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Eureka Math™ Homework Helper

2015–2016

Algebra IIModule 3

Lessons 1–6

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2015-16 M3 ALGEBRA II

Lesson 1: Integer Exponents


1. Suppose your class tried to fold an unrolled roll of paper towels, which is 11 inches wide and 92 feet long.The paper towels are approximately 0.04 inches thick.a. Complete each table, and represent the area and thickness using powers of 2.

Number of Folds

𝒏𝒏

Thickness After 𝒏𝒏 Folds

(inches)

Number of Folds

𝒏𝒏

Area on Top After 𝒏𝒏 Folds

(square inches)

0 𝟎𝟎.𝟎𝟎𝟎𝟎 = 𝟎𝟎.𝟎𝟎𝟎𝟎 ∙ 𝟐𝟐𝟎𝟎 0 𝟏𝟏𝟐𝟐𝟏𝟏𝟎𝟎𝟎𝟎 = 𝟏𝟏𝟐𝟐𝟏𝟏𝟎𝟎𝟎𝟎 ∙ 𝟐𝟐𝟎𝟎

1 𝟎𝟎.𝟎𝟎𝟎𝟎 = 𝟎𝟎.𝟎𝟎𝟎𝟎 ∙ 𝟐𝟐𝟏𝟏 1 𝟔𝟔𝟎𝟎𝟔𝟔𝟐𝟐 = 𝟏𝟏𝟐𝟐𝟏𝟏𝟎𝟎𝟎𝟎 ∙ 𝟐𝟐−𝟏𝟏

2 𝟎𝟎.𝟏𝟏𝟔𝟔 = 𝟎𝟎.𝟎𝟎𝟎𝟎 ∙ 𝟐𝟐𝟐𝟐 2 𝟑𝟑𝟎𝟎𝟑𝟑𝟔𝟔 = 𝟏𝟏𝟐𝟐𝟏𝟏𝟎𝟎𝟎𝟎 ∙ 𝟐𝟐−𝟐𝟐

3 𝟎𝟎.𝟑𝟑𝟐𝟐 = 𝟎𝟎.𝟎𝟎𝟎𝟎 ∙ 𝟐𝟐𝟑𝟑 3 𝟏𝟏𝟏𝟏𝟏𝟏𝟎𝟎 = 𝟏𝟏𝟐𝟐𝟏𝟏𝟎𝟎𝟎𝟎 ∙ 𝟐𝟐−𝟑𝟑

4 𝟎𝟎.𝟔𝟔𝟎𝟎 = 𝟎𝟎.𝟎𝟎𝟎𝟎 ∙ 𝟐𝟐𝟎𝟎 4 𝟔𝟔𝟏𝟏𝟕𝟕 = 𝟏𝟏𝟐𝟐𝟏𝟏𝟎𝟎𝟎𝟎 ∙ 𝟐𝟐−𝟎𝟎

b. Create an algebraic function that describes the thickness in inches after 𝑛𝑛 folds.

𝑻𝑻(𝒏𝒏) = 𝟎𝟎.𝟎𝟎𝟎𝟎 ∙ 𝟐𝟐𝒏𝒏, where 𝒏𝒏 is a positive integer.

c. The formula shown determines the minimum length in inches, 𝐿𝐿, of a long rectangular piece of paperof thickness 𝑇𝑇 inches needed to fold it in half 𝑛𝑛 times, always folding perpendicular to the long side.Use the formula to determine whether it is possible to fold an 11-inch by 22-inch sheet of papertowel in half 6 times.

𝐿𝐿 =𝜋𝜋𝑇𝑇6

(2𝑛𝑛 + 4)(2𝑛𝑛 − 1)

When 𝑻𝑻 = 𝟎𝟎.𝟎𝟎𝟎𝟎, the length required to fold 𝟔𝟔 times is 𝑳𝑳 = 𝝅𝝅(𝟎𝟎.𝟎𝟎𝟎𝟎)𝟔𝟔 �𝟐𝟐𝟔𝟔 + 𝟎𝟎��𝟐𝟐𝟔𝟔 − 𝟏𝟏� ≈ 𝟕𝟕𝟎𝟎.

The required length of 𝟕𝟕𝟎𝟎 inches is much longer than the 𝟐𝟐𝟐𝟐-inch sheet of paper towel. Therefore, the paper towel cannot be folded in half 𝟔𝟔 times.

I need to remember to convert the length of the roll of paper towels from feet to inches to calculate the area on top in square inches.

I see this pattern in the table.

1

© 2015 Great Minds eureka-math.org ALG II-M3-HWH-1.3.0-09.2015

Homework Helper A Story of Functions

2015-16

M3 ALGEBRA II


2. Apply the properties of exponents to rewrite each expression in the form 𝑘𝑘𝑥𝑥𝑛𝑛, where 𝑛𝑛 is a positive integer. a. (3𝑥𝑥3)(2𝑥𝑥2)−3(6𝑥𝑥)2

