## Engage NY Eureka Math Algebra 2 Module 3 Lesson 25 Answer Key

### Eureka Math Algebra 2 Module 3 Lesson 25 Opening Exercise Answer Key

Opening Exercise:

Suppose a ball is dropped from an initial height h_{0} and that each time it rebounds, its new height is 60% of its previous height.

a. What are the first four rebound heights h_{1}, h_{2}, h_{3}, and h_{4} after being dropped from a height of h_{0} = 10 ft.?

Answer:

The rebound heights are h_{1} = 6 ft., h_{2} = 3.6 ft., h_{3} = 2.16 ft., and h_{4} = 1.296 ft.

b. Suppose the initial height is A ft. What are the first four rebound heights? Fill in the following table:

Answer:

c. How is each term in the sequence related to the one that came before it?

Answer:

Each term is 0.6 times the previous term.

d. Suppose the initial height is A ft. and that each rebound, rather than being 60% of the previous height, is r times the previous height, where 0 < r < 1. What are the first four rebound heights? What is the nth rebound height?

Answer:

The rebound heights are h_{1} = Ar ft, h_{2} = Ar^{2} ft,, h_{3} = Ar^{3} ft. and h_{4} = Ar^{4} ft. The nth rebound height is h_{n} = ft.

e. What kind of sequence is the sequence of rebound heights?

Answer:

The sequence of rebounds is geometric (geometrically decreasing).

f. Suppose that we define a function f with domain the positive integers so that f(1) is the first rebound height, f(2) is the second rebound height, and continuing so that f(k) is the kth rebound height for positive integers kth. What type of function would you expect f to be?

Answer:

Since each bounce has a rebound height of r times the previous height, the function f should be exponentially decreasing.

g. On the coordinate plane below, sketch the height of the bouncing ball when A = 10 and r = 0.60, assuming that the highest points occur at x = 1, 2, 3, 4,…….

Answer:

h. Does the exponential function f(x) = 10(0.60)^{x} for real numbers x model the height of the bouncing ball? Explain how you know.

Answer:

No. Exponentialfunctions do not have the same behavior as a bouncing ball. The graph off is the smooth curve that connects the points at the “top” of the rebounds, as shown in the graph at right.

i. What does the function f(n) = 10(0.60)^{n} for integers n ≥ 0 model?

Answer:

The exponential function f(n) = 10(0.60)^{n} models the height of the rebounds for integer values of n.

### Eureka Math Algebra 2 Module 3 Lesson 25 Exercise Answer Key

Exercises:

Exercise 1.

Jane works for a video game development company that pays her a starting salary of $100 per day, and each day she works she earns $100 more than the day before.

a. How much does she earn on day 5?

Answer:

On day 5, she earns $500.

b. If you were to graph the growth of her salary for the first 10 days she worked, what would the graph look like?

Answer:

The graph would be a set of points lying on a straight line.

c. What kind of sequence is the sequence of Jane’s earnings each day?

Answer:

The sequence of her earnings is arithmetic (that is, the sequence is arithmetically increasing).

Exercise 2.

A laboratory culture begins with 1,000 bacteria at the beginning of the experiment, which we denote by time 0 hours. By time 2 hours, there are 2,890 bacteria.

a. If the number of bacteria is increasing by a common factor each hour, how many bacteria are there at time 1 hour? At time 3 hours?

Answer:

If P_{0} is the original population, the first three terms of the geometric sequence are P_{0}, P_{0}r, and P_{0},r^{2}. In this case, P_{0}, = 1000 and P_{2}, = P_{0}r^{2} = 2890, so r^{2} = 2.89 and r = 1.7. Therefore, P_{1}, = P_{0}r = 1700 and P_{3} = P_{0}r^{3} = 2890 . 1.7 = 4913.

b. Find the explicit formula for term P_{n} of the sequence in this case.

Answer:

P_{n} = P_{0}r^{n} = 1000(1.7)^{n}.

c. How would you find term P_{n + 1} if you know term P_{n}? Write a recursive formula for P_{n + 1} in terms of P_{n}.

Answer:

You would multiply the nth term by r, which in this case is 1.7. We have P_{n + 1} = 1.7 P_{n}.

d. If P_{0} is the initial population, the growth of the population P_{n} at time n hours can be modeled by the sequence P_{n} = P(n), where P is an exponential function with the following form:

P(n) = p_{0} 2^{kn} where k > 0. Find the value of k and write the function P in this form. Approximate k to four decimal places.

