lesson 5 homework practice problem solving strategies answer key

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• Lesson 5: Problem-Solving Strategies

Hi Everyone!

On this page you will find some material about Lesson 5. Read through the material below, watch the videos, and follow up with your instructor if you have questions.

Table of Contents

In this section you will find some important information about the specific resources related to this lesson:

• the learning outcomes,
• the section in the textbook,
• the homework,
• supporting video.

Learning Outcomes. (extracted from the textbook)

• State the four steps in the basic problem-solving procedure.
• Divide radical expressions.
• Solve problems using a diagram.
• Solve problems using trial and error.
• Solve problems involving money.
• Solve problems using calculation.

Topic . This lesson covers

Section 1.3: Problem-Solving Strategies

pages 29-35, ex. 1-6.

Practice Homework:

page 36: 19, 21, 25, 35, 38, 41, 43

ALEKS Assignment

Warmup Questions

These are questions on fundamental concepts that you need to know before you can embark on this lesson. Don’t skip them! Take your time to do them, and check your answer by clicking on the “Show Answer” tab.

Warmup Question 1

$$(2\sqrt{5a})(-4\sqrt{5a}).$$

Show Answer 1

$$(2\sqrt{5a})(-4\sqrt{5a})= -8\sqrt{(5a)^2}=-8(5a)=-40a$$

Warmup Question 2

$$(\sqrt 3-6)(\sqrt 3+6).$$

Show Answer 2

$$(\sqrt 3-6)(\sqrt 3+6)= (\sqrt 3)^2-6^2 = 3 – 36 = -33$$

If you are not comfortable with the Warmup Questions, don’t give up! Click on the indicated lesson for a quick catchup. A brief review will help you boost your confidence to start the new lesson, and that’s perfectly fine.

Need a review? Check Lesson 6 .

Quick Intro

This is like a mini-lesson with an overview of the main objects of study. It will often contain a list of key words, definitions and properties – all that is new in this lesson. We will use this opportunity to make connections with other concepts. It can be also used as a review of the lesson.

A Quick Intro to Division of Radicals and Rationalization

Key Words. Radicals, division of radicals, simplified form, rationalization, conjugate.

When dividing radical terms, the following property can be very helpful.

Division Property

$$\dfrac{\sqrt[n]a}{\sqrt[n]b}=\sqrt[n]{\dfrac{a}{b}}$$

If not, it may be necessary to rationalize the denominator. On Lesson 5 we listed three conditions for a radical expression to be in simplified form. The third one is:

There should be no radicals in the denominator of a fraction.

$\bigstar$ We gave $\dfrac{1}{\sqrt 2}$ as an example that fails this condition. To simplify it, we multiply both the numerator and the denominator by $\sqrt 2$.

$$\underbrace{\dfrac{1}{\sqrt 2}}_{\text{radical in the denominator}}=\dfrac{1}{\sqrt 2}\cdot\dfrac{\sqrt 2}{\sqrt 2}=\dfrac{1\cdot\sqrt 2}{\sqrt 2\cdot\sqrt 2}=\underbrace{\dfrac{\sqrt 2}{2}}_{\text{no radical in the denominator}}$$

$\bigstar$ The process of removing a radical from the denominator is called rationalization .

$\bigstar$ The key idea was to multiply the original denominator by another copy of it, since squaring eliminates the radical.

$$(\sqrt a)(\sqrt a) = (\sqrt a)^2=a.$$

$\bigstar$ But what if we have $\dfrac{1}{\sqrt 3 -6}$? Squaring $\sqrt{3}-6$ will not help (try it!). In the Warmup Question #2 we saw that

$$(\sqrt 3 -6)(\sqrt 3 +6)=-33$$

results in a number free of radical. So

$$\underbrace{\dfrac{1}{\sqrt 3 -6}}_{\text{radical in the denominator}}=\dfrac{1}{\sqrt 3 -6}\cdot\dfrac{\sqrt 3 +6}{\sqrt 3 +6}$$

$$=\dfrac{1\cdot(\sqrt 3 +6)}{(\sqrt 3 -6)\cdot(\sqrt 3 +6)}=\underbrace{-\dfrac{\sqrt 3 +6}{33}}_{\text{no radical in the denominator}}$$

$\bigstar$ This happens because the above product is a difference of squares

$$(a-b)(a+b)=a^2-b^2,$$

and squaring a single radical eliminates the radical.

$\bigstar$ We say that $a-b$ and $a+b$ are conjugates . So if the denominator is $\sqrt 3 -6$, we rationalize it by multiplying the numerator and the denominator by its conjugate $\sqrt 3+6$.