𝟑𝟑 ∙ 𝟐𝟐−𝟑𝟑 ∙ 𝟔𝟔𝟐𝟐 ∙ 𝒙𝒙𝟑𝟑+(𝟐𝟐∙−𝟑𝟑)+𝟐𝟐 =𝟑𝟑 ∙ 𝟔𝟔𝟐𝟐 ∙ 𝒙𝒙−𝟏𝟏

𝟐𝟐𝟑𝟑=𝟐𝟐𝟔𝟔𝟐𝟐𝒙𝒙

b. � 3𝑥𝑥2

5𝑥𝑥−4�−2

�𝟑𝟑𝒙𝒙𝟐𝟐

𝟏𝟏𝒙𝒙−𝟎𝟎�−𝟐𝟐

=𝟑𝟑−𝟐𝟐 ∙ 𝒙𝒙𝟐𝟐∙−𝟐𝟐

𝟏𝟏−𝟐𝟐 ∙ 𝒙𝒙−𝟎𝟎∙−𝟐𝟐=𝟑𝟑−𝟐𝟐 ∙ 𝒙𝒙−𝟎𝟎

𝟏𝟏−𝟐𝟐 ∙ 𝒙𝒙𝟎𝟎=

𝟏𝟏𝟐𝟐

𝟑𝟑𝟐𝟐 ∙ 𝒙𝒙𝟎𝟎 ∙ 𝒙𝒙𝟎𝟎=

𝟐𝟐𝟏𝟏𝟕𝟕𝒙𝒙𝟏𝟏𝟐𝟐

3. Apply the properties of exponents to verify that the statement 3

(2𝑛𝑛)3∙82𝑛𝑛

27𝑛𝑛+1=

19�23�

3𝑛𝑛 for integer values

of 𝑛𝑛 is an identity.

𝟑𝟑

(𝟐𝟐𝒏𝒏)𝟑𝟑 ∙𝟎𝟎𝟐𝟐𝒏𝒏

𝟐𝟐𝟔𝟔𝒏𝒏+𝟏𝟏=

𝟑𝟑𝟐𝟐𝟑𝟑𝒏𝒏

∙�𝟐𝟐𝟑𝟑�𝟐𝟐𝒏𝒏

(𝟑𝟑𝟑𝟑)𝒏𝒏+𝟏𝟏

=𝟑𝟑 ∙ 𝟐𝟐𝟔𝟔𝒏𝒏

𝟐𝟐𝟑𝟑𝒏𝒏 ∙ 𝟑𝟑𝟑𝟑𝒏𝒏+𝟑𝟑

=𝟐𝟐𝟑𝟑𝒏𝒏

𝟑𝟑𝟑𝟑𝒏𝒏+𝟐𝟐

=𝟐𝟐𝟑𝟑𝒏𝒏

𝟑𝟑𝟑𝟑𝒏𝒏 ∙ 𝟑𝟑𝟐𝟐

=𝟏𝟏𝟑𝟑𝟐𝟐�𝟐𝟐𝟑𝟑�𝟑𝟑𝒏𝒏

=𝟏𝟏𝟕𝟕�𝟐𝟐𝟑𝟑�𝟑𝟑𝒏𝒏

So, 𝟑𝟑

(𝟐𝟐𝒏𝒏)𝟑𝟑∙ 𝟎𝟎𝟐𝟐𝒏𝒏

𝟐𝟐𝟔𝟔𝒏𝒏+𝟏𝟏=

𝟏𝟏𝟕𝟕�𝟐𝟐𝟑𝟑�𝟑𝟑𝒏𝒏

for all integer values of 𝒏𝒏.

Rewriting each base as a power of 2 or 3 will help me to combine terms.

I can use the identities (𝑥𝑥𝑥𝑥)𝑛𝑛 = 𝑥𝑥𝑛𝑛𝑥𝑥𝑛𝑛 and

�𝑥𝑥𝑥𝑥�𝑛𝑛

= 𝑥𝑥𝑛𝑛𝑥𝑥𝑛𝑛 to simplify the expression.

I know that 1𝑥𝑥𝑎𝑎

= 𝑥𝑥−𝑎𝑎, which

allows me to write the expression with positive integer exponents.

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2015-16

M3 ALGEBRA II


Once I isolate 𝑡𝑡, I can substitute 2𝑥𝑥3 for 𝑥𝑥 to create an expression for 𝑡𝑡 in terms of 𝑥𝑥.

4. If 𝑥𝑥 = 2𝑥𝑥3, and 𝑥𝑥 = 𝑡𝑡3𝑥𝑥2, show that 𝑡𝑡 = 12𝑥𝑥7.

Since 𝒚𝒚 = 𝒕𝒕𝟑𝟑𝒙𝒙𝟐𝟐, it follows that

𝒕𝒕 = 𝒚𝒚 ∙ 𝟑𝟑𝒙𝒙𝟐𝟐 = 𝒚𝒚 ∙ 𝟑𝟑(𝟐𝟐𝒚𝒚𝟑𝟑)𝟐𝟐 = 𝒚𝒚 ∙ 𝟑𝟑 ∙ 𝟎𝟎𝒚𝒚𝟔𝟔 = 𝟏𝟏𝟐𝟐𝒚𝒚𝟔𝟔.

Therefore, if 𝒙𝒙 = 𝟐𝟐𝒚𝒚𝟑𝟑 and 𝒚𝒚 = 𝒕𝒕𝟑𝟑𝒙𝒙𝟐𝟐, then 𝒕𝒕 = 𝟏𝟏𝟐𝟐𝒚𝒚𝟔𝟔.

5. Write the first five terms of the following recursively defined sequence in the form 𝑘𝑘𝑥𝑥𝑛𝑛. 𝑎𝑎𝑛𝑛+1 = (2𝑎𝑎𝑛𝑛)−1, 𝑎𝑎1 = 𝑥𝑥, (𝑥𝑥 ≠ 0)

