eureka math lesson 3 homework 2.1 answer key

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Eureka Math Grade 2 Answer Key | Engage NY Math 2nd Grade Answer Key Solutions

Engageny math grade 2 answer key | eureka math 2nd grade answers key pdf free download.

• Eureka Math Grade 2 Module 1 Answer Key
• Eureka Math Grade 2 Module 2 Answer Key
• Eureka Math Grade 2 Module 3 Answer Key
• Eureka Math Grade 2 Module 4 Answer Key
• Eureka Math Grade 2 Module 5 Answer Key
• Eureka Math Grade 2 Module 6 Answer Key
• Eureka Math Grade 2 Module 7 Answer Key
• Eureka Math Grade 2 Module 8 Answer Key

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Eureka Math Grade 4 Module 6 Lesson 4 Answer Key

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

Answer: 2/10 = 0.2, 3/10 = 0.3, 4/10 = 0.4, 8/10 = 0.8, 6/10 = 0.6, 0.1 = 1/10, 0.2 = 2/10, 0.3 = 3/10, 0.7 = 7/10, 0.5 = 5/10, 5/10 = 0.5, 0.8 = 8/10, 7/10 = 0.7, 0.4 = 4/10, 9/10 = 0.9, 10/10 = 1, 11/10 = 1.1, 12/10 = 1.2, 15/10 = 1.5, 25/10 = 2.5, 45/10 = 4.5, 38/10 = 3.8, 1 = 10/10, 2 = 20/10, 5 = 50/10, 4 = 40/10, 4.1 = 41/10, 4.2 = 42/10, 4.6 = 46/10, 2.6 = 26/10, 3.6 = 36/10, 3.4 = 34/10, 2.3 = 23/10, 43/10 = 4.3, 20/10 = 2, 1.8 = 18/10, 34/10 = 3.4, 50/10 = 5, 4.7 = 47/10, 28/10 = 2.8, 30/10 = 3, 3.2 = 32/10, 20/10 = 2, 2.1 = 21/10.

Question 1. \(\frac{2}{10}\) = 0.2

Answer: 2/10 = 0.2.

Explanation: In the above-given question, given that, \(\frac{2}{10}\). 2/10 = 0.2.

Question 2. \(\frac{3}{10}\) = 0.3

Answer: 3/10 = 0.3.

Explanation: In the above-given question, given that, \(\frac{3}{10}\). 3/10 = 0.3.

Question 3. \(\frac{4}{10}\) = 0.4

Answer: 4/10 = 0.4.

Explanation: In the above-given question, given that, \(\frac{4}{10}\). 4/10 = 0.4.

Question 4. \(\frac{8}{10}\) = 0.8

Answer: 8/10 = 0.8.

Explanation: In the above-given question, given that, \(\frac{8}{10}\). 8/10 = 0.8.

Question 5. \(\frac{6}{10}\) = 0.6

Answer: 6/10 = 0.6.

Explanation: In the above-given question, given that, \(\frac{6}{10}\). 6/10 = 0.6.

Question 6. 0.1 = \(\frac{}{10}\)

Answer: 1/10 = 0.1.

Explanation: In the above-given question, given that, \(\frac{1}{10}\). 1/10 = 0.1.

Question 7. 0.2 = \(\frac{}{10}\)

Question 8. 0.3 = \(\frac{}{10}\)

Question 9. 0.7 = \(\frac{}{10}\)

Answer: 7/10 = 0.7.

Explanation: In the above-given question, given that, \(\frac{7}{10}\). 7/10 = 0.7.

Question 10. 0.5 = \(\frac{}{10}\)

Answer: 5/10 = 0.5.

Explanation: In the above-given question, given that, \(\frac{5}{10}\). 5/10 = 0.5.

Question 11. \(\frac{5}{10}\) = 0.5

Question 12. 0.8 = \(\frac{}{10}\)

Question 13. \(\frac{7}{10}\) = 0.7

Question 14. 0.4 = \(\frac{}{10}\)

Question 15. \(\frac{9}{10}\) = 0.9

Answer: 9/10 = 0.9.

Explanation: In the above-given question, given that, \(\frac{9}{10}\). 9/10 = 0.9.

Question 16. \(\frac{10}{10}\) = 1

Answer: 10/10 = 1.

Explanation: In the above-given question, given that, \(\frac{10}{10}\). 10/10 = 1.

Question 17. \(\frac{11}{10}\) = 1.1

Answer: 11/10 = 1.1.

Explanation: In the above-given question, given that, \(\frac{11}{10}\). 11/10 = 1.1.

Question 18. \(\frac{12}{10}\) = 1.2

Answer: 12/10 = 1.2.

Explanation: In the above-given question, given that, \(\frac{12}{10}\). 12/10 = 1.2.

