Big Ideas Math Answers Grade 6 | Big Ideas Math Textbook 6th Grade Solution Key

Big Ideas Math Answers Grade 6

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Big Ideas Math Book 6th Grade Answer Key | Big Ideas Math Answers 6th Grade Solutions Pdf

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There are several ways you can download Middle School Big Ideas Math Modeling Real Life Grade 6 Solutions PDF. You can find the direct chapter-wise links in the above sections, you just need to click on the link to get the detailed information about that complete chapter. You can also find them on google along with the solutions.

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Big Ideas Math Answers Grade 2 | Big Ideas Math Book 2nd Grade Solutions

Big Ideas Math Answers Grade 2

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  • First, find out the amount of time you have for the preparation.
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  • Prepare all the chapters without leaving any choice.
  • Take Breaks after every chapter.
  • Drink plenty of water
  • Enough sleep
  • Set the best time of the day as study time
  • Study Everyday
  • Do not read everything the night before the exam.
  • Try to complete the preparation a few days before the test.
  • Understand the concept and write the revision notes.
  • Revise early in the morning.

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The students of grade 2 can get the chapterwise Bigideas Math 6th Grade Answers on Big Ideas Math Answers.

Big Ideas Math Answers Grade 7 | Big Ideas Math Book 7th Grade Answer Key

Big Ideas Math Answers Grade 7

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Big Ideas Math Book 7th Grade Answer Key | Big Ideas Math Answers 7th Grade Solutions Pdf

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Download Common Core 2019 Curriculum Big Ideas Math Answers Free PDF from Grade K to Grade 12

Big Ideas Math Answers pdf for grade k to grade 12

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Middle School Big Ideas Math Answers

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Big Ideas Math Answers Grade K Chapter 13 Identify Measure and Compare Objects

Big Ideas Math Answers Grade K Chapter 13 Identify Measure and Compare Objects

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Big Ideas Math Book Grade K Answer Key Chapter 13 Measure and Compare Objects

Students can develop an understanding of Identify Measure and Compare Objects from here. Refer Big Ideas Math Book Grade K Answer Key Chapter 13 Measure and Compare Objects to score the maximum marks in the exams. We make solving the problems more fun so that it creates interest in students to become masters in maths. Click on the below attached and link below and get the step by step explanation.

Vocabulary

Lesson: 1 Compare Heights

Lesson: 2 Compare Lengths

Lesson: 3 Use Number to Compare Lengths

Lesson: 4 Compare Weights

Lesson: 5 Use Numbers to Compare Weights

Lesson: 6 Compare Capacities

Lesson: 7 Describe Objects by Attributes

Chapter 13: Measure and Compare Objects

Measure and Compare Objects Vocabulary

Review Words

fewer
more

Answers:
Fewer:
less amount or the small amount which is countable
more:
many or more amount which is rarely countable, In some cases uncountable
From the below diagram umbrellas are 4 which are less in amount
The water pounds are 6 when compared with umbrellas those water pounds are more

Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects v 1

Vocabulary Cards

Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects v 2
Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects v 3

Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects v 4
Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects v 5

Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects v 6
Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects v 7

Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects v 8
Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects v 9

Lesson 13.1 Compare Heights

Explore and Grow

Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects 13.1 1

Directions:
Cut out the Height Sort Cards. Compare the objects to the children shown. Then sort the cards into the categories shown.
Answer:
The children shown for heights.

Explanation:
Compared the heights of the objects and cut the high sort cards and pasted them according to their heights.

Think and Grow

Directions: Compare the heights of the objects.
• Circle the taller slide.
• Draw a line through the shorter lamp.
Are the mugs the same height? Circle the thumbs up for yes or the thumbs down for no.

Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects 13.1 2.1
Answer: circled the taller one.
Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.

step 2: line drawn through the shorter lamp

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.

Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects 13.1 2
Answer: Yes, the mugs are of the same height.

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.

Apply and Grow: Practice

Directions:
1
and 2 Circle the taller object. 3 and 4 Draw a line through the shorter object. 5 Are the lion and the giraffe the same height? Circle the thumbs up for yes or the thumbs down for no.

Question 1.
Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects 13.1 3
Answer: Circled the taller object

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.

Question 2.
Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects 13.1 4
Answer: circled the taller object

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.

Question 3.
Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects 13.1 5
Answer: line drawn through small to big

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.

Question 4.
Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects 13.1 6
Answer: line drawn through smaller one

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.

Question 5.
Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects 13.1 7
Answer: lion and giraffe are not same height.

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.

Think and Grow: Modeling Real Life

Directions:
• Draw a building that is taller than the building shown.
• Draw a building that is shorter than the building shown.

Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects 13.1 8.1
Answer:
Building drawn taller than the building shown

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.

Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects 13.1 8
Answer:
Building drawn shorter than the building shown

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.

Compare Heights Homework & Practice 13.1

Directions:
1 and 2 Circle the taller object. 3 and 4 Draw a line through the shorter object.

Question 1.
Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects 13.1 9
Answer:
Circled the taller object

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.

Question 2.
Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects 13.1 10
Answer:
Circled the taller object.

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.

Question 3.
Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects 13.1 11
Answer:
Drawn the line through the shorter object

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.

Question 4.
Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects 13.1 12
Answer:
Drawn line through the shorter object

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.

Directions:
5
and 6 Are the objects the same height? Circle the thumbs up for yes or the thumbs down for no. 7 Draw a building that is the same height as the building shown.

Question 5.
Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects 13.1 13
Answer:
No they are not same height.

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.

Question 6.
Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects 13.1 14
Answer:
The two rockets are of same height

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.

Question 7.
Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects 13.1 15
Answer:
Building drawn equal to the building shown.

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.

Lesson 13.2 Compare Lengths

Explore and Grow

Directions: Cut out the Length Sort Cards. Compare the objects to the pencil shown. Then sort the cards into the categories shown.

Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects 13.2 1
Answer:
The right giraffe is longer
the left giraffe is shorter

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.

the second picture to represent height and short from the sort out pictures
first girraffe is longer than the second one
second geraffe is shorter than the first one.

Think and Grow

Directions: Compare the lengths of the objects.
• Circle the longer surfboard.
• Draw a line through the shorter watch.
• Are the shoes the same length? Circle the thumbs up for yes or the thumbs down for no.

Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects 13.2 2.1
Answer:
Circle drawn through the longer surfboard.

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.


step 2:
line drawn through the shorter watch.

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.


Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects 13.2 2
Answer:
The shoes are not in same height

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.

Apply and Grow: Practice

Directions:
1 and 2 Circle the longer object. 3 and 4 Draw a line through the shorter object

Question 1.
Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects 13.2 3
Answer:
Circled the longer object

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.

Question 2.
Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects 13.2 4
Answer:
Circled the longer object.

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.

Question 3.
Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects 13.2 5
Answer:
Line drawn through the shorter object

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.

Question 4.
Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects 13.2 6
Answer:
Line drawn through the shorter object

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.

Think and Grow: Modeling Real Life

Directions:
• Draw a string that holds fewer beads than the string shown. Tell how you know.
• Draw a string that holds the same number of beads as the string shown. Tell how you know.

Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects 13.2 7.1
Answer:
string that holds fewer beads than the string shown
the below figure shows the small string

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.

Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects 13.2 7
Answer:
string that holds the same number of beads as the string shown
above figure shows equal strings

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.

Compare Lengths Homework & Practice 13.2

Directions:
1 and 2 Circle the longer object. 3 and 4 Draw a line through the shorter object.

Question 1.
Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects 13.2 8
Answer:
Circled the longer object

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.

Question 2.
Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects 13.2 9
Answer:
Circled the longer object

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.

Question 3.
Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects 13.2 10
Answer:
Line drawn through the shorter object

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.

Question 4.
Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects 13.2 11
Answer:
Line drawn through the shorter object

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.

Directions:
5 and 6 Are the objects the same length? Circle the thumbs up for yes or the thumbs down for no. 7 Draw a string that holds more beads than the string shown. Tell how you know.

Question 5.
Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects 13.2 12
Answer:
The above object has same length

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.

Question 6.
Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects 13.2 13
Answer:
The above objects are not equall

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.

Question 7.
Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects 13.2 14

Answer:
The below figure shows the string that holds more beads

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.

Lesson 13.3 Use Number to Compare Lengths

Explore and Grow

Directions: Build a linking cube train with 4 cubes. Circle the objects that are longer than the cube train.

Big Ideas Math Solutions Grade K Chapter 13 Measure and Compare Objects 13.3 1
Answer:
All the objects are longer.

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
the measurement or extent of something from end to end; the greater of two or the greatest of three dimensions of an object

Think and Grow

Directions: Compare the lengths of the cube trains with the given number of cubes.
• Circle the number of the train that is longer. Color to show how you know.
• Draw a line through the number of the train that is shorter. Color to show how you know.
• Are the cube trains the same length? Circle the thumbs up for yes or the thumbs down for no. Color to show how you know.

Big Ideas Math Solutions Grade K Chapter 13 Measure and Compare Objects 13.3 2
Answer:
The longer number train is circled and colored.

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
the measurement or extent of something from end to end; the greater of two or the greatest of three dimensions of an object

Answer:
Line drawn through the shorter train.

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
the measurement or extent of something from end to end; the greater of two or the greatest of three dimensions of an object

 

Big Ideas Math Solutions Grade K Chapter 13 Measure and Compare Objects 13.3 3
Answer:
Both the trains are equal.

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
the measurement or extent of something from end to end; the greater of two or the greatest of three dimensions of an object

Apply and Grow: Practice

Directions:
Compare the lengths of the cube trains with the given number of cubes. 1 and 2 Circle the number of the train that is longer. Color to show how you know. 3 Draw a line through the number of the train that is shorter. Color to show how you know.

Question 1.
Big Ideas Math Solutions Grade K Chapter 13 Measure and Compare Objects 13.3 4

Answer:
Circled the number of the train that is longer

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
the measurement or extent of something from end to end; the greater of two or the greatest of three dimensions of an object

Question 2.
Big Ideas Math Solutions Grade K Chapter 13 Measure and Compare Objects 13.3 5
Answer:
Circled the train which is longer and colored

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
the measurement or extent of something from end to end; the greater of two or the greatest of three dimensions of an object

Question 3.
Big Ideas Math Solutions Grade K Chapter 13 Measure and Compare Objects 13.3 6
Answer:
Line drawn through the shorter train

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
the measurement or extent of something from end to end; the greater of two or the greatest of three dimensions of an object

Think and Grow: Modeling Real Life

Directions: Each car on a roller coaster holds 2 people.
• Do the roller-coaster trains hold the same number of people? Circle the thumbs up for yes or the thumbs down for no. Tell how you know.
• Circle the roller-coaster train that holds more people. Tell how you know

Big Ideas Math Solutions Grade K Chapter 13 Measure and Compare Objects 13.3 7

Answer:
No, The roller-coaster trains does not  hold the same number of people.
the first roller -coaster contains two seats that means 4 people and the second has 5 seats with 10 people
second has more number of seats and more people

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
the measurement or extent of something from end to end; the greater of two or the greatest of three dimensions of an object

Answer: Circled the roller-coaster train that holds more people. The first roller-coaster has 4 seats with 8 people
and the second roller-coaster has 6 seats with 12 people.

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
the measurement or extent of something from end to end; the greater of two or the greatest of three dimensions of an object

Big Ideas Math Solutions Grade K Chapter 13 Measure and Compare Objects 13.3 8

Use Number to Compare Lengths Homework & Practice 13.3

Directions:
1 and 2 Compare the lengths of the cube trains with the given number of cubes. Circle the number of the train that is longer. Color to show how you know.

Question 1.
Big Ideas Math Solutions Grade K Chapter 13 Measure and Compare Objects 13.3 9
Answer:
Circled the number of the train that is longer

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
the measurement or extent of something from end to end; the greater of two or the greatest of three dimensions of an object

Question 2.
Big Ideas Math Solutions Grade K Chapter 13 Measure and Compare Objects 13.3 10
Answer:
Circled the longer train and colored.

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
the measurement or extent of something from end to end; the greater of two or the greatest of three dimensions of an object

Directions:
3 Compare the lengths of the cube trains with the given number of cubes. Draw a line through the number of the train that is shorter. Color to show how you know. 4 Compare the lengths of the cube trains with the given number of cubes. Are the cube trains the same length? Circle the thumbs up for yes or the thumbs down for no. Color to show how you know. 5 Each car on a roller coaster holds 2 people. Draw a line through the roller-coaster train that holds fewer people. Tell how you know.

Question 3.
Big Ideas Math Solutions Grade K Chapter 13 Measure and Compare Objects 13.3 11
Answer:
Line through the number train that is shorter.

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
the measurement or extent of something from end to end; the greater of two or the greatest of three dimensions of an object

Question 4.
Big Ideas Math Solutions Grade K Chapter 13 Measure and Compare Objects 13.3 12
Answer:
No, The both trains are not equal

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
the measurement or extent of something from end to end; the greater of two or the greatest of three dimensions of an object

Question 5.
Big Ideas Math Solutions Grade K Chapter 13 Measure and Compare Objects 13.3 13
Answer:
Line drawn through the roller-coaster which contains fewer people

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
the measurement or extent of something from end to end; the greater of two or the greatest of three dimensions of an object

Lesson 13.4 Compare Weights

Explore and Grow

Directions: Cut out the Weight Sort Cards. Compare the objects to the lion shown. Then sort the cards into the categories shown.

Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects 13.4 1
Answer:
The elephant is heavier and the mouse is lighter.

Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight

Think and Grow

Directions: Compare the weights of the objects.
• Circle the heavier object.
• Draw a line through the lighter object.
• Are the markers the same weight? Circle the thumbs up for yes or the thumbs down for no.

Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects 13.4 2
Answer:
stapler is heavier than measuring tape

Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight

step 2:
Line through lighter object

Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight

Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects 13.4 3
Answer:
Yes, the markers are same weight.

Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight

Apply and Grow: Practice

Directions:
1 and 2 Circle the heavier object. 3 and 4 Draw a line through the lighter object. 5 Are the objects the same weight? Circle the thumbs up for yes or the thumbs down for no.

Question 1.
Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects 13.4 4
Answer:
crayon box is heavier than one crayon

Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight

Question 2.
Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects 13.4 5
Answer:
circled the heavier object

Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight

Question 3.
Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects 13.4 6
Answer:
Line drawn through the lighter object.

Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight

Question 4.
Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects 13.4 7
Answer:
line drawn through the lighter object.

Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight

Question 5.
Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects 13.4 8
Answer:
Brick is heavier than the tissue box
so, they are not equal

Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight

Think and Grow: Modeling Real Life

Directions:
• Circle the object you can carry. Tell why you can carry the object.
• Circle the object you cannot carry. Tell why you cannot carry the object.

Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects 13.4 9
Answer:
The folder is lighter in weight in can be easily carried

Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight

Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects 13.4 10
Answer:
The bus cannot be carried because it heavier in weight

Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight

Compare Weights Homework & Practice 13.4

Directions:
1 and 2 Circle the heavier object. 3 and 4 Draw a line through the lighter object.

Question 1.
Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects 13.4 11
Answer: The chair is heavier than the glue

Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight

Question 2.
Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects 13.4 12
Answer:
The line drawn through the file

Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight

Question 3.
Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects 13.4 13
Answer:
Line drawn through the lighter object

Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight

Question 4.
Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects 13.4 14
Answer:
Line drawn through the lighter object.

Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight

Directions:
5 and 6 Are the objects the same weight? Circle the thumbs up for yes or the thumbs down for no. 7 Circle the object you can carry. Tell why you can carry the object.

Question 5.
Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects 13.4 15
Answer:
Both the objects are same weight

Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight

Question 6.
Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects 13.4 16
Answer:
Both the objects are not in equal weight

Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight

Question 7.
Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects 13.4 17
Answer:
We can carry the violin because it is lighter in weight

Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight

Lesson 13.5 Use Numbers to Compare Weights

Explore and Grow

Directions: Hold some counting bears in one hand and a different amount of counting bears in your other hand. Place the groups of bears on the correct buckets on the scale.

Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects 13.5 1

Think and Grow

Directions:
• Compare the weights of the groups of linking cubes. Match each group of linking cubes with the correct side of the balance.
• Compare the weights of the groups of linking cubes. Match each group of linking cubes with the correct side of the balance.
• Circle the number of linking cubes that makes the balance scale even.

Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects 13.5 2
Answer:
Each group of linking measures show the correct sides

Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight

Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects 13.5 3
Answer:
Does not show the equal measures

Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight

Apply and Grow: Practice

Directions:
1 – 3 Compare the weights of the groups of linking cubes. Match each group of linking cubes with the correct side of the balance scale.

Question 1.

Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects 13.5 4
Answer:
when weight is more the machine goes down and which has less weight it comes up.

Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight

Question 2.
6Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects 13.5 5
Answer:
The three side should be up as it has less weight and 6 should be down as it has more weight.

Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight

Question 3.
Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects 13.5 6
Answer:
10 should be down as it has the more weight and 3 + 1 =4 should be up as it has less weight.

Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight

Think and Grow: Modeling Real Life

Directions:
• Circle the basket that is lighter. Tell how you know.
• Circle the basket that is heavier. Tell how you know.

Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects 13.5 7
Answer:
First basket is lighter in weight because number of bottles in first basket is less than the second one

Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight

Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects 13.5 8
Answer:
The first one is heavier because the number of bottle in the first basket is more than the second basket.

Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight

Use Numbers to Compare Weights Homework & Practice 13.5

Directions:
1 and 2 Compare the weights of the groups of linking cubes. Match each group of linking cubes with the correct side of the balance scale.

Question 1.
Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects 13.5 9
Answer:
As 5 has more linking cubes than 1, 5 should be down and 1 should be up.

Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight

Question 2.
Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects 13.5 10
Answer:
The weighing measure shows correct measures.

Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight

Directions:
3 Compare the weights of the groups of linking cubes. Match each group of linking cubes with the correct side of the balance scale. 4 Circle the Circle the basket number of linking cubes that makes the balance scale even. 5 Circle the basket that is heavier. Tell how you know.

Question 3.
Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects 13.5 11
Answer:
2 + 2 = 4 is lighter and 5 is heavier,
5 will be down and and 4 will be up

Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight

Question 4.
Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects 13.5 12
Answer: Line drawn through the correct measures

Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight

Question 5.

Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects 13.5 13
Answer:
The first basket is heavier than the second one

Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight

Lesson 13.6 Compare Capacities

Explore and Grow

Directions: Cut out the Capacity Sort Cards. Compare the objects to the bucket shown. Then sort the cards into the categories shown.

Big Ideas Math Solutions Grade K Chapter 13 Measure and Compare Objects 13.6 1

Think and Grow

Directions: Compare the capacities of the objects.
• Circle the object that holds more.
• Draw a line through the object that holds less.
• Do the recycling bins hold the same amount? Circle the thumbs up for yes or the thumbs down for no.

Big Ideas Math Solutions Grade K Chapter 13 Measure and Compare Objects 13.6 2
Answer: The cup contains the more

Explanation:
The maximum amount that something can contain
The maximum is that holds more
and the minimum that holds less

Big Ideas Math Solutions Grade K Chapter 13 Measure and Compare Objects 13.6 3

Answer:
The recycling bins hold the same amount

Explanation:
The maximum amount that something can contain
The maximum is that holds more
and the minimum that holds less

Apply and Grow: Practice

Directions:
1 and 2 Circle the object that holds more. 3 and 4 Draw a line through the object that holds less. 5 Do the milk containers hold the same amount? Circle the thumbs up for yes or the thumbs down for no.