𝒂𝒂𝟏𝟏 = 𝒚𝒚 𝒂𝒂𝟐𝟐 = (𝟐𝟐𝒂𝒂𝟏𝟏)−𝟏𝟏 = (𝟐𝟐𝒚𝒚)−𝟏𝟏 = 𝟐𝟐−𝟏𝟏𝒚𝒚−𝟏𝟏 =

𝟏𝟏𝟐𝟐𝒚𝒚

−𝟏𝟏

𝒂𝒂𝟑𝟑 = (𝟐𝟐𝒂𝒂𝟐𝟐)−𝟏𝟏 = �𝟐𝟐 ∙𝟏𝟏𝟐𝟐𝒚𝒚�

−𝟏𝟏= �

𝟏𝟏𝒚𝒚�

−𝟏𝟏= 𝒚𝒚

𝒂𝒂𝟎𝟎 = (𝟐𝟐𝒂𝒂𝟑𝟑)−𝟏𝟏 = (𝟐𝟐𝒚𝒚)−𝟏𝟏 = 𝟐𝟐−𝟏𝟏𝒚𝒚−𝟏𝟏 =𝟏𝟏𝟐𝟐𝒚𝒚

−𝟏𝟏

𝒂𝒂𝟏𝟏 = (𝟐𝟐𝒂𝒂𝟎𝟎)−𝟏𝟏 = �𝟐𝟐 ∙𝟏𝟏𝟐𝟐𝒚𝒚�

−𝟏𝟏= �

𝟏𝟏𝒚𝒚�

−𝟏𝟏= 𝒚𝒚

The first five terms are 𝒚𝒚, 𝟏𝟏𝟐𝟐𝒚𝒚−𝟏𝟏, 𝒚𝒚,

𝟏𝟏𝟐𝟐𝒚𝒚−𝟏𝟏, 𝒚𝒚.

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M3 ALGEBRA II

Lesson 2: Base 10 and Scientific Notation

I can also rewrite this quotient as 0.7 × 10−2, which I can rewrite as 7 × 10−3.

Rewriting 10−3 using the property 𝑎𝑎𝑥𝑥+𝑦𝑦 = 𝑎𝑎𝑥𝑥 ∙𝑎𝑎𝑦𝑦 will produce two numerical expressions with the same order of magnitude.


1. Write the following numbers used in these statements in scientific notation. (Note: These numbers have been rounded.) a. The boiling point of silver is 2435 Kelvin.

𝟐𝟐.𝟒𝟒𝟒𝟒𝟒𝟒 × 𝟏𝟏,𝟎𝟎𝟎𝟎𝟎𝟎 = 𝟐𝟐.𝟒𝟒𝟒𝟒𝟒𝟒 × 𝟏𝟏𝟎𝟎𝟒𝟒

b. The Van der Waals radius of a carbon atom is 0.00000000017 meters.

𝟏𝟏.𝟕𝟕 × 𝟏𝟏𝟎𝟎−𝟏𝟏𝟎𝟎

2. Write the following numbers in decimal form. (Note: These numbers have been rounded.) a. The mass of Mars is 6.42 × 1023 kg.

𝟔𝟔𝟒𝟒𝟐𝟐,𝟎𝟎𝟎𝟎𝟎𝟎,𝟎𝟎𝟎𝟎𝟎𝟎,𝟎𝟎𝟎𝟎𝟎𝟎,𝟎𝟎𝟎𝟎𝟎𝟎,𝟎𝟎𝟎𝟎𝟎𝟎,𝟎𝟎𝟎𝟎𝟎𝟎,𝟎𝟎𝟎𝟎𝟎𝟎

b. The longest wavelength for waves classified as ultraviolet is 4 × 10−7meters.

𝟎𝟎.𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟒𝟒

3. Perform the following calculations without rewriting the numbers in decimal form.

a. (6.5 × 10−2) − (4.7 × 10−3)

�𝟔𝟔.𝟒𝟒 × 𝟏𝟏𝟎𝟎−𝟐𝟐� − �𝟒𝟒.𝟕𝟕 × 𝟏𝟏𝟎𝟎−𝟒𝟒� = �𝟔𝟔.𝟒𝟒 × 𝟏𝟏𝟎𝟎−𝟐𝟐� − �𝟒𝟒.𝟕𝟕 × 𝟏𝟏𝟎𝟎−𝟏𝟏 × 𝟏𝟏𝟎𝟎−𝟐𝟐�

= �𝟔𝟔.𝟒𝟒 × 𝟏𝟏𝟎𝟎−𝟐𝟐� − �𝟎𝟎.𝟒𝟒𝟕𝟕 × 𝟏𝟏𝟎𝟎−𝟐𝟐� = 𝟔𝟔.𝟎𝟎𝟒𝟒 × 𝟏𝟏𝟎𝟎−𝟐𝟐

b. 3.5 × 108

5 × 1010

𝟒𝟒.𝟒𝟒 × 𝟏𝟏𝟎𝟎𝟖𝟖

𝟒𝟒 × 𝟏𝟏𝟎𝟎𝟏𝟏𝟎𝟎=�𝟒𝟒𝟒𝟒 × 𝟏𝟏𝟎𝟎−𝟏𝟏� × 𝟏𝟏𝟎𝟎𝟖𝟖

𝟒𝟒 × 𝟏𝟏𝟎𝟎𝟏𝟏𝟎𝟎=𝟒𝟒𝟒𝟒 × (𝟏𝟏𝟎𝟎−𝟏𝟏 × 𝟏𝟏𝟎𝟎𝟖𝟖)

𝟒𝟒 × 𝟏𝟏𝟎𝟎𝟏𝟏𝟎𝟎= 𝟕𝟕 × 𝟏𝟏𝟎𝟎−𝟒𝟒

I remember that a positive, decimal number is written in scientific notation if it is expressed as a product 𝑑𝑑 × 10𝑛𝑛, where 𝑑𝑑 is a finite decimal number so that 1 ≤ 𝑑𝑑 < 10, and 𝑛𝑛 is an integer.

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M3 ALGEBRA II


I know that given an average speed 𝑟𝑟, the distance is given by 𝑑𝑑 = 𝑟𝑟 ∙ 𝑡𝑡, so

𝑡𝑡 = 𝑑𝑑𝑟𝑟.