Question 19. \(\frac{15}{10}\) = 1.5.

Answer: 15/10 = 1.5.

Explanation: In the above-given question, given that, \(\frac{15}{10}\). 15/10 = 1.5.

Question 20. \(\frac{25}{10}\) = 2.5.

Answer: 25/10 = 2.5.

Explanation: In the above-given question, given that, \(\frac{25}{10}\). 25/10 = 2.5.

Question 21. \(\frac{45}{10}\) = 4.5.

Answer: 45/10 = 4.5.

Explanation: In the above-given question, given that, \(\frac{45}{10}\). 45/10 = 4.5.

Question 22. \(\frac{38}{10}\) = 3.8.

Answer: 38/10 = 3.8.

Explanation: In the above-given question, given that, \(\frac{38}{10}\). 38/10 = 3.8.

Question 23. 1 = \(\frac{}{10}\)

Question 24. 2 = \(\frac{}{10}\)

Answer: 20/10 = 2.

Explanation: In the above-given question, given that, \(\frac{20}{10}\). 20/10 = 2.

Question 25. 5 = \(\frac{}{10}\)

Answer: 50/10 = 5.

Explanation: In the above-given question, given that, \(\frac{50}{10}\). 50/10 = 5.

Question 26. 4 = \(\frac{}{10}\)

Answer: 40/10 = 4.

Explanation: In the above-given question, given that, \(\frac{40}{10}\). 40/10 = 4.

Question 27. 4.1 = \(\frac{}{10}\)

Answer: 41/10 = 4.1.

Explanation: In the above-given question, given that, \(\frac{41}{10}\). 41/10 = 4.1.

Question 28. 4.2 = \(\frac{}{10}\)

Answer: 42/10 = 4.2.

Explanation: In the above-given question, given that, \(\frac{42}{10}\). 42/10 = 4.2.

Question 29. 4.6 = \(\frac{}{10}\)

Answer: 46/10 = 4.6.

Explanation: In the above-given question, given that, \(\frac{46}{10}\). 46/10 = 4.6.

Question 30. 2.6 = \(\frac{}{10}\)

Answer: 26/10 = 2.6.

Explanation: In the above-given question, given that, \(\frac{26}{10}\). 26/10 = 2.6.

Question 31. 3.6 = \(\frac{}{10}\)

Answer: 36/10 = 3.6.

Explanation: In the above-given question, given that, \(\frac{36}{10}\). 36/10 = 3.6.

Question 32. 3.4 = \(\frac{}{10}\)

Answer: 34/10 = 3.4.

Explanation: In the above-given question, given that, \(\frac{34}{10}\). 34/10 = 3.4.

Question 33. 2.3 = \(\frac{}{10}\)

Answer: 23/10 = 2.3.

Explanation: In the above-given question, given that, \(\frac{23}{10}\). 23/10 = 2.3.

Question 34. 4\(\frac{3}{10}\) = 4.3.

Answer: 43/10 = 4.3.

Explanation: In the above-given question, given that, \(\frac{43}{10}\). 43/10 = 4.3.

Question 35. \(\frac{20}{10}\) = 2.

Question 36. 1.8 = \(\frac{}{10}\)

Answer: 18/10 = 1.8.

Explanation: In the above-given question, given that, \(\frac{18}{10}\). 18/10 = 1.8.

Question 37. 3\(\frac{4}{10}\) = 3.4.

Question 38. \(\frac{50}{10}\) = 5.

Question 39. 4.7 = \(\frac{}{10}\)

Answer: 47/10 = 4.7.

Explanation: In the above-given question, given that, \(\frac{47}{10}\). 47/10 = 4.7.

Question 40. 2\(\frac{8}{10}\) = 2.8.

Answer: 28/10 = 2.8.

Explanation: In the above-given question, given that, \(\frac{28}{10}\). 28/10 = 2.8.

Question 41. \(\frac{30}{10}\) = 3.

Answer: 30/10 = 3.

Explanation: In the above-given question, given that, \(\frac{30}{10}\). 30/10 = 3.

Question 42. 3.2 = \(\frac{}{10}\)

Answer: 32/10 = 3.2.

Explanation: In the above-given question, given that, \(\frac{32}{10}\). 32/10 = 3.2.

Question 43. \(\frac{20}{10}\) = 2.

Question 44. 2.1 = \(\frac{}{10}\)

Answer: 21/10 = 2.1.

Explanation: In the above-given question, given that, \(\frac{21}{10}\). 21/10 = 2.1.

