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## 8.E: Homework Chapter 8 Answer Key

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General Questions

1. 3 pizzas, limited by the cheese

Mole-to-Mole Conversions

3. 0.28 moles Cl 2(g)

5. 3.0 moles NH 3

7. 2.4 moles NaCl (aq)

9. 2 NO (g) + 5H 2(g) → 2NH 3(g) + 2H 2 O (g)

2.18 moles NH 3

11. 2.30 moles aluminum chloride

13. Cu 2 O (s) + C (s) → 2Cu (s) + CO (g)

3.2 moles carbon

Mass-to-Mass Conversions

15. 31.3 g NaCl

17. 25.1 g CH 3 OH

19. 2 Al (s) + 3Cl 2(g) → 2AlCl 3(s)

0.56 g AlCl 3

21. 2 Fe (s) + 3S (l) → Fe 2 S 3(s)

6.0 g Fe 2 S 3

23. 94.2 g NO 2

25. 6CO 2(g) + 6H 2 O (l) → 6O 2(g) + C 6 H 12 O 6(aq)

3.8 g C 6 H 12 O 6

4.1 g O 2

Limiting Reactant, Theoretical Yield, and Percent Yield

27. Limiting reactant: 1.4 mol Cl 2

2.8 mol KCl

29. 5.54 g H 2 O

18.0 g NaCl

Limiting reactant: 12.3 g NaOH

31. Limiting reactant: 0.56 mol Al

33. Limiting reactant: 1.4 g Mg

35. Limiting reactant: 21.88 g Cl 2

Theoretical yield: 22.5 g HCl

37. 4AB + C 2 → 2(AB) 2 C

Limiting reactant: 34.5 g AB

39.2 g C 2

42.9 g (AB) 2 C

39. C 6 H 12 O 6 + 6O 2 → 6CO 2 + 6H 2 O

a. Limiting reactant: 10.5 mol O 2

b. 0 Limiting reactant: 0.45 mol O 2

c. 2 Limiting reactant: 13.5 g O 2

d. 4 Limiting reactant: 45.654 g C 6 H 12 O 6

41. SnO 2 + 2H 2 → Sn + 2H 2 O

Limiting reactant: 5.6 mol H 2

3.3 x 10^2 g Sn, 11 g H 2 , 1.0 x 10^2 g H 2 O, 90. g SnO 2

Challenge Problems

43. 6CO 2 + 6H 2 O → C 6 H 12 O 6 + 6O 2

57.5 g C 6 H 12 O 6

45. 81% CO 2

57% H 2 O

90% NaC 2 H 3 O 2

Limiting reactant: 0.99 g NaHCO 3

0.52 g CO 2 , 0.21 g H 2 O, 0.97 g NaC 2 H 3 O 2

47. Mg(OH) 2(s) + 2HCl (aq) → MgCl 2(aq) + 2H 2 O (l)

3.6 g Mg(OH) 2

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## Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series

Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series helps you to get a grip on the concepts from surface level to a deep level. Finish your homework or assignments in time by solving questions from B ig Ideas Math Book Algebra 2 Ch 8 Sequences and Series here. Access the user-friendly solutions provided for all the concepts of Chapter 8 Sequences and Series from Big Ideas Math Algebra 2 Textbooks here for free of cost.

Improve your performance in the final exams by practicing the Big Ideas Math Algebra 2 Answers Ch 8 Sequences and Series on a daily basis. BIM Algebra 2 Chapter 8 Sequences and Series Solution Key is given by subject experts adhering to the Latest Common Core Curriculum. Sequences and Series Big Ideas Math Algebra 2 Chapter 8 Answer Key encourages students and teachers to learn math in a simple and fun learning way.

## Big Ideas Math Book Algebra 2 Answer Key Chapter 8 Sequences and Series

Check out Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series aligned as per the Big Ideas Math Textbooks. Use the below available links for learning the Topics of BIM Algebra 2 Chapter 8 Sequences and Series easily and quickly. You just need to tap on them and avail the underlying concepts in it and score better grades in your exams. The Solutions covered here include Questions from Chapter Tests, Review Tests, Cumulative Practice, Cumulative Assessments, Exercise Questions, etc.

- Sequences and Series Maintaining Mathematical Proficiency – Page 407
- Sequences and Series Mathematical Practices – Page 408
- Lesson 8.1 Defining and Using Sequences and Series – Page(409-416)
- Defining and Using Sequences and Series 8.1 Exercises – Page(414-416)
- Lesson 8.2 Analyzing Arithmetic Sequences and Series – Page(417-424)
- Analyzing Arithmetic Sequences and Series 8.2 Exercises – Page(422-424)
- Lesson 8.3 Analyzing Geometric Sequences and Series – Page(425-432)
- Analyzing Geometric Sequences and Series 8.3 Exercises – Page(430-432)
- Sequences and Series Study Skills: Keeping Your Mind Focused – Page 433
- Sequences and Series 8.1 – 8.3 Quiz – Page 434
- Lesson 8.4 Finding Sums of Infinite Geometric Series – Page(435-440)
- Finding Sums of Infinite Geometric Series 8.4 Exercises – Page(439-440)
- Lesson 8.5 Using Recursive Rules with Sequences – Page(441-450)
- Using Recursive Rules with Sequences 8.5 Exercises – Page(447-450)
- Sequences and Series Performance Task: Integrated Circuits and Moore s Law – Page 451
- Sequences and Series Chapter Review – Page(452-454)
- Sequences and Series Chapter Test – Page 455
- Sequences and Series Cumulative Assessment – Page(456-457)