4. The distance from the Sun to Neptune is about 4.5 × 109 km. The speed of light is approximately 3.0 × 108 meters per second. Approximately how long does it take light from the Sun to reach Neptune?

Since 𝟒𝟒.𝟒𝟒 × 𝟏𝟏𝟎𝟎𝟗𝟗 𝐤𝐤𝐤𝐤 𝟒𝟒.𝟎𝟎 × 𝟏𝟏𝟎𝟎𝟖𝟖 𝐤𝐤𝐬𝐬

= 𝟒𝟒.𝟒𝟒 × 𝟏𝟏𝟎𝟎𝟏𝟏𝟐𝟐 𝐤𝐤 𝟒𝟒.𝟎𝟎 × 𝟏𝟏𝟎𝟎𝟖𝟖 𝐤𝐤𝐬𝐬

= 𝟏𝟏.𝟒𝟒 × 𝟏𝟏𝟎𝟎𝟒𝟒 s,

it takes light approximately 𝟒𝟒.𝟐𝟐 hours to reach Neptune from the Sun.

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2015-16

M3 ALGEBRA II

Lesson 3: Rational Exponents—What are 212 and 2

13?

Lesson 3: Rational Exponents—What are 𝟐𝟐𝟏𝟏𝟐𝟐 and 𝟐𝟐

𝟏𝟏𝟑𝟑?

1. Rewrite in radical form. If the number is rational, write it without using radicals or exponents.

a. 2723 b. �1

2�−13

𝟐𝟐𝟐𝟐𝟐𝟐𝟑𝟑 = �√𝟐𝟐𝟐𝟐𝟑𝟑 �

𝟐𝟐= 𝟑𝟑𝟐𝟐 = 𝟗𝟗 �𝟏𝟏

𝟐𝟐�−𝟏𝟏𝟑𝟑 = 𝟐𝟐

𝟏𝟏𝟑𝟑 = √𝟐𝟐𝟑𝟑

2. Rewrite the following expressions in exponent form.

a. √643

√𝟔𝟔𝟒𝟒𝟑𝟑 = �𝟔𝟔𝟒𝟒�𝟏𝟏𝟑𝟑 = 𝟔𝟔

𝟒𝟒𝟑𝟑

b. �√6�3

�√𝟔𝟔�𝟑𝟑

= �𝟔𝟔𝟏𝟏𝟐𝟐�

𝟑𝟑= 𝟔𝟔

𝟑𝟑𝟐𝟐

3. Use the graph of 𝑓𝑓(𝑥𝑥) = 3𝑥𝑥 shown to the right to estimate the following powers of 3.

a. 345

𝟑𝟑𝟒𝟒𝟓𝟓 = 𝟑𝟑𝟎𝟎.𝟖𝟖, so 𝟑𝟑

𝟒𝟒𝟓𝟓 ≈ 𝟐𝟐.𝟒𝟒

b. 31.2

𝟑𝟑𝟏𝟏.𝟐𝟐 ≈ 𝟑𝟑.𝟐𝟐

I can rewrite an exponential expression

𝑎𝑎𝑚𝑚𝑛𝑛 as √𝑎𝑎𝑚𝑚𝑛𝑛 or �√𝑎𝑎𝑛𝑛 �

𝑚𝑚.

I know that √6 = 612

because the root index of a square root is 2.

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2015-16

M3 ALGEBRA II

Lesson 3: Rational Exponents—What are 212 and 2

13?

4. Rewrite each expression in the form 𝑘𝑘𝑥𝑥𝑛𝑛, where 𝑘𝑘 is a real number, 𝑥𝑥 is a positive real number, and 𝑛𝑛 is rational.

a. 16

√25𝑥𝑥6

𝟏𝟏𝟔𝟔√𝟐𝟐𝟓𝟓𝒙𝒙𝟔𝟔

=𝟏𝟏𝟔𝟔

�(𝟓𝟓𝒙𝒙𝟑𝟑)𝟐𝟐=𝟏𝟏𝟔𝟔𝟓𝟓𝒙𝒙𝟑𝟑

=𝟏𝟏𝟔𝟔𝟓𝟓𝒙𝒙−𝟑𝟑

b. �64𝑥𝑥4�−12

�𝟔𝟔𝟒𝟒𝒙𝒙𝟒𝟒�−𝟏𝟏𝟐𝟐

= �𝒙𝒙𝟒𝟒

𝟔𝟔𝟒𝟒�

𝟏𝟏𝟐𝟐

=𝒙𝒙𝟐𝟐

√𝟔𝟔𝟒𝟒=𝒙𝒙𝟐𝟐

𝟖𝟖=𝟏𝟏𝟖𝟖𝒙𝒙𝟐𝟐

5. Find a value of 𝑥𝑥 for which 𝑥𝑥3

2 = 125.

𝒙𝒙𝟑𝟑𝟐𝟐 = 𝟏𝟏𝟐𝟐𝟓𝟓

�√𝒙𝒙�𝟑𝟑

= 𝟏𝟏𝟐𝟐𝟓𝟓

��√𝒙𝒙�𝟑𝟑�𝟏𝟏𝟑𝟑 = 𝟏𝟏𝟐𝟐𝟓𝟓

𝟏𝟏𝟑𝟑

√𝒙𝒙 = √𝟏𝟏𝟐𝟐𝟓𝟓𝟑𝟑

√𝒙𝒙 = 𝟓𝟓 𝒙𝒙 = 𝟐𝟐𝟓𝟓

I can rewrite 25𝑥𝑥6 as (5𝑥𝑥3)2.