Answer: 1/10 = 0.1, 2/10 = 0.2, 3/10 = 0.3, 7/10 = 0.7, 5/10 = 0.5, 0.2 = 2/10, 0.3 = 3/10, 0.4 = 4/10, 0.8 = 8/10, 0.6 = 6/10, 4/10 = 0.4, 0.9 = 9/10, 6/10 = 0.6, 0.5 = 5/10, 9/10 = 0.9, 10/10 = 1, 11/10 = 1.1, 12/10 = 1.2, 17/10 = 1.7, 27/10 = 2.7, 47/10 = 4.7, 34/10 = 3.4, 1 = 10/10, 2 = 20/10, 4 = 40/10, 3 = 30/10, 3.1 = 31/10, 3.2 = 32/10, 3.6 = 36/10, 1.6 = 16/10, 2.6 = 26/10, 4.2 = 42/10, 2.5 = 25/10, 34/10 = 3.4, 20/10 = 2, 4.6 = 46/10, 24/10 = 2.4, 40/10 = 4, 2.3 = 23/10, 30/10 = 3, 4.1 = 41/10.

Question 1. \(\frac{1}{10}\) = 0.1.

Question 2. \(\frac{2}{10}\) = 0.2.

Question 3. \(\frac{3}{10}\) = 0.3.

Question 4. \(\frac{7}{10}\) = 0.7.

Question 5. \(\frac{5}{10}\) = 0.5.

Question 6. 0.2 = \(\frac{}{10}\)

Question 7. 0.3 = \(\frac{}{10}\)

Question 8. 0.4 = \(\frac{}{10}\)

Question 9. 0.8 = \(\frac{}{10}\)

Question 10. 0.6 = \(\frac{}{10}\)

Question 11. \(\frac{4}{10}\) = 0.4

Question 12. 0.9 = \(\frac{}{10}\)

Question 13. \(\frac{6}{10}\) = 0.6.

Question 14. 0.5 = \(\frac{}{10}\)

Question 15. \(\frac{9}{10}\) = 0.9.

Question 16. \(\frac{10}{10}\) =1.

Question 17. \(\frac{11}{10}\) = 1.1.

Question 18. \(\frac{12}{10}\) = 1.2.

Question 19. \(\frac{17}{10}\) = 1.7.

Answer: 17/10 = 1.7.

Explanation: In the above-given question, given that, \(\frac{17}{10}\). 17/10 = 1.7.

Question 20. \(\frac{27}{10}\) = 2.7.

Answer: 27/10 = 2.7.

Explanation: In the above-given question, given that, \(\frac{27}{10}\). 27/10 = 2.7.

Question 21. \(\frac{47}{10}\) = 4.7.

Question 22. \(\frac{34}{10}\) = 3.4.

Question 25. 4 = \(\frac{}{10}\)

Question 26. 3 = \(\frac{}{10}\)

Question 27. 3.1 = \(\frac{}{10}\)

Answer: 31/10 = 3.1.

Explanation: In the above-given question, given that, \(\frac{31}{10}\). 31/10 = 3.1.

Question 28. 3.2 = \(\frac{}{10}\)

Question 29. 3.6 = \(\frac{}{10}\)

Question 30. 1.6 = \(\frac{}{10}\)

Answer: 16/10 = 1.6.

Explanation: In the above-given question, given that, \(\frac{16}{10}\). 16/10 = 1.6.

Question 31. 2.6 = \(\frac{}{10}\)

Question 32. 4.2 = \(\frac{}{10}\)

Question 33. 2.5 = \(\frac{}{10}\)

Question 34. 3\(\frac{4}{10}\) = 3.4.

Question 35. \(\frac{50}{10}\) = 5.

Question 36. 1.7 = \(\frac{}{10}\)

Question 37. 4\(\frac{3}{10}\) = 4.3.

Question 38. \(\frac{20}{10}\) = 2.

Question 39. 4.6 = \(\frac{}{10}\)

Question 40. 2\(\frac{4}{10}\) =2.4.

Answer: 24/10 = 2.4.

Explanation: In the above-given question, given that, \(\frac{24}{10}\). 24/10 = 2.4.

Question 41. \(\frac{40}{10}\) = 4.

Question 42. 2.3 = \(\frac{}{10}\)

Question 43. \(\frac{30}{10}\) = 3.

Question 44. 4.1 = \(\frac{}{10}\)

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

Answer: a.The length of the shaded part of the meter stick in centimeters = 1cm. b. 1/10 fraction of a meter is 1 centimeter. c. The length of the shaded portion of the meter stick = 1/10. d. decimal form = 0.1. e. 1/10 fraction of a meter is 10 centimeters.

Explanation: In the above-given question, given that, 1 meter = 10 centimeters. a.The length of the shaded part of the meter stick in centimeters = 1cm. b. 1/10 fraction of a meter is 1 centimeter. c. The length of the shaded portion of the meter stick = 1/10. d. decimal form = 0.1. e. 1/10 fraction of a meter is 10 centimeters.

