Solution Sets
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Using Bar Models to Solve Equations
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Solving Simple Equations with a Pan BalanceÑPart 1
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Solving Simple Equations with a Pan BalanceÑPart 2
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Combining Like Terms
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Generating Equivalent Expressions and Equations
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T-Shirt Cost Estimates
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Reversing Operations
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Building and Solving Equations with the Pan-Balance Model
Home Link 6-10
Comparing Multiple Strategies for Solving Equations
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Unit 6 Progress Check
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Engage ny eureka math 6th grade module 4 lesson 11 answer key, eureka math grade 6 module 4 lesson 11 example answer key.
How many fives are in the model? Answer: 5
How many threes are In the model? Answer: 2
What does the expression represent in words? Answer: The sum of two groups of five and two groups of three
What expression could we write to represent the model? Answer: 2 × 5 + 2 × 3
b. Use the new model and the previous model to answer the next set of questions.
How many threes are in the model? Answer: 2
What does the expression represent in words? Answer: Two groups of the sum of five and three
What expression could we write to represent the model? Answer: (5 + 3) + (5 + 3) or 2(5 + 3)
Is the model in part (a) equivalent to the model in part (b)? Answer: Yes, because both expressions have two 5’s and two 3’s. Therefore, 2 × 5 + 2 × 3 = 2(5 + 3).
d. What relationship do we see happening on either side of the equal sign? Answer: On the left-hand side, 2 is being multiplied by 5 and then by 3 before adding the products together. On the right-hand side, the 5 and 3 are added first and then multiplied by 2.
e. In Grade 5 and in Module 2 of this year, you have used similar reasoning to solve problems. What Is the name of the property that is used to say that 2(5 + 3) is the same as 2 × 5 + 2 × 3? Answer: The name of the property is the distributive property.
What does the model represent in words? Answer: a plus a plus b plus b, two a’s plus two b’s, two times a plus two times b
What does 2a mean? Answer: 2a means that there are 2 a’s or 2 × a.
How many a’s are in the model? Answer: 2
How many b’s are in the model? Answer: 2
What expression could we write to represent the model? Answer: 2a + 2b
How many a’s are in the expression? Answer: 2
How many b’s are in the expression? Answer: 2
What expression could we write to represent the model? Answer: (a + b) + (a + b) = 2(a + b)
Are the two expressions equivalent? Answer: Yes. Both models include 2 a’s and 2 b’s. Therefore, 2a + 2b = 2(a + b).
Use GCF and the distributive property to write equivalent expressions. 1. 3f + 3g = __________ Answer: 3(f + g)
What is the question asking us to do? Answer: We need to rewrite the expression as an equivalent expression in factored form, which means the expression is written as the product of factors. The number outside of the parentheses is the GCF.
How would Problem 1 look if we expanded each term? Answer: 3 ∙ f + 3 ∙ g
What is the GCF in Problem 1? Answer: 3
How can we use the GCF to rewrite this expression? Answer: 3 goes on the outside, and f + g will go inside the parentheses. 3(f + g)
2. 6x + 9y = __________ Answer: 3(2x + 3y)
How would Problem 2 look if we expanded each term? Answer: 2 ∙ 3 ∙ x + 3 ∙ 3 ∙ y
What is the GCF in Problem 2? Answer: The GCF is 3.
How can we use the GCF to rewrite this expression? Answer: I will factor out the 3 from both terms and place it in front of the parentheses. I will place what is left in the terms inside the parentheses: 3(2x + 3y).
3. 3c + 11c = _________ Answer: c(3 + 11)
Is there a greatest common factor in Problem 3? Answer: Yes. When I expand, I can see that each term has a common factor c. 3 ∙ c + 11 ∙ c
Rewrite the expression using the distributive property. Answer: c(3 + 11)
4. 24b + 8 = _________ Answer: 8(3b + 1)
Explain how you used GCF and the distributive property to rewrite the expression in Problem 4. Answer: I first expanded each term. I know that 8 goes into 24, so I used it in the expansion. 2 ∙ 2 ∙ 2 ∙ 3 ∙ b + 2 ∙ 2 ∙ 2 I determined that 2 ∙ 2 ∙ 2, or 8, is the common factor. So, on the outside of the parentheses I wrote 8, and on the inside I wrote the leftover factor, 3b + 1 ∙ 8(3b + 1)
Why is there a 1 in the parentheses? Answer: When I factor out a number, lam leaving behind the other factor that multiplies to make the original number. In this case, when I factor out an 8 from 8, I am left with a 1 because 8 × 1 = 8.
