6.5 math homework answers

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Texas Go Math Grade 5 Lesson 6.5 Answer Key Use Multiplication

Refer to our Texas Go Math Grade 5 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 5 Lesson 6.5 Answer Key Use Multiplication.

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What information am I given? I need to use the size of each __________ of sandwich and the number of ___________ she cuts.

Plan What s my plan or strategy? I can _______________ to organize the information from the problem. Then I can use the organized information to find ___________________

So, Erica has __________ one-third-size sandwich pieces. Answer: Read I need to find the number of 1/3 size sandwich pieces does Eric has after she cuts 6 sandwiches into thirds. I need to use the size of each piece of sandwich and the number of sandwiches she cuts. Plan I can draw a diagram to organize the information from the problem. Then I can use the organized information to find the number of 1/3 size sandwich pieces does Eric has after she cuts 6 sandwiches into thirds.

Math Talk Mathematical Processes

Explain how you can use multiplication to check your answer. Answer: I can multiply the quotient and the divisor to see if the product is equal to the dividend.

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Roberto is cutting 3 blueberry pies into halves to give to his neighbors. How many neighbors will get a \(\frac{1}{2}\)-size pie piece? Read What do I need to find?

What information am I given?

Plan What is my plan or strategy?

So, 6 neighbors will get a 1/2 -size pie piece.

Explain how the diagram you drew for the division problem helps you write a multiplication sentence. Answer: Since Roberto is cutting 3 pies, my diagram needs to show 3 circles to represent the pies. I can divide each of the circles into halves. To find the total number of halves in the 3 circles, I can multiply the number of halves in each circle by the number of circles.

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Problem Solving

Question 4. H.O.T. Use Tools Brianna has a sheet of paper that is 6 feet long. She cuts the length of paper into sixths and then cuts the length of each of these \(\frac{1}{6}\)-pieces into thirds. How many pieces does she have? How many inches long is each piece? Answer:

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Texas Go Math Grade 5 Lesson 6.5 Homework and Practice Answer Key

Question 1. Julian wants to put a border around two sides of his rectangular-shaped garden. He has a 12-foot piece of lumber that will be cut into \(\frac{1}{3}\)-foot pieces. How many \(\frac{1}{3}\)-foot pieces can Julian cut from the lumber? Answer: 12 ÷ 1/3 = 12 x 3 = 36 Julian can cut 36-foot pieces from the lumber. Explanation: Julian wants to put a border around two sides of his rectangular-shaped garden. He has a 12-foot piece of lumber that will be cut into 1/3-foot pieces. Divide 12 by 1/3 the result is 36. So, Julian can cut 36-foot pieces from the lumber.

Go Math Lesson 6.5 Answer Key Homework 5th Grade Question 2. The camp counselors bought a 6-pound bag of raisins for an afternoon snack for the campers. If the counselors package the raisins into \(\frac{1}{8}\)-pound individual servings, how many individual servings can they make? Answer: 6 ÷ 1/8 = 6 x 8 = 48 They can make 48 individual servings. Explanation: The camp counselors bought a 6-pound bag of raisins for an afternoon snack for the campers. The counselors package the raisins into 1/8-pound individual servings. Divide 6 by 1/8 the result is 48. So, they can make 48 individual servings.

Question 3. Kim has 5 yards of denim. She cuts each yard of denim into thirds. How many \(\frac{1}{3}\)-yard pieces of denim does she have? Answer: 5 ÷ 1/3 = 5 x 3 = 15 She have 15 pieces with 1/3-yard pieces of denim. Explanation: Kim has 5 yards of denim. She cuts each yard of denim into thirds. Divide 5 by 1/3 the result is 15. She have 15 pieces with 1/3-yard pieces of denim.

Question 4. If it takes Fran \(\frac{1}{5}\) of an hour to paint one section of the fence, how many sections can she paint in 2 hours? Answer: 1/5 x 2 = 2/5 She can paint 2/5 sections in 2 hours. Explanation: Fran takes 1/5 of an hour to paint one section of the fence. Multiply 1/5 with 2 the product is 2/5. So, she can paint 2/5 sections in 2 hours.

Question 5. You plan to sell pumpkin pie slices at the fall festival. If you cut 4 pies into \(\frac{1}{8}\)-size pieces, how many slices will you have to sell? Describe your strategy for solving the problem. Answer: 4 ÷ 1/8 = 4 x 8 = 32 I have to sell 32 slices. Explanation: I plan to sell pumpkin pie slices at the fall festival. If I cut 4 pies into 1/8-size pieces. Divide 4 by 1/8 the result is 32. So, I have to sell 32 slices.