Question 2. Fill in the blanks. a. 1 tenth = __10__ hundredths b. \(\frac{1}{10}\)m = \(\frac{}{10}\)m c. \(\frac{2}{10}\)m = \(\frac{20}{}\)m

Answer: a. 1 tenth = 10 hundredths. b. 1/10 = 10/10. c. 2/10 = 20/100.

Explanation: In the above-given question, given that, a. 1 tenth = 10 hundredths. b. 1/10 = 10/10. c. 2/10 = 20/100.

Answer: 0.24 m.

Answer: 0.38 m.

Answer: 0.84 m.

Explanation: In the above-given question, given that, \(\frac{8}{10}\)m + \(\frac{4}{100}\)m = \(\frac{84}{100}\)m = 0.84m 8/10 = 0.8. 4/100 = 0.04. 0.8 + 0.04 = 0.84 m

Answer: 0.17 m.

Explanation: In the above-given question, given that, \(\frac{1}{10}\)m + \(\frac{7}{100}\)m = \(\frac{17}{100}\)m = 0.17m 1/10 = 0.1. 7/100 = 0.07. 0.1 + 0.07 = 0.17 m

Answer: 0.19 m.

Explanation: In the above-given question, given that, \(\frac{1}{10}\)m + \(\frac{9}{100}\)m = \(\frac{19}{100}\)m = 0.19m 1/10 = 0.1. 9/100 = 0.09. 0.1 + 0.09 = 0.19 m.

Question 5. Draw a number bond, pulling out the tenths from the hundredths as in Problem 3. Write the total as the equivalent decimal. a. \(\frac{19}{100}\)m b. \(\frac{28}{100}\)m c. \(\frac{77}{100}\) m d. \(\frac{94}{100}\)m

Answer: a. 0.19 m.

Answer: b. 0.28 m.

Answer: c. 0.77 m.

Answer: 0.94 m.

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

Answer: 0.6 m.

Explanation: In the above-given question, given that, \(\frac{6}{10}\)m  = 0.6m 6/10 = 0.6.

Question 2. Draw a number bond, pulling out the tenths from the hundredths. Write the total as the equivalent decimal. a. \(\frac{62}{100}\)m b. \(\frac{27}{100}\)

Answer: a.0.62 m.

Answer: b.0.27 m.

Eureka Math Grade 4 Module 6 Lesson 4 Homework Answer Key

Answer: a.The length of the shaded part of the meter stick in centimeters = 3cm. b. 1/30 fraction of a meter is 3 centimeters. c. The length of the shaded portion of the meter stick = 1/30. d. decimal form = 0.03. e. 1/30 fraction of a meter is 30 centimeters.

Explanation: In the above-given question, given that, 1 meter = 100 centimeters. a.The length of the shaded part of the meter stick in centimeters = 3cm. b. 1/30 fraction of a meter is 3 centimeters. c. The length of the shaded portion of the meter stick = 1/30. d. decimal form = 0.03. e. 1/30 fraction of a meter is 30 centimeters.

Question 2. Fill in the blanks. a. 5 tenths = __50__ hundredths b. \(\frac{5}{10}\)m = \(\frac{}{100}\)m c. \(\frac{4}{10}\)m = \(\frac{40}{}\) m

Answer: a. 5 tenth = 50 hundredths. b. 5/10 = 50/100. c. 4/10 = 40/100.

Explanation: In the above-given question, given that, a. 5 tenth = 50 hundredths. b. 5/10 = 50/10. c. 4/10 = 40/100.

Answer: 0.46 m.

Answer: 0.15 m.

Explanation: In the above-given question, given that, \(\frac{1}{10}\)m + \(\frac{5}{100}\)m = \(\frac{15}{100}\)m = 0.15m 1/10 = 0.1. 5/100 = 0.05. 0.1 + 0.05 = 0.15 m

Answer: 0.41 m.

Question 5. Draw a number bond, pulling out the tenths from the hundredths, as in Problem 3 of the Homework. Write the total as the equivalent decimal. a. \(\frac{23}{100}\)m b. \(\frac{38}{100}\)m c. \(\frac{82}{100}\)m d. \(\frac{76}{100}\)m

Answer: a.0.23 m.

Answer: b. 0.38 m.

Answer: c. 0.82 m.

Answer: d. 0.76 m.

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Eureka Math Algebra 2 Module 3 Lesson 6 Answer Key

Engage ny eureka math algebra 2 module 3 lesson 6 answer key, eureka math algebra 2 module 3 lesson 6 exercise answer key.