How is this related to the first two examples? Answer: In the first two examples, we saw that we could rewrite the expressions by thinking about groups. We can either think of 24b + 8 as 8 groups of 3b and 8 groups of 1 or as 8 groups of the sum of 3b + 1. This shows that 8(3b) + 8(1) = 8(3b + 1)is the some as 24b + 8.
Exercise 1. Apply the distributive property to write equivalent expressions. a. 7x + 7y Answer: 7(x + y)
b. 15g + 20h Answer: 5(3g + 4h)
c. 18m + 42n Answer: 6(3m + 7n)
d. 30a + 39b Answer: 3(10a + 13b)
e. 11f + 15f Answer: f(11 + 15)
f. 18h + 13h Answer: h(18 + 13)
g. 55m + 11 Answer: 11(5m + 1)
h. 7 + 56y Answer: 7(1 + 8y)
2. Evaluate each of the expressions below. a. 6x + 21 y and 3(2x + 7y) x = 3 and y = 4 Answer: 6(3) + 21(4) 3(23 + 74) 18 + 84 3(6 + 28) 102 3(34) 102 102
b. 5g + 7g and g(5 + 7) g = 6 Answer: 5(6) + 7(6) 6(5 + 7) 30 + 42 6(12) 72 72
c. 14x + 2 and 2(7x + 1) x = 10 Answer: 14(10) + 2 2(7.10 + 1) 140 + 2 2(70 + 1) 142 2(71) 142 142
d. Explain any patterns that you notice in the results to parts (a) – c). Answer: Both expressions in parts (a) – (c) evaluated to the same number when the indicated value was substituted for the variable. This shows that the two expressions are equivalent for the given values.
e. What would happen if other values were given for the variables? Answer: Because the two expressions in each part are equivalent, they evaluate to the same number, no matter what value is chosen for the variable.
How can use you use your knowledge of GCF and the distributive property to write equivalent expressions? Answer: We can use our knowledge of GCF and the distributive property to change expressions from standard form to factored form.
Find the missing value that makes the two expressions equivalent. 4x + 12y ___(x + 3y) 35x + 50y ___(7x + 10y) 18x + 9y ___(2x + y) 32x + 8y ___(4x + y) 100x + 700y ___(x + 7y) Answer: 4x + 12y 4 (x + 3y) 35x + 50y 5(7x + 10y) 18x + 9y 9(2x + y) 32x + 8y 8(4x + y) 100x + 700y 100(x + 7y)
Explain how you determine the missing number. Answer: I would expand each term and determine the greatest common factor. The greatest common factor is the number that is placed on the blank line.
Question 2. Use greatest common factor and the distributive property to write equivalent expressions in factored form for the following expressions. a. 4d + 12e Answer: 4(d + 3e) or 4(1d + 3e)
b. 18x + 30y Answer: 6(3x + 5y)
c. 21a + 28y Answer: 7(3a + 4y)
d. 24f + 56g Answer: 8(3f + 7g)
Use greatest common factor and the distributive property to write equivalent expressions in factored form.
Question 1. 2x + 8y Answer: 2(x + 4y)
Question 2. 13ab + 15 ab Answer: ab(13 + 15)
Question 3. 20g + 24h Answer: 4(5g + 6h)
Greatest Common Factor – Round 1 Directions: Determine the greatest common factor of each pair of numbers.