## Sequences and Series Maintaining Mathematical Proficiency

Solve the equation. Check your solution(s). Question 4. 7x + 3 = 31 Answer: 7x+3=31 7x=31-3 7x=28 x=28/7 x=4

## Sequences and Series Mathematical Practices

Mathematically proficient students consider the available tools when solving a mathematical problem.

Monitoring Progress

Use a spreadsheet to help you answer the question. Question 1. A pilot flies a plane at a speed of 500 miles per hour for 4 hours. Find the total distance flown at 30-minute intervals. Describe the pattern. Answer:

Question 2. A population of 60 rabbits increases by 25% each year for 8 years. Find the population at the end of each year. Describe the type of growth. Answer: Given, A population of 60 rabbits increases by 25% each year for 8 years. Year 1 of 8: 75 Year 2 of 8: 94 Year 3 of 8: 117 Year 4 of 8: 146 Year 5 of 8: 183 Year 6 of 8: 229 Year 7 of 8: 286 Year 8 of 8 (Final year): 357

Question 3. An endangered population has 500 members. The population declines by 10% each decade for 80 years. Find the population at the end of each decade. Describe the type of decline. Answer:

Question 4. The top eight runners finishing a race receive cash prizes. First place receives $200, second place receives $175, third place receives $150, and so on. Find the fifth through eighth place prizes. Describe the type of decline. Answer:

## Lesson 8.1 Defining and Using Sequences and Series

Essential Question How can you write a rule for the nth term of a sequence? A sequence is an ordered list of numbers. There can be a limited number or an infinite number of terms of a sequence. a 1 , a 2 , a 3 , a 4 , . . . , a n , . . .Terms of a sequence Here is an example. 1, 4, 7, 10, . . . , 3 n-2 , . . .

EXPLORATION 1

Communicate Your Answer

Question 2. How can you write a rule for the nth term of a sequence? Answer: nth term of a sequence a n = a 1 + (n-1)(d) Question 3. What do you notice about the relationship between the terms in (a) an arithmetic sequence and (b) a geometric sequence? Justify your Answer: An arithmetic sequence has a constant difference between each consecutive pair of terms. This is similar to the linear functions that have the form y=mx +b. A geometric sequence has a constant ratio between each pair of consecutive terms.

## Defining and Using Sequences and Series 8.1 Exercises

Question 2. COMPLETE THE SENTENCE In a sequence, the numbers are called __________ of the sequence. Answer: In a sequence, the numbers are called the terms of the sequence.

Monitoring Progress and Modeling with Mathematics

Maintaining Mathematical Proficiency

## Lesson 8.2 Analyzing Arithmetic Sequences and Series

Essential Question How can you recognize an arithmetic sequence from its graph? In an arithmetic sequence, the difference of consecutive terms, called the common difference, is constant. For example, in the arithmetic sequence 1, 4, 7, 10, . . . , the common difference is 3.

EXPLORATION 2

Question 3. How can you recognize an arithmetic sequence from its graph? Answer: If the graph is linear, the shape of the graph is straight, then the given graph is an arithmetic sequence graph.

Question 4. Find the sum of the terms of each arithmetic sequence. a. 1, 4, 7, 10, . . . , 301 b. 1, 2, 3, 4, . . . , 1000 c. 2, 4, 6, 8, . . . , 800 Answer: a. tn = a + (n – 1)d 301 = 4 + (n – 1)3 301 = 4 + 3n – 3 301 = 3n + 1 3n = 300 n = 300/3 n = 100 b. 1000 = 2 + (n – 1)1 1000 = 2 + n – 1 1000 = n + 1 n = 999 c. 800 = 4 + (n – 1)2 800 = 4 + 2n – 2 800 = 2 + 2n 798 = 2n n = 399

## Analyzing Arithmetic Sequences and Series 8.2 Exercises

Question 34. a 8 = −15, a 17 = −78 Answer:

Question 36. a 12 = −38, a 19 = −73 Answer: 12 + 38 + 19 + 73 = 142

Question 38. a 12 = 9, a 27 = 15 Answer:

Question 48. \(\sum_{i=1}^{26}\)(4i + 7) Answer: Step1: Find the first and last terms. a1 = 4(1) + 7 = 11. a26 = 4(26) + 7 = 111. Step2: Find the sum S29 = 29(11 + 111/2) = 29(61) S29 = 1,769

Question 52. \(\sum_{i=1}^{39}\)(−4.1 + 0.4i ) Answer: Step1: Find the first and last terms a1 = -4.1 + 0.4(1) = -3.7 a39 = -4.1 + 0.4(39) = 11.5 Step2: Find the sum S39 = 39(-3.7 + 11.5/2) = 39(3.9) S39 = 152.1