Exercises 1 – 3:

Exercise 1. Assume that there is initially 1 cm of water in the tank, and the height of the water doubles every 10 seconds. Write an equation that could be used to calculate the height H(t) of the water in the tank at any time t. Answer: The height of the water at time t seconds con be modeled by H(t) = 2 t/10

Exercise 2. How would the equation in Exercise 1 change if…

a. the initial depth of water in the tank was 2 cm? Answer: H(t) = 2 . 2 t/10

b. the initial depth of water in the tank was cm? Answer: H(t) = \(\frac{1}{2}\) . 2 t/10

c. the initial depth of water in the tank was 10 cm? Answer: H(t) = 10 . 2 t/10

d. the initial depth of water in the tank was A cm, for some positive real number A? Answer: H(t) = A . 2 t/10

Exercise 3. How would the equation in Exercise 2, part (d), change if…

a. the height tripled every ten seconds? Answer: H(t) = A . 3 t/10

b. the height doubled every five seconds? Answer: H(t) = A . 2 t/5

c. the height quadrupled every second? Answer: H(t) = A . 4 t

d. the height halved every ten seconds? Answer: H(t) = A . (0.5 t/10 )

Example 1. Consider two identical water tanks, each of which begins with a height of water 1 cm and fills with water at a different rate. Which equations can be used to calculate the height of water in each tank at time t? Use H 1 for tank 1 and H 2 for tank 2.

a. If both tanks start filling at the same time, which one fills first? Answer: Tank 2 fills first because the level is rising more quickly.

b. We want to know the average rate of change of the height of the water in these tanks over an interval that starts at a fixed time T as they are filling up. What is the formula for the average rate of change of a function f on an interval [a, b]? Answer: \(\frac{f(b)-f(a)}{b-a}\)

c. What is the formula for the average rate of change of the function H 1 on an interval [a, b]? Answer: \(\frac{\boldsymbol{H}_{1}(\boldsymbol{b})-\boldsymbol{H}_{1}(\boldsymbol{a})}{\boldsymbol{b}-\boldsymbol{a}}\)

d. Let’s calculate the average rate of change of the function H 1 on the interval [T, T + 0. 1], which is an interval one-tenth of a second long starting at an unknown time T. Answer:

Exercises 4 – 8:

On average, over the time interval [T, T + 0. 1], the water in tank 2 rises at a rate of approximately 1.16123H 2 (T) centimeters per second.

On overage, over the time interval [T, T + 0.01], the water in tank 1 rises at a rate of approximately 0.693387H 1 (T) centimeter per second.

Over the time interval [T, T + 0.001], the water in tank 2 rises at an average rate of approximately 1. 09922H 2 (T) centimeters per second.

Exercise 6. In Exercise 5, the average rate of change of the height of the water in tank 1 on the interval [T, T + 0.001] can be described by the expression c 1 2T, and the average rate of change of the height of the water in tank 2 on the interval [T, T + 0. 001] can be described by the expression c 2 3T• What are approximate values of c 1 and c 2 ? Answer: c 1 ≈ 0.69339 and c 2 ≈ 1. 09922.

Exercise 7. As an experiment, let’s look for a value of b so that if the height of the water can be described by H(t) = b t , then the expression for the average rate of change on the interval [T, T + 0. 001] is 1 . H(T).

a. Write out the expression for the average rate of change of H(t) = b t on the interval [T, T + 0.001]. Answer: \(\frac{\boldsymbol{H}_{b}(\boldsymbol{T}+\mathbf{0 . 0 0 1})-\boldsymbol{H}_{b}(\boldsymbol{T})}{\mathbf{0 . 0 0 1}}\)

c. Now we want to find the value of b that satisfies the equation you found in part (b), but we do not have a way to explicitly solve this equation. Look back at Exercise 6; which two consecutive integers have b between them? Answer: We are looking for the base of the exponent that produces a rate of change on a small interval near t that is 1. H(t). When that base is 2, the value of the rate is roughly 0.69H (t). When the base is 3, the value of the rate is roughly 1.1H. Since 0.69 < 1 < 1.1, the base we are looking for is somewhere between 2 and 3.

d. Use your calculator and a guess-and-check method to find an approximate value of b to 2 decimal places. Answer: Students may choose to use a table such as a table shown below. Make sure that students are maintaining enough decimal places of b 0.001 to determine which value is closest to 0.001.

Then b ≈ 2.72.

Exercise 8. Verify that for the value of b found in Exercise 7, \(\frac{H_{b}(T+0.001)-H_{b}(T)}{0.001}\) ≈ H b (T), where H b (T) = b T Answer:

When the height of the water increases by a factor of 2.72 units per second, the height at any time is equal to the rate of change of height at that time.