Answer:

We know that P(n) = 1000(1.7)^{n} and 1.7 = 2^{log2(1.7)} with k = log_{2}(1.7) = \(\frac{\log (1.7)}{\log (2)}\) ≈ 0.7655.

Thus, we can express P in the form: P(n) = 1000(2^{0.7655n}).

e. Use the function in part (d) to determine the value of t when the population of bacteria has doubled.

Answer:

We need to solve 2000 = 1000(2^{0.7655t}), which happens when the exponent is 1.

0.7655t = 1

t = \(\frac{1}{0.7655}\)

t ≈ 1.306

This population doubles in roughly 1.306 hours, which is about 1 hour and 18 minutes.

f. If P_{0} is the initial population, the growth of the population P at time t can be expressed in the following

form: P(n) = P_{0}e^{kn}, where k > 0. Find the value of k, and write the function P in this form. Approximate k to four decimal places.

Answer:

Substituting in the formula for t = 2, we get 2890 = 1000e^{2k}. Solving for k, we get k = ln(1.7) ≈ 0.5306. Thus, we can express P in the form: P(t) = 1000(e^{o.53o6t}).

g. Use the formula in part (d) to determine the value of t when the population of bacteria has doubled.

Answer:

Substituting in the formula with k = 0.5306, we get 2000 = 1000e^{0.5306t}. Solving for t, we get

t = \(\frac{\ln (2)}{0.5306}\) ≈ 1.306, which is the same value we found in part (e).

Exercise 3.

The first term a_{0} of a geometric sequence is -5, and the common ratio r is -2.

a. What are the terms a_{0}, a_{1}, and a_{2}?

Answer:

a_{0} = -5

a_{1} = 10

a_{2} = -20

b. Find a recursive formula for this sequence.

Answer:

The recursive formulais a_{n + 1} = -2a_{n}, with a_{0} = -5.

c. Find an explicit formula for this sequence.

Answer:

The explicit formula is a_{n} = -5(-2)^{n}, for n ≥ 0.

d. What is term a_{9}?

Answer:

Using the explicit formula, we find: a_{9} = (-5) (-2)^{9} = 2560.

e. What is term a10?

Answer:

One solution is to use the explicit formula: a_{10} = (-5) . (-2)^{10} = -5120.

Another solution is to use the recursive formula: a_{10} = a_{9} . (-2) = -5120.

Exercise 4.

Term a_{4} of a geometric sequence is 5.8564, and term a_{5} is -6.44204.

a. What is the common ratio r?

Answer:

We have r = \(\frac{-6.44204}{5.8564}\) = -1.1. The common ratio is -1.1.

b. What is term a_{0}?

Answer:

From the definition of a geometric sequence, a_{4} = a_{0}r^{4} = 5.8564, so a_{0} = \(\frac{5.8564}{(-1.1)^{4}}=\frac{5.8564}{1.4641}\) = 4.

c. Find a recursive formula for this sequence.

Answer:

The recursive formulais a_{n + 1} = -1. 1(a_{n}) with a_{0} = 4.

d. Find an explicit formula for this sequence.

Answer:

The explicit formulais a_{n} = 4(1.1)^{n}, for n ≥ 0.

Exercise 5.

The recursive formula for a geometric sequence is a_{n + 1} = 3.92(a_{n}) with a_{0} = 4.05. Find an explicit formula for this sequence.

Answer:

The common ratio is 3.92, and the initial value is 4.05, so the explicit formula is a_{n} = 4.05(3.92)^{n} for n ≥ 0.

Exercise 6.

The explicit formula for a geometric sequence is a_{n} = 147(2.1)^{3n}. Find a recursive formula for this sequence.

Answer:

First, we rewrite the sequence as a_{n} = 147(2.1^{3})^{n} = 147(9.261)^{n}. We then see that the common ratio is 9.261, and the initial value is 147, so the recursive formula is a_{n + 1} = (9.261)a_{n} with a_{0} = 147.

### Eureka Math Algebra 2 Module 3 Lesson 25 Problem Set Answer Key

Question 1.