Question 1.
Big Ideas Math Solutions Grade K Chapter 13 Measure and Compare Objects 13.6 4

Answer:
The bag contains the more amount

Explanation:
The maximum amount that something can contain
The maximum is that holds more
and the minimum that holds less

Question 2.
Big Ideas Math Solutions Grade K Chapter 13 Measure and Compare Objects 13.6 5
Answer:
The container contains more amount

Explanation:
The maximum amount that something can contain
The maximum is that holds more
and the minimum that holds less

Question 3.
Big Ideas Math Solutions Grade K Chapter 13 Measure and Compare Objects 13.6 6

Answer:
second figure holds the less

Explanation:
The maximum amount that something can contain
The maximum is that holds more
and the minimum that holds less

Question 4.
Big Ideas Math Solutions Grade K Chapter 13 Measure and Compare Objects 13.6 7
Answer:
line drawn through the object which contain less

Explanation:
The maximum amount that something can contain
The maximum is that holds more
and the minimum that holds less

Question 5.
Big Ideas Math Solutions Grade K Chapter 13 Measure and Compare Objects 13.6 8
Answer:
The milk container does not contain the equal amount

Explanation:
The maximum amount that something can contain
The maximum is that holds more
and the minimum that holds less

Think and Grow: Modeling Real Life

Directions:
• You are going camping. Circle the backpack that can hold all of your camping supplies. Tell how you know.
• You are going to school. Circle the bag that cannot hold all of your school supplies. Tell how you know.

Big Ideas Math Solutions Grade K Chapter 13 Measure and Compare Objects 13.6 9
Answer:
Circled the backpack that holds all of your camping supplies.
The second bag holds all the things.

Explanation:
The maximum amount that something can contain
The maximum is that holds more
and the minimum that holds less

 

Answer:
Circled the bag that cannot hold all of your school supplies.

Explanation:
The maximum amount that something can contain
The maximum is that holds more
and the minimum that holds less

Compare Capacities Homework & Practice 13.6

Directions: 1 and 2 Circle the object that holds more. 3 and 4 Draw a line through the object that holds less.

Question 1.
Big Ideas Math Solutions Grade K Chapter 13 Measure and Compare Objects 13.6 11
Answer:
The second spoon contains the more amount.

Explanation:
The maximum amount that something can contain
The maximum is that holds more
and the minimum that holds less

Question 2.
Big Ideas Math Solutions Grade K Chapter 13 Measure and Compare Objects 13.6 12
Answer:
The Big bucket contains the more amount

Explanation:
The maximum amount that something can contain
The maximum is that holds more
and the minimum that holds less

Question 3.
Big Ideas Math Solutions Grade K Chapter 13 Measure and Compare Objects 13.6 13
Answer:
Line Drawn through the shorter object

Explanation:
The maximum amount that something can contain
The maximum is that holds more
and the minimum that holds less

Question 4.
Big Ideas Math Solutions Grade K Chapter 13 Measure and Compare Objects 13.6 14
Answer:
Line drawn through the shorter object

Explanation:
The maximum amount that something can contain
The maximum is that holds more
and the minimum that holds less

Directions:
5 and 6 Do the objects hold the same amount? Circle the thumbs up for yes or the thumbs down for no. 7 Your class is going on a field trip. Circle the vehicle that can hold your class. Tell how you know.

Question 5.
Big Ideas Math Solutions Grade K Chapter 13 Measure and Compare Objects 13.6 15
Answer:
Both the basket contains the same amount

Explanation:
The maximum amount that something can contain
The maximum is that holds more
and the minimum that holds less

Question 6.
Big Ideas Math Solutions Grade K Chapter 13 Measure and Compare Objects 13.6 16
Answer:
The objects does not contain the same amount

Explanation:
The maximum amount that something can contain
The maximum is that holds more
and the minimum that holds less

Question 7.
Big Ideas Math Solutions Grade K Chapter 13 Measure and Compare Objects 13.6 17
Answer:
The bus can hold all the children

Explanation:
The maximum amount that something can contain
The maximum is that holds more
and the minimum that holds less

Lesson 13.7 Describe Objects by Attributes

Explore and Grow

Directions: Cut out the Measurable Attribute Sort Cards. Place the objects that you can measure using length or height into the length or height box. Then place the objects that you can measure using weight into the weight box. Then place the objects that you can measure using capacity into the capacity box.

Big Ideas Math Solutions Grade K Chapter 13 Measure and Compare Objects 13.7 1

Think and Grow

Directions: Circle the measurable attributes of the object

Big Ideas Math Solutions Grade K Chapter 13 Measure and Compare Objects 13.7 2
Answer:
cubes are the measuring attribute

Explanation:
The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers

 

Big Ideas Math Solutions Grade K Chapter 13 Measure and Compare Objects 13.7 3
Answer:
weighing machine is the measuring attribute

The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers

Big Ideas Math Solutions Grade K Chapter 13 Measure and Compare Objects 13.7 4
Answer:
Capacity is the measuring attribute

The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers

Apply and Grow: Practice

Directions:
1– 4 Circle the measurable attributes of the object. 5 Circle the objects that have capacity as an attribute.

Question 1.
Big Ideas Math Solutions Grade K Chapter 13 Measure and Compare Objects 13.7 5
Answer:
cubes are the measuring attribute

The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers

Question 2.
Big Ideas Math Solutions Grade K Chapter 13 Measure and Compare Objects 13.7 6
Answer:
weighing is the measuring attribute

The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers

Question 3.
Big Ideas Math Solutions Grade K Chapter 13 Measure and Compare Objects 13.7 7
Answer:
cubes are the measuring attribute

The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers

Question 4.
Big Ideas Math Solutions Grade K Chapter 13 Measure and Compare Objects 13.7 8
Answer:
cubes are the measuring attribute

The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers

Question 5.
Big Ideas Math Solutions Grade K Chapter 13 Measure and Compare Objects 13.7 9
Answer:
The glue is the capacity attribute

The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers

Think and Grow: Modeling Real Life

Directions:
• Draw an object that has capacity as an attribute.
• Draw an object that does not have capacity as an attribute.

Big Ideas Math Solutions Grade K Chapter 13 Measure and Compare Objects 13.7 10

Answer:
Jar is the capacity object
pencil is not the capacity object

Explanation:
The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers

Describe Objects by Attributes Homework & Practice 13.7

Directions:
1 – 3 Circle the measurable attributes of the object.

Question 1.
Big Ideas Math Solutions Grade K Chapter 13 Measure and Compare Objects 13.7 11
Answer:
Cubes are the measuring object

Explanation:
The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers

Question 2.
Big Ideas Math Solutions Grade K Chapter 13 Measure and Compare Objects 13.7 12
Answer:
cubes are the measuring object

Explanation:
The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers

Question 3.
Big Ideas Math Solutions Grade K Chapter 13 Measure and Compare Objects 13.7 13
Answer:
cube is the measuring object

Explanation:
The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers

Directions:
4 Circle the measurable attributes of the broccoli. 5 Circle the objects that have length as an attribute. 6 Draw an object that has length as an attribute. 7 Draw an object that has weight as an attribute.

Question 4.
Big Ideas Math Solutions Grade K Chapter 13 Measure and Compare Objects 13.7 14
Answer:
Weighing is the measuring attribute

Explanation:
The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers

Question 5.
Big Ideas Math Solutions Grade K Chapter 13 Measure and Compare Objects 13.7 15
Answer:
fork, tree, toy helicopter are length attributes

The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers

Question 6.
Big Ideas Math Solutions Grade K Chapter 13 Measure and Compare Objects 13.7 16
Answer:
fork is the length attribute

The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers

Question 7.
Big Ideas Math Solutions Grade K Chapter 13 Measure and Compare Objects 13.7 17
Answer:

The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers

water bottle is the capacity attribute

Measure and Compare Objects Performance Task

Directions: You use one bucket to collect rainwater on Monday and a different bucket to collect rainwater on Tuesday. On Monday, you collect 1 less than 7 fluid ounces of rainwater. On Tuesday, you collect 1 more than 3 fluid ounces of rainwater. 1 Circle the number on each bucket that shows the amount of rainwater you collect. Then circle the day that you collect more rainwater. 2 Draw a bucket for Wednesday that is taller and holds more water than Monday’s bucket. 3 The amount of rainwater you collect on Wednesday is the same as the amount you collect in all on Monday and Tuesday. Write an addition sentence to tell how much rainwater you collect on Wednesday.

Question 1.
Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects 1
Answer:

Question 2.
Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects 2

Answer: the diagram is shown below.

Question 3.
Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects 3
Answer:
1 less than 7 fluid ounces of rainwater = 6
1 more than 3 fluid ounces of rainwater = 4

The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers

Measure and Compare Objects Activity

Measurement Boss

Directions: Each player flips a Measurement Boss Card and places it on the page. Compare the objects based on the attribute of the card. The player with the object that is longer, taller, heavier, or holds more takes both cards. Repeat until all cards have been used.

Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects 4

Measure and Compare Objects Chapter Practice

Directions:
1 and 2 Circle the taller object. 3 Draw a line through the shorter object. 4 Are the crayons the same length? Circle the thumbs up for yes or the thumbs down for no. 5 Draw a string that holds the same number of beads as the string shown. Tell how you know.

13.1 Compare Heights

Question 1.
Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects chp 1

Answer:
The door is taller than the chair

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.

Question 2.
Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects chp 2
Answer:
The dog is taller than the cat

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.

13.2 Compare Lengths

Question 3.
Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects chp 3
Answer:
Line drawn through the shorter object

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.

Question 4.
Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects chp 4
Answer: second crayon is shorter.

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.

Question 5.
Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects chp 5
Answer:
string drawn with equal length.

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.

Directions:
6 and 7 Compare the lengths of the cube trains that have the given number of cubes. Circle the number of the train that is longer. Color to show how you know. 8 Draw a line through the lighter object. Are the footballs the same weight? 9 Circle the thumbs up for yes or the thumbs down for no.

13.3 Use Numbers to Compare Lengths

Question 6.
Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects chp 6
Answer:
7 is the bigger train than the 5

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
the measurement or extent of something from end to end; the greater of two or the greatest of three dimensions of an object

 

Question 7.
Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects chp 7
Answer:
10 train is lengthier than the 9

Explanation:
Great or more than average height,
especially (with reference to an object) relative to width and height
measuring a specified distance from top to bottom.
the measurement or extent of something from end to end; the greater of two or the greatest of three dimensions of an object

13.4 Compare Weights

Question 8.
Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects chp 8
Answer:
Line drawn through the lighter object

Question 9.
Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects chp 9
Answer: Yes, the foot balls are same weight

Directions:
10 Compare the weights of the groups of linking cubes. Match each group of linking cubes with the correct side of the balance scale. 11 Circle the bag that is lighter. Tell how you know. 12 and 13 Draw a line through the object that holds less.

13.5 Use Numbers to Compare Weights

Question 10.
Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects chp 10
Answer: 4 comes down and 2 goes up

Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight
.

Question 11.
Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects chp 11
Answer: The bag that contains 3 balls are lighter.

Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight

13.6 Compare Capacities

Question 12.
Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects chp 12
Answer: Line drawn through the lighter object.

Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight

Question 13.
Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects chp 13
Answer: Line drawn through the lighter object.

Explanation:
heavier of great weight; difficult to lift or move
of great density; thick or substantial
lighter than the expected item or which weighs less weight

Directions:
14 Do the water bottles hold the same amount? Circle the thumbs up for yes or the thumbs down for no. 15 – 17 Circle the measurable attributes of the object. 18 Circle the objects that have capacity as an attribute.

Question 14.
Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects chp 14
Answer: The water bottles does not hold the same amount of water.

13.7 Describe Objects by Attributes

Question 15.
Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects chp 15
Answer:
Cubes are the measuring attribute of length.

Explanation:
The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers

Question 16.
Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects chp 16
Answer:
capacity is the measuring attribute.

Explanation:
The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers

Question 17.
Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects chp 17
Answer:
It can be measured with cubes length attribute.

Explanation:
The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers

Question 18.
Big Ideas Math Answers Grade K Chapter 13 Measure and Compare Objects chp 18
Answer:
Circled the measuring attribute.

Explantion:
The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers

Measure and Compare Objects Cumulative Practice

Directions:
Shade the circle next to the answer. 1 Which group has a yellow pencil that is longer than the red pencil? 2 Which five frame shows how many sharks are in the picture? 3 Which group has all rectangles?
Question 1.
Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects cp 1
Answer:
Second picture has yellow longer.

Explanation:
The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers

Question 2.
Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects cp 2
Answer:
There are 5 sharks in the picture.

Explanation:
The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers

Question 3.
Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects cp 3
Answer:
The third image shows all rectangles

Explantion:
The attributes we measure come in two main types, quantitative and categorical. Quantitative attributes are attributes we measure using numbers. For example, your height is a quantitative attribute. On the other hand, a categorical attribute is something measured without using numbers

Directions:
Shade the circle next to the answer. 4 Which subtraction sentence not tells how many geese are left? 5 Which shape is a solid shape? 6 Which solid shape does not stack or slide?

Question 4.
Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects cp 4
Answer:
the first substraction sentence shows

Explanation:
There are 8 birds 3 are flying and 5 are sitting
and the picture shows 8 – 3 = 5

Question 5.
Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects cp 5
Answer:
The pyramid shows the solid shape

Explanation:
Solid Shapes. Objects that occupy space are called solid shapes. Their surfaces are called faces. Faces meet at edges and edges meet at vertices. Some examples of solid shapes: Cone, Cuboid, Sphere, Cylinder, Cube

Question 6.
Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects cp 6
Answer:
the box will not slide or stack

Explanation:
Solid Shapes. Objects that occupy space are called solid shapes. Their surfaces are called faces. Faces meet at edges and edges meet at vertices. Some examples of solid shapes: Cone, Cuboid, Sphere, Cylinder, Cube

Directions:
7 Circle the objects that have capacity as an attribute. 8 Circle the object that looks like a cylinder that is above the ball. 9 Find the number of dots on each domino. Write each number. Draw a line through the number that is less than the other number.

Question 7.
Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects cp 7
Answer: The aquarium is capacity attribute.

Explanation:
Capacity Attributes means any and all current or future defined characteristics, certificates, tag, credits, ancillary service attributes, or accounting construct

Question 8.
Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects cp 8
Answer:
Nothing shows like a cylinder above the ball.

Question 9.
Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects cp 9
Answer:
Line drawn through the less number.

Explanation:
10 + 5 = 15
8 + 10 = 18
line drawn through the dice

Directions:
10 Trace the shapes that are hexagons. Write the number of sides and the number of vertices of a hexagon. 11 You have 10 apples. Classify the apples into 2 categories. Circle the groups. Then complete the number bond to match your picture. 12 Draw a larger triangle that can be formed by the 2 triangles shown.

Question 10.
Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects cp 10
Answer:
there are two hexagons of 6 sides and 6 vertices.

Explanation:
a plane figure with six straight sides and angles

Question 11.

Answer:
The apples are made into 2 groups of 5 each.

Explanation:
the total ten apples are made in to 2 groups 5 in each group
3 are red and big 2 are yellow and small
3 are red and small  and 2 are yellow and small.

Question 12.
Big Ideas Math Answer Key Grade K Chapter 13 Measure and Compare Objects cp 12
Answer:
Drawn a big triangle with the help of two triangles

Explanation:
A triangle has 3 sides and 3 vertices
with equal length

Final Words:

Hope the solutions given on this page are helpful for you guys. Make your children ready by making them practice for the test using Big Ideas Math Kth Grade Solution Key Ch 13 Measure and Compare Objects. Get the solutions for each and every question with the simple tricks from Big Ideas Math Answer Grade K. Keep in touch with us to get the answers for all Big Ideas Math Grade K Chapters.

Big Ideas Math Answers Grade K Chapter 4 Compare Numbers to 10

Big Ideas Math Answers Grade K Chapter 4 Compare Numbers to 10

Download pdf link of Big Ideas Math Answers Grade K Chapter 4 Compare Numbers to 10 is available here. So, the students of elementary school can Download Big Ideas Math Book Grade K Solution Key Chapter 4 Compare Numbers to 10 for free of cost. This Compare Numbers to 10 chapter includes the topics like Vocabulary, Compare Groups to 10 by Matching, Compare Groups to 10 by Counting, Classify Objects into Categories, Classify and Compare by Counting, and so on. Big Ideas Math Answers Grade K Chapter 4 Compare Numbers to 10 helps you to secure the highest marks in the exams and also to improve your math skills.

Big Ideas Math Book Grade K Answer Key Chapter 4 Compare Numbers to 10

Check out the links given in the below section before you start practicing the problems. You can perform well in practice tests, Assessment tests, and chapter tests by referring to our Bigideas Math Book Grade K Answer Key Chapter 4 Compare Numbers to 10. Also, it helps you to complete your homework without any delay by using the BIM Grade K Answers for Chapter 4 Compare Numbers to 10.

Vocabulary

Lesson: 1 Compare Groups to 10 by Matching

Lesson: 2 Compare Groups to 10 by Counting

Lesson: 3 Compare Numbers to 10

Lesson: 4 Classify Objects into Categories

Lesson: 5 Classify and Compare by Counting

Chapter: 4 – Compare Numbers to 10

Compare Numbers to 10 Vocabulary

Big Ideas Math Answers Grade K Chapter 4 Compare Numbers to 10 1
Big Ideas Math Answers Grade K Chapter 4 Compare Numbers to 10 2

Directions:
Count the number of dog toys in front of each dog. Write each number. Circle the number that is greater than the other number.
Answer:
The Number of Dog Toys are greater than the number Dogs.

Explanation:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Compare-Numbers-to-10-Compare Numbers to 10 Vocabulary

Number of Dogs = 2
Number of Dog Toys = 5

Chapter 4 Vocabulary Cards

Big Ideas Math Answers Grade K Chapter 4 Compare Numbers to 10 3
Big Ideas Math Answers Grade K Chapter 4 Compare Numbers to 10 4

Lesson 4.1 Compare Groups to 10 by Matching

Explore and Grow
Big Ideas Math Answer Key Grade K Chapter 4 Compare Numbers to 10 5

Directions:
Use cubes to show the number of letters. Use cubes to show the number of mailboxes. Tell which group has more and which group has less.

Answer:
The count of Letters is greater than the count of mailboxes.

Explanation:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Compare-Numbers-to-10-Lesson-4.1-Compare-Groups-to-10-by-Matching1

Think and Grow

Big Ideas Math Answer Key Grade K Chapter 4 Compare Numbers to 10 6
Big Ideas Math Answer Key Grade K Chapter 4 Compare Numbers to 10 6.1

Answer:
The number of fire extensions is greater than the count of caps, cameras, and minks.

Explanation:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Compare-Numbers-to-10-Lesson-4.1-Think and Grow
a-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Compare-Numbers-to-10-Lesson-4.1-Think and Grow.

The number of caps = Seven or 7.
The number of fire extension = Nine or 9.
A number of cameras = Five or 5.
The number of mink = Six or 6.
Directions:
Draw lines between the objects in each group.

  • Is the number of toothbrushes equal to the number of toothpaste tubes? Circle the thumbs up for yes or the thumbs down for no.
  • Circle the group that is greater in number than the other group.
  • Draw a line through the group that is less in number than the other group.
    Directions:
    Draw lines between the objects in each group.
    1 Is the number of paint palettes equal to the number of paintbrushes? Circle the thumbs up for yes or the thumbs down for no. 2 Circle the group that is greater in number than the other group. 3 Draw a line through the group that is less in number than the other group.