Eureka Math Algebra 2 Module 3 Lesson 6 Problem Set Answer Key

Question 1. The product 4 . 3 . 2 . 1 is called 4 factorial and is denoted by 4!. Then 10! = 10 . 9 . 8 . 7 . 6 . 5 . 4 . 3 . 2 . 1, and for any positive integer n, n! = n(n – 1)(n – 2) ……… 3 . 2 . 1.

b. Evaluate the sum \(1+\frac{1}{1 !}\). Answer: 2

c. Evaluate the sum \(1+\frac{1}{1 !}+\frac{1}{2 !}\) Answer: 2.5

d. Use a calculator to approximate the sum \(1+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !}\) to 7 decimal places. Do not round the fractions before evaluating the sum. Answer: \(\frac{8}{3}\) ≈ 2.6666667

e. Use a calculator to approximate the sum \(1+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !}+\frac{1}{4 !}\) to 7 decimal places. Do not round the fractions before evaluating the sum. Answer: \(\frac{65}{24}\) ≈ 2.7083333

f. Use a calculator to approximate sums of the form \(1+\frac{1}{1 !}+\frac{1}{2 !}+\cdots+\frac{1}{k !}\) to 7 decimal places for k = 5, 6, 7,8,9, 10. Do not round the fractions before evaluating the sums with a calculator. Answer: If k = 5, the sum is \(\frac{163}{60}\) ≈ 2.1766667.

If k = 6, the sum is \(\frac{1957}{720}\) ≈ 2.7180556.

1f k = 7,the sum is \(\frac{685}{252}\) ≈ 2. 7182540.

If k = 8, the sum is \(\frac{109 601}{40320}\) ≈ 2.7182788.

If k = 9, the sum is \(\frac{98 461}{36 288}\) ≈ 2.7182815.

If k = 10, the sum is \(\frac{9 864 101}{3 628 800}\) ≈ 2.7182818.

g. Make a conjecture about the sums \(1+\frac{1}{1 !}+\frac{1}{2 !}+\cdots+\frac{1}{k !}\) for positive integers k as k increases in size. Answer: It seems that as k gets larger, the sums \(1+\frac{1}{1 !}+\frac{1}{2 !}+\cdots+\frac{1}{k !}\) get closer to e.

h. Would calculating terms of this sequence ever yield an exact value of e? Why or why not? Answer: No. The number e is irrational, so it cannot be written as a quotient of integers. Any finite sum \(1+\frac{1}{1 !}+\frac{1}{2 !}+\cdots+\frac{1}{k !}\) can be expressed as a single rational number with denominator k!, so the sums are all rational numbers. However, the more terms that are calculated, the closer to e the sum becomes, so these sums provide better and better rational number approximations of e.

Question 2. Consider the sequence given by a n = (1 + \(\frac{1}{n}\)) n where n ≥ 1 is an integer.

a. Use your calculator to approximate the first 5 terms of this sequence to 7 decimal places. Answer: a 1 = (1 + \(\frac{1}{1}\)) 1 = 2 a 2 = (1 + \(\frac{1}{2}\)) 2 = 2.25 a 3 = (1 + \(\frac{1}{3}\)) 3 ≈ 2.3703704 a 4 = (1 + \(\frac{1}{4}\)) 4 ≈ 2.4414063 a 5 = (1 + \(\frac{1}{5}\)) 5 = 2.4883200

b. Does it appear that this sequence settles near a particular value? Answer: No, the numbers get bigger, but we cannot tell if it keeps getting bigger or settles on or near a particular value.

c. Use a calculator to approximate the following terms of this sequence to 7 decimal places. Answer: i. a 100 = 2.7081383 ii. a 1000 = 2.7169239 iii. a 10,000 = 2.7181459 iv. a 100,000 = 2.7182682 v. a 1,000,000 = 2.7182805 vi. a 100,000 = 2.7182816 vii. a 100,000,000 = 2.7182818

d. Does it appear that this sequence settles near a particular value? Answer: Yes, it appears that as n gets really large (at least 100,000,000), the terms a n of the sequence settle near the value of e.

e. Compare the results of this exercise with the results of Problem 1. What do you observe? Answer: It took about 10 terms of the sum in Problem 1 to see that the sum settled at the value e, but it takes 100,000,000 terms of the sequence in this problem to see that the sum settles at the value e.

Question 3. If x = 5a 4 and a = 2e 3 , express x in terms of e, and approximate to the nearest whole number. Answer: If x = 5a 4 and a = 2e 3 , then x = 5(2e 3 ) 4 . Rewriting the right side in an equivalent form gives x = 80e 12 ≈ 13020383.

Question 4. If a = 2b 3 and b = –\(\frac{1}{2}\)e -2 , express a in terms of e, and approximate to four decimal places. Answer: If a = 2b 3 and b = –\(\frac{1}{2}\)e -2 , then a = 2 (-\(\frac{1}{2}\) e -2 ) 3 . Rewriting the right side in an equivalent form gives a = –\(\frac{1}{4}\)e -6 ≈ -0.0006.