Video Lesson

Many times the mini-lesson will not be enough for you to start working on the problems. You need to see someone explaining the material to you. In the video you will find a variety of examples, solved step-by-step – starting from a simple one to a more complex one. Feel free to play them as many times as you need. Pause, rewind, replay, stop… follow your pace!

A description of the video

In the video you will see the following radical expressions.

• $\dfrac{3\sqrt x+1}{2\sqrt x}$
• $\dfrac{3\sqrt 5 + 1}{2\sqrt 7 -3}$

Try Questions

Now that you have read the material and watched the video, it is your turn to put in practice what you have learned. We encourage you to try the Try Questions on your own. When you are done, click on the “Show answer” tab to see if you got the correct answer.

Try Question 1

Simplify $$\dfrac{\sqrt{15}}{5\sqrt{20}}.$$

$$\dfrac{\sqrt{15}}{5\sqrt{20}} = \dfrac{\sqrt{15}}{5\sqrt{20}}\cdot \dfrac{\sqrt{20}}{\sqrt{20}}$$

$$= \dfrac{\sqrt{15}\sqrt{20}}{5\cdot 20} = \dfrac{\sqrt{15} \cdot 2\sqrt{5}}{5\cdot 20}$$

$$= \dfrac{\sqrt {5\cdot 15}}{5\cdot 10} = \dfrac{5\sqrt {3}}{50} = \dfrac{\sqrt 3}{10}$$

Try Question 2

Simplify $$\dfrac{3}{4+2\sqrt 5}.$$

$$\dfrac{3}{4+2\sqrt 5}=\dfrac{3}{4+2\sqrt 5}\cdot \dfrac{4-2\sqrt 5}{4-2\sqrt 5}$$

$$= \dfrac{3(4-2\sqrt 5)}{(4+2\sqrt 5)(4-2\sqrt 5)}= \dfrac{12-6\sqrt 5}{16-20}$$

$$=\dfrac{12-6\sqrt 5}{-4} = \dfrac{-6+3\sqrt 5}{2} $$

Try Question 3

Rationalize and simplify

$$\dfrac{1+3\sqrt 2}{1-\sqrt 2}+\sqrt 2.$$

Show Answer 3

$$\dfrac{1+3\sqrt 2}{1-\sqrt 2}+\sqrt 2= \dfrac{1+3\sqrt 2}{1-\sqrt 2}\cdot\dfrac{1+\sqrt 2}{1+\sqrt 2}+\sqrt 2$$

$$=\dfrac{(1+3\sqrt 2)(1+\sqrt 2)}{(1-\sqrt 2)(1+\sqrt 2)} +\sqrt 2 = \dfrac{1+\sqrt 2+3\sqrt 2 +3\cdot 2}{1-2}+\sqrt 2 $$

$$=\dfrac{7+4\sqrt 2}{-1}+\sqrt 2 = -7-4\sqrt 2+\sqrt 2 = -7-3\sqrt 2$$

You should now be ready to start working on the homework problems. Doing the homework is an essential part of learning. It will help you practice the lesson and reinforce your knowledge.

It is time to do the homework on WeBWork:

RationalizeDenominators

When you are done, come back to this page for the Exit Questions.

Exit Questions

After doing the WeBWorK problems, come back to this page. The Exit Questions include vocabulary checking and conceptual questions. Knowing the vocabulary accurately is important for us to communicate. You will also find one last problem. All these questions will give you an idea as to whether or not you have mastered the material. Remember: the “Show Answer” tab is there for you to check your work!

• What is the goal in rationalizing the denominator? 
• Why does the ‘conjugate’ play a role in accomplishing this?

$\bigstar$ Simplify

(a) $\dfrac{3-3\sqrt{3a}}{4\sqrt{8a}}$

(b) $\dfrac{\sqrt{5}+3}{4-\sqrt 5}$

Show Answer

(a) $$\dfrac{3-3\sqrt{3a}}{4\sqrt{8a}}=\dfrac{3-3\sqrt{3a}}{4\cdot 2 \sqrt{2a}} $$

$$= \dfrac{3-3\sqrt{3a}}{8\sqrt{2a}}\cdot\dfrac{\sqrt{2a}}{\sqrt{2a}} =\dfrac{(3-3\sqrt{3a}) \sqrt{2a}}{8\cdot 2a} $$

$$= \dfrac{3\sqrt{2a}-3\sqrt{6a^2}}{16a} =\dfrac{3\sqrt{2a}-3a\sqrt 6}{16a}$$

(b) $$\dfrac{\sqrt{5}+3}{4-\sqrt 5}= \dfrac{\sqrt{5}+3}{4-\sqrt 5}\cdot \dfrac{4+\sqrt 5}{4+\sqrt 5} $$

$$= \dfrac{(\sqrt{5}+3)(4+\sqrt 5)}{(4-\sqrt 5)(4-\sqrt 5)}=\dfrac{4\sqrt{5}+5+12+3\sqrt 5}{16-5}$$

$$ =\dfrac{7\sqrt 5+17}{11}$$

Need more help?