Go Math 5th Grade Lesson 6.5 Homework Answer Key Question 6. Marcel has 6 strips of paper. Each strip is 1 foot long. He folds each strip into \(\frac{1}{4}\)-foot sections. Describe how you can draw a diagram to find the number of sections Marcel made when he folded the paper strips. Answer:

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Answer Key 6.5

• [latex]6p-42[/latex]
• [latex]32k^2+16k[/latex]
• [latex]12x+6[/latex]
• [latex]18n^3+21n^2[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrrrrr} &8n&+&8&& \\ \times &4n&+&6&& \\ \hline &32n^2&+&32n&& \\ &&&48n&+&48 \\ \hline &32n^2&+&80n&+&48 \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrrrrr} &x&-&4&& \\ \times &2x&+&1&& \\ \hline &2x^2&-&8x&& \\ &&&x&-&4 \\ \hline &2x^2&-&7x&-&4 \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrrrrr} &7b&-&5&& \\ \times &8b&+&3&& \\ \hline &56b^2&-&40b&& \\ &&+&21b&-&15 \\ \hline &56b^2&-&19b&-&15 \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrrrrr} &4r&+&8&& \\ \times &r&+&8&& \\ \hline &4r^2&+&8r&& \\ &&+&32r&+&64 \\ \hline &4r^2&+&40r&+&64 \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrrrrr} &5v&-&2&& \\ \times &3v&-&4&& \\ \hline &15v^2&-&6v&& \\ &&-&20v&+&8 \\ \hline &15v^2&-&26v&+&8 \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrrrrr} &a&-&8&& \\ \times &6a&+&4&& \\ \hline &6a^2&-&48a&& \\ &&+&4a&-&32 \\ \hline &6a^2&-&44a&-&32 \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrrrrr} &6x&-&4y&& \\ \times &5x&+&y&& \\ \hline &30x^2&-&20xy&& \\ &&+&6xy&-&4y^2 \\ \hline &30x^2&-&14xy&-&4y^2 \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrrrrr} &8u&-&7v&& \\ \times &2u&+&3v&& \\ \hline &16u^2&-&14uv&& \\ &&+&24uv&-&21v^2 \\ \hline &16u^2&+&10uv&-&21v^2 \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrrrrr} &8x&+&3y&& \\ \times &7x&+&5y&& \\ \hline &56x^2&+&21xy&& \\ &&+&40xy&+&15y^2 \\ \hline &56x^2&+&61xy&+&15y^2 \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrrrrr} &a&-&3b&& \\ \times &5a&+&8b&& \\ \hline &5a^2&-&15ab&& \\ &&+&8ab&-&24b^2 \\ \hline &5a^2&-&7ab&-&24b^2 \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrrrrrrr} &6r^2&-&r&+&5&& \\ \times &&&r&-&7&& \\ \hline &6r^3&-&r^2&+&5r&& \\ &&-&42r^2&+&7r&-&35 \\ \hline &6r^3&-&43r^2&+&12r&-&35 \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrrrrrrr} &4x^2&+&3x&+&5&& \\ \times &&&4x&+&8&& \\ \hline &16x^3&+&12x^2&+&20x&& \\ &&+&32x^2&+&24x&+&40 \\ \hline &16x^3&+&44x^2&+&44x&+&40 \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrrrrrrr} &2n^2&-&2n&+&5&& \\ \times &&&6n&-&4&& \\ \hline &12n^3&-&12n^2&+&30n&& \\ &&-&8n^2&+&8n&-&20 \\ \hline &12n^3&-&20n^2&+&38n&-&20 \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrrrrrrr} &4b^2&+&4b&+&4&& \\ \times &&&2b&-&3&& \\ \hline &8b^3&+&8b^2&+&8b&& \\ &&-&12b^2&-&12b&-&12 \\ \hline &8b^3&-&4b^2&-&4b&-&12 \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrrrrrrr} &6x^2&-&7xy&+&4y^2&& \\ \times &&&6x&+&3y&& \\ \hline &36x^3&-&42x^2y&+&24xy^2&& \\ &&+&18x^2y&-&21xy^2&+&12y^3 \\ \hline &36x^3&-&24x^2y&+&3xy^2&+&12y^3 \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrrrrrrr} &7m^2&+&6mn&+&4n^2&& \\ \times &&&3m&-&2n&& \\ \hline &21m^3&+&18m^2n&+&12mn^2&& \\ &&-&14m^2n&-&12mn^2&-&8n^3 \\ \hline &21m^3&+&4m^2n&-&8n^3&& \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrrrrrrrrr} &6n^2&-&5n&+&6&&&& \\ \times&8n^2&+&4n&+&6&&&& \\ \hline &48n^4&-&40n^3&+&48n^2&&&& \\ &&+&24n^3&-&20n^2&+&24n&& \\ &&&&+&36n^2&-&30n&+&36 \\ \hline &48n^4&-&16n^3&+&64n^2&-&6n&+&36 \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrrrrrrrrr} &7a^2&-&6a&+&1&&&& \\ \times&2a^2&+&6a&+&3&&&& \\ \hline &14a^4&-&12a^3&+&2a^2&&&& \\ &&+&42a^3&-&36a^2&+&6a&& \\ &&&&+&21a^2&-&18a&+&3 \\ \hline &14a^4&+&30a^3&-&13a^2&-&12a&+&3 \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrrrrrrrrr} &3k^2&+&3k&+&6&&&& \\ \times&5k^2&+&3k&+&3&&&& \\ \hline &15k^4&+&15k^3&+&30k^2&&&& \\ &&+&9k^3&+&9k^2&+&18k&& \\ &&&&+&9k^2&+&9k&+&18 \\ \hline &15k^4&+&24k^3&+&48k^2&+&27k&+&18 \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrrrrrrrrr} &6u^2&+&4uv&+&3v^2&&&& \\ \times &7u^2&+&8uv&-&6v^2&&&& \\ \hline &42u^4&+&28u^3v&+&21u^2v^2&&&& \\ &&+&48u^3v&+&32u^2v^2&+&24uv^3&& \\ &&&&-&36u^2v^2&-&24uv^3&-&18v^4 \\ \hline &42u^4&+&76u^3v&+&17u^2v^2&-&18v^4&& \\ \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrrrrrrrrrrrrr} &2n^3&-&8n^2&+&3n&+&6&&&&&& \\ \times&n^3&-&6n^2&-&2n&+&3&&&&&& \\ \hline &2n^6&-&8n^5&+&3n^4&+&6n^3&&&&&& \\ &&-&12n^5&+&48n^4&-&18n^3&-&36n^2&&&& \\ &&&&-&4n^4&+&16n^3&-&6n^2&-&12n&& \\ &&&&&&+&6n^3&-&24n^2&+&9n&+&18 \\ \hline &2n^6&-&20n^5&+&47n^4&+&10n^3&-&66n^2&-&3n&+&18 \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrrrrrrrrrrrrr} &a^3&+&2a^2&+&3a&+&3&&&&&& \\ \times&a^3&+&2a^2&-&4a&+&1&&&&&& \\ \hline &a^6&+&2a^5&+&3a^4&+&3a^3&&&&&& \\ &&+&2a^5&+&4a^4&+&6a^3&+&6a^2&&&& \\ &&&&-&4a^4&-&8a^3&-&12a^2&-&12a&& \\ &&&&&&+&a^3&+&2a^2&+&3a&+&3 \\ \hline &a^6&+&4a^5&+&3a^4&+&2a^3&-&4a^2&-&9a&+&3 \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrrrrr} &2x&+&1&& \\ \times &3x&-&4&& \\ \hline &6x^2&+&3x&& \\ &&-&8x&-&4 \\ \hline &3(6x^2&-&5x&-&4) \\ &18x^2&-&15x&-&12 \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrrrrr} &2x&-&3&& \\ \times &x&-&4&& \\ \hline &2x^2&-&3x&& \\ &&-&8x&+&12 \\ \hline &5(2x^2&-&11x&+&12) \\ &10x^2&-&55x&+&60 \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrrrrr} &4x&-&5&& \\ \times &2x&+&1&& \\ \hline &8x^2&-&10x&& \\ &&+&4x&-&5 \\ \hline &3(8x^2&-&6x&-&5) \\ &24x^2&-&18x&-&15 \end{array}[/latex]
• [latex]\phantom{a}[/latex] [latex]\begin{array}[t]{rrrrrr} &2x&-&6&& \\ \times &4x&+&1&& \\ \hline &8x^2&-&24x&& \\ &&+&2x&-&6 \\ \hline &2(8x^2&-&22x&-&6) \\ &16x^2&-&44x&-&12 \end{array}[/latex]

Intermediate Algebra Copyright © 2020 by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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Eureka Math Grade 6 Module 5 Lesson 4 Answer Key

Engage ny eureka math grade 6 module 5 lesson 4 answer key, eureka math grade 6 module 5 lesson 4 exercise answer key.