Convert the following recursive formulas for sequences to explicit formulas.

a. a_{n + 1} = 4.2 + a_{n} with a_{0} = 12

Answer:

a_{n} = 12 + 4 . 2n for n ≥ 0.

b. a_{n + 1} = 4. 2a_{n} with a_{0} = 12

Answer:

a_{n} = 12(4.2)^{n} for n ≥ 0.

c. a_{n + 1} = √5a^{n} with a_{0} = 2

Answer:

a_{n} = 2(√5)^{n} for n ≥ 0

d. a_{n + 1} = √5 + a^{n} with _{0} = 2

Answer:

a_{n} = 2 + n√5 for n ≥ 0

e. a_{n + 1} = πa_{n} with a_{0} = π

Answer:

a_{n} = π(π)^{n} = π^{n + 1} for n ≥ 0.

Question 2.

Convert the following explicit formulas for sequences to recursive formulas.

a. a_{n} = \(\frac{1}{5}\)(3^{n})for n ≥ 0

Answer:

a_{n + 1} = 3a_{n} with a_{0} = \(\frac{1}{5}\)

b. a_{n} = 16 – 2n for n ≥ 0

Answer:

a_{n + 1} = a_{n} – 2 with a_{0} = 16

c. a_{n} = 16(\(\frac{1}{2}\))^{n} for n ≥ 0

Answer:

a_{n + 1}= \(\frac{1}{2}\)a_{n} with a_{0} = 16

d. a_{n} = 71 – \(\frac{6}{7}\)n for n ≥ 0

Answer:

a_{n + 1} = a_{n} – \(\frac{6}{7}\) with a_{0} = 71

e. a_{n} = 190(1.03)^{n} for n ≥ 0

Answer:

a_{n + 1} = 1.03a_{n} with a_{0} = 190

Question 3.

If a geometric sequence has a_{1} = 256 and a_{8} = 512, find the exact value of the common ratio r.

Answer:

The recursive formulais a_{n + 1} = a_{n} . r, so we have

a_{8} = a_{7}(r)

= a_{6}(r^{2})

= a_{5}(r^{3})

.

.

= a_{1}(r^{7})

512 = 256(r^{7})

2 = r^{7}

r = \(\sqrt[7]{2}\)

Question 4.

If a geometric sequence has a_{2} = 495 and a_{6} = 311, approximate the value of the common ratio r to four decimal places.

Answer:

The recursi ve formulais _{n + 1} = a_{n} . r, so we have

a_{6} = a_{5}(r)

= a_{4}(r^{2})

= a_{3}(r^{3})

= a_{2}(r^{4})

311 = 495(r^{4})

r^{4} = \(\frac{311}{495}\)

r = \(\sqrt[4]{\frac{311}{495}}\) ≈ 0.8903

Question 5.

Find the difference between the terms a_{10} of an arithmetic sequence and a geometric sequence, both of which begin at term a_{0} and have a_{2} = 4 and a_{4} = 12.

Answer:

Arithmetic: The explicit formula has the form a_{n} = a_{0} + nd, so a_{2} = a_{0} + 2d and a_{4} = a_{0} + 4d. Then a_{4} – a_{2} = 12 – 4 = 8 and a_{4} – a_{2} = (a_{0} + 4d) – (a_{0} + 2d), so that 8 = 2d and d = 4, we know that a_{0} = a_{2} – 2d = 4 – 8 = -4. So, the explicit formula for thos airthmetic sequence is a_{n} = -4 + 4n. we then know that a_{10} = -4 + 40 = 36

Geometric: The explicit formula has the form a_{n} = a_{0}(r^{n}), so a_{2} = a_{0}(r^{2}) and a_{4} = a_{0}(r^{4}), so \(\frac{a_{4}}{a_{2}}\)r^{2} and \(\frac{a_{4}}{a_{2}}=\frac{12}{4}\) = 3. Thus, r^{2} = 3, so r = ±√3. Since r^{2} = 3, we have a_{2} = 4 = a_{0}(r^{2}), so that a_{0} = \(\frac{4}{3}\). Then the a_{10} = \(\frac{4}{3}\)(±√3)<sup>10</sup> = \(\frac{4}{3}\)(3^{5}) = 4(3^{4}) = 324.

Thus, the difference between the terms a_{10} of these two sequence is 324 – 36 = 288.

Question 6.

Given the geometric sequence defined by the following values of a_{0} and r, find the value of n so that a_{n} has the specified value.

a. a_{0} = 64, r = \(\frac{1}{2}\), a_{n} = 2

Answer:

The explicit formula for this geometric sequence is a_{n} = 64(\(\frac{1}{2}\))^{n} and a_{n} = 2.