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2015-16

M3 ALGEBRA II

Lesson 4: Properties of Exponents and Radicals


1. Evaluate each expression for 𝑎𝑎 = 8 and 𝑏𝑏 = 16.

a. �√𝑎𝑎3 + √𝑏𝑏�2

�√𝟖𝟖𝟑𝟑 + √𝟏𝟏𝟏𝟏�𝟐𝟐

= (𝟐𝟐 + 𝟒𝟒)𝟐𝟐 = 𝟑𝟑𝟏𝟏

b. �𝑎𝑎23 − 𝑏𝑏

32�−1

�𝟖𝟖𝟐𝟐𝟑𝟑 − 𝟏𝟏𝟏𝟏

𝟑𝟑𝟐𝟐�

−𝟏𝟏= ��√𝟖𝟖𝟑𝟑 �

𝟐𝟐− �√𝟏𝟏𝟏𝟏�

𝟑𝟑�−𝟏𝟏

= �𝟐𝟐𝟐𝟐 − 𝟒𝟒𝟑𝟑�−𝟏𝟏 = (−𝟏𝟏𝟔𝟔)−𝟏𝟏 = −𝟏𝟏𝟏𝟏𝟔𝟔

2. Rewrite each expression so that each term is in the form 𝑘𝑘𝑥𝑥𝑛𝑛, where 𝑘𝑘 is a real number, 𝑥𝑥 is a positive real number, and 𝑛𝑛 is a rational number.

a. 2𝑥𝑥3+𝑥𝑥2−√𝑥𝑥

√𝑥𝑥3

𝟐𝟐𝒙𝒙𝟑𝟑 + 𝒙𝒙𝟐𝟐 − √𝒙𝒙√𝒙𝒙𝟑𝟑 =

𝟐𝟐𝒙𝒙𝟑𝟑 + 𝒙𝒙𝟐𝟐 − 𝒙𝒙𝟏𝟏𝟐𝟐

𝒙𝒙𝟏𝟏𝟑𝟑

= 𝟐𝟐𝒙𝒙𝟖𝟖𝟑𝟑 + 𝒙𝒙

𝟓𝟓𝟑𝟑 − 𝒙𝒙

𝟏𝟏𝟏𝟏

b. �3𝑥𝑥23 ∙ −2𝑥𝑥

43�−2

�𝟑𝟑𝒙𝒙𝟐𝟐𝟑𝟑 ∙ −𝟐𝟐𝒙𝒙

𝟒𝟒𝟑𝟑�

−𝟐𝟐= �𝟑𝟑 ∙ −𝟐𝟐 ∙ 𝒙𝒙

𝟐𝟐𝟑𝟑 ∙ 𝒙𝒙

𝟒𝟒𝟑𝟑�

−𝟐𝟐

= �−𝟏𝟏𝒙𝒙𝟐𝟐𝟑𝟑+

𝟒𝟒𝟑𝟑�

−𝟐𝟐

= �−𝟏𝟏𝒙𝒙𝟐𝟐�−𝟐𝟐

= �𝟏𝟏

−𝟏𝟏𝒙𝒙𝟐𝟐�𝟐𝟐

=𝟏𝟏

𝟑𝟑𝟏𝟏𝒙𝒙𝟒𝟒

=𝟏𝟏𝟑𝟑𝟏𝟏

𝒙𝒙−𝟒𝟒

I can write these exponential expressions in radical form using the identities learned in Lesson 3.

I remember that

𝑏𝑏𝑚𝑚

𝑏𝑏𝑛𝑛= 𝑏𝑏𝑚𝑚 ∙ 𝑏𝑏−𝑛𝑛 = 𝑏𝑏𝑚𝑚−𝑛𝑛.

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2015-16

M3 ALGEBRA II


3. Determine whether the following numerical expression is real or complex. If it is real, determine whether it is rational or irrational and positive or negative.

�823 − 8−

23�

−12

�𝟖𝟖𝟐𝟐𝟑𝟑 − 𝟖𝟖−

𝟐𝟐𝟑𝟑�

−𝟏𝟏𝟐𝟐= �𝟒𝟒 −

𝟏𝟏𝟒𝟒�−𝟏𝟏𝟐𝟐

= �𝟏𝟏𝟓𝟓𝟒𝟒�−𝟏𝟏𝟐𝟐

= �𝟒𝟒𝟏𝟏𝟓𝟓�𝟏𝟏𝟐𝟐

=√𝟒𝟒√𝟏𝟏𝟓𝟓

=𝟐𝟐√𝟏𝟏𝟓𝟓

Thus, �𝟖𝟖𝟐𝟐𝟑𝟑 − 𝟖𝟖−

𝟐𝟐𝟑𝟑�−𝟏𝟏𝟐𝟐

is real, positive, and irrational.

4. Show that 26 ∙ 5𝑛𝑛 − 5𝑛𝑛+2 = 5𝑛𝑛 for any integer 𝑛𝑛.

𝟐𝟐𝟏𝟏 ∙ 𝟓𝟓𝒏𝒏 − 𝟓𝟓𝒏𝒏+𝟐𝟐 = 𝟐𝟐𝟏𝟏 ∙ 𝟓𝟓𝒏𝒏 − 𝟓𝟓𝟐𝟐 ∙ 𝟓𝟓𝒏𝒏 = 𝟓𝟓𝒏𝒏�𝟐𝟐𝟏𝟏 − 𝟓𝟓𝟐𝟐� = 𝟓𝟓𝒏𝒏(𝟏𝟏) = 𝟓𝟓𝒏𝒏

Thus, for any integer 𝒏𝒏, 𝟐𝟐𝟏𝟏 ∙ 𝟓𝟓𝒏𝒏 − 𝟓𝟓𝒏𝒏+𝟐𝟐 = 𝟓𝟓𝒏𝒏.

5. Rewrite the following radical expression as an equivalent exponential expression in which each variable occurs no more than once.