Question 62. USING EQUATIONS Find the value of n. a. \(\sum_{i=1}^{n}\)(3i + 5) = 544 Answer: \(\sum_{i=1}^{n}\)(3i + 5) = 544 3 \(\sum_{i=1}^{n}\)(i + 5n) = 544 3n(n + 1)/2 + 5n = 544 (3n² + 13n)/2 + 5n = 544 3n² + 13n – 1088 = 0 (3n + 64) (n – 17) = 0 n = 17 n = -64/3 n = -64/3 is a negative value. So, it is not possible n = 17 Thus the value of n is 17.

b. \(\sum_{i=1}^{n}\)(−4i − 1) = −1127 Answer: \(\sum_{i=1}^{n}\)(−4i − 1) = −1127 -4(n)(n + 1)/2 – n = -1127 2n(n + 1) + n = 1127 2n² + 3n – 1127 = 0 (n – 23) (2n + 49) = 0 n = 23 n = -49/2 n = -49/2 is a negatuve value. So, it is not possible n = 23.

c. \(\sum_{i=5}^{n}\)(7 + 12i) = 455 Answer: \(\sum_{i=5}^{n}\)(7 + 12i) = 455 (7 + 12(5)) + (7 + 12(6)) + ….(7 + 12n) = 455 7n – 28 + 6n² + 6n – 120 = 455 6n² + 13n – 603 = 0 (n – 9) (6n + 67) = 0 n = 9 or n = -67/6 n = -67/6 is a negatuve value. So, it is not possible n = 9.

d. \(\sum_{i=3}^{n}\)(−3 − 4i) = −507 Answer: \(\sum_{i=3}^{n}\)(−3 − 4i) = −507 (-3 – 4(3)) + (-3 – 4(4)) + …. + (-3 – 4n) = -507 -3(n – 2) – 4(n – 2)(3 + n)/2 = -507 -3(n – 2) – 2(n – 2) (n + 3) = 507 3n – 6 + 2n² + 2n – 12 = 507 2n² + 5n – 525 = 0 (n – 15)(2n + 35) = 0 n = 15 or n = -35/2 n = -35/2 is a negatuve value. So, it is not possible n = 15.

Question 64. CRITICAL THINKING The expressions 3 −x, x, and 1 − 3x are the first three terms in an arithmetic sequence. Find the value of x and the next term in the sequence. Answer: Consider 3 – x, x, 1 – 3x are in A.P. x – (3 – x) = x – 3x – x x – 3 + x = 1 – 4x 2x – 3 = 1 – 4x 2x + 4x = 1 + 3 6x = 4 x = 2/3 The next term is 3 – x, x, 1 – 3x a4 = a + 3d 3 – x + 3(2x – 3) 3 – x + 6x – 9 -6 + 5x -6 × 5 × (2/3) -6 + 10/3 -18 + 10/3 a4 = -8/3 The value of x is 2/3 and next term in the sequence is -8/3.

Simplify the expression. Question 66. \(\frac{7}{7^{1 / 3}}\) Answer: \(\frac{7}{7^{1 / 3}}\) 7/7 -3 7 × 7 × 7 × 7 = 2401

Question 68. \(\left(\frac{9}{49}\right)^{1 / 2}\) Answer: \(\left(\frac{9}{49}\right)^{1 / 2}\) √(9/49) = 3/7

Tell whether the function represents exponential growth or exponential decay. Then graph the function. Question 70. y= 2e x Answer:

Question 72. y = 3e -x Answer:

## Lesson 8.3 Analyzing Geometric Sequences and Series

Essential Question How can you recognize a geometric sequence from its graph? In a geometric sequence, the ratio of any term to the previous term, called the common ratio, is constant. For example, in the geometric sequence 1, 2, 4, 8, . . . , the common ratio is 2.

Question 4. Find the sum of the terms of each geometric sequence. a. 1, 2, 4, 8, . . . , 8192 b. 0.1, 0.01, 0.001, 0.0001, . . . , 10 -10 Answer: a. tn = ar n-1 tn = 8192, a = 1 and r = 2 8192 = 1 × 2 n-1 2 13 = 2 n-1 n = 14 Sn = a(r n – 1) × 1/r – 1 Sn = 1(16384 – 1) × 1/2-1 Sn = 16383 b. 0.1, 0.01, 0.001, 0.0001, . . . , 10 -10 an = 10^-10 r = 0.01/0.1 = 1/10 10 -10 = 1 . (1/10) n-1 (1/10) 10 = 1/10 n-1 10 = n – 1 n – 1 = 10 n = 11 Sn = a1/1 – r Sn = 0.1/0.9 Sn = 1/9

## Analyzing Geometric Sequences and Series 8.3 Exercises

Question 2. WRITING How can you determine whether a sequence is geometric from its graph? Answer: Graph of a geometric sequence behaves like graph of exponential function.

Question 30. a 5 = 1, r = −\(\frac{1}{5}\) Answer:

## Sequences and Series Study Skills: Keeping Your Mind Focused

8.1–8.3 What Did You Learn?