2 = 64(\(\frac{1}{2}\))^{n}

\(\frac{1}{32}\) = (\(\frac{1}{2}\))^{n}

(\(\frac{1}{2}\))^{5} = (\(\frac{1}{2}\))^{n}

n = 5

Thus, a_{5} = 2.

b. a_{0} = 13, r = 3, a_{n} = 85293

Answer:

The explicit formula for this geometric sequence is a_{n} = 13(3)^{n}, and we have a_{n} = 85293

13(3)^{n} = 85293

3^{n} = 6561

3^{n} = 3^{8}

n = 8

Thus a_{8} = 85293

c.a_{0} = 6.7, r = 1.9, a_{n} = 7804.8

Answer:

The explicit formula for this geometric sequence is a_{n} = 6.7(1.9)^{n}, and we have a_{n} = 7804.8

6.7(1.9)^{n} = 7804.8

(1.9)^{n} = 1164.9

n log(1.9) = log(1164.9)

n = \(\frac{\log (1164.9)}{\log (1.9)}\) = 11

Thus, a_{11} = 7804.8

d. a_{0}= 10958, r = 0.7, a_{n} = 25.5

Answer:

The explicit formula for this geometric sequence is a_{n} =10958(0.7)^{n}, and we have a_{n} = 25.5

10958(0.7)^{n} = 25.5

log(10958) + nlog(0.7) = log(25.5)

n = \(\frac{\log (25.5)-\log (10958)}{\log (0.7)}\)

n = 17

Thus, a_{17} = 25.5

Question 7.

Jenny planted a sunflower seedling that started out 5 cm tall, and she finds that the average daily growth is 3.5 cm.

a. Find a recursive formula for the height of the sunflower plant on day n.

Answer:

h_{n + 1} = 3.5 + h_{n} with h_{0} = 5.

b. Find an explicit formula for the height of the sunflower plant on day n ≥ 0.

Answer:

h = 5 + 3.5n

Question 8.

Kevin modeled the height of his son (in inches) at age n years for n = 2,3, …, 8 by the sequence h_{n} = 34 + 3.2(n – 2). Interpret the meaning of the constants 34 and 3.2 in his model.

Answer:

At age 2, Kevin’s son was 34 inches tall, and between the ages of 2 and 8 he grew at a rate of 3.2 inches per year.

Question 9.

Astrid sells art prints through an online retailer. She charges a flat rate per order for an order processing fee, sales tax, and the same price for each print. The formula for the cost of buying n prints is given by P, = 4.5 + 12.6n.

a. Interpret the number 4. 5 in the context of this problem.

Answer:

The number 4.5 represents a $4. 50 order processing fee.

b. Interpret the number 12.6 in the context of this problem.

Answer:

The number 12.6 represents the cost of each print, including the sales tax.

c. Find a recursive formula for the cost of buying n prints.

Answer:

P_{n} = 12.6 + P_{n – 1} with P_{1} = 17.10

(Notice that it makes no sense to start the sequence with n = 0, since that would mean you need to pay the processing fee when you do not place an order.)

Question 10.

A bouncy ball rebounds to 90% of the height of the preceding bounce. Craig drops a bouncy ball from a height of 20 feet.

a. Write out the sequence of the heights h_{1}, h_{2}, h_{3}, and h_{4} of the first four bounces, counting the initial height as h_{0} = 20.

Answer:

h_{1} = 18

h_{2} = 16.2

h_{3} = 14.58

h_{4} = 13.122

b. Write a recursive formula for the rebound height of a bouncy ball dropped from an initial height of 20 feet.

Answer:

h_{n + 1} = 0.9 h_{n} with h_{0} = 20

c. Write an explicit formula for the rebound height of a bouncy ball dropped from an initial height of 20 feet.

Answer:

h_{n} = 20(0.9)^{n}; for n ≥ 0

d. How many bounces does it take until the rebound height is under 6 feet?

Answer:

20(0.9^{n} < 6

n log(0.9) < log(6) – log(20) n > \(\frac{\log (6)-\log (20)}{\log (0.9)}\)

n > 11.42

So, it takes 12 bounces for the bouncy ball to rebound under 6 feet.

e. Extension: Find a formula for the minimum number of bounces needed for the rebound height to be under y, feet, for a real number 0 < y < 20.

Answer:

20(0.9)^{n} < y

n log(0.9) < log(y) – log(20) n > \(\frac{\log (y)-\log (20)}{\log (0.9)}\)

Rounding this up to the next integer with the ceiling function, it takes |\(\frac{\log (y)-\log (20)}{\log (0.9)}\)| bounces for the bouncy ball to rebound under y feet.