Apply and Grow: Practice

Question 1.
Big Ideas Math Answer Key Grade K Chapter 4 Compare Numbers to 10 7

Answer:
The number of paintbrushes is greater than the number of paint palettes.

Explanation:

Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Compare-Numbers-to-10-Lesson-4.1-Apply-and-Grow-Practice1
Number of paintbrushes = Seven or 7
The number of paint palettes = Six or 6.

Question 2.
Big Ideas Math Answer Key Grade K Chapter 4 Compare Numbers to 10 8

Answer:
The number of wooden spoons is greater than the number of chef cap.

Explanation:

Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Compare-Numbers-to-10-Lesson-4.1-Apply-and-Grow-Practice2

The number of wooden spoons = Ten or 10.
The number of chef cap = Seven or 7.

Question 3.
Big Ideas Math Answer Key Grade K Chapter 4 Compare Numbers to 10 9

Answer:
The number of thermometers is greater than the number of nurse caps.

Explanation:

Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Compare-Numbers-to-10-Lesson-4.1-Apply-and-Grow-Practice3

The number of thermometers = Eight or 8.
The number of nurse caps = Four or 4.

Directions:

  • Draw nails on the board so that the number of nails is greater than the number of hammers. Draw lines between the objects in each group to show that you are correct.
  • Draw nails on the board so that the number of nails is less than the number of hammers. Draw lines between the objects in each group to show that you are correct.

Think and Grow: Modeling Real Life

Big Ideas Math Answer Key Grade K Chapter 4 Compare Numbers to 10 10

Answer:
a. The number of nails on the board are greater than the number of hampers.

b. The number of hampers are greater than the number of nails on the board.

Explanation:
a.
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Compare-Numbers-to-10-Lesson-4.1-Think-and-Grow-Modeling-Real-Life
The number of nails = Seventeen or 17.
The number of hampers = Fifteen or 15.
b.
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Compare-Numbers-to-10-Lesson-4.1-Think-and-Grow-Modeling-Real-Life

A number of nails = Twelve or 12.
Number of hampers = Fifteen or 15.

Compare Groups to 10 by Matching Homework & Practice 4.1

Question 1.
Big Ideas Math Answer Key Grade K Chapter 4 Compare Numbers to 10 11
Answer:
The Number of Nurse caps are greater than the number of chakras.

Explanation:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Compare-Numbers-to-10-Lesson-4.1-Compare-Groups-to-10-by-Matching-Homework-&-Practice 4.1-1.

The number of nurse caps = Eight or 8.
The number of wooden chakras = Six or 6.

Directions:
1 Draw lines between the objects in each group. Circle the group that is greater in number than the other group.
2 Draw lines between the objects in each group. Draw a line through the group that is less in number than the other group. 3 Draw wands on the cloth so that the number of wands is equal to the number of magician hats. Draw lines between the objects in each group to show that you are correct.

Question 2.
Big Ideas Math Answer Key Grade K Chapter 4 Compare Numbers to 10 12

Answer:
The number of Screw drives are greater than the Number of Caps.

Explanation:
Compare Groups to 10 by Matching Homework & Practice 4.1

Number of Caps = Seven or 7.
The number of Screw drives = Nine or 9.

Question 3.
Big Ideas Math Answer Key Grade K Chapter 4 Compare Numbers to 10 13

Answer:
The number of magical hats is equal to the number of wands.

Explanation:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Compare-Numbers-to-10-Lesson-4.1-Compare-Groups-to-10-by-Matching-Homework-&-Practice 4.1-3
The number of magical hats = Six or 6.
The number of wands = Six or 6.

Lesson 4.2 Compare Groups to 10 by Counting

Explore and Grow

Big Ideas Math Answer Key Grade K Chapter 4 Compare Numbers to 10 14

Directions:
Use cubes to show the number of penguins. Use cubes to show the number of eggs. Tell which group has more and which group has less.

Answer:
The number of Penguins is comparatively more than the number of Eggs.

Explanation:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Compare-Numbers-to-10-Lesson-Lesson-4.2-Compare-Groups-to-10-by-Counting-EXPLORE

Number of Penguins = Eight or 8.
Number of Eggs = Six or 6.

Directions:
Count the objects in each group. Write each number.

  • Is the number of green dots equal to the number of purple dots? Circle the thumbs up for yes or the thumbs down for no.
  • Compare the numbers of blackberries and bears. Circle the number that is greater than the other number.
  • Compare the numbers of bears and fish. Draw a line through the number that is less than the other number.

Think and Grow

Big Ideas Math Answer Key Grade K Chapter 4 Compare Numbers to 10 15

Answer:
a. No, the number of Green dots is not equal to the number of Purple dots.
b. The number of Blueberries is greater than the number of Bears.
c. The number of Bears is lesser than the number of Fishes.

Explanation:
a. Number of Green dots = 50
Number of Purple dots = 200

b.
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Compare-Numbers-to-10-Lesson-4.2-Think and Grow.

Number of Blueberries = Ten or 10.
Number of Bears = Four or 4.

c.
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Compare-Numbers-to-10-Lesson-4.2-Think and Grow1.

Number of Bears = Four or 4.
Number of Fishes = Seven or 7.

Directions:
Count the objects in each group. Write each number. 1 Is the number of panda bears equal to the number of bamboo sticks? Circle the thumbs up for yes or the thumbs down for no. 2 Circle the number that is greater than the other number. 3 Draw a line through the number that is less than the other number.

Apply and Grow: Practice

Question 1.
Big Ideas Math Answer Key Grade K Chapter 4 Compare Numbers to 10 16

Answer:
The number of Bears is equal to the number of Bamboo Sticks.

Explanation:

Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Compare-Numbers-to-10-Lesson-4.2-Apply and Grow-Practice

Number of Bears = Six or 6.
The number of Bamboo Sticks = Six or 6.

Question 2.
Big Ideas Math Answer Key Grade K Chapter 4 Compare Numbers to 10 17

Answer:
The number of Leaves is greater than the number of Koala Bears.

Explanation:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Compare-Numbers-to-10-Lesson-4.2-Apply and Grow-Practice2
The number of Leaves = Ten or 10.
The number of Koala Bear = Seven or 7.

Question 3.
Big Ideas Math Answer Key Grade K Chapter 4 Compare Numbers to 10 18

Answer:
The number of Leaves are greater than the number of Kangaroos.

Explanation:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Compare-Numbers-to-10-Lesson-4.2-Apply and Grow-Practice3
Number of Kangaroos = Eight or 8.
Number of Leaves = Nine or 9.

Directions:

  • Draw bananas hanging from the tree so that the number of bananas is greater than the number of monkeys. Write the number of each object. Circle the number that is greater than the other number.
  • Draw bananas hanging from the tree so that the number of bananas is less than the number of monkeys. Write the number of each object. Draw a line through the number that is less than the other number.


Think and Grow: Modeling Real Life

Big Ideas Math Answer Key Grade K Chapter 4 Compare Numbers to 10 19

Answer:
a. The number of Bananas are greater than the number of Monkeys.
b. The number of Bananas are lesser than the number of Monkeys.

Explanation:
a. Number of Bananas = Eight or 8.
Number of Monkeys = Seven or 7
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Compare-Numbers-to-10-Compare Numbers to 10 Vocabulary

b. Number of Bananas = Six or 6.
Number of Monkeys = Nine or 9.

Compare Groups to 10 by Counting Homework & Practice 4.2

Question 1.
Big Ideas Math Answer Key Grade K Chapter 4 Compare Numbers to 10 20

 

Answer:
The number of Rabbits are Equal to the number of Carrots.

Explanation:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Compare Groups to 10 by Counting Homework & Practice 4.2.1

Number of Rabbits = Eight or 8.
Number of Carrots = Eight or 8.

Question 2.
Big Ideas Math Answer Key Grade K Chapter 4 Compare Numbers to 10 21

Answer:
The number of  Trees are greater than the number of Giraffes.
Explanation:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Compare Groups to 10 by Counting Homework & Practice 4.2.2

Number of Giraffes = Six or 6.
Number of Trees = Nine or 9.

Directions:
1 Count the objects in each group. Write each number. Is the number of yellow counters equal to the number of red counters? Circle the thumbs up for yes or the thumbs down for no.
2 Count the objects in each group. Write each number. Circle the number that is greater than the other number. 3 Draw hay bales in the pen so that the number of hay bales is less than the number of donkeys. Write the number of each object. Draw a line through the number that is less than the other number.

Question 3.
Big Ideas Math Answer Key Grade K Chapter 4 Compare Numbers to 10 22
Answer:
The Number of Donkeys are greater than the number of hay bales.

Explanation:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Compare Groups to 10 by Counting Homework & Practice 4.2.3

Number of Donkeys = 8
Number of Hay Bales = 5

Lesson 4.3 Compare Numbers to 10

Explore and Grow

Big Ideas Math Answer Key Grade K Chapter 4 Compare Numbers to 10 23
Answer:
The Number of Yellow boxes in the cubes are greater than the number of Blue boxes in the cubes.

Explanation:

Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Lesson 4.3

Number of Blue boxes in the cubes = Six or 6
Number of Yellow boxes in the cubes = Nine or 9

Directions:
Which number is greater than the other number? Which number is less than the other number? Use counters to show how you know.

Think and Grow

Big Ideas Math Answer Key Grade K Chapter 4 Compare Numbers to 10 24
Answer:
The number of Yellow Stars above the line are greater than the number of  Gray Stars below the line.

Explanation:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Lesson 4.3-Think and Grow

Number of Yellow Stars above the line = Ten or 10.
The number of Gray Stars below the line = Seven or 7.

Big Ideas Math Answer Key Grade K Chapter 4 Compare Numbers to 10 25
Answer:
The number of Smiles below the line are greater than the number of Smiles above are line.

Explanation:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Lesson 4.3-Think and Grow..
Number of Smiles above are line = One or 1.
Number of Smiles below the line = Six or 6.

Directions:
Compare the numbers.

  • Are the numbers equal? Circle the thumbs up for yes or the thumbs down for no. Draw to show how you know.
  • Circle the number that is greater than the other number. Draw to show how you know.
  • Draw a line through the number that is less than the other number. Draw to show how you know.

Apply and Grow: Practice

Question 1.
Big Ideas Math Answer Key Grade K Chapter 4 Compare Numbers to 10 26
Answer:
The number of Blue Plus symbols above the line are greater than the number of Green Plus symbols above the line.

Explanation:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Lesson 4.3-Apply-and-Grow-Practice-1
Number of Green Plus symbols above the line = Five or 5
The number of Blue Plus symbols above the line = Nine or 9.

Question 2.
Big Ideas Math Answer Key Grade K Chapter 4 Compare Numbers to 10 27
Answer:
The number of Yellow Moon above the line are greater than the number of Blue Moon.

Explanation:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Lesson 4.3-Apply-and-Grow-Practice-1.

Number of Yellow Moon above the line = Eight or 8
The number of Blue Moon below the line = Six or 6.

Question 3.

Big Ideas Math Answer Key Grade K Chapter 4 Compare Numbers to 10 28
Answer:
The number of Blue arrows above the line are greater than the number of Black arrows below the line.

Explanation:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Lesson 4.3-Apply-and-Grow-Practice-2.

The number of Blue arrows above the line = Nine or 9.
The number of Black arrows below the line = Eight or 8.

Directions:
Compare the numbers. 1 Are the numbers equal? Circle the thumbs up for yes or the thumbs down for no. Draw to show how you know. 2 Circle the number that is greater than the other number. Draw to show how you know. 3 Draw a line through the number that is less than the other number. Draw to show how you know.

Think and Grow: Modeling Real Life

Big Ideas Math Answer Key Grade K Chapter 4 Compare Numbers to 10 29

Big Ideas Math Answer Key Grade K Chapter 4 Compare Numbers to 10 30
Answer:
a. The number of Flowers I have are greater than the number of Flowers my Friend has.
b. The number of Flowers my friend has are lesser than the number of Flowers I have.

Explanation:
a. Number of Flowers I have = Seven or 7.
Number of Flowers my Friend has = Six or 6.
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Lesson 4.3-Think and Grow-Modeling Real Life
b. Number of Flowers I have = Ten or 10.
Number of Flowers my Friend has = Nine or 9.
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Lesson 4.3-Think and Grow-Modeling Real Life.

Directions

  • You have a number of flowers that is greater than 6 and less than 8. Your friend has a number of flowers that is one more than 5. Write and draw how many flowers you each have. Circle the number that is greater than the other number.
  • You have a number of flowers that is one more than 9. Your friend has a number of flowers that is one less than 10. Write and draw how many flowers you each have. Draw a line through the number that is less than the other number.

Compare Numbers to 10 Homework & Practice 4.3

Question 1.
Big Ideas Math Answer Key Grade K Chapter 4 Compare Numbers to 10 31

Answer:
The number of clouds above the line are greater than the number of clouds below the line.

Explanation:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Lesson 4.3-Compare Numbers to 10 Homework & Practice 4.3.1
Number of clouds above the line = 10
Number of clouds below the line = 5

Question 2.
Big Ideas Math Answers Grade K Chapter 4 Compare Numbers to 10 32
Answer:
The number of Triangles above the line are greater than the number of Triangles below the line.

Explanation:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Lesson 4.3-Compare Numbers to 10 Homework & Practice 4.3.2
Number of Triangles above the line = Nine or 9.
Number of Triangles below the line = Seven or 7.

Question 3.
Big Ideas Math Answers Grade K Chapter 4 Compare Numbers to 10 33
Answer:
Number of Marbles I have are greater than the number of Marbles my Friend has.

Explanation:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Lesson 4.3-Compare Numbers to 10 Homework & Practice 4.3.3
Number of Marbles I have = Six or 6.
Number of Marbles my Friend has = Five or 5.


Directions:
1
Compare the numbers. Draw a line through the number that is less than the other number. Draw to show how you know.
2 Compare the numbers. Are the numbers equal? Circle the thumbs up for yes or the thumbs down for no. Draw to show how you know. 3 You have a number of marbles that is greater than 5 and less than 7. Your friend has a number of marbles that is one more than 4. Write and draw how many marbles you each have. Circle the number that is greater than the other number.

Lesson 4.4 Classify Objects into Categories

Explore and Grow

Big Ideas Math Answer Key Grade K Chapter 4 Compare Numbers to 10 34

Directions:
Place a yellow counter on each tree that is yellow. Place a red counter on each tree that is not yellow.

Answer:
The number of Red counters on the Trees are greater than Number of Yellow counter on the Trees

Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Lesson 4.4-Classify-Objects-into-Categories-
Answer:
Number of Yellow counter on the Trees = Four or 4.
Number of Red counters on the Trees = Eight or 8.

Directions:

  • Circle the objects that are green. Cross out the objects that are not green.
  • Circle the animals that are brown. Cross out the animals that are not brown.
  • Circle the animals that have wings. Cross out the animals that do not have wings.

Think and Grow

Big Ideas Math Answers Grade K Chapter 4 Compare Numbers to 10 35

Big Ideas Math Answers Grade K Chapter 4 Compare Numbers to 10 36

Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Lesson 4.4-Think and Grow

Directions:
1 Circle the pigs that have spots. Cross out the pigs that do not have spots.
2 Circle the blocks that have letters. Cross out the blocks that do have letters.
3 Circle the animals that have tails. Cross out the animals that do not have tails.

Apply and Grow: Practice

Question 1.
Big Ideas Math Answer Key Grade K Chapter 4 Compare Numbers to 10 37

Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Lesson 4.4-Apply-and-Grow-Practice1

Question 2.
Big Ideas Math Answers Grade K Chapter 4 Compare Numbers to 10 38
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Lesson 4.4-Think and Grow

Question 3.
Big Ideas Math Answer Key Grade K Chapter 4 Compare Numbers to 10 39
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Lesson 4.4-Think and Grow

 

Directions:

  • Circle the dogs that are black. Cross out the dogs that are not black.
  • Classify the dogs another way. Circle the dogs that belong in the new category. Cross out the dogs that do not belong in the new category. Tell how you classified the dogs.

Think and Grow: Modeling Real Life

Big Ideas Math Answers Grade K Chapter 4 Compare Numbers to 10 40
Big Ideas Math Answers Grade K Chapter 4 Compare Numbers to 10 40.1

Answer:
a.
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Lesson 4.4-Think-and-Grow-Modeling-Real-Life

b.
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Lesson 4.4-Think-and-Grow-Modeling-Real-Life..

The New type of Dogs appearance is completely different from normal dogs like different nose, eyes, tail and many more.

Classify Objects into Categories Homework & Practice 4.4

Question 1.
Big Ideas Math Answers Grade K Chapter 4 Compare Numbers to 10 41
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Classify Objects into Categories Homework & Practice 4.4.1

Question 2.
Big Ideas Math Answers Grade K Chapter 4 Compare Numbers to 10 42

Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Classify Objects into Categories Homework & Practice 4.4.2

Directions:
1 Circle the cars that are blue. Cross out the cars that are not red. 2 Circle the toys that are balls. Cross out the toys that are not balls.
3 Circle the animals that have stripes. Cross out the animals that do not have stripes. 4 Circle the animals that have 2 legs. Cross out the animals that do not have 2 legs. 5 Classify the books into 2 categories. Circle the books in one category. Cross out the books in the other category. Tell how you classied the books.

Question 3.
Big Ideas Math Answers Grade K Chapter 4 Compare Numbers to 10 43

Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Classify Objects into Categories Homework & Practice 4.4.3

Question 4.
Big Ideas Math Answers Grade K Chapter 4 Compare Numbers to 10 44

Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Classify Objects into Categories Homework & Practice 4.4.4

Question 5.
Big Ideas Math Answers Grade K Chapter 4 Compare Numbers to 10 45

Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Classify Objects into Categories Homework & Practice 4.4.5
The two types of Books  are:
1. Closed book
2. Open Book

Lesson 4.5 Classify and Compare by Counting

Explore and Grow

Big Ideas Math Answers Grade K Chapter 4 Compare Numbers to 10 46
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Lesson 4.5-Classify and Compare by Counting-Explore and Grow

Directions:
Classify the animals into the categories shown. Write the marks in the chart. Count the marks and write the numbers to tell how many animals are in each category. Circle the number that is greater than the other number.

Think and Grow

Big Ideas Math Answers Grade K Chapter 4 Compare Numbers to 10 47

Big Ideas Math Answers Grade K Chapter 4 Compare Numbers to 10 48
Answer:

Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Lesson 4.5-Classify-and-Compare by Counting-Think and Grow...

Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Lesson 4.5-Classify-and-Compare-by-Counting-Think-and-Grow...

Directions:
1 and 2 Classify the animals into the categories shown. Write the marks in the chart. Count the marks and write the numbers to tell how many animals are in each category. Draw a line through the number that is less than the other number

Apply and Grow: Practice

Big Ideas Math Answers Grade K Chapter 4 Compare Numbers to 10 49

Question 1.
Big Ideas Math Answers Grade K Chapter 4 Compare Numbers to 10 50

Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Lesson 4.5-Apply-and-Grow-Practice1

Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Lesson-4.5-Apply-and-Grow-Practice1
Number of Four legs Animals = Five or 5.
Number of  not Four legs Animals = Eight or 8.

Question 2.
Big Ideas Math Answers Grade K Chapter 4 Compare Numbers to 10 51
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Lesson-4.5-Apply-and-Grow-Practice2

Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Lesson-4.5-Apply-and-Grow-Practice2..
Number of

.

Directions:

  • Classify the buttons into the categories shown. Write the marks in the chart. Count the marks and write the numbers to tell how many buttons are in each category. Circle the number that is greater than the other number.
  • Classify the buttons by color. Count the number of buttons in each category. Write each number. Circle the numbers that are equal.