Core Concepts Section 8.1Sequences, p. 410 Series and Summation Notation, p. 412 Formulas for Special Series, p. 413

Section 8.2 Rule for an Arithmetic Sequence, p. 418 The Sum of a Finite Arithmetic Series, p. 420

Section 8.3 Rule for a Geometric Sequence, p. 426 The Sum of a Finite Geometric Series, p. 428

Mathematical Practices Question 1. Explain how viewing each arrangement as individual tables can be helpful in Exercise 29 on page 415. Answer:

Question 2. How can you use tools to find the sum of the arithmetic series in Exercises 53 and 54 on page 423? Answer:

Question 3. How did understanding the domain of each function help you to compare the graphs in Exercise 55 on page 431? Answer:

Study Skills: Keeping Your Mind Focused

## Sequences and Series 8.1 – 8.3 Quiz

## Lesson 8.4 Finding Sums of Infinite Geometric Series

Essential Question How can you find the sum of an infinite geometric series?

Writing a Conjecture Work with a partner. Look back at the infinite geometric series in Exploration 1. Write a conjecture about how you can determine whether the infinite geometric series a 1 + a 1 r + a 1 r 2 + a 1 r 3 +. . .has a finite sum.

EXPLORATION 3

Writing a Formula Work with a partner. In Lesson 8.3, you learned that the sum of the first n terms of a geometric series with first term a 1 and common ratio r≠ 1 is S n = a 1 \(\left(\frac{1-r^{n}}{1-r}\right)\) When an infinite geometric series has a finite sum, what happens to r n as n increases? Explain your reasoning. Write a formula to find the sum of an infinite geometric series. Then verify your formula by checking the sums you obtained in Exploration 1.

Question 4. How can you find the sum of an infinite geometric series? Answer: There is an equation for it, Sum = a1(1 – r) a1 = the first term of the series r = rate of change

Question 5. Find the sum of each infinite geometric series, if it exists. a. 1 + 0.1 + 0.01 + 0.001 + 0.0001 +. . . b. \(2+\frac{4}{3}+\frac{8}{9}+\frac{16}{27}+\frac{32}{81}+\cdots\) Answer: Given, 1 + 0.1 + 0.01 + 0.001 + 0.0001 +. . . a1 = 1 |r| < 1, the series does have a limit given by formula of limit or sum of an infinite geometric series S∞ = a1/1-r S∞ = 1/1 – 0.1 = 1/0.9 = 1.11… b. \(2+\frac{4}{3}+\frac{8}{9}+\frac{16}{27}+\frac{32}{81}+\cdots\) r = a2/a1 r = 4/3/2 r = 2/3 Each ratio is 2/3, so the sequence is geometric a1 = 2 and r = 2/3 Sn = a1/1 – r S = 2/(1-2/3) S = 6 The sum of infinite geometric series S = 6.

## Finding Sums of Infinite Geometric Series 8.4 Exercises

Vocabulary and Core Concept Check Question 1. COMPLETE THE SENTENCE The sum S n of the first n terms of an infinite series is called a(n) ________. Answer:

## Lesson 8.5 Using Recursive Rules with Sequences

Essential Question How can you define a sequence recursively?A recursive rule gives the beginning term(s) of a sequence and a recursive equation that tells how a n is related to one or more preceding terms.

Question 4. How can you define a sequence recursively? Answer: A recursive sequence is also called the recurrence sequence it is a sequence of numbers indexed by an integer and generated by solving a recurrence equation.

Question 5. Write a recursive rule that is different from those in Explorations 1–3. Write the first six terms of the sequence. Then graph the sequence and classify it as arithmetic, geometric, or neither. Answer:

## Using Recursive Rules with Sequences 8.5 Exercises

Question 4. a 1 = 1 a n = a n-1 − 5 Answer: a 1 = 1 a n = a n-1 − 5 a2 = a1 – 5 = 1-5 = -4 a3 = a2 – 5 = -4 – 5 = -9 a4 = a3 – 5 = -9 – 5 = -14 a5 = a4 – 5 = -14 – 5 = -19 a6 = a5 – 5 = -19 – 5 = -24

Question 6. f(0) = 10 f(n) = \(\frac{1}{2}\)f(n− 1) Answer: f(0) = 10 f(n) = \(\frac{1}{2}\)f(n− 1) f(1) = \(\frac{1}{2}\)f(0) = 1/2 × 10 = 5 f(2) = \(\frac{1}{2}\)f(1) = 1/2 × 5 = 5/2 f(3) = \(\frac{1}{2}\)f(2) = 1/2 × 5/2 = 5/4 f(4) = \(\frac{1}{2}\)f(3) = 1/2 × 5/4 = 5/8 f(5) = \(\frac{1}{2}\)f(4) = 1/2 × 5/8 = 5/16