�162𝑥𝑥3𝑦𝑦12𝑧𝑧94

�𝟏𝟏𝟏𝟏𝟐𝟐𝒙𝒙𝟑𝟑𝒚𝒚𝟏𝟏𝟐𝟐𝒛𝒛𝟗𝟗𝟒𝟒 = �𝟐𝟐 ∙ 𝟖𝟖𝟏𝟏 ∙ 𝒙𝒙𝟑𝟑 ∙ 𝒚𝒚𝟏𝟏𝟐𝟐 ∙ 𝒛𝒛𝟗𝟗�𝟏𝟏𝟒𝟒

= 𝟐𝟐𝟏𝟏𝟒𝟒 ⋅ 𝟖𝟖𝟏𝟏

𝟏𝟏𝟒𝟒 ⋅ �𝒙𝒙𝟑𝟑�

𝟏𝟏𝟒𝟒 ⋅ �𝒚𝒚𝟏𝟏𝟐𝟐�

𝟏𝟏𝟒𝟒 ⋅ �𝒛𝒛𝟗𝟗�

𝟏𝟏𝟒𝟒

= 𝟐𝟐𝟏𝟏𝟒𝟒 ∙ 𝟑𝟑 ∙ 𝒙𝒙

𝟑𝟑𝟒𝟒 ∙ 𝒚𝒚𝟑𝟑 ∙ 𝒛𝒛

𝟗𝟗𝟒𝟒

6. Use properties of exponents to find two integers that are upper and lower estimates of the value of 250.8.

𝟔𝟔.𝟓𝟓 < 𝟔𝟔.𝟖𝟖 < 𝟏𝟏

𝟐𝟐𝟓𝟓𝟔𝟔.𝟓𝟓 < 𝟐𝟐𝟓𝟓𝟔𝟔.𝟖𝟖 < 𝟐𝟐𝟓𝟓𝟏𝟏

√𝟐𝟐𝟓𝟓 < 𝟐𝟐𝟓𝟓𝟔𝟔.𝟖𝟖 < 𝟐𝟐𝟓𝟓𝟏𝟏

𝟓𝟓 < 𝟐𝟐𝟓𝟓𝟔𝟔.𝟖𝟖 < 𝟐𝟐𝟓𝟓

I can use the same process I used in Problem 1 to evaluate these exponential expressions.

I know 81 = 34, so

8114 = (34)

14 = 3.

Since 25 is a perfect square, I know √25 is an integer, and I can write √25 as 250.5.

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2015-16

M3 ALGEBRA II

Lesson 5: Irrational Exponents—What are 2√2 and 2𝜋𝜋?

Lesson 5: Irrational Exponents—What Are 𝟐𝟐√𝟐𝟐 and 𝟐𝟐𝝅𝝅?

1. Use properties of exponents to rewrite the following expressions as a number or an exponential expression with only one exponent.

a. �21+√3�1−√3

�𝟐𝟐𝟏𝟏+√𝟑𝟑�𝟏𝟏−√𝟑𝟑

= 𝟐𝟐(𝟏𝟏+√𝟑𝟑)(𝟏𝟏−√𝟑𝟑) = 𝟐𝟐𝟏𝟏−𝟑𝟑 = 𝟐𝟐−𝟐𝟐 = 𝟏𝟏𝟒𝟒

b. 53+√2 ⋅ 53−√2

𝟓𝟓𝟑𝟑+√𝟐𝟐 ⋅ 𝟓𝟓𝟑𝟑−√𝟐𝟐 = 𝟓𝟓��𝟑𝟑+√𝟐𝟐�+�𝟑𝟑−√𝟐𝟐�� = 𝟓𝟓𝟔𝟔 = 𝟏𝟏𝟓𝟓𝟔𝟔𝟐𝟐𝟓𝟓

2. Between what two integer powers of 4 does 4√7 lie?

Since 𝟒𝟒 < 𝟕𝟕 < 𝟗𝟗, it follows that 𝟐𝟐 < √𝟕𝟕 < 𝟑𝟑, so 𝟒𝟒𝟐𝟐 < 𝟒𝟒√𝟕𝟕 < 𝟒𝟒𝟑𝟑.

3. Follow the steps below to approximate the number 4√7. Use the approximation √7 ≈ 2.645751.

a. Find a sequence of four intervals that contain √7 whose endpoints get successively closer to √7.

𝟐𝟐 < √𝟕𝟕 < 𝟑𝟑 𝟐𝟐.𝟔𝟔 < √𝟕𝟕 < 𝟐𝟐.𝟕𝟕 𝟐𝟐.𝟔𝟔𝟒𝟒 < √𝟕𝟕 < 𝟐𝟐.𝟔𝟔𝟓𝟓 𝟐𝟐.𝟔𝟔𝟒𝟒𝟓𝟓 < √𝟕𝟕 < 𝟐𝟐.𝟔𝟔𝟒𝟒𝟔𝟔

b. Find a sequence of four intervals that contain 4√7 whose endpoints get successively closer to 4√7. Write your intervals in the form 4𝑟𝑟 < 4√7 < 4𝑠𝑠 for rational numbers 𝑟𝑟 and 𝑠𝑠.

𝟒𝟒𝟐𝟐 < 𝟒𝟒√𝟕𝟕 < 𝟒𝟒𝟑𝟑 𝟒𝟒𝟐𝟐.𝟔𝟔 < 𝟒𝟒√𝟕𝟕 < 𝟒𝟒𝟐𝟐.𝟕𝟕 𝟒𝟒𝟐𝟐.𝟔𝟔𝟒𝟒 < 𝟒𝟒√𝟕𝟕 < 𝟒𝟒𝟐𝟐.𝟔𝟔𝟓𝟓 𝟒𝟒𝟐𝟐.𝟔𝟔𝟒𝟒𝟓𝟓 < 𝟒𝟒√𝟕𝟕 < 𝟒𝟒𝟐𝟐.𝟔𝟔𝟒𝟒𝟔𝟔

I recognize �1 + √3��1 − √3� as the factored form of the difference of two squares: (𝑎𝑎 + 𝑏𝑏)(𝑎𝑎 − 𝑏𝑏) = 𝑎𝑎2 − 𝑏𝑏2.