Think and Grow: Modeling Real Life

Big Ideas Math Answers Grade K Chapter 4 Compare Numbers to 10 52

Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Lesson 4.5-Think and Grow-Modeling Real Life....

  • Directions:
    1 Classify the animals into the categories shown. Write the marks in the chart. Count the marks and write the numbers to tell how many animals are in each category. Circle the number that is greater than the other number.

Classify and Compare by Counting Homework & Practice 4.5

Question 1.
Big Ideas Math Answers Grade K Chapter 4 Compare Numbers to 10 53

Answer:

Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Classify and Compare by Counting Homework & Practice 4.5.1

Directions:
2 Classify the frogs into the categories shown. Write the marks in the chart. Count the marks and write the numbers to tell how many frogs are in each category. Draw a line through the number that is less than the other number. 3 Classify the frogs by color. Count the number of frogs in each category. Write each number. Circle the numbers that are equal.

Big Ideas Math Answers Grade K Chapter 4 Compare Numbers to 10 54

Question 2.
Big Ideas Math Solutions Grade K Chapter 4 Compare Numbers to 10 55

Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Classify and Compare by Counting Homework & Practice 4.5.2

Question 3.
Big Ideas Math Solutions Grade K Chapter 4 Compare Numbers to 10 56
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Classify and Compare by Counting Homework & Practice 4.5.3.

 

Compare Numbers to 10 Performance Task

Big Ideas Math Solutions Grade K Chapter 4 Compare Numbers to 10 57

Question 1.
Big Ideas Math Solutions Grade K Chapter 4 Compare Numbers to 10 58
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-Compare Numbers to 10 Performance Task1

Question 2.
Big Ideas Math Solutions Grade K Chapter 4 Compare Numbers to 10 59
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-4.5-Compare Numbers to 10 Performance Task-2.3

Directions:
1 Count the toys in each group. Write each number. Is the number of spring toys equal to the number of balls? Circle the thumbs up for yes or the thumbs down for no. 2 Classify the toys into the categories shown. Write the marks in the chart. Count the marks and write the numbers to tell how many toys are in each category. Circle the number that is greater than the other number. 3 More balls are delivered in a box. The number of balls in the box is less than the number of balls in the picture. Draw the balls in the box. Write the number of balls in the box.

Compare Numbers to 10 Activity

Toss and Compare

Big Ideas Math Solutions Grade K Chapter 4 Compare Numbers to 10 60

Directions:
Take turns tossing a counter onto the board. If the counter lands on Newton or Descartes, choose any number from 0 to 10. Write the numbers on your Toss and Compare Numbers from 0 to 10 Recording Sheet. Circle the number that is greater than the other number. Circle both numbers if they are equal. Repeat this process until you fill your sheet.

Compare Numbers to 10 Chapter Practice

4.1 Compare Groups to 1o by Matching

Question 1.
Big Ideas Math Solutions Grade K Chapter 4 Compare Numbers to 10 61

Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-4.5-4.1 Compare Groups to 1o by Matching1

Question 2.
Big Ideas Math Solutions Grade K Chapter 4 Compare Numbers to 10 62

Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-4.5-4.1 Compare Groups to 1o by Matching2

Question 3.
Big Ideas Math Solutions Grade K Chapter 4 Compare Numbers to 10 63

Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-4.5-4.1 Compare Groups to 1o by Matching3

Directions:
Draw lines between the objects in each group.
1 Is the number of red cherries equal to the number of black cherries? Circle the thumbs up for yes or the thumbs down for no. 2 Circle the group that is greater in number than the other group. 3 Draw a line through the group that is less in number than the other group.

4.2 Compare Groups to 1o by Counting

Question 4.
Big Ideas Math Solutions Grade K Chapter 4 Compare Numbers to 10 64
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-4.5-4.1 Compare Groups to 1o by Matching4

Question 5.
Big Ideas Math Solutions Grade K Chapter 4 Compare Numbers to 10 65
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-4.5-4.1 Compare Groups to 1o by Matching5

Question 6.
Big Ideas Math Solutions Grade K Chapter 4 Compare Numbers to 10 66
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-4.5-4.1 Compare Groups to 1o by Matching6.

Directions:
Count the objects in each group. Write each number.
4 Is the number of squirrels equal to the number of acorns? Circle the thumbs up for yes or the thumbs down for no. 5 Circle the number that is greater than the other number. 6 Draw a line through the number that is less than the other number.

4.3 Compare Numbers to 10

Question 7.
Big Ideas Math Solutions Grade K Chapter 4 Compare Numbers to 10 67
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-4.5-4.3 Compare Numbers to 10.1

Question 8.
Big Ideas Math Solutions Grade K Chapter 4 Compare Numbers to 10 68
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-4.5-4.3 Compare Numbers to 10.2

Question 9.
Big Ideas Math Solutions Grade K Chapter 4 Compare Numbers to 10 69

Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-4.5-4.3 Compare Numbers to 10.9

Directions:
Compare the numbers. 7 Are the numbers equal? Circle the thumbs up for yes or the thumbs down for no. 8 Circle the number that is greater than the other number. Draw to show how you know. 9 Draw a line through the number that is less than the other number. Draw to show how you know.

4.4 Classify Objects into Categories

Question 10.
Big Ideas Math Answer Key Grade K Chapter 4 Compare Numbers to 10 70
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-4.5-4.3 Compare Numbers to 10.10

4.5 Classify and Compare by Counting

Question 11.
Big Ideas Math Answer Key Grade K Chapter 4 Compare Numbers to 10 71
Answer:

Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-4.5-4.3 Compare Numbers to 10.11.

Directions:
10 Circle the buttons that have 4 holes. Cross out the buttons that do not have 4 holes. 11 Classify the animals into the categories shown. Write the marks in the chart. Count the marks and write the numbers to tell how many animals are in each category. Circle the number that is greater than the other number.

Question 1.
Big Ideas Math Answer Key Grade K Chapter 4 Compare Numbers to 10 72
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-4.5-Compare Numbers to 10 Cumulative Practice 1 - 4.1.

Question 2.
Big Ideas Math Answer Key Grade K Chapter 4 Compare Numbers to 10 73
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-4.5-Compare Numbers to 10 Cumulative Practice 1 - 4.2.

Question 3.
Big Ideas Math Answer Key Grade K Chapter 4 Compare Numbers to 10 74
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-4.5-Compare Numbers to 10 Cumulative Practice 1 - 4.3.

Directions:
Shade the circle next to the answer. 1 Which five frame shows the number of puzzle pieces? 2 Which basket has 0 apples? 3 Which number is greater than 8?

Question 4.
Big Ideas Math Answer Key Grade K Chapter 4 Compare Numbers to 10 75
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-4.5-Compare Numbers to 10 Cumulative Practice 1 - 4.4.
Question 5.
Big Ideas Math Answer Key Grade K Chapter 4 Compare Numbers to 10 76
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-4.5-Compare Numbers to 10 Cumulative Practice 1 - 4.5.

Question 6.
Big Ideas Math Answer Key Grade K Chapter 4 Compare Numbers to 10 77
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-4.5-Compare Numbers to 10 Cumulative Practice 1 - 4.6.

Directions:
Shade the circle next to the answer. 4 Which block belongs in the group? 5 Which ticket shows the numbers 1 to 5 in order? 6 Which group does not have 7 balls?

Question 7.
Big Ideas Math Answer Key Grade K Chapter 4 Compare Numbers to 10 78
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-4.5-Compare Numbers to 10 Cumulative Practice 1 - 4.7.

Question 8.
Big Ideas Math Answer Key Grade K Chapter 4 Compare Numbers to 10 79

Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-4.5-Compare Numbers to 10 Cumulative Practice 1 - 4.8.

Question 9.
Big Ideas Math Answer Key Grade K Chapter 4 Compare Numbers to 10 80

Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-4.5-Compare Numbers to 10 Cumulative Practice 1 - 4.9.

 

Directions
7 Count the flowers. Color the boxes to show how many. 8 Count the salamanders. Say the number. Write the number. 9 Draw lines between the objects in each group. Is the number of detective hats equal to the number of magnifying glasses? Circle the thumbs up for yes or the thumbs down for no.

Question 10.
Big Ideas Math Answer Key Grade K Chapter 4 Compare Numbers to 10 81
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-4.5-Compare Numbers to 10 Cumulative Practice 1 - 4.10.

Question 11.
Big Ideas Math Answer Key Grade K Chapter 4 Compare Numbers to 10 82
Answer:
The number of Frogs are greater than the Number of Lotus.

Explanation:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-4.5-Compare Numbers to 10 Cumulative Practice 1 - 4.11.

Number of Frogs = Five or 5.
Number of Lotus = Three or 3.

Question 12.
Big Ideas Math Solutions Grade K Chapter 4 Compare Numbers to 10 83
Answer:
Big-Ideas-Math-Book-Grade-K-Answer-Key-Chapter-4-4.5-Compare Numbers to 10 Cumulative Practice 1 - 4.12.

Directions:
10 Count the objects in each group. Write each number. Circle the number that is greater than the other number. 11 Draw lines between the objects in each group. Draw a line through the group that is less in number than the other group. 12 Draw 4 pieces of string. Is the number of pieces of string equal to the number of buttons? Circle the thumbs up for yes or the thumbs down for no.

Final words:

After practicing the problems test yourself by solving the problems given at the end of the chapter. The solutions are prepared from Big Ideas Math – Modeling Real Life Grade K Student Edition Set. Even teachers can make use of the Big Ideas Math Answers Grade K Chapter 4 Compare Numbers to 10. Bookmark our page to get the free pdf of all Big Ideas Math Grade K Answers for Chapters.

Big Ideas Math Answers Grade 1 Chapter 10 Measure and Compare Lengths

Big Ideas Math Answers Grade 1 Chapter 10

Students can find the exact solution for every question of Big Ideas Math Answers Grade 1 Chapter 10 Measure and Compare Lengths. So, the students who are searching for the Grade 1 Solution Key can get them on the Big Ideas Math Answer page. In order to excel in the exams, you have to choose the best material. Thus Download Big Ideas Math Answers Grade 1 Chapter 10 Measure and Compare Lengths from here and start practicing the problems.

Big Ideas Math 1st Grade Answer Key Chapter 10 Measure and Compare Lengths

Here we can discuss different topics on Measuring and Comparing, Order Objects by Length, Compare Lengths Indirectly, Measure Lengths, Measure More Lengths, Solve Compare Problems Involving Length, and so on. Compare the topic you learned here with the real time and understand the topic deeply. Those topics were being set up by the mathematical experts as indicated by the latest edition. Scroll down this page to get the answers to all the inquiries. Click on the links to look at the subjects shrouded in this chapter Measuring and comparing lengths.

Lesson 1 Order Objects by Length

Lesson 2 Compare Lengths Indirectly

Lesson 3 Measure Lengths

Lesson 4 Measure More Lengths

Lesson 5 Solve Compare Problems Involving Length

Performance Task

Measure and Compare Lengths Vocabulary

Organize It

Review Words:
longer
shorter

Use the review words to complete the graphic organizer.
Big Ideas Math Answer Key Grade 1 Chapter 10 Measure and Compare Lengths 1
Answer:
The first image longer and the second image is shorter.

Explanation:
Big-Ideas-Math-Answer-Key-Grade-1-Chapter-10-Measure-and-Compare-Lengths-1

Define It

Use your vocabulary cards to identify the words. Find each word in the word search.
Big Ideas Math Answer Key Grade 1 Chapter 10 Measure and Compare Lengths 2
Answer:
Big-Ideas-Math-Answer-Key-Grade-1-Chapter-10-Measure-and-Compare-Lengths-2

Lesson 10.1 Order Objects by Length

Explore and Grow

Draw an object that is shorter than the pencil and longer than the crayon.

Big Ideas Math Answer Key Grade 1 Chapter 10 Measure and Compare Lengths 3

Big Ideas Math Answer Key Grade 1 Chapter 10 Measure and Compare Lengths 4
Answer:
Big-Ideas-Math-Answer-Key-Grade-1-Chapter-10-Measure-and-Compare-Lengths-5
Chalk is smaller than crayon.

Show and Grow

Question 1.
Order from longest to shortest.
Big Ideas Math Answer Key Grade 1 Chapter 10 Measure and Compare Lengths 5
_____________, _____________, _____________
Answer:
Purple, Pink, Blue.

Explanation:
In the above image, the purple color tube is the longest after that the pink color tube is the longest, and the blue color tube is the shortest.

Question 2.
Order from shortest to longest.
Big Ideas Math Answer Key Grade 1 Chapter 10 Measure and Compare Lengths 6
_____________, _____________, _____________
Answer:
Yellow, Green, Black.

Explanation:
In the above image, the yellow brush is the longest, and then the green brush is the longest. The black brush is the shortest.

Apply and Grow: Practice

Question 3.
Order from longest to shortest.
Big Ideas Math Answer Key Grade 1 Chapter 10 Measure and Compare Lengths 7
_____________, _____________, _____________
Answer:
Purple crayon, Red crayon, Green crayon.

Explanation:
In the above image, the Purple crayon is the longest, and then the red crayon is the longest. The green crayon is the shortest.

Question 4.
Order from shortest to longest.
Big Ideas Math Answer Key Grade 1 Chapter 10 Measure and Compare Lengths 8
_____________, _____________, _____________
Answer:
The order from shortest to longest is, Pink is the shortest, green is the largest, and blue is the longest.

Explanation:
In the above image, we can see that the order from shortest to longest is, Pink is the shortest, green is the largest, and blue is the longest.

Question 5.
YOU BE THE TEACHER
Your friend ordered from shortest to longest. Is your friend correct? Explain.
Big Ideas Math Answer Key Grade 1 Chapter 10 Measure and Compare Lengths 9
  Yellow   ,          green  ,           red   
Answer:
No, my friend is not correct.

Explanation:
No, my friend is not correct. He represented the order from longest to shortest. In the above image, the red chill is the shortest, and the green chili is the longest. The black brush is the shortest.

Think and Grow: Modeling Real Life

Your yarn is longer than Newton’s. Descartes’s is longer than Newton’s and shorter than yours. Who has the longest yarn?
Big Ideas Math Answer Key Grade 1 Chapter 10 Measure and Compare Lengths 10
Draw a picture:
Big Ideas Math Answer Key Grade 1 Chapter 10 Measure and Compare Lengths 11
Who has the longest yarn?
You             Newton      Descartes
Answer:
My yarn is the longest yarn.

Explanation:
Big-Ideas-Math-Answers-1st-Grade-1-Chapter-10-Measure-and-Compare-Lengths-img-1
As my yarn is longer than Newton’s and Descartes’s is longer than Newton’s and shorter than mine. So the longest yarn is mine, and the largest yarn is Descartes. The shortest yarn is Newton’s.

Show and Grow

Question 6.
Descartes’s pencil is shorter than Newton’s. Yours is shorter than Newton’s and longer than Descartes’s. Who has the shortest pencil?
Draw a picture:
Big Ideas Math Answer Key Grade 1 Chapter 10 Measure and Compare Lengths 12
Who has the shortest pencil?
Descartes        Newton           You
Answer: Your pencil is shorted among the three.

Order Objects by Length Practice 10.1

Order from longest to shortest.

Question 1.
Big Ideas Math Answer Key Grade 1 Chapter 10 Measure and Compare Lengths 13
_____________, _____________, _____________
Answer:
The order of the bats from longest to the shortest is
Bat 2, Bat 1, Bat 3.

Explanation:
In the above image, we can see the longest bat is bat 2 and then the largest bat is bat 1. The shortest bat is bat 3.

Question 2.
Big Ideas Math Answer Key Grade 1 Chapter 10 Measure and Compare Lengths 14
_____________, _____________, _____________
Answer:
The order of the colors from longest to shortest is
Gold, Blue, Red.

Explanation:
In the above image, we can see that the longest color is gold and then the largest color is blue. The shortest color is red.

Question 3.
Order from shortest to longest.
Big Ideas Math Answer Key Grade 1 Chapter 10 Measure and Compare Lengths 15
_____________, _____________, _____________
Answer:
The order from shortest to longest is
Vine 3, Vine 1, Vine 2.

Explanation:
In the above image, we can see that the shortest vine is vine 3 after that the largest vine is vine 1 and the longest vine is Vine 2.

Question 4.
DIG DEEPER!
Use the clues to match. The red pencil is longer than the yellow pencil. The shortest pencil is blue.
Big Ideas Math Answer Key Grade 1 Chapter 10 Measure and Compare Lengths 16
Answer:
The order from shortest to longest is
Blue pencil, Yellow pencil, Red pencil.

Explanation:
Big-Ideas-Math-Answer-Key-Grade-1-Chapter-10-Measure-and-Compare-Lengths-16
As by the given clue, the Blue pencil is the shortest pencil, the Yellow pencil is the largest pencil, and the Red pencil is the longest pencil.

Question 5.
Modeling Real Life
Your jump rope is longer than Newton’s. Descartes’s is longer than Newton’s and shorter than yours. Who has the longest jump rope?
Big Ideas Math Answer Key Grade 1 Chapter 10 Measure and Compare Lengths 17
Who has the longest jump rope?
You           Newton        Descartes
Answer:
My rope is the longest jump rope.

Explanation:
As my jump rope is longer than Newton’s and Descartes’s jump rope longer than Newton’s and Newton’s jump rope is shorter than yours. So my jump rope is Longer, Newton’s jump rope is the Largest, and the shortest rope is Descartes.

Review & Refresh

Compare.

Question 6.
25 ○ 52
Answer:
25 > 52.

Explanation:
Given that to compare 25 and 52. So we can see that 52 is greater than 25.

Question 7.
41 ○ 44
Answer:
41 > 44.

Explanation:
Given that to compare 41 and 44. So we can see that 44 is greater than 41.

Lesson 10.2 Compare Lengths Indirectly

Explore and Grow

Use string to compare the keys. Which key is longer?
How do you know?

Big Ideas Math Answers 1st Grade 1 Chapter 10 Measure and Compare Lengths 18

Big Ideas Math Answers 1st Grade 1 Chapter 10 Measure and Compare Lengths 19
Answer:
Key 1 is longer than Key 2.

Explanation:
In the above image, we can see that key 1 is longer than key 2.

Show and Grow

Question 1.
Circle the longer object.
Big Ideas Math Answers 1st Grade 1 Chapter 10 Measure and Compare Lengths 20
Answer:
The Pen is longer than Eraser.

Explanation:
Big-Ideas-Math-Answers-1st-Grade-1-Chapter-10-Measure-and-Compare-Lengths-20
In the above image, we can see that the Pen is longer than the Easer.

Question 2.
Draw a line through the shorter object.
Big Ideas Math Answers 1st Grade 1 Chapter 10 Measure and Compare Lengths 21
Answer:

Explanation:

Apply and Grow: Practice

Question 3.
Draw a line through the shorter object.
Big Ideas Math Answers 1st Grade 1 Chapter 10 Measure and Compare Lengths 22
Answer:
Object 2 is shorter than object 1.

Explanation:
Big-Ideas-Math-Answers-1st-Grade-1-Chapter-10-Measure-and-Compare-Lengths-22
In the above image, we can see that object two is shorter than object 1.

Question 4.
Circle the longer object.
Big Ideas Math Answers 1st Grade 1 Chapter 10 Measure and Compare Lengths 23
Answer:
The spoon is longer than the brush.

Explanation:
Big-Ideas-Math-Answers-1st-Grade-1-Chapter-10-Measure-and-Compare-Lengths-23
We can see in the above image, that the spoon is longer than the brush. So we will round off the spoon.