Question 11.

Show that when a quantity a_{0} = A is increased by x%, its new value is a_{1} = A (1 + \(\frac{x}{100}\)) If this quantity is again increased by x%, what is its new value a_{2}? If the operation is performed n times in succession, what is the final value of the quantity a_{n}?

Answer:

We know that x% of a number A is represented by \(\frac{x}{100}\)A. Thus, when a_{0} = A is increased by X%, the new quantity

a_{1} = A + \(\frac{x}{100}\)A = A(1 + \(\frac{x}{100}\))

If we increase it again by x%, we have

a_{2} = a_{1} + \(\frac{x}{100}\) a_{1}

= (1 + \(\frac{x}{100}\))a_{1}

= (1 + \(\frac{x}{100}\))(1 + \(\frac{x}{100}\))a_{0}

= (1 + \(\frac{x}{100}\))^{2} a_{0}

If we repeat this operation n times, we find that

a_{n} = (1 + \(\frac{x}{100}\))^{n} a_{0}.

Question 12.

When Eli and Daisy arrive at their cabin in the woods in the middle of winter, the interior temperature is 40°F.

a. Eli wants to turn up the thermostat by 2°F every 15 minutes. Find an explicit formula for the sequence that represents the thermostat settings using Eli’s plan.

Answer:

Let n represent the number of 15-minute increments. The function E(n) = 40 + 2n models the thermostat settings using Eli’s plan.

b. Daisy wants to turn up the thermostat by 4% every 15 minutes. Find an explicit formula for the sequence that represents the thermostat settings using Daisy’s plan.

Answer:

Let n represent the number of 15-minute increments. The function D(n) = 40(1.04)^{n} models the thermostat settings using Daisy’s plan.

c. Which plan gets the thermostat to 60°F most quickly?

Answer:

Making a table of values, we see that Eli’s plan sets the thermostat to 60°F first.

d. Which plan gets the thermostat to 72°F most quickly?

Answer:

Continuing the table of values from part (c), we see that Daisy’s plan sets the thermostat to 72°F first.

Question 13.

In nuclear fission, one neutron splits an atom causing the release of two other neutrons, each of which splits an atom and produces the release of two more neutrons, and so on.

a. Write the first few terms of the sequence showing the numbers of atoms being split at each stage after a single atom splits. Use a_{0} = 1.

Answer:

a_{0} = 1, a_{1} = 2, a_{2} = 4, a_{3} = 8.

b. Find the explicit formula that represents your sequence in part (a).

Answer:

a_{n} = 2^{n}

c. If the interval from one stage to the next is one-millionth of a second, write an expression for the number of atoms being split at the end of one second.

Answer:

At the end of one-second n = 1000000, s0 2^{1000000} atoms are being split.

d. If the number from part (c) were written out, how many digits would it have?

Answer:

The number of digits in a number x is given by rounding up log(x) to the next largest integer; that is, by the ceiling of log(x), [log(x)]. Thus, there are [Iog(2^{1000000})] digits.

Since log(2^{1000000}) = 1000000 log(2) ≈ 301030, there are 301030 digits in the number 2^{1000000}.

### Eureka Math Algebra 2 Module 3 Lesson 25 Exit Ticket Answer Key

Question 1.

Every year, Mikhail receives a 3% raise in his annual salary. His starting annual salary was $40,000.

a. Does a geometric or arithmetic sequence best model Mikhail’s salary in year n? Explain how you know.

Answer:

Because Mikhail’s salary increases by a multiple of itself each year, a geometric sequence is an appropriate model.

b. Find a recursive formula for a sequence, S, which represents Mikhail’s salary in year n.

Answer:

Mikhail’s annual salary can be represented by the sequence S_{n + 1} = 1.03 S_{n}. with S_{0} = $40,000.

Question 2.

Carmela’s annual salary in year n can be modeled by the recursive sequence C_{n + 1} = 1.05 C_{n} where C_{0} = $75,000.

a. What does the number 1.05 represent in the context of this problem?

Answer:

The number 1.05 represents the growth rate of her salary with time; it indicates that she is receiving a 5% raise each year.

b. What does the number $75, 000 represent in the context of this problem?

Answer:

Carmela’s starting annual salary was $75, 000, before she earned any raises.

c. Find an explicit formula for a sequence that represents Carmela’s salary.

Answer:

Carmela’s salary can be represented by the sequence C_{n} = $75,000 (1.05)^{n}.