Line 2: m = − 2 ; m = − 2 ; Parallel

y = − 0.2 x + 21 y = − 0.2 x + 21

More than 250

y = − 3 00 x + 11 , 5 00 y = − 3 00 x + 11 , 5 00

• ⓑ 100 students per year
• ⓒ P ( t ) = 1 00 t + 17 00 P ( t ) = 1 00 t + 17 00

Extrapolation

y = − 1.294 x + 49.412 ;   r = − 0.974 y = − 1.294 x + 49.412 ;   r = − 0.974

Practice Test

y = −1.5x − 6

y = −2x − 1

Perpendicular

(−7, 0); (0, −2)

y = −0.25x + 12

Slope = −1 and y-intercept = 6

y = 875x + 10,625

• ⓑ dropped an average of 46.875, or about 47 people per year
• ⓒ y = −46.875t + 1250

In early 2018

y = 0.00455x + 1979.5

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