Question 5.
DIG DEEPER
Which object is longer? Explain.
Big Ideas Math Answers 1st Grade 1 Chapter 10 Measure and Compare Lengths 24
Answer:
The book’s shelf is longer than the key.

Explanation:
Big-Ideas-Math-Answers-1st-Grade-1-Chapter-10-Measure-and-Compare-Lengths-24
As we can see in the above image, that the books shelf is longer than the key. So we will circle the books shelf.

Think and Grow: Modeling Real Life

A green crayon is shorter than a blue crayon. The blue crayon is shorter than a yellow crayon. Is the green crayon longer than or shorter than the yellow crayon?
Big Ideas Math Answers 1st Grade 1 Chapter 10 Measure and Compare Lengths 25
Draw a picture:
Big Ideas Math Answers 1st Grade 1 Chapter 10 Measure and Compare Lengths 26
Longer              Shorter
Answer:
The green crayon is shorter than the yellow crayon.

Explanation:
Big-Ideas-Math-Answers-1st-Grade-1-Chapter-10-Measure-and-Compare-Lengths-img-1-1
Given that green crayon is shorter than blue crayon, and blue crayon is shorter than yellow crayon. So green crayon is shorter than yellow crayon.

Show and Grow

Question 6.
A yellow ribbon is longer than a pink ribbon. The pink ribbon is longer than a blue ribbon. Is the yellow ribbon longer than or shorter than the blue ribbon?
Big Ideas Math Answers 1st Grade 1 Chapter 10 Measure and Compare Lengths 27
Draw a picture:
Big Ideas Math Answers 1st Grade 1 Chapter 10 Measure and Compare Lengths 28
Longer            Shorter
Answer:
The Yellow ribbon is longer than the blue ribbon.

Explanation:
Big-Ideas-Math-Answers-1st-Grade-1-Chapter-10-Measure-and-Compare-Lengths- img 2
Given that yellow ribbon is longer than pink ribbon, and the pink ribbon is longer than a blue ribbon. So yellow ribbon is longer than the blue ribbon.

Compare Lengths Indirectly Practice 10.2

Question 1.
Circle the longer object.
Big Ideas Math Answers 1st Grade 1 Chapter 10 Measure and Compare Lengths 29
Answer:
Object 1 is longer than object 2.

Explanation:
In the above image, we can see that object one is longer than image two. So we will circle object one.

Question 2.
Draw a line through the shorter object.
Big Ideas Math Answers 1st Grade 1 Chapter 10 Measure and Compare Lengths 30
Answer:
Object two is shorter than object one.

Explanation:
Big-Ideas-Math-Answers-1st-Grade-1-Chapter-10-Measure-and-Compare-Lengths-30
In the above image, we can see that object two is shorter than object one. So we will circle object two.

Question 3.
DIG DEEPER
Use the clues to match.
The blue string is longer than the orange string.
The purple string is shorter than the orange string.
Big Ideas Math Answers 1st Grade 1 Chapter 10 Measure and Compare Lengths 31
Answer:
The order of the strings from longest to shortest is
Blue string, Orange string, Purple string.

Explanation:
Big-Ideas-Math-Answers-1st-Grade-1-Chapter-10-Measure-and-Compare-Lengths-31
Given that the blue string is longer than the orange string, and the purple string is shorter than the orange string.
So the order of the strings is the blue string is longer, the orange string is the largest and the purple string is the shortest.

Question 4.
Modeling Real Life
A kayak is shorter than a canoe. The canoe is shorter than a paddle board. Is the kayak longer than or shorter than the paddle board?

Longer             Shorter
Answer:
The Kayak is shorter than the paddle board.

Explanation:
Given that the Kayak is shorter than a canoe, and the canoe is shorter than a paddle board. So the Kayak is shorter than the paddle board.

Review & Refresh

Question 5.
Circle the objects that have capacity as an attribute.
Big Ideas Math Answers 1st Grade 1 Chapter 10 Measure and Compare Lengths 32
Answer:
The objects which have the capacity to store are circled in the below image.

Explanation:
Big-Ideas-Math-Answers-1st-Grade-1-Chapter-10-Measure-and-Compare-Lengths-32
The circled items have the capacity to store. As we can see that the lid jar is used to store any items, and chalk box is used to store chalk pieces, and glass is used to store water or any liquids, and a glue bottle is used to store glue. So these items have the capacity to store. And the remaining things do not have that much capacity to store.

Lesson 10.3 Measure Lengths

Explore and Grow

Find and measure the objects shown in your classroom.

Big Ideas Math Answers 1st Grade 1 Chapter 10 Measure and Compare Lengths 33

Answer:
The length of the table is about two colored red tiles.

Explanation:
The table is measured by using the colored red tile. As each tile is equal to one unit. So the length of the table is two colored red tiles, which means two units. We must take each tile without gaps or any overlaps between them.

Big Ideas Math Answers 1st Grade 1 Chapter 10 Measure and Compare Lengths 34

Answer:
The length of the pencil is about one tile.

Explanation:
The pencil is measured by using the colored red tile. As each tile is equal to one unit. So the length of the pencil is one colored red tile, which means one unit. We must take each tile without gaps or any overlaps between them.

Show and Grow

Measure

Question 1.
Big Ideas Math Answers 1st Grade 1 Chapter 10 Measure and Compare Lengths 35

about ___________ color tile
Answer:
The length of the object is about one tile.

Explanation:
The object is measured by using the colored tile. As each tile is equal to one unit. So the length of the object is about one colored tile, which means one unit. We must take each tile without gaps or any overlaps between them.

Question 2.
Big Ideas Math Answers 1st Grade 1 Chapter 10 Measure and Compare Lengths 36

about ___________ color tile

Answer:
The length of the object is about five colored tiles.

Explanation:
The object is measured by using the colored tile. As each tile is equal to one unit. So the length of the object is about five colored tiles, which means five units. We must take each tile without gaps or any overlaps between them.

Apply and Grow: Practice

Measure

Question 3.
Big Ideas Math Answers 1st Grade 1 Chapter 10 Measure and Compare Lengths 37

about ___________ color tile
Answer:
The length of the object is about four colored tiles.

Explanation:
The object is measured by using the colored tile. As each tile is equal to one unit. So the length of the object is about four colored tiles, which means four units. We must take each tile without gaps or any overlaps between them.

Question 4.
Big Ideas Math Answers 1st Grade 1 Chapter 10 Measure and Compare Lengths 38

about ___________ color tile
Answer:
The length of the object is about three colored tiles.

Explanation:
The object is measured by using the colored tile. As each tile is equal to one unit. So the length of the object is about three colored tiles, which means three units. We must take each tile without gaps or any overlaps between them.

Question 5.
MP Precision
Which picture shows the correct way to measure the straw?
Big Ideas Math Answers 1st Grade 1 Chapter 10 Measure and Compare Lengths 39
Answer:
The red color tiles are the correct way to measure the straw.

Explanation:
In the above image, we can see that the correct way to measure the straw is the red color tiles.

Think and Grow: Modeling Real Life

Will the scissors fit inside a pencil case that is 7 color tiles long?
Big Ideas Math Answers 1st Grade 1 Chapter 10 Measure and Compare Lengths 40
Circle:      Yes     No
Tell how you know:
Answer:
Yes, the scissors can fit inside a pencil case that is 7 color tiles long.

Explanation:
Yes, the scissors can fit inside a pencil case. As the length of the scissors is four colored tiles. As the scissors are measured by using the colored tile. And each tile is equal to one unit. So the length of the scissors is about four colored tiles, which means four units. And the pencil case is 7 colored tiles long. So the scissors can fit. We must take each tile without gaps or any overlaps between them.

Show and Grow

Question 6.
Will the cell phone fit inside a case that is 5 color tiles long?
Big Ideas Math Answers 1st Grade 1 Chapter 10 Measure and Compare Lengths 41
Circle:     Yes     No
Tell how you know:
Answer:
No, the cell phone cannot fit inside a case of 5 color tiles long.

Explanation:
No, the cell phone cannot fit inside a case of 5 color tiles long. As the length of the cell phone is six colored tiles. And the cell phone is measured by using the colored tile. And each tile is equal to one unit. So the length of the cell phone is about six colored tiles, which means six units. And cannot fit inside 5 colored tiles long. We must take each tile without gaps or any overlaps between them.

Measure Lengths Practice 10.3

Measure.

Question 1.
Big Ideas Math Answers 1st Grade 1 Chapter 10 Measure and Compare Lengths 42
about ___________ color tile
Answer:
The length of the object is about four colored tiles.

Explanation:
The object is measured by using the colored tile. As each tile is equal to one unit. So the length of the object is about four colored tiles, which means four units. We must take each tile without gaps or any overlaps between them.

Question 2.
Big Ideas Math Answers 1st Grade 1 Chapter 10 Measure and Compare Lengths 43
about ___________ color tile
Answer:
The length of the object is about two colored tiles.

Explanation:
The object is measured by using the colored tile. As each tile is equal to one unit. So the length of the object is about two colored tiles, which means two units. We must take each tile without gaps or any overlaps between them.

Question 3.
Big Ideas Math Answers 1st Grade 1 Chapter 10 Measure and Compare Lengths 44
about ___________ color tile
Answer:
The length of the glue stick is about three colored tiles.

Explanation:
The glue stick is measured by using the colored tile. As each tile is equal to one unit. So the length of the glue stick is about three colored tiles, which means three units. We must take each tile without gaps or any overlaps between them.

Question 4.
MP Reasoning
The green yarn is about 3 color tiles long. How long is the blue yarn?
Big Ideas Math Answers 1st Grade 1 Chapter 10 Measure and Compare Lengths 45
about ___________ color tile
Answer:
The length of the blue yarn is about six color tiles.

Explanation:
Given that the length of the green yarn is three color tiles long, so we should find the length of the blue yarn. To measure the length of the blue yarn, we will take the colored tiles without gaps or any overlaps between them. As each tile is equal to one unit. So the length of the blue yarn is about six colored tiles, which means six units.

Question 5.
Modeling Real Life
Will the gift card fit inside an envelope that is 8 color tiles long?
Big Ideas Math Answers 1st Grade 1 Chapter 10 Measure and Compare Lengths 46
Circle:      Yes      No
Tell how you know:
Answer:
Yes, the gift can fit inside an envelope that is 8 color tiles long.

Explanation:
Yes, the gif card can fit inside an envelope that is 8 color tiles long. As the length of the gift card is four colored tiles. And the gift card is measured by using the colored tile. And each tile is equal to one unit. So the length of the gift is about four colored tiles, which means four units. And the envelope is 8 colored tiles long. So the gift card can fit. We must take each tile without gaps or any overlaps between them.

Review & Refresh

Question 6.
Complete the fact family.
7 + 3 = _________                                     __________ – 3 = 7
_________ + _________ = _________             _________ – 7 = _________
Answer:
7 + 3 is 10,
3 + 7 is 10
10 – 3 is 7,
10 –  7 is 3.

Explanation:
As a fact family represents a group of math facts, equations, which are created by using the same set of digits. This fact family defines the relation between the three numbers which are involved. In addition or subtraction in a fact family, there will be four addition and subtractions will be created using these three numbers. So
7 + 3 is 10,
3 + 7 is 10
10 – 3 is 7,
10 –  7 is 3.

Lesson 10.4 Measure More Lengths

Explore and Grow

Find and measure the objects shown in your classroom two ways. What do you notice?
Big Ideas Math Solutions Grade 1 Chapter 10 Measure and Compare Lengths 47

Answer:
The length of the table is about two colored red tiles and the length of the table which is measured using a paper clip is one paper clip.

Explanation:
The table is measured by using the colored red tile. As each tile is equal to one unit. So the length of the table is two colored red tiles, which means two units. We must take each tile without gaps or any overlaps between them. And to measure the table with the paper clip, we will place the paper clip without gaps or overlaps between them. And each paper clip is equal to about one unit. So the length of the table using the paper clip is one paper clip, which means one unit. We have noticed that if we take a colored tile, then the length of the table will be two color tiles. And if we take a paper clip then the length of the table will be one paper clip. So the length is doubled while measuring with color tile than the paper clip.

Big Ideas Math Solutions Grade 1 Chapter 10 Measure and Compare Lengths 48
Answer:
The length of the pencil is about two colored red tiles and the length of the pencil which is measured using a paper clip is one paper clip.

Explanation:
The pencil is measured by using the colored red tile. As each tile is equal to one unit. So the length of the table is two colored red tiles, which means two units. We must take each tile without gaps or any overlaps between them. And to measure the table with the paper clip, we will place the paper clip without gaps or overlaps between them. And each paper clip is equal to about one unit. So the length of the pencil using the paper clip is one paper clip, which means one unit. We have noticed that if we take a colored tile, then the length of the table will be two color tiles. And if we take a paper clip then the length of the pencil will be one paper clip. So the length is doubled while measuring with color tile than the paper clip.

Show and Grow

Measure

Question 1.
Big Ideas Math Solutions Grade 1 Chapter 10 Measure and Compare Lengths 49
about __________ color tiles about __________ paper clips
Answer:
The length of the brush is about four colored tiles and the length of the brush which is measured using a paper clip is two paper clips.

Explanation:
The brush is measured by using the colored tile. As each tile is equal to one unit. So the length of the table is two colored tiles, which means two units. We must take each tile without gaps or any overlaps between them. And to measure the brush with the paper clip, we will place the paper clip without gaps or overlaps between them. And each paper clip is equal to about one unit. So the length of the brush using the paper clip is one paper clip, which means one unit. We have noticed that if we take a colored tile, then the length of the table will be two color tiles. And if we take a paper clip then the length of the brush will be one paper clip. So the length is doubled while measuring with color tile than the paper clip.

Question 2.
Big Ideas Math Solutions Grade 1 Chapter 10 Measure and Compare Lengths 50
about __________ color tiles about __________ paper clips
Answer:
The length of the skating board is about two colored tiles and the length of the pencil which is measured using a paper clip is one paper clip.

Explanation:
The skating board is measured by using the colored tile. As each tile is equal to one unit. So the length of the table is two colored tiles, which means two units. We must take each tile without gaps or any overlaps between them. And to measure the skating board with the paper clip, we will place the paper clip without gaps or overlaps between them. And each paper clip is equal to about one unit. So the length of the skating board using the paper clip is one paper clip, which means one unit. We have noticed that if we take a colored tile, then the length of the skating board will be two color tiles. And if we take a paper clip then the length of the pencil will be one paper clip. So the length is doubled while measuring with color tile than the paper clip.

Apply and Grow: Practice

Measure.

Question 3.
Big Ideas Math Solutions Grade 1 Chapter 10 Measure and Compare Lengths 51
about __________ color tiles about __________ paper clips
Answer:
The length of the object is about six colored tiles and the length of the pencil which is measured using a paper clip is three paper clips.

Explanation:
The object is measured by using the colored tile. As each tile is equal to one unit. So the length of the object is six colored tiles, which means six units. We must take each tile without gaps or any overlaps between them. And to measure the table with the paper clip, we will place the paper clip without gaps or overlaps between them. And each paper clip is equal to about one unit. So the length of the object using the paper clip is three paper clips, which means three units. We have noticed that if we take a colored tile, then the length of the object will be six color tiles. And if we take a paper clip then the length of the pencil will be three paper clips. So the length is doubled while measuring with color tile than the paper clip.

Question 4.
Big Ideas Math Solutions Grade 1 Chapter 10 Measure and Compare Lengths 52
about __________ color tiles about __________ paper clips
Answer:

The length of the sunscreen lotion bottle is about four colored tiles and the length of the sunscreen lotion bottle which is measured using a paper clip is two paper clips.

Explanation:
The sunscreen lotion bottle is measured by using the colored tile. As each tile is equal to one unit. So the length of the sunscreen lotion bottle is four colored tiles, which means four units. We must take each tile without gaps or any overlaps between them. And to measure the sunscreen lotion bottle with the paper clip, we will place the paper clip without gaps or overlaps between them. And each paper clip is equal to about one unit. So the length of the sunscreen lotion bottle using the paper clip is two paper clips, which means two units. We have noticed that if we take a colored tile, then the length of the sunscreen lotion bottle will be four color tiles. And if we take a paper clip then the length of the sunscreen lotion bottle will be two paper clips. So the length is doubled while measuring with color tile than the paper clip.

Question 5.
YOU BE THE TEACHER
Your friend says the pencil is more paper clips long than color tiles. Is your friend correct? Explain.
Big Ideas Math Solutions Grade 1 Chapter 10 Measure and Compare Lengths 53
Answer:
No, my friend is not correct.

Explanation:
No, my friend is not correct. As we can see from the above problems that while measuring with the paper clip we need fewer paper clips than the color tiles. And the length is doubled while measuring with color tile than the paper clip. So my friend is not correct.

Think and Grow: Modeling Real Life

Your guitar is 33 color tiles long. Is your guitar more than or less than 33 paper clips long?
Big Ideas Math Solutions Grade 1 Chapter 10 Measure and Compare Lengths 54
Circle:      more than 33         less than 33
Tell how you know:
Answer:
My guitar will be less than 33 paper clips.

Explanation:
My guitar will be less than 33 paper clips. As we can see from the above problems that while measuring with the paper clip we need fewer paper clips than the color tiles. And the length is doubled while measuring with color tile than the paper clip. So there will be fewer paper clips than the color tiles.

Show and Grow

Question 6.
Your mailbox is 11 paper clips long. Is your mailbox more than or less than 11 color tiles long?
Big Ideas Math Solutions Grade 1 Chapter 10 Measure and Compare Lengths 55
Circle:       more than 11         less than 11
Tell how you know:
Answer:
My mailbox will have more than 11 color tiles long.

Explanation:
My mailbox will have more than 11 color tiles. As we can see from the above problems that while measuring with the paper clip we need fewer paper clips than the color tiles. And the length is doubled while measuring with color tile than the paper clip. So there will be more color tiles than the paper clips.

Measure More Lengths Practice 10.4

Measure

Question 1.
Big Ideas Math Solutions Grade 1 Chapter 10 Measure and Compare Lengths 56
about __________ color tiles about __________ paper clips
Answer:
The length of the rocket is about six colored tiles and the length of the pencil which is measured using a paper clip is three paper clips.

Explanation:
The rocket is measured by using the colored tile. As each tile is equal to one unit. So the length of the rocket is six colored tiles, which means six units. We must take each tile without gaps or any overlaps between them. And to measure the rocket with the paper clip, we will place the paper clip without gaps or overlaps between them. And each paper clip is equal to about one unit. So the length of the rocket using the paper clip is three paper clips, which means three units. We have noticed that if we take a colored tile, then the length of the rocket will be six color tiles. And if we take a paper clip then the length of the pencil will be three paper clips. So the length is doubled while measuring with color tile than the paper clip.

Question 2.
YOU BE THE TEACHER
Your friend says the marker is more color tiles long than paper clips. Is your friend correct? Explain.
Big Ideas Math Solutions Grade 1 Chapter 10 Measure and Compare Lengths 57
Answer:
Yes, the marker has more color tiles long than paper clips. So my friend is correct.

Explanation:
Yes, the marker has more color tiles long than paper clips. As we can see from the above problems that while measuring with the paper clip we need fewer paper clips than the color tiles. And the length is doubled while measuring with color tile than the paper clip. So there will be more color tiles than the paper clips for the marker. So my friend is correct.

Question 3.
Modeling Real Life
Your folder is 15 color tiles long. Is your folder more than or less than 15 paper clips long?
Big Ideas Math Solutions Grade 1 Chapter 10 Measure and Compare Lengths 58
Circle:       more than 15         less than 15
Tell how you know:

Answer:
My folder will be less than 15 paper clips.