Question 1. GCF of 10 and 50 Answer: 10
Question 2. GCF of 5 and 35 Answer: 5
Question 3. GCF of 3 and 12 Answer: 3
Question 4. GCF of 8 and 20 Answer: 4
Question 5. GCF of 15 and 35 Answer: 5
Question 6. GCF of 10 and 75 Answer: 5
Question 7. GCF of 9 and 30 Answer: 3
Question 8. GCF of 15 and 33 Answer: 3
Question 9. GCF of 12 and 28 Answer: 4
Question 10. GCF of 16 and 40 Answer: 8
Question 11. GCF of 24 and 32 Answer:8 Question 12. GCF of 35 and 49 Answer: 7
Question 13. GCF of 45 and 60 Answer: 15
Question 14. GCF of 48 and 72 Answer: 24
Question 15. GCF of 50 and 42 Answer: 2
Question 16. GCF of 45 and 72 Answer: 9
Question 17. GCF of 28 and 48 Answer: 4
Question 18. GCF of 44 and 77 Answer: 11
Question 19. GCF of 39 and 66 Answer: 3
Question 20. GCF of 64 and 88 Answer: 8
Question 21. GCF of 42 and 56 Answer: 14
Question 22. GCF of 28 and 42 Answer: 14
Question 23. GCF of 13 and 91 Answer: 13
Question 24. GCF of 16 and 84 Answer: 4
Question 25. GCF of 36 and 99 Answer: 9
Question 26. GCF of 39 and 65 Answer: 13
Question 27. GCF of 27 and 87 Answer: 3
Question 28. GCF of 28 and 70 Answer: 14
Question 29. GCF of 29 and 91 Answer: 13
Question 30. GCF of 34 and 51 Answer: 17
Greatest Common Factor – Round 2 Directions: Determine the greatest common factor of each pair of numbers.
Question 1. GCF of 20 and 80 Answer: 20
Question 2. GCF of 10 and 70 Answer: 10
Question 3. GCF of 9 and 36 Answer: 9
Question 4. GCF of 12 and 24 Answer: 12
Question 5. GCF of 15 and 45 Answer: 15
Question 6. GCF of 10 and 95 Answer: 5
Question 7. GCF of 9 and 45 Answer: 9
Question 8. GCF of 18 and 33 Answer: 3
Question 9. GCF of 12 and 32 Answer: 4
Question 10. GCF of 16 and 56 Answer: 8
Question 11. GCF of 40 and 7 Answer: 8
Question 12. GCF of 35 and 63 Answer: 7
Question 13. GCF of 30 and 75 Answer: 15
Question 14. GCF of 42 and 72 Answer: 6
Question 15. GCF of 30 and 28 Answer: 2
Question 16. GCF of 33 and 99 Answer: 33
Question 17. GCF of 38 and 76 Answer: 38
Question 18. GCF of 26 and 65 Answer: 13
Question 19. GCF of 39 and 48 Answer: 3
Question 20. GCF of 72 and 88 Answer: 8
Question 21. GCF of 21 and 56 Answer: 7
Question 22. GCF of 28 and 52 Answer: 4
Question 23. GCF of 51 and 68 Answer: 17
Question 24. GCF of 48 and 84 Answer: 12
Question 25. GCF of 21 and 63 Answer: 21
Question 26. GCF of 64 and 80 Answer: 16
Question 27. GCF of 36 and 90 Answer: 18
Question 28. GCF of 28 and 98 Answer: 14
Question 29. GCF of 39 and 91 Answer: 13
Question 30. GCF of 38 and 95 Answer: 19
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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
Eureka Math Grade 6 Module 4 Lesson 23 Example Answer Key. For each equation or inequality your teacher displays, write the equation or inequality and then substitute 3 for every x. Determine if the equation or inequality results in a true number sentence or a false number sentence. Display 5 + x = 8.
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Eureka Math Grade 6 Module 4 Answer Key | Engage NY Math 6th Grade Module 4 Answer Key March 31, 2021 / By Sachin
Engage NY Eureka Math Grade 6 Module 4 Lesson 24 Answer Key Eureka Math Grade 6 Module 4 Lesson 24 Opening Exercise Answer Key Opening Exercise: State whether each number sentence is true or false.
Your solution's ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. Question: Algebra II A 6.4 Synthetic Division Worksheet Directions: Divide the polynomials using synthetic divisio (standard form ). If one of the terms is missing, you must (x^ (2)+5x+1)-: (x+3)
Eureka Math Grade 6 Module 4 Lesson 11 Example Answer Key a. Use the model to answer the following questions. How many fives are in the model? Answer: 5 How many threes are In the model? Answer: 2 What does the expression represent in words? Answer: The sum of two groups of five and two groups of three What expression could we write to represent the model? Answer: 2 × 5 + 2 × 3
Question: 6.4 Exercise 4Given the following circuit, find I1,I2,I3,Va, and Vb. Present your calculation steps andcheck them out by performing the simulation.Figure 1.21: Find I1,I2,I3,Va, and Vb6.4.1 CalculationNotes:Explanations, formulas, and equations are expected rather than only results.The whole circuit equivalent resistance: Req=I1=I2=I3=Va=.Vb=.