Opening Exercise:

Draw and label the altitude of each triangle below.

Exploratory Challenge/Exercises 1 – 5:

Question 1. Use rectangle X and the triangle with the altitude inside (triangle X) to show that the area formula for the triangle is A = \(\frac{1}{2}\) × base × height.

a. Step One: Find the area of rectangle X. Answer: A = 3 in. × 2.5 in. = 7.5 in 2

b. Step Two: What is half the area of rectangle X? Answer: Half of the area of the rectangle is 7.5 in 2 ÷ 2= 3.75 in 2

c. Step Three: Prove, by decomposing triangle X, that it is the same as half of rectangle X. Please glue your decomposed triangle onto a separate sheet of paper. Glue it into rectangle X. What conclusions can you make about the triangle’s area compared to the rectangle’s area? Answer: Because the triangle fits inside half of the rectangle, we know the triangle’s area is half of the rectangle’s area.

Question 2. Use rectangle Y and the triangle with a side that ¡s the altitude (triangle Y) to show the area formula for the triangle is A = \(\frac{1}{2}\) × base × height.

a. Step One: Find the area of rectangle Y. Answer: A = 3 in. × 3 in. = 9 in 2

b. Step Two: What is half the area of rectangle Y? Answer: Half the area of the rectangle is 9 in 2 ÷ 2 = 4.5 in 2 .

c. Step Three: Prove, by decomposing triangle Y, that it is the same as half of rectangle Y. Please glue your decomposed triangle onto a separate sheet of paper. Glue it into rectangle Y. What conclusions can you make about the triangle’s area compared to the rectangle’s area? Answer: The right triangle also fits in exactly half of the rectangle, so the triangle’s area is • Students may struggle once again half the size of the rectangle’s area. with this step since they have yet to see an obtuse

Question 3. Use rectangle Z and the triangle with the altitude outside (triangle Z) to show that the area angle. formula for the triangle is A = \(\frac{1}{2}\) × base × height.

a. Step One: Find the area of rectangle Z. Answer: A= 3 in. × 2.5 in.=7.5 in 2

b. Step Two: What is half the area of rectangle Z? Answer: Half of the area of the rectangle is 7.5 in 2 ÷2 = 3.75 in 2 .

c. Step Three: Prove, by decomposing triangle Z, that it is the same as half of rectangle Z. Please glue your decomposed triangle onto a separate sheet of paper. Glue it into rectangle Z. What conclusions can you make about the triangle’s area compared to the rectangle’s area? Answer: Similar to the other two triangles, when the altitude is outside the triangle, the area of the triangle is exactly half of the area of the rectangle.

Question 4. When finding the area of a triangle, does it matter where the altitude is located? Answer: It does not matter where the altitude is located. To find the area of a triangle, the formula is always A = \(\frac{1}{2}\) × base × height.

Question 5. How can you determine which part of the triangle is the base and which is the height? Answer: The base and the height of any triangle form a right angle because the altitude is always perpendicular to the base.

Exercises 6 – 8:

Calculate the area of each triangle. Figures are not drawn to scale.

Question 8. Draw three triangles (acute, right, and obtuse) that have the same area. Explain how you know they have the same area. Answer: Answers will vary.

Eureka Math Grade 6 Module 5 Lesson 4 Problem Set Answer Key

Calculate the area of each figure below. Figures are not drawn to scale.

Which student calculated the area correctly? Explain why the other student is not correct. Answer: Donovan calculated the area correctly. Although Darnell did use the altitude of the triangle, he used the length between the altitude and the base rather than the length of the actual base.

a. Determine the area of the larger triangle if it has a height of 12.2 m. Answer: A = \(\frac{1}{2}\) (10.14 m) (12.2 m) = 61.854 m 2

b. Let A be the area of the unshaded (white) triangle in square meters. Write and solve an equation to determine the value of A, using the areas of the larger triangle and the gray triangle. Answer: 40.325m 2 + A = 61.854m 2 40.325m 2 + A – 40.325 m 2 = 61.854 m 2 – 40.325 m 2 A = 21.529 m 2

Eureka Math Grade 6 Module 5 Lesson 4 Exit Ticket Answer Key

Find the area of each triangle. Figures are not drawn to scale.

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