Explanation:
My folder will be less than 15 paper clips. As we can see from the above problems that while measuring with the paper clip we need fewer paper clips than the color tiles. And the length is doubled while measuring with color tile than the paper clip. So there will be fewer than 15 paper clips than the color tiles for my folder.

Review & Refresh

Question 4.
8 tigers swim.
5 tigers leave.
How many tigers are left?
___________ – ___________ = ___________ tigers
Answer:
the number of tigers left is
8 – 5= 3 tigers.

Explanation:
The number of tigers swim is 8 and the number of tigers which leave is 5 tigers. So the number of tigers left is
8 – 5= 3 tigers.

Question 5.
You have 6 pencils.
You lose 2 pencils.
How many pencils are left?
___________ – ___________ = ___________ pencils
Answer:
the remaining pencils are
6 – 2= 4 pencils.

Explanation:
The number of pencils I have is 6 pencils, and the number of pencils lost is 2 pencils. So the remaining pencils are
6 – 2= 4 pencils.

Lesson 10.5 Solve Compare Problems Involving Length

Explore and Grow

Draw a line that is 2 color tiles longer than the pencil.

Big Ideas Math Answer Key Grade 1 Chapter 10 Measure and Compare Lengths 59

Draw a line that is 2 color tiles shorter than the pencil.

Big Ideas Math Answer Key Grade 1 Chapter 10 Measure and Compare Lengths 59
Answer:

Show and Grow

Question 1.
Your lunch box is 6 paper clips long. Your friend’s is 3 paper clips long. How many paper clips longer is your lunch box?
Big Ideas Math Answer Key Grade 1 Chapter 10 Measure and Compare Lengths 60
Big Ideas Math Answer Key Grade 1 Chapter 10 Measure and Compare Lengths 61
__________ paper clips
Answer:
My lunch box is three paper clips longer than my friend’s and my friend’s lunch box is three times shorter than my friend’s lunch box.

Explanation:
Big-Ideas-Math-Answers-1st-Grade-1-Chapter-10-Measure-and-Compare-Lengths-84-1
Given that my lunch box is six paper clips long and my friend’s lunch box is three paper clips long. So there will be three paper clips longer than my friend’s. And my friend’s lunch box is three paper clips shorter than my lunch box.

Apply and Grow: Practice

Question 2.
Your scarf is 10 paper clips long. Your friend’s is 7 paper clips long. How many paper clips longer is your scarf?
Big Ideas Math Answer Key Grade 1 Chapter 10 Measure and Compare Lengths 62
Big Ideas Math Answer Key Grade 1 Chapter 10 Measure and Compare Lengths 63
Answer:
Myscarf is three paper clips longer than my friend’s and my friend’s scarf is three times shorter than my friend’s scarf.

Eplanation:
Big-Ideas-Math-Answers-1st-Grade-1-Chapter-10-Measure-and-Compare-Lengths-84-2
Given that my scarf is ten paper clips long and my friend’s scarf is seven paper clips long. So there will be three paper clips longer than my friend’s. And my friend’s scarf is three paper clips shorter than my scarf.

Question 3.
Your marker is 6 color tiles long. Your friend’s is 7 color tiles long. How many tiles shorter is your marker?
Big Ideas Math Answer Key Grade 1 Chapter 10 Measure and Compare Lengths 64
Answer:
My marker is one tile shorter than my friend’s marker.

Explanation:
Big-Ideas-Math-Answer-Key-Grade-1-Chapter-10-Measure-and-Compare-Lengths-64
Given that my marker is 6 color tiles long and my friend’s marker is 7 color tiles. So my marker is 7 – 6 which is one color tile shorter than my friend’s. And my friend’s marker is one color tile longer than mine.

Question 4.
MP Reasoning
Your pencil is 4 color tiles long. Your
friend’s is 2 color tiles long. Complete the sentences.

Your pencil is ________ color tiles ________ than your friend’s.
Your friend’s pencil is ________ color tiles ________ than yours.
Answer:
Your pencil is two color tiles longer than your friend’s.
Your friend’s pencil is two color tiles shorter than yours.

Explanation:
Given that my pencil is 4 color tiles long and my friend’s pencil is 2 color tiles. So my pencil is 4 – 2 which is two color tiles longer than my friend’s. And my friend’s pencil is two color tiles shorter than mine.

Think and Grow: Modeling Real Life

Your friend’s paper chain is 6 paper clips shorter than yours. Your chain is 12 paper clips long. How long is your friend’s?
Big Ideas Math Answer Key Grade 1 Chapter 10 Measure and Compare Lengths 65
Model:
Big Ideas Math Answer Key Grade 1 Chapter 10 Measure and Compare Lengths 66
Equation:

____________ paper clips long
Answer:
My paper chain is six color tiles longer than my friend’s.

Explanation:
Big-Ideas-Math-Answer-Key-Grade-1-Chapter-10-Measure-and-Compare-Lengths-66
Given that my paper chain is 12 color tiles long and my friend’s paper chain is 6 color tiles. So my paper chain is 12 – 6 which is six color tile shorter than my friend’s. And my friend’s paper chain is six color tile longer than mine.

Show and Grow

Question 5.
Your paper airplane is 9 color tiles shorter than your friend’s. Your friend’s paper airplane is 16 color tiles long. How long is yours?
Big Ideas Math Answer Key Grade 1 Chapter 10 Measure and Compare Lengths 67
Model:
Big Ideas Math Answer Key Grade 1 Chapter 10 Measure and Compare Lengths 68
Answer:
Equation:

____________ color titles long
Answer:
My airplane is seven color tiles longer than my friend.

Explanation:
Big-Ideas-Math-Answer-Key-Grade-1-Chapter-10-Measure-and-Compare-Lengths-68
Given that my airplane is 9 color tiles long and my friend’s airplane is 16 color tiles. So my marker is 16 – 9 which is seven color tile shorter than my friend’s. And my friend’s airplane is seven color tile longer than mine.

Solve Compare Problems Involving Length Practice 10.5

Question 1.
Your backpack is 15 paper clips long. Your friend’s is 12 paper clips long. How many paper clips longer is your backpack?
Big Ideas Math Answer Key Grade 1 Chapter 10 Measure and Compare Lengths 69
Big Ideas Math Answer Key Grade 1 Chapter 10 Measure and Compare Lengths 70
Answer:
Three paper clips longer than my friend’s

Explanation:
Big-Ideas-Math-Answer-Key-Grade-1-Chapter-10-Measure-and-Compare-Lengths-61
Given that my backpack is 15 paper clips long and my friend’s backpack is 12 paper clips long. So there will be three paper clips longer than my friend’s.

Question 2.
MP Reasoning
Your baseball miff is 8 paper clips long. Your friend’s is 7 paper clips long. Complete the sentences.
Big Ideas Math Answer Key Grade 1 Chapter 10 Measure and Compare Lengths 71
Your friend’s baseball mitt is ________ paper clip _________ than yours.
Your baseball mitt is _________ paper clip _________ than your friends.
Answer:
Your friend’s baseball mitt is 1 paper clip shorter than yours.
Your baseball mitt is 1 paper clip longer than your friends.

Explanation:
Given that my baseball miff is 8 paper clips long and my friend’s baseball miff is 7 paper clips. So my baseball miff is 8 – 7 which is 1 paper clip longer than my friend’s. And my friend’s baseball miff is 1 paper clip shorter than mine.

Question 3.
Modeling Real Life
Your desk is 7 paper clips longer than your friend’s. Your friend’s desk is 14 paper clips long. How long is yours?
Big Ideas Math Answer Key Grade 1 Chapter 10 Measure and Compare Lengths 72
___________ paper clips long
Answer:
7 paper clips long.

Explanation:
Big-Ideas-Math-Answer-Key-Grade-1-Chapter-10-Measure-and-Compare-Lengths-61-1
As my desk is 7 paper clips longer than my friend’s and my friend’s desk is 14 paper clips long. So my desk will be 14 – 7= 7 paper clips long.

Review & Refresh

Use the picture to complete the number bond.

Question 4.
Big Ideas Math Answer Key Grade 1 Chapter 10 Measure and Compare Lengths 73
Answer:
By number bond, the total number of flowers is 5 + 3= 8 flowers.

Explanation:
Big-Ideas-Math-Answer-Key-Grade-1-Chapter-10-Measure-and-Compare-Lengths-73
This number bond explains to us how the numbers are joined together and how they break down into a certain number of parts. In the above image, we can see that there are five yellow flowers and three red flowers. By number bond, we will add all the flowers. So the total number of flowers is 5 + 3= 8 flowers. Now they break down by five yellow flowers in one circle and three red flowers in another circle.

Question 5.
Big Ideas Math Answer Key Grade 1 Chapter 10 Measure and Compare Lengths 74
Answer:
By number bond, the total number of cans is 3 + 3= 8 cans.

Explanation:
Big-Ideas-Math-Answer-Key-Grade-1-Chapter-10-Measure-and-Compare-Lengths-74

This number bond explains to us how the numbers are joined together and how they break down into a certain number of parts. In the above image, we can see that there are three green cans and three red cans. By number bond, we will add all the cans. So the total number of cans is 3 + 3= 6 flowers. Now they break down by three green cans in one circle and three red cans in another circle.

Measure and Compare Lengths Performance Task

Big Ideas Math Answers 1st Grade 1 Chapter 10 Measure and Compare Lengths 75

Question 1.
Use a piece of string to compare the routes from your house to the library, the post office, and the school.Order the routes from shortest to longest.
____________, ____________, ____________
Answer:
The order from shortest to longest is the post office, the library, and the school.

Explanation:
The route from my house to the library is three meters and the route from my house to the post office is two meters and the route from my house to the school is four meters. So the order from shortest to longest is the post office, the library, and the school.

Question 2.
Use a piece of string to measure the different routes from your house to your friend’s house. Color the route you would use to ride your bike to your friend’s house.
Answer:

Question 3.
a. The bakery is farther from your house than the pool. The park is closer to your house than the pool. Which place is closest to your house?
Park             Bakery              Pool
Answer:

b. Label the park, bakery, and pool on the map.
Answer:

Measure and Compare Lengths Chapter Practice

Order Objects by Length Homework & Practice 10.1

Question 1.
Order from longest to shortest.
Big Ideas Math Answers 1st Grade 1 Chapter 10 Measure and Compare Lengths 76
____________, _____________, ______________
Answer:
The order from longest to shortest is
Shark, Lobster, Fish.

Explanation:
In the above image, the longest is the shark, the largest is the lobster and the shortest is the fish.

Question 2.
MP Problem Solving
A green snake is shorter than a black snake. A brown snake is shorter than a black snake. Which snake is the longest?
green        block         brown
Answer:
The black snake is the longest.

Explanation:
Given that a green snake is shorter than a black snake and a brown snake is shorter than a black snake. So the longest snake is a black snake.

Compare Lengths Indirectly Homework & Practice 10.2

Question 3.
Circle the longer object.
Big Ideas Math Answers 1st Grade 1 Chapter 10 Measure and Compare Lengths 77
Answer:
We will circle the second image.

Explanation:
Big-Ideas-Math-Answers-1st-Grade-1-Chapter-10-Measure-and-Compare-Lengths-77
In the above image, the longest object is the second image. So we will circle the second image.

Question 4.
Circle the longer Object.
Big Ideas Math Answers 1st Grade 1 Chapter 10 Measure and Compare Lengths 78
Answer:
We will circle the second image.

Explanation:
Big-Ideas-Math-Answers-1st-Grade-1-Chapter-10-Measure-and-Compare-Lengths-78
In the above image, the longest object is the second image. So we will circle the second image.

Question 5.
Draw a line through the shorter object.
Big Ideas Math Answers 1st Grade 1 Chapter 10 Measure and Compare Lengths 79
Answer:
The first image is the shorter image.

Explanation:
Big-Ideas-Math-Answers-1st-Grade-1-Chapter-10-Measure-and-Compare-Lengths-79
In the above image, the shorter object is the first image. So we will circle the first image.

Measure Lengths Homework & Practice 10.3

Measure

Question 6.
Big Ideas Math Answers 1st Grade 1 Chapter 10 Measure and Compare Lengths 80
about ___________ color tiles
Answer:
The length of the object is about four colored tiles.

Explanation:
The object is measured by using the colored tile. As each tile is equal to one unit. So the length of the object is about four colored tiles, which means four units. We must take each tile without gaps or any overlaps between them.

Question 7.
Big Ideas Math Answers 1st Grade 1 Chapter 10 Measure and Compare Lengths 81
about ___________ color tiles
Answer:
The length of the object is about five colored tiles.

Explanation:
The object is measured by using the colored tile. As each tile is equal to one unit. So the length of the object is about five colored tiles, which means five units. We must take each tile without gaps or any overlaps between them.

Measure More Lengths Homework & Practice 10.4

Measure

Question 8.
Big Ideas Math Answers 1st Grade 1 Chapter 10 Measure and Compare Lengths 82
about ___________ color tiles
about ___________ paper clips
Answer:
The length of the object is about four colored tiles and the length of the object which is measured using a paper clip is two paper clips.

Explanation:
The object is measured by using the colored tile. As each tile is equal to one unit. So the length of the object is four colored tiles, which means four units. We must take each tile without gaps or any overlaps between them. And to measure the object with the paper clip, we will place the paper clip without gaps or overlaps between them. And each paper clip is equal to about one unit. So the length of the object using the paper clip is two paper clips, which means two units. We have noticed that if we take a colored tile, then the length of the object will be four color tiles. And if we take a paper clip then the length of the pencil will be two paper clips. So the length is doubled while measuring with color tile than the paper clip.

Question 9.
Modeling Real Life
Your hockey stick is 18 paper clips long. Is your hockey stick more than or less than 18 color tiles long?
Big Ideas Math Answers 1st Grade 1 Chapter 10 Measure and Compare Lengths 83
Circle: more than 18 less than 18
Tell how you know:
Answer:
My hockey stick has more than 18 color tiles.

Explanation:
My hockey stick has more than 18 color tiles. As we can see from the above problems that while measuring with the paper clip we need fewer paper clips than the color tiles. And the length is doubled while measuring with color tile than the paper clip. So there will be more than 18 color tiles than the paper clips for the hockey stick.

Solve Compare Problems Involving Length Homework & Practice 10.5

Question 10.
Your water bottle is 5 paper clips long. Your Friend’s is 4 paper clips long. How many paper clips longer is your water bottle?
Big Ideas Math Answers 1st Grade 1 Chapter 10 Measure and Compare Lengths 84
Answer:
One paper clip long.

Explanation:
Big-Ideas-Math-Answers-1st-Grade-1-Chapter-10-Measure-and-Compare-Lengths-84
Given that my water bottle is 5 paper clips long and my friend’s water bottle is 4 paper clips long. So there will be one paper clip longer than my friend’s.

Question 11.
Your bookshelf is 19 color tiles long. Your friend’s is 15 color tiles long. How many tiles longer is your bookshelf?
Big Ideas Math Answers 1st Grade 1 Chapter 10 Measure and Compare Lengths 85
Big Ideas Math Answers 1st Grade 1 Chapter 10 Measure and Compare Lengths 86
Answer:
My bookshelf is four tiles longer.

Explanation:
Big-Ideas-Math-Answers-1st-Grade-1-Chapter-10-Measure-and-Compare-Lengths-86
Given that my bookshelf is 19 paper clips long and my friend’s bookshelf is 15 paper clips long. So there will be four paper clips longer than my friend’s.

Question 12.
MP Reasoning
Your pencil is 6 color tiles long. Your friend’s is 3 color tiles long. Complete the sentences.

Your pencil is ____________ color tiles ____________ than your friend’s.
Your friend’s pencil is ____________ color tiles ____________ than yours.
Answer:
Your pencil is three color tiles longer than your friend’s.
Your friend’s pencil is three color tiles shorter than yours.

Explanation:
As my pencil is six color tiles long, and my friend’s pencil is three color tiles long. So my pencil will be three color tiles longer than my friend’s pencil. And my friend’s pencil is three color tiles shorter than my pencil.

Conclusion:

Hope the information regarding Big Ideas Math Answers Grade 1 Chapter 10 Measure and Compare Lengths is beneficial for you. If you have any doubts please post the comments in the below-mentioned comment box. Share this pdf link with your friends and help them to overcome the difficulties in maths.

Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions

Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions

For better preparation and good quality of education, we have shared the Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions for high school students. Provided BIM Math Book Algebra 2 Solution Key are aligned topic-wise as per the latest common core 2019 curriculum Big Ideas Math Textbook. So, students can easily understand the concepts of chapter 1 linear functions.

Linear functions topics are very hard to solve for high school students but it can be easy by practicing the questions covered in the Solution Key of Big Ideas Math Book Algebra 2 Chapter 1 Linear Functions. Also, you can download all topics BIM Algebra 2 Ch 1 Linear Functions Answer Key simply by clicking the available links below.

Big Ideas Math Book Algebra 2 Answer Key Chapter 1 Linear Functions

Step by Step Solutions explained for all the questions in the BIM Algebra 2 Ch 1 Solution key are easy to learn & understand as they are prepared by the professional subject experts. After referring to the topic wise Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions, every high school student can become a pro in the concepts and gain better subject knowledge. So, click on the respective links and start preparing each topic of the Big Ideas math book Algebra 2 Chapter 1 Linear Functions efficiently.

Linear Functions Maintaining Mathematical Proficiency

Evaluate.

Question 1.
5 • 23 + 7

Question 2.
4 – 2(3 + 2)2

Question 3.
48 ÷ 42 + \(\frac{3}{5}\)

Question 4.
50 ÷ 52 • 2

Question 5.
\(\frac{1}{2}\)(22+ 22)

Question 6.
\(\frac{1}{6}\)(6 + 18) – 22

Graph the transformation of the figure.

Question 7.
Translate the rectangle 1 unit right and 4 units up.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1

Question 8.
Reflect the triangle in the y-axis. Then translate 2 units left.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 2

Question 9.
Translate the trapezoid 3 units down. Then reflect in the x-axis.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 3

Question 10.
ABSTRACT REASONING Give an example to show why the order of operations is important when evaluating a numerical expression. Is the order of transformations of figures important? Justify your answer.

Linear Functions Maintaining Mathematical Practices

Monitoring Progress

Use a graphing calculator to graph the equation using the standard viewing window and a square viewing window. Describe any differences in the graphs.

Question 1.
y = 2x – 3

Question 2.
y = | x + 2 |

Question 3.
y = -x2 + 1

Question 4.
y = \(\sqrt{x-1}\)

Question 5.
y = x3 – 2

Question 6.
y = 0.25x3

Determine whether the viewing window is square. Explain.

Question 7.
-8 ≤ x ≤ 8, -2 ≤ y ≤ 8

Question 8.
-7 ≤ x ≤ 8, -2 ≤ y ≤ 8

Question 9.
-6 ≤ x ≤ 9, -2 ≤ y ≤ 8

Question 10.
-2 ≤ x≤ 2, -3 ≤ y ≤ 3

Question 11.
-4 ≤ x ≤ 5, -3 ≤ y ≤ 3

Question 12.
-4 ≤ x ≤ 4, -3 ≤ y ≤ 3

Lesson 1.1 Parent Functions and Transformations

Essential Question

What are the characteristics of some of the basic parent functions?

EXPLORATION 1
Identifying Basic Parent Functions
Work with a partner.
Graphs of eight basic parent functions are shown below. Classify each function as constant, linear, absolute value, quadratic, square root, cubic, reciprocal, or exponential. Justify your reasoning.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 4

Communicate Your Answer

Question 2.
What are the characteristics of some of the basic parent functions?

Question 3.
Write an equation for each function whose graph is shown in Exploration 1. Then use a graphing calculator to verify that your equations are correct.