Question 58. MODELING WITH MATHEMATICS You borrow $10,000 to build an extra bedroom onto your house. The loan is secured for 7 years at an annual interest rate of 11.5%. The monthly payment is $173.86. a. Find the balance after the fourth payment. b. Find the amount of the last payment. Answer: Given, You borrow $10,000 to build an extra bedroom onto your house. The loan is secured for 7 years at an annual interest rate of 11.5%. The monthly payment is $173.86. 0.115/12 = 0.0096 an = 1.0096 an-1 an-1 is the balance before payment

So that balance after the 4th payment will be = $9684.05 and balance after 85 payment is 173.86 – 159.49 = 14.37

Question 62. DRAWING CONCLUSIONS You sprain your ankle and your doctor prescribes 325 milligrams of an anti-in ammatory drug every 8 hours for 10 days. Sixty percent of the drug is removed from the bloodstream every 8 hours. a. Write a recursive rule for the amount of the drug in the bloodstream after n doses. Answer: Given, You sprain your ankle and your doctor prescribes 325 milligrams of an anti-in ammatory drug every 8 hours for 10 days. Sixty percent of the drug is removed from the bloodstream every 8 hours. an = 0.4 an-1 + 325 a1 = 325

b. The value that a drug level approaches after an extended period of time is called the maintenance level. What is the maintenance level of this drug given the prescribed dosage? Answer: A doctor prescribes 325 milligram of an anti-inflammatory drug every 8 hours for 10 days and 60% of the drug is removed from the bloodstream in every 8 hours. Substitute n = 30 in the above recursive rule and simplify to get the final answer. a30 = 541.66

c. How does doubling the dosage affect the maintenance level of the drug? Justify your answer. Answer: an = 0.4 an-1 + 650 for n > 1 It is seen that after n = 12, the same value of 1083.33 is repeating. This implies that the maintenance level is 1083.33 Also, the maintenance level is 1083.33 Since 1083.33/541.6 ≈ 2, the maintenance level doubles when the dose is doubled.

Question 68. MAKING AN ARGUMENT Your friend says it is impossible to write a recursive rule for a sequence that is neither arithmetic nor geometric. Is your friend correct? Justify your answer. Answer:

Question 72. 2\(\sqrt{52}\) − 5 = 15 Answer: 2\(\sqrt{52}\) − 5 = 15 Squaring on both sides 4 × 52 – 25 = 15 208 – 25 = 15 183 ≠ 15

Question 74. 2\(\sqrt [ 3 ]{ x }\) − 13 = −5 Answer: 2\(\sqrt [ 3 ]{ x }\) − 13 = −5 Cubing on both sides 8 × x – 2197 = -125 8x = 2197 – 125 8x = 2072 x = 259

## Sequences and Series Performance Task: Integrated Circuits and Moore s Law

8.4–8.5What Did You Learn?

Core Vocabulary partial sum, p. 436 explicit rule, p. 442 recursive rule, p. 442

Core Concepts Section 8.4 Partial Sums of Infinite Geometric Series, p. 436 The Sum of an Infinite Geometric Series, p. 437

Section 8.5 Evaluating Recursive Rules, p. 442 Recursive Equations for Arithmetic and Geometric Sequences, p. 442 Translating Between Recursive and Explicit Rules, p. 444

Mathematical Practices Question 1. Describe how labeling the axes in Exercises 3–6 on page 439 clarifies the relationship between the quantities in the problems. Answer:

Question 2. What logical progression of arguments can you use to determine whether the statement in Exercise 30 on page 440 is true? Answer:

Question 3. Describe how the structure of the equation presented in Exercise 40 on page 448 allows you to determine the starting salary and the raise you receive each year. Answer:

Question 4. Does the recursive rule in Exercise 61 on page 449 make sense when n= 5? Explain your reasoning. Answer:

Performance Task: Integrated Circuits and Moore s Law

## Sequences and Series Chapter Review

8.1 Defining and Using Sequences and Series (pp. 409–416)

8.2 Analyzing Arithmetic Sequences and Series (pp. 417–424)

Question 12. Find the sum \(\sum_{i=1}^{36}\)(2 + 3i) . Answer:

8.3 Analyzing Geometric Sequences and Series (pp. 425–432)

8.4 Finding Sums of Infinite Geometric Series (pp. 435–440)

8.5 Using Recursive Rules with Sequences (pp. 441–450)

Write the first six terms of the sequence. Question 22. a 1 = 7, a n = a n-1 + 11 Answer:

Question 23. a 1 = 6, a n = 4a n-1 Answer: Given that a1 = 6, an = 4an-1 If n= 2. a2 = 4a2-1 a2 = 4a1 a2 = 4(6) = 24. a3 = 4(24) = 96 a4 = 4(96) = 384 a5 = 4(384) =1,536 a6 = 4( 1,536) = 6,144

Question 24. f(0) = 4, f(n) = f(n − 1) + 2n Answer: Given that, f(0) = 4 and f(n) = f(n-1) + 2n If n = 1. f(1) = f(1-1) + 2(1) = f(0) + 2 = 4 + 1 = 5 n = 2 f(2) = f(2-1) + 2(2) = 5 + 4 f(2) = 9. n = 3 f(3) = f(3-1) + 2(3) f(3) = f(2) + 6 = 9 + 6 f(3) = 15. f(4) = f(4-1) + 2(4) f(4) = f(3) + 8 = 15 + 8 f(4) = 23. f(5) = f(5-1) + 2(5) = f(4) + 10 = 23 + 10 f(5) = 33. f(6) = f(6-1) + 2(6) = f(5) + 12 = 33 + 12 f(6) = 45