I can use the property 𝑏𝑏𝑥𝑥 ⋅ 𝑏𝑏𝑦𝑦 = 𝑏𝑏𝑥𝑥+𝑦𝑦 to rewrite the given expression.

Using the given approximation for √7, I can select values of 𝑟𝑟 and 𝑠𝑠 that create smaller intervals that contain 4√7.

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2015-16

M3 ALGEBRA II

Lesson 5: Irrational Exponents—What are 2√2 and 2𝜋𝜋?

c. Use a calculator to find estimates to four decimal places of the endpoints of the intervals in part (b).

𝟏𝟏𝟔𝟔.𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎 < 𝟒𝟒√𝟕𝟕 < 𝟔𝟔𝟒𝟒.𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎 𝟑𝟑𝟔𝟔.𝟕𝟕𝟓𝟓𝟕𝟕𝟑𝟑 < 𝟒𝟒√𝟕𝟕 < 𝟒𝟒𝟐𝟐.𝟐𝟐𝟐𝟐𝟒𝟒𝟑𝟑 𝟑𝟑𝟕𝟕.𝟕𝟕𝟓𝟓𝟒𝟒𝟐𝟐 < 𝟒𝟒√𝟕𝟕 < 𝟑𝟑𝟗𝟗.𝟑𝟑𝟗𝟗𝟔𝟔𝟔𝟔 𝟑𝟑𝟗𝟗.𝟏𝟏𝟐𝟐𝟒𝟒𝟓𝟓 < 𝟒𝟒√𝟕𝟕 < 𝟑𝟑𝟗𝟗.𝟏𝟏𝟕𝟕𝟕𝟕𝟕𝟕

d. Is 4√7 rational or irrational? Explain your response.

We cannot tell whether the decimal expansion of 𝟒𝟒√𝟕𝟕 will have a repeating pattern, so we cannot determine whether it is rational or irrational based on the estimates from part (c).

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2015-16

M3 ALGEBRA II

Lesson 6: Euler’s Number, 𝑒𝑒

Lesson 6: Euler’s Number, 𝒆𝒆

1. For any positive integer 𝑛𝑛, 𝑛𝑛 factorial is defined by 𝑛𝑛! = 𝑛𝑛(𝑛𝑛 − 1)(𝑛𝑛 − 2)⋯3 ∙ 2 ∙ 1. Given that 9! = 9 ∙ 8 ∙ 7 ∙ 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 362,880, calculate 10! and 11!.

𝟏𝟏𝟏𝟏! = 𝟏𝟏𝟏𝟏 ∙ 𝟗𝟗 ∙ 𝟖𝟖 ∙ 𝟕𝟕 ∙ 𝟔𝟔 ∙ 𝟓𝟓 ∙ 𝟒𝟒 ∙ 𝟑𝟑 ∙ 𝟐𝟐 ∙ 𝟏𝟏 = 𝟏𝟏𝟏𝟏 ∙ 𝟗𝟗! = 𝟏𝟏𝟏𝟏 ∙ 𝟑𝟑𝟔𝟔𝟐𝟐,𝟖𝟖𝟖𝟖𝟏𝟏 = 𝟑𝟑,𝟔𝟔𝟐𝟐𝟖𝟖,𝟖𝟖𝟏𝟏𝟏𝟏

𝟏𝟏𝟏𝟏! = 𝟏𝟏𝟏𝟏 ∙ 𝟏𝟏𝟏𝟏 ∙ 𝟗𝟗 ∙ 𝟖𝟖 ∙ 𝟕𝟕 ∙ 𝟔𝟔 ∙ 𝟓𝟓 ∙ 𝟒𝟒 ∙ 𝟑𝟑 ∙ 𝟐𝟐 ∙ 𝟏𝟏 = 𝟏𝟏𝟏𝟏 ∙ 𝟏𝟏𝟏𝟏! = 𝟏𝟏𝟏𝟏 ∙ 𝟑𝟑,𝟔𝟔𝟐𝟐𝟖𝟖,𝟖𝟖𝟏𝟏𝟏𝟏 = 𝟑𝟑𝟗𝟗,𝟗𝟗𝟏𝟏𝟔𝟔,𝟖𝟖𝟏𝟏𝟏𝟏

2. If 𝑎𝑎 = 5𝑏𝑏2 and 𝑏𝑏 = − 14 𝑒𝑒

3, express 𝑎𝑎 in terms of 𝑒𝑒, and approximate the value of 𝑎𝑎 to four decimal places.

𝒂𝒂 = 𝟓𝟓�−𝟏𝟏𝟒𝟒 𝒆𝒆

𝟑𝟑�𝟐𝟐

= 𝟓𝟓� 𝟏𝟏𝟏𝟏𝟔𝟔 𝒆𝒆𝟔𝟔� ≈ 𝟏𝟏𝟐𝟐𝟔𝟔.𝟏𝟏𝟕𝟕𝟏𝟏𝟓𝟓

I know that (𝑛𝑛 + 1)! = (𝑛𝑛 + 1) ∙ 𝑛𝑛!, and I can use this to calculate the values of the factorial expressions.

I can substitute − 14 𝑒𝑒

3 for 𝑏𝑏 into the equation 𝑎𝑎 = 5𝑏𝑏2 to express 𝑎𝑎 in terms of 𝑒𝑒.