1.1 Lesson

Monitoring Progress

Question 1.
Identify the function family to which g belongs. Compare the graph of g to the graph of its parent function.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 5

Graph the function and its parent function. Then describe the transformation.

Question 2.
g(x) = x + 3

Question 3.
h(x) = (x – 2)2

Question 4.
n(x) = – | x |

Graph the function and its parent function. Then describe the transformation.

Question 5.
g(x) = 3x

Question 6.
h(x) = \(\frac{3}{2}\)x2

Question 7.
c(x) = 0.2|x|

Use a graphing calculator to graph the function and its parent function. Then describe the transformations

Question 8.
h(x) = –\(\frac{1}{4}\)x + 5

Question 9.
d(x) = 3(x – 5)2 – 1

Question 10.
The table shows the amount of fuel in a chainsaw over time. What type of function can you use to model the data? When will the tank be empty?
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 6

Parent Functions and Transformations 1.1 Exercises

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
The function f(x) = x2 is the ______ of f(x) = 2x2 – 3.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 1

Question 2.
DIFFERENT WORDS, SAME QUESTION Which is different? Find “both” answers.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 7
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–6, identify the function family to which f belongs. Compare the graph of f to the graph of its parent function.

Question 3.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 8
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 3

Question 4.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 9
Answer:

Question 5.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 10
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 5

Question 6.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 11
Answer:

Question 7.
MODELING WITH MATHEMATICS
At 8:00 A.M., the temperature is 43°F. The temperature increases 2°F each hour for the next 7 hours. Graph the temperatures over time t (t = 0 represents 8:00 A.M.). What type of function can you use to model the data? Explain.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 7

Question 8.
MODELING WITH MATHEMATICS
You purchase a car from a dealership for $10,000. The trade-in value of the car each year after the purchase is given by the function f(x) = 10,000 – 250x2. What type of function models the trade-in value?
Answer:

In Exercises 9–18, graph the function and its parent function. Then describe the transformation.

Question 9.
g(x) = x + 4
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 9

Question 10.
f(x) = x – 6
Answer:

Question 11.
f(x) = x2 – 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 11

Question 12.
h(x) = (x+ 4)2
Answer:

Question 13.
g(x) = | x – 5 |
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 13

Question 14.
f(x) = 4 + | x |
Answer:

Question 15.
h(x) = -x2
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 15
Answer:

Question 16.
g(x) = -x
Answer:

Question 17.
f(x) = 3
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 17

Question 18.
f(x) = -2
Answer:

In Exercises 19–26, graph the function and its parent function. Then describe the transformation.

Question 19.
f(x) = \(\frac{1}{3}\)x
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 19

Question 20.
g(x) = 4x
Answer:

Question 21.
f(x) = 2x2

Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 21

Question 22.
h(x) = \(\frac{1}{3}\)x2

Answer:

Question 23.
h(x) = \(\frac{3}{4}\)x
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 23

Question 24.
g(x) = \(\frac{4}{3}\)x
Answer:

Question 25.
h(x) = 3 | x |
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 25

Question 26.
f(x) = \(\frac{1}{2}\) | x |
Answer:

In Exercises 27–34, use a graphing calculator to graph the function and its parent function. Then describe the transformations.

Question 27.
f(x) = 3x + 2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 27

Question 28.
h(x) = -x + 5
Answer:

Question 29.
h(x) = -3 | x | – 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 29

Question 30.
f(x) = \(\frac{3}{4}\) | x | + 1
Answer:

Question 31.
g(x) = \(\frac{1}{2}\)x2 – 6
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 31

Question 32.
f(x) = 4x2 – 3
Answer:

Question 33.
f(x) = -(x + 3)2 + \(\frac{1}{4}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 33

Question 34.
g(x) = – | x – 1 | – \(\frac{1}{2}\)
Answer:

ERROR ANALYSIS In Exercises 35 and 36, identify and correct the error in describing the transformation of the parent function.

Question 35.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 12
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 35

Question 36.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 13
Answer:

MATHEMATICAL CONNECTIONS In Exercises 37 and 38, find the coordinates of the figure after the transformation.

Question 37.
Translate 2 units down.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 13.1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 37

Question 38.
Reflect in the x-axis.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 15
Answer:

USING TOOLS In Exercises 39–44, identify the function family and describe the domain and range. Use a graphing calculator to verify your answer.

Question 39.
g(x) = | x + 2 | – 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 39

Question 40.
h(x) = | x – 3 | + 2
Answer:

Question 41.
g(x) = 3x + 4
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 41

Question 42.
f(x) = -4x + 11
Answer:

Question 43.
f(x) = 5x2 – 2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 43

Question 44.
f(x) = -2x2 + 6
Answer:

Question 45.
MODELING WITH MATHEMATICS The table shows the speeds of a car as it travels through an intersection with a stop sign. What type of function can you use to model the data? Estimate the speed of the car when it is 20 yards past the intersection.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 16
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 45

Question 46.
THOUGHT PROVOKING In the same coordinate plane, sketch the graph of the parent quadratic function and the graph of a quadratic function that has no x-intercepts. Describe the transformation(s) of the parent function.
Answer:

Question 47.
USING STRUCTURE Graph the functions f(x) = | x – 4 | and g(x) = | x | – 4. Are they equivalent? Explain.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 47

Question 48.
HOW DO YOU SEE IT? Consider the graphs of f, g, and h.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 17
a. Does the graph of g represent a vertical stretch or a vertical shrink of the graph of f? Explain your reasoning.
b. Describe how to transform the graph of f to obtain the graph of h.
Answer:

Question 49.
MAKING AN ARGUMENT Your friend says two different translations of the graph of the parent linear function can result in the graph of f(x) = x – 2. Is your friend correct? Explain.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 49

Question 50.
DRAWING CONCLUSIONS A person swims at a constant speed of 1 meter per second. What type of function can be used to model the distance the swimmer travels? If the person has a 10-meter head start, what type of transformation does this represent? Explain.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 17.1
Answer:

Question 51.
PROBLEM SOLVING You are playing basketball with your friends. The height (in feet) of the ball above the ground t seconds after a shot is released from your hand is modeled by the function f(t) = -16t2 + 32t + 5.2.
a. Without graphing, identify the type of function that models the height of the basketball.
b. What is the value of t when the ball is released from your hand? Explain your reasoning.
c. How many feet above the ground is the ball when it is released from your hand? Explain.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 51

Question 52.
MODELING WITH MATHEMATICS The table shows the battery lives of a computer over time. What type of function can you use to model the data? Interpret the meaning of the x-intercept in this situation.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 18
Answer:

Question 53.
REASONING Compare each function with its parent function. State whether it contains a horizontal translation, vertical translation, both, or neither. Explain your reasoning.
a. f(x) = 2 | x | – 3
b. f(x) = (x – 8)2
c. f(x) = | x + 2 | + 4
d. f(x) = 4x2

Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 53

Question 54.
CRITICAL THINKING
Use the values -1, 0, 1, and 2 in the correct box so the graph of each function intersects the x-axis. Explain your reasoning.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 19
Answer:

Maintaining Mathematical Proficiency

Determine whether the ordered pair is a solution of the equation. (Skills Review Handbook)

Question 55.
f(x) = | x + 2 |; (1, -3)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 55

Question 56.
f(x) = | x | – 3; (-2, -5)
Answer:

Question 57.
f(x) = x – 3; (5, 2)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 57

Question 58.
f(x) = x – 4; (12, 8)
Answer:

Find the x-intercept and the y-intercept of the graph of the equation. (Skills Review Handbook)

Question 59.
y = x
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 59

Question 60.
y = x + 2
Answer:

Question 61.
3x + y = 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 61

Question 62.
x – 2y = 8
Answer:

Lesson 1.2 Transformations of Linear and Absolute Value Functions

Essential Question

How do the graphs of y = f(x) + k, y = f(x – h), and y = -f(x) compare to the graph of the parent function f?

EXPLORATION 1
Transformations of the Parent Absolute Value Function
Work with a partner.
Compare the graph of the function
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 20
y = | x | + k Transformation
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 20.1
to the graph of the parent function
f(x) = | x |.

EXPLORATION 2
Transformations of the Parent Absolute Value Function
Work with a partner.
Compare the graph of the function
y = | x – h | Transformation
to the graph of the parent function
f(x) = | x |. Parent function
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 22

EXPLORATION 3
Transformation of the Parent Absolute Value Function
Work with a partner.
Compare the graph of the function
y = – | x | Transformation
to the graph of the parent function
f(x) = | x | Parent function
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 23

Communicate Your Answer

Question 4.
Transformation How do the graphs of y = f (x) + k, y = f(x – h), and y = -f(x) compare to the graph of the parent function f?

Question 5.
Compare the graph of each function to the graph of its parent function f. Use a graphing calculator to verify your answers are correct.
a. y = \([\sqrt{x}/latex] – 4
b. y = [latex][\sqrt{x + 4}/latex]
c. y = –[latex][\sqrt{x}/latex]
d. y = x2 + 1
e. y = (x – 1)2
f. y = -x2

1.2 Lesson

Monitoring Progress

Write a function g whose graph represents the indicated transformation of the graph of f. Use a graphing calculator to check your answer.

Question 1.
f(x) = 3x; translation 5 units up

Question 2.
f(x) = | x | – 3; translation 4 units to the right

Question 3.
f(x) = – | x + 2 | – 1; reflection in the x-axis

Question 4.
f(x) = [latex]\frac{1}{2}\)x+ 1; reflection in the y-axis

Write a function g whose graph represents the indicated transformation of the graph of f. Use a graphing calculator to check your answer.

Question 5.
f(x) = 4x+ 2; horizontal stretch by a factor of 2

Question 6.
f(x) = | x | – 3; vertical shrink by a factor of \(\frac{1}{3}\)

Question 7.
Let the graph of g be a translation 6 units down followed by a reflection in the x-axis of the graph of f(x) = | x |. Write a rule for g. Use a graphing calculator to check your answer.

Question 8.
WHAT IF? In Example 5, your revenue function is f(x) = 3x. How does this affect your profit for 100 downloads?

Transformations of Linear and Absolute Value Functions 1.2 Exercises

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
The function g(x) = | 5x |- 4 is a horizontal ___________ of the function f(x) = | x | – 4.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.2 Question 1

Question 2.
WHICH ONE DOESN’T BELONG? Which transformation does not belong with the other three? Explain your reasoning.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 24
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–8, write a function g whose graph represents the indicated transformation of the graph of f. Use a graphing calculator to check your answer.

Question 3.
f(x) = x – 5; translation 4 units to the left
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.2 Question 3

Question 4.
f(x) = x + 2; translation 2 units to the right
Answer:

Question 5.
f(x) = | 4x + 3 | + 2; translation 2 units down
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.2 Question 5

Question 6.
f(x) = 2x – 9; translation 6 units up
Answer:

Question 7.
f(x) = 4 – | x + 1 |
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 25
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.2 Question 7

Question 8.
f(x) = | 4x | + 5
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 26
Answer:

Question 9.
WRITING Describe two different translations of the graph of f that result in the graph of g.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 27
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.2 Question 9

Question 10.
PROBLEM SOLVING You open a café. The function f(x) = 4000x represents your expected net income (in dollars) after being open x weeks. Before you open, you incur an extra expense of $12,000. What transformation of f is necessary to model this situation? How many weeks will it take to pay off the extra expense?
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 28
Answer:

In Exercises 11–16, write a function g whose graph represents the indicated transformation of the graph of f. Use a graphing calculator to check your answer.

Question 11.
f(x) = -5x+ 2; reflection in the x-axis
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.2 Question 11

Question 12.
f(x) = \(\frac{1}{2}\)x – 3; reflection in the x-axis
Answer:

Question 13.
f(x) = | 6x | – 2; reflection in the y-axis
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.2 Question 13

Question 14.
f(x) = | 2x – 1 | + 3; reflection in the y-axis
Answer:

Question 15.
f(x) = -3 + | x – 11 |; reflection in the y-axis
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.2 Question 15

Question 16.
f(x) = -x+ 1; reflection in the y-axis

In Exercises 17–22, write a function g whose graph represents the indicated transformation of the graph of f. Use a graphing calculator to check your answer.

Question 17.
f(x) = x + 2; vertical stretch by a factor of 5
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.2 Question 17

Question 18.
f(x) = 2x+ 6; vertical shrink by a factor of \(\frac{1}{2}\)
Answer:

Question 19.
f(x) = | 2x | + 4; horizontal shrink by a factor of \(\frac{1}{2}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.2 Question 19

Question 20.
f(x) = | x+ 3 | ; horizontal stretch by a factor of 4
Answer:

Question 21.
f(x) = -2 | x – 4 | + 2
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 29
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.2 Question 21

Question 22.
f(x) = 6 – x
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 30
Answer:

ANALYZING RELATIONSHIPS In Exercises 23–26, match the graph of the transformation of f with the correct equation shown. Explain your reasoning.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 31

Question 23.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 32
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.2 Question 23

Question 24.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 33
Answer:

Question 25.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 34
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.2 Question 25

Question 26.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 35
Answer:
A. y = 2f(x)
B. y = f(2x)
C. y = f(x + 2)
D. y = f(x) + 2

In Exercises 27–32, write a function g whose graph represents the indicated transformations of the graph of f.

Question 27.
f(x) = x; vertical stretch by a factor of 2 followed by a translation 1 unit up
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.2 Question 27

Question 28.
f(x) = x; translation 3 units down followed by a vertical shrink by a factor of \(\frac{1}{3}\)
Answer:

Question 29.
f(x) = | x | ; translation 2 units to the right followed by a horizontal stretch by a factor of 2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.2 Question 29

Question 30.
f(x) = | x |; reflection in the y-axis followed by a translation 3 units to the right
Answer:

Question 31.
f(x) = | x |
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 36
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.2 Question 31

Question 32.
f(x) = | x |
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 37
Answer:

ERROR ANALYSIS In Exercises 33 and 34, identify and correct the error in writing the function g whose graph represents the indicated transformations of the graph of f.

Question 33.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 38
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.2 Question 33

Question 34.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 39
Answer:

Question 35.
MAKING AN ARGUMENT Your friend claims that when writing a function whose graph represents a combination of transformations, the order is not important. Is your friend correct? Justify your answer.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.2 Question 35

Question 36.
MODELING WITH MATHEMATICS During a recent period of time, bookstore sales have been declining. The sales (in billions of dollars) can be modeled by the function f(t) = –\(\frac{7}{5}\)t + 17.2, where t is the number of years since 2006. Suppose sales decreased at twice the rate. How can you transform the graph of f to model the sales? Explain how the sales in 2010 are affected by this change.
Answer:

MATHEMATICAL CONNECTIONS For Exercises 37–40, describe the transformation of the graph of f to the graph of g. Then find the area of the shaded triangle.

Question 37.
f(x) = | x – 3 |
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 40
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.2 Question 37

Question 38.
f(x) = – | x | – 2
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 41
Answer:

Question 39.
f(x) = -x + 4
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 42
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.2 Question 39

Question 40.
f(x) = x – 5
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 43
Answer:

Question 41.
ABSTRACT REASONING The functions f(x) = mx + b and g(x) = mx + c represent two parallel lines.
a. Write an expression for the vertical translation of the graph of f to the graph of g.
b. Use the definition of slope to write an expression for the horizontal translation of the graph of f to the graph of g.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.2 Question 41

Question 42.
HOW DO YOU SEE IT? Consider the graph of f(x) = mx + b. Describe the effect each transformation has on the slope of the line and the intercepts of the graph.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 44
a. Reflect the graph of f in the y-axis.
b. Shrink the graph of f vertically by a factor of \(\frac{1}{3}\).
c. Stretch the graph of f horizontally by a factor of 2.
Answer:

Question 43.
REASONING The graph of g(x) = -4 |x | + 2 is a reflection in the x-axis, vertical stretch by a factor of 4, and a translation 2 units down of the graph of its parent function. Choose the correct order for the transformations of the graph of the parent function to obtain the graph of g. Explain your reasoning.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.2 Question 43

Question 44.
THOUGHT PROVOKING You are planning a cross-country bicycle trip of 4320 miles. Your distance d (in miles) from the halfway point can be modeled by d = 72 |x – 30 |, where x is the time (in days) and x = 0 represents June 1. Your plans are altered so that the model is now a right shift of the original model. Give an example of how this can happen. Sketch both the original model and the shifted model.
Answer:

Question 45.
CRITICAL THINKING Use the correct value 0, -2, or 1 with a, b, and c so the graph of g(x) = a|x – b | + c is a reflection in the x-axis followed by a translation one unit to the left and one unit up of the graph of f(x) = 2 |x – 2 | + 1. Explain your reasoning.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.2 Question 45

Maintaining Mathematical Proficiency

Evaluate the function for the given value of x. (Skills Review Handbook)

Question 46.
f(x) = x + 4; x = 3
Answer:

Question 47.
f(x) = 4x – 1; x = -1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.2 Question 47

Question 48.
f(x) = -x + 3; x = 5
Answer:

Question 49.
f(x) = -2x – 2; x = -1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.2 Question 49

Create a scatter plot of the data. (Skills Review Handbook)

Question 50.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 45
Answer:

Question 51.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 46
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.2 Question 51

Linear Functions Study Skills Taking Control of Your Class Time

1.1 – 1.2 What Did You Learn?

Core Vocabulary
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 47

Core Concepts

Section 1.1
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 48

Section 1.2
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 49

Mathematical Practices

Question 1.
How can you analyze the values given in the table in Exercise 45 on page 9 to help you determine what type of function models the data?

Question 2.
Explain how you would round your answer in Exercise 10 on page 16 if the extra expense is $13,500.

Study Skills

Taking Control of Your Class Time

Question 1.
Sit where you can easily see and hear the teacher, and the teacher can see you.

Question 2.
Pay attention to what the teacher says about math, not just what is written on the board.

Question 3.
Ask a question if the teacher is moving through the material too fast.

Question 4.
Try to memorize new information while learning it.

Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 50

Question 5.
Ask for clarification if you do not understand something.

Question 6.
Think as intensely as if you were going to take a quiz on the material at the end of class.

Question 7.
Volunteer when the teacher asks for someone to go up to the board.

Question 8.
At the end of class, identify concepts or problems for which you still need clarification.

Question 9.
Use the tutorials at BigIdeasMath.com for additional help.

Linear Functions 1.1-1.2 Quiz

Identify the function family to which g belongs. Compare the graph of the function to the graph of its parent function. (Section 1.1)

Question 1.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 51

Question 2.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 52

Question 3.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 53

Graph the function and its parent function. Then describe the transformation. (Section 1.1)

Question 4.
f(x) = \(\frac{3}{2}\)

Question 5.
f(x) = 3x

Question 6.
f(x) = 2(x – 1)2

Question 7.
f(x) = – | x + 2 | – 7

Question 8.
f(x) = \(\frac{1}{4}\)x2 + 1

Question 9.
f(x) = –\(\frac{1}{2}\)x – 4

Write a function g whose graph represents the indicated transformation of the graph of f. (Section 1.2)

Question 10.
f(x) = 2x + 1; translation 3 units up

Question 11.
f(x) = -3 | x – 4 | ; vertical shrink by a factor of \(\frac{1}{2}\)

Question 12.
f(x) = 3 | x + 5 |; reflection in the x-axis

Question 13.
f(x) = \(\frac{1}{3}\)x – \(\frac{2}{3}\) ; translation 4 units left

Write a function g whose graph represents the indicated transformations of the graph of f. (Section 1.2)

Question 14.
Let g be a translation 2 units down and a horizontal shrink by a factor of \(\frac{2}{3}\) of the graph of f(x) =x.