Write a recursive rule for the sequence. Question 25. 9, 6, 4, \(\frac{8}{3}\), \(\frac{16}{9}\), . . . Answer:

Question 26. 2, 2, 4, 12, 48, . . . Answer: Given that the sequence is 2, 2, 4, 12, 48. First, assume that, a1 = 2, a2 = 2 = 1 x 2 = 1 x a1. a3 = 4 = 2 x 2 = 2 x a2. a4 = 12 = 3 x 4 = 3 x a3. a5 = 48 = 4 x 12 = 4 x a4. We can conclude that an = (n-1) x an-1 The recursive rule for the sequence is a1 = 2, an = (n-1) x an-1.

Question 27. 7, 3, 4, −1, 5, . . . Answer: Given that the sequence is 7, 3, 4, -1, 5. Here a1 = 7, a2 = 3, a3 = 4, a5 = -1, a6 = 5. Therefore, the recursive rule for the sequence is an = an-2 – an-1.

Question 28. Write a recursive rule for a n = 105 (\(\frac{3}{5}\)) n−1 . Answer: Given that, an = 105(3/5) n−1 . Compare the given equation with the nth term an = a1 x r n−1 .. The value of a1 is 105 and the constant ratio r = 3/5. Substitute r in the above equation. an = r x a n−1 . an = 3/5 x a n−1 . Hence the recursive equation is an = 3/5 x a n−1 .

Write an explicit rule for the sequence. Question 29. a 1 = −4, a n = a n-1 + 26 Answer: Given that, a1 = -4, an = an-1 + 26. Let us consider n = 2. a2 = a2-1 + 26 = a1 + 26 = -4 + 26 = 22. a3 = a3-1 + 26 = a2 + 26 = 22 + 26 = 48. a4 = a4-1 + 26 = a3 + 26 = 48 + 26 = 74. a5 = a5-1 + 26 = a4 + 26 = 74 + 26 = 100. a6 = a6-1 + 26 = a5 + 26 = 100 + 26 = 126.

Question 30. a 1 = 8, a n = −5a n-1 Answer: Given that, a1 = 8, an = -5an-1. Let us consider n = 2. a2 = -5(a2-1) = -5a1 = -5(8) = 40. a3 = -5(a3-1) = -5a2 = -5(40) = -200. a4 = -5(a4-1) = -5a3 = -5(-200) = 1000. a5 = -5(a5-1) = -5a4 = -5(1000) = -5000. a6 = -5(a6-1) = -5a5 = -5(-5000) = 25,000.

Question 31. a 1 = 26, a n = \(\frac{2}{5}\)a n-1 . Answer: Given that, a1 = 26, an = 2/5 (an-1) Let us consider n = 2 a2 = 2/5 (a2-1) = 2/5 (a1) = 2/5 x 26 = 10.4 a3 = 2/5 (a3-1) = 2/5 (a2) = 2/5 x 10.4 = 4.16 a4 = 2/5 (a4-1) = 2/5 (a3) = 2/5 x 4.16 = 1.664 a5 = 2/5 (a5-1) = 2/5 (a4) = 2/5 x 1.664 = 0.6656 a6 = 2/5 (a6-1) = 2/5 (a5) = 2/5 x 0.6656 = 0.26624.

Question 32. A town’s population increases at a rate of about 4% per year. In 2010, the town had a population of 11,120. Write a recursive rule for the population P n of the town in year n. Let n = 1 represent 2010. Answer:

## Sequences and Series Chapter Test

Question 3. \(\sum_{k=1}^{\infty}\)2(0.8) k−1 Answer:

## Sequences and Series Cumulative Assessment

Question 2. You take out a loan for $16,000 with an interest rate of 0.75% per month. At the end of each month, you make a payment of $300. a. Write a recursive rule for the balance an of the loan at the beginning of the nth month. b. How much do you owe at the beginning of the 18th month? c. How long will it take to pay off the loan? d. If you pay $350 instead of $300 each month, how long will it take to pay off the loan? How much money will you save? Explain. Answer:

Question 7. Classify the solution(s) of each equation as real numbers, imaginary numbers, or pure imaginary numbers. Justify your answers. a. x + \(\sqrt{-16}\) = 0 b. (11 – 2i) – (-3i + 6) = 8 + x c. 3x 2 – 14 = -20 d. x 2 + 2x = -3 e. x 2 = 16 f. x 2 – 5x – 8 = 0 Answer:

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Similarities: Both types of bonds result from overlap of atomic orbitals on adjacent atoms and contain a maximum of two electrons. Differences: σ bonds are stronger and result from end-to-end overlap and all single bonds are σ bonds; π bonds between the same two atoms are weaker because they result from side-by-side overlap, and multiple bonds contain one or more π bonds (in addition to a σ bond).