Question 5. If x = 3e 4 and e = \(\frac{s}{2 x^{3}}\) show that s = 54e 13 , and approximate sto the nearest whole number. Answer: Rewrite the equation e = \(\frac{s}{2 x^{3}}\) to isolate the variable s. e = \(\frac{s}{2 x^{3}}\) 2x 3 e = s

By the substitution property, if s = 2x 3 e and x = 3e 4 , then s = 2(3e 4 ) 3 e. Rewriting the right side in an equivalent form gives s = 2 . 27e 12 . e = 54e 13 ≈ 23890323.

Question 6. The following graph shows the number of barrels of oil produced by the Glenn Pool well in Oklahoma from 1910 to 1916.

a. Estimate the average rate of change of the amount of oil produced by the well on the interval [1910, 1916], and explain what that number represents. Answer: Student responses will vary based on how they read the points on the graph. Over the interval [1910, 1916], the average rate of change is roughly \(\frac{300-3200}{1916-1910}=-\frac{2900}{6}\) ≈ -483.33. This says that the production of the well decreased by an average of about 483 barrels of oil each year between 1910 and 1916.

b. Estimate the average rate of change of the amount of oil produced by the well on the interval [1910, 1913], and explain what that number represents. Answer: Student responses will vary based on how they read the points on the graph. Over the interval [1910, 19131], the average rate of change is roughly \(\frac{800-3200}{1913-1910}=-\frac{2400}{3}\) = -800. This says that the production of the well decreased by an average of about 800 barrels of oil per year between 1910 and 1913.

c. Estimate the average rate of change of the amount of oil produced by the well on the interval [1913, 1916], and explain what that number represents. Answer: Student responses will vary based on how they read the points on the graph. Over the interval [1913, 1916], the average rate of change is roughly \(\frac{300-800}{1916-1913}=-\frac{500}{3}\) ≈ -166.67

This says that the production of the well decreased by on an average of about 166.67 barrels of oil per year between 1913 and 1916.

d. Compare your results for the rates of change in oil production in the first half and the second half of the time period in question in parts (b) and (c). What do those numbers say about the production of oil from the well? Answer: The production dropped much more rapidly in the first three years than it did in the second three years. Looking at the graph, it looks like the oil in the well might be running out, so less and less can be extracted each year.

e. Notice that the average rate of change of the amount of oil produced by the well on any interval starting and ending in two consecutive years is always negative. Explain what that means in the context of oil production. Answer: Because the average rate of change of oil production over a one-year period is always negative, the well is producing less oil each year than it did the year before.

Question 7. The following table lists the number of hybrid electric vehicles (HEVs) sold in the United States between 1999 and 2013.

a. During which one-year interval is the average rate of change of the number of HEVs sold the largest? Explain how you know. Answer: The average rate of change of the number of HEVs sold is largest during [2011,2012] because of the number of HEVssold increases by the largest amount between those two years.

b. Calculate the average rate of change of the number of HEVs sold on the interval [2003,2004], and explain what that number represents. Answer: On the interval [2003,2004], the average rate of change in sales of HEVs is \(\frac{84199-47600}{2004-2003}\) which is 36,599. This means that during this one-year period, HEVs were selling at a rate of 36, 599 vehicles per year.

c. Calculate the average rate of change of the number of HEVs sold on the interval [2003, 2008], and explain what that number represents. Answer: On the interval [2003, 2008], the average rate of change in sales of HEVs is \(\frac{312,386-47,600}{2008-2003}\), which is 52,957.2. This means that during this five-year period, HEVs were selling at an average rate of 52,957 vehicles per year.

d. What does it mean if the average rate of change of the number of HEVs sold is negative? Answer: If the average rate of change of the vehicles sold is negative, then the sales are declining. This means that fewer cards were sold than in the previous year.

Question 8. The formula for the area of a circle of radius r can be expressed as a function A(r) = πr 2 .

a. Find the average rate of change of the area of a circle on the interval [4, 5]. Answer: \(\frac{A(5)-A(4)}{5-4}=\frac{25 \pi-16 \pi}{1}\) = 9π

b. Find the average rate of change of the area of a circle on the interval [4, 4.1]. Answer: \(\frac{A(4.1)-A(4)}{4.1-4}=\frac{16.81 \pi-16 \pi}{0.1}\) = 8.1π

c. Find the average rate of change of the area of a circle on the interval [4, 4.01]. Answer: \(\frac{A(4.01)-A(4)}{4.01-4}=\frac{16.0801 \pi-16 \pi}{0.01}\) = 8.01π

d. Find the average rate of change of the area of a circle on the interval [4, 4.001]. Answer: \(\frac{A(4.001)-A(4)}{4.001-4}=\frac{16.008001 \pi-16 \pi}{0.001}\) = 8.001π

e. What is happening to the average rate of change of the area of the circle as the interval gets smaller and smaller? Answer: The average rate of change of the area of the circle appears to be getting close to 8π.