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2015-16

M3 ALGEBRA II


I know the average rate of change of a function 𝑓𝑓 on an interval [𝑎𝑎, 𝑏𝑏] is the number 𝑓𝑓(𝑏𝑏)−𝑓𝑓(𝑎𝑎)

𝑏𝑏−𝑎𝑎.

3. The following table lists the population of Austin, Texas, between 2010 and 2014.

Year Population of Austin, Texas 2010 816,022 2011 839,714 2012 865,571 2013 887,124 2014 912,791

Source: U.S. Census Bureau, 𝟐𝟐𝟏𝟏𝟏𝟏𝟒𝟒

a. Calculate the average rate of change of the population of Austin on the interval [2011, 2014], and explain what that number represents.

Average rate of change in population of Austin on [𝟐𝟐𝟏𝟏𝟏𝟏𝟏𝟏,𝟐𝟐𝟏𝟏𝟏𝟏𝟒𝟒]: 𝟗𝟗𝟏𝟏𝟐𝟐,𝟕𝟕𝟗𝟗𝟏𝟏−𝟖𝟖𝟑𝟑𝟗𝟗,𝟕𝟕𝟏𝟏𝟒𝟒

𝟐𝟐𝟏𝟏𝟏𝟏𝟒𝟒−𝟐𝟐𝟏𝟏𝟏𝟏𝟏𝟏= 𝟐𝟐𝟒𝟒, 𝟑𝟑𝟓𝟓𝟗𝟗

This means that during this three-year period, the population of Austin increased at a rate of 𝟐𝟐𝟒𝟒,𝟑𝟑𝟓𝟓𝟗𝟗 people per year.

b. Compare your results for the average rate of change in population on the interval [2010, 2012] with that on the interval [2012, 2014]. What do those numbers say about the population of Austin?

Average rate of change in population of Austin on [𝟐𝟐𝟏𝟏𝟏𝟏𝟏𝟏,𝟐𝟐𝟏𝟏𝟏𝟏𝟐𝟐]: 𝟖𝟖𝟔𝟔𝟓𝟓,𝟓𝟓𝟕𝟕𝟏𝟏−𝟖𝟖𝟏𝟏𝟔𝟔,𝟏𝟏𝟐𝟐𝟐𝟐

𝟐𝟐𝟏𝟏𝟏𝟏𝟐𝟐−𝟐𝟐𝟏𝟏𝟏𝟏𝟏𝟏= 𝟐𝟐𝟒𝟒, 𝟕𝟕𝟕𝟕𝟒𝟒. 𝟓𝟓

Average rate of change in population of Austin on [𝟐𝟐𝟏𝟏𝟏𝟏𝟐𝟐,𝟐𝟐𝟏𝟏𝟏𝟏𝟒𝟒]: 𝟗𝟗𝟏𝟏𝟐𝟐,𝟕𝟕𝟗𝟗𝟏𝟏−𝟖𝟖𝟔𝟔𝟓𝟓,𝟓𝟓𝟕𝟕𝟏𝟏

𝟐𝟐𝟏𝟏𝟏𝟏𝟒𝟒−𝟐𝟐𝟏𝟏𝟏𝟏𝟐𝟐= 𝟐𝟐𝟑𝟑, 𝟔𝟔𝟏𝟏𝟏𝟏

These numbers mean that the average rate of increase in population was greater from 2010 to 2012 than it was from 2012 to 2014. The fact that the average rates of change are positive on both intervals indicates that the population increased from 2010 to 2012 and from 2012 to 2014, but it increased more quickly on the interval from 2010 to 2012.

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2015-16

M3 ALGEBRA II


I know the average rate of change of 𝑆𝑆 on the interval

[6, 6 + ℎ] is 𝑆𝑆(6+ℎ)−𝑆𝑆(6)

(6+ℎ)−6.

4. The formula for the surface area of a sphere of radius 𝑟𝑟 can be expressed as a function 𝑆𝑆(𝑟𝑟) = 4𝜋𝜋𝑟𝑟2.

a. Find the average rate of change of the surface area of the sphere on the interval [6, 6 + ℎ] for some

small positive number ℎ.

𝑺𝑺(𝟔𝟔 + 𝒉𝒉) − 𝑺𝑺(𝟔𝟔)(𝟔𝟔 + 𝒉𝒉) − 𝟔𝟔

=𝟒𝟒𝟒𝟒(𝟔𝟔 + 𝒉𝒉)𝟐𝟐 − 𝟒𝟒𝟒𝟒�𝟔𝟔𝟐𝟐�

𝒉𝒉

=�𝟑𝟑𝟔𝟔 + 𝟏𝟏𝟐𝟐𝒉𝒉+ 𝒉𝒉𝟐𝟐�𝟒𝟒𝟒𝟒 − (𝟑𝟑𝟔𝟔)𝟒𝟒𝟒𝟒

𝒉𝒉

=𝟏𝟏𝒉𝒉 �

𝟏𝟏𝟐𝟐𝒉𝒉 + 𝒉𝒉𝟐𝟐�𝟒𝟒𝟒𝟒

= (𝟏𝟏𝟐𝟐 + 𝒉𝒉)𝟒𝟒𝟒𝟒

b. What happens to the average rate of change of the surface area of the sphere on the interval [6, 6 + ℎ] as ℎ → 0?

As 𝒉𝒉 → 𝟏𝟏, 𝟏𝟏𝟐𝟐 + 𝒉𝒉 → 𝟏𝟏𝟐𝟐 + 𝟏𝟏, so as 𝒉𝒉 gets smaller, the average rate of change approaches (𝟏𝟏𝟐𝟐 + 𝟏𝟏)𝟒𝟒𝟒𝟒, which is 𝟒𝟒𝟖𝟖𝟒𝟒.

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