The specific average bond distance is the distance with the lowest energy. At distances less than the bond distance, the positive charges on the two nuclei repel each other, and the overall energy increases.

Bonding: One σ bond and one π bond. The s orbitals are filled and do not overlap. The p orbitals overlap along the axis to form a σ bond and side-by-side to form the π bond.

No, two of the p orbitals (one on each N) will be oriented end-to-end and will form a σ bond.

Hybridization is introduced to explain the geometry of bonding orbitals in valance bond theory.

There are no d orbitals in the valence shell of carbon.

trigonal planar, sp 2 ; trigonal pyramidal (one lone pair on A) sp 3 ; T-shaped (two lone pairs on A sp 3 d , or (three lone pairs on A) sp 3 d 2

(a) Each S has a bent (109°) geometry, sp 3

(b) Bent (120°), sp 2

(c) Trigonal planar, sp 2

(d) Tetrahedral, sp 3

(a) XeF 2 (b)

(c) linear (d) sp 3 d

(b) P atoms, trigonal pyramidal; S atoms, bent, with two lone pairs; Cl atoms, trigonal pyramidal; (c) Hybridization about P, S, and Cl is, in all cases, sp 3 ; (d) Oxidation states P +1, S − 1 1 3 , − 1 1 3 , Cl +5, O –2. Formal charges: P 0; S 0; Cl +2: O –1

Phosphorus and nitrogen can form sp 3 hybrids to form three bonds and hold one lone pair in PF 3 and NF 3 , respectively. However, nitrogen has no valence d orbitals, so it cannot form a set of sp 3 d hybrid orbitals to bind five fluorine atoms in NF 5 . Phosphorus has d orbitals and can bind five fluorine atoms with sp 3 d hybrid orbitals in PF 5 .

A triple bond consists of one σ bond and two π bonds. A σ bond is stronger than a π bond due to greater overlap.

(b) The terminal carbon atom uses sp 3 hybrid orbitals, while the central carbon atom is sp hybridized. (c) Each of the two π bonds is formed by overlap of a 2 p orbital on carbon and a nitrogen 2 p orbital.

(a) sp 2 ; (b) sp ; (c) sp 2 ; (d) sp 3 ; (e) sp 3 ; (f) sp 3 d ; (g) sp 3

(a) sp 2 , delocalized; (b) sp , localized; (c) sp 2 , delocalized; (d) sp 3 , delocalized

Each of the four electrons is in a separate orbital and overlaps with an electron on an oxygen atom.

(a) Similarities: Both are bonding orbitals that can contain a maximum of two electrons. Differences: σ orbitals are end-to-end combinations of atomic orbitals, whereas π orbitals are formed by side-by-side overlap of orbitals. (b) Similarities: Both are quantum-mechanical constructs that represent the probability of finding the electron about the atom or the molecule. Differences: ψ for an atomic orbital describes the behavior of only one electron at a time based on the atom. For a molecule, ψ represents a mathematical combination of atomic orbitals. (c) Similarities: Both are orbitals that can contain two electrons. Differences: Bonding orbitals result in holding two or more atoms together. Antibonding orbitals have the effect of destabilizing any bonding that has occurred.

An odd number of electrons can never be paired, regardless of the arrangement of the molecular orbitals. It will always be paramagnetic.

Bonding orbitals have electron density in close proximity to more than one nucleus. The interaction between the bonding positively charged nuclei and negatively charged electrons stabilizes the system.

The pairing of the two bonding electrons lowers the energy of the system relative to the energy of the nonbonded electrons.

(a) H 2 bond order = 1, H 2 + H 2 + bond order = 0.5, H 2 − H 2 − bond order = 0.5, strongest bond is H 2 ; (b) O 2 bond order = 2, O 2 2+ O 2 2+ bond order = 3; O 2 2− O 2 2− bond order = 1, strongest bond is O 2 2+ ; O 2 2+ ; (c) Li 2 bond order = 1, Be 2 + Be 2 + bond order = 0.5, Be 2 bond order = 0, strongest bond is Li 2 Li 2 ;(d) F 2 bond order = 1, F 2 + F 2 + bond order = 1.5, F 2 − F 2 − bond order = 0.5, strongest bond is F 2 + ; F 2 + ; (e) N 2 bond order = 3, N 2 + N 2 + bond order = 2.5, N 2 − N 2 − bond order = 2.5, strongest bond is N 2

(a) H 2 ; (b) N 2 ; (c) O; (d) C 2 ; (e) B 2

Yes, fluorine is a smaller atom than Li, so atoms in the 2 s orbital are closer to the nucleus and more stable.

N 2 has s-p mixing, so the π orbitals are the last filled in N 2 2+ . N 2 2+ . O 2 does not have s-p mixing, so the σ p orbital fills before the π orbitals.

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Lesson 8.1.1 b: b: f: 700 yes yes c: c: 480 no yes d: 1080 d: yes h: no 8-7. 8-9. 8-10. 8-11. 8-12. a: 1100 a: no e: no b: The measure of an exterior angle of a triangle equals the sum of the measures of its remote interior angles. c: a + b + c = 1800 (the sum of the interior angles of a triangle is 1800), x + c = 1800

a) 30 seconds b) 30 m. Chapter 8 Solutions. These answers are to be used to check against your solutions. Your homework should show all of your work, not just the answers! Section 8.3 - Vectors in 3-D. 12. ˛˚⃑ =√26. 13. ˛˚⃑ =√69.