Don’t wait too long to do the following.

• Watch the additional video resources.
• Talk to your instructor.
• Form a study group.
• Visit a tutor. For more information, check the tutoring page .

Lessons Menu

• Lesson 1: Applications of Linear Equations
• Lesson 2: Ratio, Proportion, and Variation
• Lesson 3: The Nature of Mathematical Reasoning
• Lesson 4: Estimation and Interpreting Graphs
• Lesson 6: Statement and Quantifiers
• Lesson 7: Measures of Length: Converting Units and the Metric System
• Lesson 8: Measures of Area, Volume, and Capacity
• Lesson 9: Measures of Weight & Temperature
• Lesson 10: Percents
• Lesson 11: Simple Interest
• Lesson 12: Compound Interest
• Lesson 13: Basic Concepts of Probability
• Lesson 14: Tree Diagrams, Tables and Sample Spaces
• Lesson 15: Gathering and Organizing Data/Picturing Data
• Lesson 16: Measures of Average
• Lesson 17: Measures of Variation
• Lesson 18: Measures of Position
• Lesson 19: The Normal Distribution/Applications of the Normal Distribution
• Lesson 20: Correlation and Regression Analysis
• Lesson 21: Points, Lines, Planes & Angles
• Lesson 22: Triangles
• Lesson 23: Polygons and Perimeter/Areas of Polygons and Circles

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Chapter 2, Lesson 5: Solving Equations with the Variable on Each Side

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McGraw Hill My Math Grade 5 Chapter 1 Lesson 5 Answer Key Understand Place Value

All the solutions provided in McGraw Hill Math Grade 5 Answer Key PDF Chapter 1 Lesson 5 Understand Place Value  will give you a clear idea of the concepts.

McGraw-Hill My Math Grade 5 Answer Key Chapter 1 Lesson 5 Understand Place Value

Use models to describe the relationship between the value of the digits in the decimal 0.77 and their place-value position.

Talk About It

The decimal model shows 0.033. Use this decimal to answer Exercises 1—4.

Question 1. What is the value of the 3 in the hundredth place? Answer:

Question 3. The digit in the hundredth place is how many times as much as the digit in the thousandth place? Answer: The digit in the hundredth place is 10 times bigger than the digit in the thousandth place. 300 x 10 = 3000

Practice It

Use each model to write a decimal ¡n standard form and word form. Then complete each sentence.

Question 9. Micah asked for directions to a local park. He was told to go 0.33 miles south to get to the park. The value of the digit in the tenths place is how many times as much as the value of the digit in the hundredth place? Answer: The above-given miles: 0.33 The value present in the tenth place is 3 from the tenth place, we need to check the value of the digit in the hundredth place. The value of the digit in the tenth place is 10 times as much as the value of the digit in the hundredth place.

Question 10. Mathematical PRACTICE 2 Use Number Sense In science class, Abigail’s frog weighed 0.88 kilograms. What is the value of the digit in the tenth place? Answer: The above-given decimal value is 0.88 kg Here asked, the value present in the tenth place. The value of the digit in the tenths place is 0.8

Write About It

Question 13. How are the place-value positions to either side of a particular number related? Answer: The place value positions to either side of a number are related because they are 10 times smaller or ten times bigger – a number in the hundreds column is ten times bigger than the same number in the tens column, for example. Each place has a value of 10 times the place to its right.

McGraw Hill My Math Grade 5 Chapter 1 Lesson 5 My Homework Answer Key

Problem Solving

Question 3. Michael bought 0.44 pounds of sliced turkey. What is the value of the digit in the hundredth place? _____________________ The value of the digit in the hundredth place is how many times as much as the value of the digit in the tenth place? _____________________ Answer: The above-given decimal: 0.44 The value of the digit in the hundredth place is 0.04 The value of the digit in the hundredth place is 10 times as much as the value of the digit in the tenth place.

Question 4. A small piece of metal weighs 0.77 grams. What is the value of the digit in the tenth place? _____________________ The digit in the tenth place is how many times as much as the value of the digit in the hundredth place? _____________________ Answer: The above-given decimal: 0.77 The value of the digit in the tenths place is 0.7 The digit in the tenths place is 1/10 times as the value of the digit in the hundredth place.

Question 5. Mathematical PRACTICE 2 Use Number Sense Write a decimal where the value of a digit is \(\frac{1}{10}\) as much as a digit in another place. Answer: The above-given fraction is 1/10 In decimal form, we can write it as 0.1 The other place would be the hundredth place. The value of the digit in the hundredth place is 0.01

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