f. Find the average rate of change of the area of a circle on the interval [4, 4 + h] for some small positive number h. Answer: \(\frac{A(4+h)-A(4)}{(4+h)-4}\) = \(\frac{(4+h)^{2} \pi-16 \pi}{h}\) = \(\frac{\left(16+8 h+h^{2}\right) \pi-16 \pi}{h}\) = \(\frac{1}{h}\) (8h + h 2 )π

= \(\frac{1}{h}\) . h(8 + h)π = (8 + h)π

g. What happens to the average rate of change of the area of the circle on the interval [4, 4 + h] as h → 0? Does this agree with your answer to part (d)? Should it agree with your answer to part (e)? Answer: As h → 0, 8 + h → 8, so as h gets smaller, the average rate of change approaches 8. This agrees with my response to part (e), and it should because as h → 0, the Interval [4, 4 + h] gets smaller.

i. What happens to the average rate of change of the area of the circle on the interval [r 0 , r 0 + h] as h → 0? Do you recognize the resulting formula? Answer: As h → 0, the expression for the average rate of change becomes 2πr 0 , which is the circumference of the circle with radius r 0 .

Question 9. The formula for the volume of a sphere of radius r can be expressed as a function V(r) = \(\frac{4}{3}\)πr 3 . As you work through these questions, you will see the pattern develop more clearly if you leave your answers in the form of a coefficient times π. Approximate the coefficient to five decimal places.

a. Find the average rate of change of the volume of a sphere on the interval [2, 3]. Answer: \(\frac{V(3)-V(2)}{3-2}=\frac{\frac{4}{3} \cdot 27 \pi-\frac{4}{3} \cdot 8 \pi}{1}\) = \(\frac{4}{3}\) . 19π ≈ 25.33333π

b. Find the average rate of change of the volume of a sphere on the interval [2, 2. 1]. Answer: \(\frac{V(2.1)-V(2)}{2.1-2}=\frac{\frac{4}{3} \pi\left(2.1^{3}-8\right)}{0.1}\) ≈ 16.81333π

c. Find the average rate of change of the volume of a sphere on the interval [2, 2. 01]. Answer: \(\frac{V(2.01)-V(2)}{2.01-2}=\frac{\frac{4}{3} \pi\left(2.01^{3}-8\right)}{0.01}\) ≈ 16.08013π

d. Find the average rate of change of the volume of a sphere on the interval [2, 2.001]. Answer: \(\frac{V(2.001)-V(2)}{2.001-2}=\frac{\frac{4}{3} \pi\left(2.001^{3}-8\right)}{0.001}\) ≈ 16.00800π

e. What is happening to the average rate of change of the volume of a sphere as the interval gets smaller and smaller? Answer: The average rate of change of the volume of the sphere appears to be getting close to 16π.

g. What happens to the average rate of change of the volume of a sphere on the interval [2, 2 + h] as h → 0? Does this agree with your answer to part (e)? Should it agree with your answer to part (e)? Answer: As h → 0, the value of the polynomial 12 + 6h + h 2 approaches 12. Then the average rate of change approaches \(\frac{4 \pi}{3}\) . 12 = 16. This agrees with my response to part (e), and it should because as h → 0, the interval [2, 2 + h] gets smaller.

i. What happens to the average rate of change of the volume of a sphere on the interval [r 0 , r 0 + h] as h → 0? Do you recognize the resulting formula? Answer: As h → 0, the expression for the average rate of change becomes 4πr 0 2 , which is the surface area of the sphere with radius r 0 .

Eureka Math Algebra 2 Module 3 Lesson 6 Exit Ticket Answer Key

Question 1. Suppose that water is entering a cylindrical water tank so that the initial height of the water is 3 cm and the height of the water doubles every 30 seconds. Write an equation of the height of the water at time t seconds. Answer: H(t) = 3(\(2^{\frac{t}{30}}\))

Question 2. Explain how the number e arose in our exploration of the average rate of change of the height of the water in the water tank. Answer: We first noticed that if the water level in the tank was doubling every second, then the average rate of change of the height of the water was roughly 0.69 times the height of the water at that time. And if the water level in the tank was tripling every second, then the average rate of change of the height of the water was roughly 1.1 times the height of the water at that time.

When we went looking for a base b so that the average rate of change of the height of the water was 1.0 times the height of the water at that time, we found that the base was roughly e. Calculating the average rate of change over shorter intervals gave a better approximation of e.

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