8.5 Section Exercises. 1. a is the real part, b is the imaginary part, and i = √− 1. 3. Polar form converts the real and imaginary part of the complex number in polar form using x = rcosθ and y = rsinθ. 5. zn = rn(cos(nθ) + isin(nθ)) It is used to simplify polar form when a number has been raised to a power.

13th Edition • ISBN: 9780321884077 George B Thomas Jr, Joel D. Hass, Maurice D. Weir. 11,924 solutions. Get your Pearson Math homework done with Quizlet! Browse through thousands of step-by-step solutions to end-of-chapter questions from the most popular Pearson Math textbooks. It's never been a better time to #LearnOn.

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Georgia Milestones Assessment System Test Prep: Grade 3 English Language Arts Literacy (ELA) Practice Workbook and Full-length Online Assessments: GMAS Study Guide. Lumos Learning. 3. 2017. ACT Aspire Test Prep: 3rd Grade Math Practice Workbook and Full-length Online Assessments: ACT Aspire Study Guide. Lumos Learning.

Answer is (0.502, 0.538). Yes, this interval does not fall below 0.50, so we can conclude that at least half of all American adults believe that major sports programs corrupt education - but we do so with only 75 percent confidence. 133. CL = 0.95; α = 1 - 0.95 = 0.05; α 2 α 2 = 0.025; zα 2 z α 2 = 1.96.

8.E: Homework Chapter 8 Answer Key is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts. Back to top 8.E: Homework Chapter 8

Access MindTap Economics for Mankiw's Principles of Macroeconomics, 8th Edition, [Instant Access] 8th Edition Chapter 8 solutions now. Our solutions are written by Chegg experts so you can be assured of the highest quality!

Everyone must do the homework for this Chapter!! You can chose to do the book homework above or the AP Classroom problems! After ch. 8 regular homework policy applies. Notes, Homework, and Reviews ... chapter 8 - Use these to help guide you while you're reading chapter 8! 1/10/20 - Practice 8.2 KEY - Quiz Monday!! 1/17/20 - Practice 8.3 KEY 1 ...

We provide step-by-step solutions that help you understand and learn how to solve for the answer. Comprehending how to calculate the answer is where the true learning begins. Armed with this knowledge, you can apply it to other textbook problems and be better prepared to succeed on test day.

Precalculus with Limits. 4th Edition • ISBN: 9781337271059 Larson, Paul Battaglia. 11,584 solutions. Get your Cengage Learning Math homework done with Quizlet! Browse through thousands of step-by-step solutions to end-of-chapter questions from the most popular Cengage Learning Math textbooks. It's never been a better time to #LearnOn.

Problem Solving REAL WORLD 9. There are 12 students in a jewelry-making class and 8 sets of charms. What fraction of a set of charms will each student get? Each student will get of a set. P173 Each friend will get cheesecakes. Chapter 8. Lesson Check (CC.5.NF.3) 1. Eight friends share 4 bunches of grapes equally.

February 15, 2021 / By Prasanna. Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series helps you to get a grip on the concepts from surface level to a deep level. Finish your homework or assignments in time by solving questions from B ig Ideas Math Book Algebra 2 Ch 8 Sequences and Series here. Access the user-friendly solutions ...

Email your homework to your parent or tutor for free; ... Answer Keys . Chapter 8:Functions and Inequalities;Lesson 1:Function Tables. Please share this page with your friends on FaceBook. Use Math Toole Complete each function table. Question 14 (request help) Question 15 ...

Similarities: Both types of bonds result from overlap of atomic orbitals on adjacent atoms and contain a maximum of two electrons. Differences: σ bonds are stronger and result from end-to-end overlap and all single bonds are σ bonds; π bonds between the same two atoms are weaker because they result from side-by-side overlap, and multiple bonds contain one or more π bonds (in addition to a ...

Homework Practice Workbook -07-660291-5 978--07-660291-9 Spanish Version Homework Practice Workbook -07-660294-X 978--07-660294- Answers For Workbooks The answers for Chapter 8 of these workbooks can be found in the back of this Chapter Resource Masters booklet. ... ents up to twenty of the key vocabulary terms from the chapter. Students ...

Answer Key - Chapter 25 (31.0K) Answer Key - Chapter 26 (36.0K) To learn more about the book this website supports, please visit its Information Center .

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Go Math Grade K Answer Key. Chapter 1 Represent, Count, and Write Numbers 0 to 5. Chapter 2 Compare Numbers to 5. Chapter 3 Represent, Count, and Write Numbers 6 to 9. Chapter 4 Represent and Compare Numbers to 10. Chapter 5 Addition. Chapter 6 Subtraction. Chapter 7 Represent, Count, and Write 11 to 19. Chapter 8 Represent, Count, and Write 20 ...