Question 15.
Let g be a translation 9 units down followed by a reflection in the y-axis of the graph of f(x) = x.

Question 16.
Let g be a reflection in the x-axis and a vertical stretch by a factor of 4 followed by a translation 7 units down and 1 unit right of the graph of f(x) = | x |.

Question 17.
Let g be a translation 1 unit down and 2 units left followed by a vertical shrink by a factor of \(\frac{1}{2}\) of the graph of f(x) = | x |.

Question 18.
The table shows the total distance a new car travels each month after it is purchased. What type of function can you use to model the data? Estimate the mileage after 1 year. (Section 1.1)
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 54

Question 19.
The total cost of an annual pass plus camping for x days in a National Park can be modeled by the function f(x) = 20x+ 80. Senior citizens pay half of this price and receive an additional $30 discount. Describe how to transform the graph of f to model the total cost for a senior citizen. What is the total cost for a senior citizen to go camping for three days? (Section 1.2)

Lesson 1.3 Modeling with Linear Functions

Essential Question
How can you use a linear function to model and analyze a real-life situation?

EXPLORATION 1
Modeling with a Linear Function
Work with a partner.
A company purchases a copier for $12,000. The spreadsheet shows how the copier depreciates over an 8-year period.
a. Write a linear function to represent the value V of the copier as a function of the number t of years.
b. Sketch a graph of the function. Explain why this type of depreciation is called straight line depreciation.
c. Interpret the slope of the graph in the context of the problem.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 55

EXPLORATION 2
Modeling with Linear Functions
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 56
Work with a partner. Match each description of the situation with its corresponding graph. Explain your reasoning.
a. A person gives $20 per week to a friend to repay a $200 loan.
b. An employee receives $12.50 per hour plus $2 for each unit produced per hour.
c. A sales representative receives $30 per day for food plus $0.565 for each mile driven.
d. A computer that was purchased for $750 depreciates $100 per year.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 56.1

Communicate Your Answer

Question 3.
How can you use a linear function to model and analyze a real-life situation?

Question 4.
Use the Internet or some other reference to find a real-life example of straight line depreciation.
a. Use a spreadsheet to show the depreciation.
b. Write a function that models the depreciation.
c. Sketch a graph of the function.

1.3 Lesson

Monitoring Progress

Question 1.
The graph shows the remaining balance y on a car loan after making x monthly payments. Write an equation of the line and interpret the slope and y-intercept. What is the remaining balance after 36 payments?
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 57

Question 2.
WHAT IF? Maple Ridge charges a rental fee plus a $10 fee per student. The total cost is $1900 for 140 students. Describe the number of students that must attend for the total cost at Maple Ridge to be less than the total costs at the other two venues. Use a graph to justify your answer.

Question 3.
The table shows the humerus lengths (in centimeters) and heights (in centimeters) of several females.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 58
a. Do the data show a linear relationship? If so, write an equation of a line of fit and use it to estimate the height of a female whose humerus is 40 centimeters long.
b. Use the linear regression feature on a graphing calculator to find an equation of the line of best fit for the data. Estimate the height of a female whose humerus is 40 centimeters long. Compare this height to your estimate in part (a).

Modeling with Linear Functions 1.3 Exercises

Question 1.
COMPLETE THE SENTENCE The linear equation y = \(\frac{1}{2}\)x + 3 is written in ____________ form.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.3 Question 1

Question 2.
VOCABULARY A line of best fit has a correlation coefficient of -0.98. What can you conclude about the slope of the line?
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–8, use the graph to write an equation of the line and interpret the slope.

Question 3.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 59
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.3 Question 3

Question 4.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 60
Answer:

Question 5.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 61
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.3 Question 5

Question 6.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 62
Answer:

Question 7.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 63
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.3 Question 7

Question 8.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 64
Answer:

Question 9.
MODELING WITH MATHEMATICS Two newspapers charge a fee for placing an advertisement in their paper plus a fee based on the number of lines in the advertisement. The table shows the total costs for different length advertisements at the Daily Times. The total cost y (in dollars) for an advertisement that is x lines long at the Greenville Journal is represented by the equation y = 2x + 20. Which newspaper charges less per line? How many lines must be in an advertisement for the total costs to be the same?
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 65
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.3 Question 9.1
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.3 Question 9.2

Question 10.
PROBLEM SOLVING While on vacation in Canada, you notice that temperatures are reported in degrees Celsius. You know there is a linear relationship between Fahrenheit and Celsius, but you forget the formula. From science class, you remember the freezing point of water is 0°C or 32°F, and its boiling point is 100°C or 212°F.
a. Write an equation that represents degrees Fahrenheit in terms of degrees Celsius.
b. The temperature outside is 22°C. What is this temperature in degrees Fahrenheit?
c. Rewrite your equation in part (a) to represent degrees Celsius in terms of degrees Fahrenheit.
d. The temperature of the hotel pool water is 83°F. What is this temperature in degrees Celsius?
Answer:

ERROR ANALYSIS In Exercises 11 and 12, describe and correct the error in interpreting the slope in the context of the situation.

Question 11.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 66
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.3 Question 11

Question 12.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 67
Answer:

In Exercises 13–16, determine whether the data show a linear relationship. If so, write an equation of a line of fit. Estimate y when x = 15 and explain its meaning in the context of the situation.

Question 13.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 68
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.3 Question 13.1
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.3 Question 13.2

Question 14.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 69
Answer:

Question 15.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 70
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.3 Question 15.1
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.3 Question 15.2

Question 16.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 71
Answer:

Question 17.
MODELING WITH MATHEMATICS The data pairs (x, y) represent the average annual tuition y (in dollars) for public colleges in the United States x years after 2005. Use the linear regression feature on a graphing calculator to find an equation of the line of best fit. Estimate the average annual tuition in 2020. Interpret the slope and y-intercept in this situation.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 72
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.3 Question 17

Question 18.
MODELING WITH MATHEMATICS The table shows the numbers of tickets sold for a concert when different prices are charged. Write an equation of a line of fit for the data. Does it seem reasonable to use your model to predict the number of tickets sold when the ticket price is $85? Explain.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 73
Answer:

USING TOOLS In Exercises 19–24, use the linear regression feature on a graphing calculator to find an equation of the line of best fit for the data. Find and interpret the correlation coefficient.

Question 19.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 74
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.3 Question 19

Question 20.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 75
Answer:

Question 21.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 76
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.3 Question 21

Question 22.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 77
Answer:

Question 23.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 78
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.3 Question 23

Question 24.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 79
Answer:

Question 25.
OPEN-ENDED Give two real-life quantities that have
(a) a positive correlation,
(b) a negative correlation, and
(c) approximately no correlation. Explain.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.3 Question 25

Question 26.
HOW DO YOU SEE IT? You secure an interest-free loan to purchase a boat. You agree to make equal monthly payments for the next two years. The graph shows the amount of money you still owe.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 80
a. What is the slope of the line? What does the slope represent?
b. What is the domain and range of the function? What does each represent?
c. How much do you still owe after making payments for 12 months?
Answer:

Question 27.
MAKING AN ARGUMENT A set of data pairs has a correlation coefficient r = 0.3. Your friend says that because the correlation coefficient is positive, it is logical to use the line of best fit to make predictions. Is your friend correct? Explain your reasoning.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.3 Question 27

Question 28.
THOUGHT PROVOKING Points A and B lie on the line y = -x + 4. Choose coordinates for points A, B, and C where point C is the same distance from point A as it is from point B. Write equations for the lines connecting points A and C and points B and C.
Answer:

Question 29.
ABSTRACT REASONING If x and y have a positive correlation, and y and z have a negative correlation, then what can you conclude about the correlation between x and z? Explain.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.3 Question 29

Question 30.
MATHEMATICAL CONNECTIONS Which equation has a graph that is a line passing through the point (8, -5) and is perpendicular to the graph of y = -4x + 1?
A. y = \(\frac{1}{4}\)x – 5
B. y = -4x + 27
C. y = –\(\frac{1}{4}\)x – 7
D. y = \(\frac{1}{4}\)x – 7
Answer:

Question 31.
PROBLEM SOLVING You are participating in an orienteering competition. The diagram shows the position of a river that cuts through the woods. You are currently 2 miles east and 1 mile north of your starting point, the origin. What is the shortest distance you must travel to reach the river?
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 81
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.3 Question 31.1
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.3 Question 31.2

Question 32.
ANALYZING RELATIONSHIPS Data from North American countries show a positive correlation between the number of personal computers per capita and the average life expectancy in the country.
a. Does a positive correlation make sense in this situation? Explain.
b. Is it reasonable to conclude that giving residents of a country personal computers will lengthen their lives? Explain.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 82
Answer:

Maintaining Mathematical Proficiency

Solve the system of linear equations in two variables by elimination or substitution. (Skills Review Handbook)

Question 33.
3x + y = 7
-2x – y = 9
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.3 Question 33

Question 34.
4x + 3y = 2
2x – 3y = 1
Answer:

Question 35.
2x + 2y = 3
x = 4y – 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.3 Question 35

Question 36.
y = 1 + x
2x + y = -2
Answer:

Question 37.
\(\frac{1}{2}\)x + 4y = 4
2x – y = 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.3 Question 37.1
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.3 Question 37.2

Question 38.
y = x – 4
4x + y = 26
Answer:

Lesson 1.4 Solving Linear Systems

Essential Question
How can you determine the number of solutions of a linear system?
A linear system is consistent when it has at least one solution. A linear system is inconsistent when it has no solution.

EXPLORATION 1
Recognizing Graphs of Linear Systems
Work with a partner.
Match each linear system with its corresponding graph. Explain your reasoning. Then classify the system as consistent or inconsistent.
a. 2x – 3y = 3
-4x + 6y = 6
Big Ideas Math Algebra 2 Answer Key Chapter 1 Linear Functions 83

b. 2x – 3y = 3
x + 2y = 5
Big Ideas Math Algebra 2 Answer Key Chapter 1 Linear Functions 84

c. 2x – 3y = 3
-4x + 6y = 6
Big Ideas Math Algebra 2 Answer Key Chapter 1 Linear Functions 85

EXPLORATION 2
Solving Systems of Linear Equations
Work with a partner.
Solve each linear system by substitution or elimination. Then use the graph of the system below to check your solution.
a. 2x + y = 5
x – y = 1
Big Ideas Math Algebra 2 Answer Key Chapter 1 Linear Functions 86

b. x+ 3y = 1
-x + 2y = 4
Big Ideas Math Algebra 2 Answer Key Chapter 1 Linear Functions 87

c. x + y = 0
3x + 2y = 1
Big Ideas Math Algebra 2 Answer Key Chapter 1 Linear Functions 88

Communicate Your Answer

Question 3.
How can you determine the number of solutions of a linear system?

Big Ideas Math Algebra 2 Answer Key Chapter 1 Linear Functions 89
Question 4.
Suppose you were given a system of three linear equations in three variables. Explain how you would approach solving such a system.

Question 5.
Apply your strategy in Question 4 to solve the linear system.
Big Ideas Math Algebra 2 Answer Key Chapter 1 Linear Functions 90

1.4 Lesson

Monitoring Progress

Question 1.
x – 2y + z = -11
3x + 2y – z = 7
-x + 2y + 4z = -9

Question 2.
x + y – z = -1
4x + 4y – 4z = -2
3x + 2y + z = 0

Question 3.
x + y + z = 8
x – y + z = 8
2x + y + 2z = 16

Question 4.
In Example 3, describe the solutions of the system using an ordered triple in terms of y.

Question 5.
WHAT IF? On the first day, 10,000 tickets sold, generating $356,000 in revenue. The number of seats sold in Sections A and B are the same. How many lawn seats are still available?

Solving Linear Systems 1.4 Exercises

Vocabulary and Core Concept Check

Question 1.
VOCABULARY The solution of a system of three linear equations is expressed as a(n)__________.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 1

Question 2.
WRITING Explain how you know when a linear system in three variables has infinitely many solutions.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–8, solve the system using the elimination method.

Question 3.
x + y – 2z = 5
-x + 2y + z = 2
2x + 3y – z = 9
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 3.1
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 3.2

Question 4.
x + 4y – 6z = -1
2x – y + 2z = -7
-x + 2y – 4z = 5
Answer:

Question 5.
2x + y – z = 9
-x + 6y + 2z = -17
5x + 7y + z = 4
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 5.1
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 5.2

Question 6.
3x + 2y – z = 8
-3x + 4y + 5z = -14
x – 3y + 4z = -14

Answer:

Question 7.
2x + 2y + 5z = -1
2x – y + z = 2
2x + 4y – 3z = 14
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 7.1
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 7.2

Question 8.
3x + 2y – 3z = -2
7x – 2y + 5z = -14
2x + 4y + z = 6
Answer:

ERROR ANALYSIS In Exercises 9 and 10, describe and correct the error in the first step of solving the system of linear equations.

Big Ideas Math Algebra 2 Answer Key Chapter 1 Linear Functions 90.1

Question 9.
Big Ideas Math Algebra 2 Answer Key Chapter 1 Linear Functions 91
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 9

Question 10.
Big Ideas Math Algebra 2 Answer Key Chapter 1 Linear Functions 92
Answer:

In Exercises 11–16, solve the system using the elimination method.

Question 11.
3x – y + 2z = 4
6x – 2y + 4z = -8
2x – y + 3z = 10
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 11

Question 12.
5x + y – z = 6
x + y + z = 2
12x + 4y = 10
Answer:

Question 13.
x + 3y – z = 2
x + y – z = 0
3x + 2y – 3z = -1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 13.1

Question 14.
x + 2y – z = 3
-2x – y + z = -1
6x – 3y – z = -7
Answer:

Question 15.
x + 2y + 3z = 4
-3x + 2y – z = 12
-2x – 2y – 4z = -14
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 15.1

Question 16.
-2x – 3y + z = -6
x + y – z = 5
7x + 8y – 6z = 31
Answer:

Question 17.
MODELING WITH MATHEMATICS Three orders are placed at a pizza shop. Two small pizzas, a liter of soda, and a salad cost $14; one small pizza, a liter of soda, and three salads cost $15; and three small pizzas, a liter of soda, and two salads cost $22. How much does each item cost?
Big Ideas Math Algebra 2 Answer Key Chapter 1 Linear Functions 93
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 17.1
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 17.2
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 17.3

Question 18.
MODELING WITH MATHEMATICS Sam’s Furniture Store places the following advertisement in the local newspaper. Write a system of equations for the three combinations of furniture. What is the price of each piece of furniture? Explain.
Big Ideas Math Algebra 2 Answer Key Chapter 1 Linear Functions 94
Answer:

In Exercises 19–28, solve the system of linear equations using the substitution method.

Question 19.
-2x + y + 6z = 1
3x + 2y + 5z = 16
7x + 3y – 4z = 11
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 19.1
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 19.2

Question 20.
x – 6y – 2z = -8
-x + 5y + 3z = 2
3x – 2y – 4z = 18
Answer:

Question 21.
x + y + z = 4
5x + 5y + 5z = 12
x – 4y + z = 9
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 21.1

Question 22.
x + 2y = -1
-x + 3y + 2z = -4
-x + y – 4z = 10
Answer:

Question 23.
2x – 3y + z = 10
y + 2z = 13
z = 5
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 23.1

Question 24.
x = 4
x + y = -6
4x – 3y + 2z = 26
Answer:

Question 25.
x + y – z = 4
3x + 2y + 4z = 17
-x + 5y + z = 8
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 25.1
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 25.2

Question 26.
2x – y – z = 15
4x + 5y + 2z = 10
-x – 4y + 3z = -20
Answer:

Question 27.
4x + y + 5z = 5
8x + 2y + 10z = 10
x – y – 2z = -2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 27.1
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 27.2

Question 28.
x + 2y – z = 3
2x + 4y – 2z = 6
-x – 2y + z = -6
Answer:

Question 29.
PROBLEM SOLVING The number of left-handed people in the world is one-tenth the number of right-handed people. The percent of right-handed people is nine times the percent of left-handed people and ambidextrous people combined. What percent of people are ambidextrous?
Big Ideas Math Algebra 2 Answer Key Chapter 1 Linear Functions 95
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 29.1
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 29.2
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 29.3

Question 30.
MODELING WITH MATHEMATICS Use a system of linear equations to model the data in the following newspaper article. Solve the system to find how many athletes finished in each place.
Big Ideas Math Algebra 2 Answer Key Chapter 1 Linear Functions 96
Answer:

Question 31.
WRITING Explain when it might be more convenient to use the elimination method than the substitution method to solve a linear system. Give an example to support your claim.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 31

Question 32.
REPEATED REASONING Using what you know about solving linear systems in two and three variables, plan a strategy for how you would solve a system that has four linear equations in four variables.
Answer:

MATHEMATICAL CONNECTIONS In Exercises 33 and 34, write and use a linear system to answer the question.

Question 33.
The triangle has a perimeter of 65 feet. What are the lengths of sides ℓ, m, and n?
Big Ideas Math Algebra 2 Answer Key Chapter 1 Linear Functions 97
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 33.1
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 33.2

Question 34.
What are the measures of angles A, B, and C?
Big Ideas Math Algebra 2 Answer Key Chapter 1 Linear Functions 97.1
Answer:

Question 35.
OPEN-ENDED Consider the system of linear equations below. Choose nonzero values for a, b, and c so the system satisfies the given condition. Explain your reasoning.
x + y + z = 2
ax + by + cz = 10
x – 2y + z = 4
a. The system has no solution.
b. The system has exactly one solution.
c. The system has infinitely many solutions.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 35

Question 36.
MAKING AN ARGUMENT A linear system in three variables has no solution. Your friend concludes that it is not possible for two of the three equations to have any points in common. Is your friend correct? Explain your reasoning.
Answer:

Question 37.
PROBLEM SOLVING A contractor is hired to build an apartment complex. Each 840-square-foot unit has a bedroom, kitchen, and bathroom. The bedroom will be the same size as the kitchen. The owner orders 980 square feet of tile to completely cover the floors of two kitchens and two bathrooms. Determine how many square feet of carpet is needed for each bedroom.
Big Ideas Math Algebra 2 Answer Key Chapter 1 Linear Functions 98
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 37.1
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 37.2

Question 38.
THOUGHT PROVOKING Does the system of linear equations have more than one solution? Justify your answer.
4x + y + z = 0
2x + \(\frac{1}{2}\)y – 3z = 0
-x – \(\frac{1}{4}\)y – z = 0
Answer:

Question 39.
PROBLEM SOLVING A florist must make 5 identical bridesmaid bouquets for a wedding. The budget is $160, and each bouquet must have 12 flowers. Roses cost $2.50 each, lilies cost $4 each, and irises cost $2 each. The florist wants twice as many roses as the other two types of flowers combined.
a. Write a system of equations to represent this situation, assuming the florist plans to use the maximum budget.
b. Solve the system to find how many of each type of flower should be in each bouquet.
c. Suppose there is no limitation on the total cost of the bouquets. Does the problem still have exactly one solution? If so, find the solution. If not, give three possible solutions.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 39.1
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 39.2
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 39.3

Question 40.
HOW DO YOU SEE IT? Determine whether the system of equations that represents the circles has no solution, one solution, or infinitely many solutions. Explain your reasoning.
Big Ideas Math Algebra 2 Answer Key Chapter 1 Linear Functions 99
Answer:

Question 41.
CRITICAL THINKING Find the values of a, b, and c so that the linear system shown has (-1, 2, -3) as its only solution. Explain your reasoning.
x + 2y – 3z = a
– x – y + z = b
2x + 3y – 2z = c
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 41

Question 42.
ANALYZING RELATIONSHIPS Determine which arrangement(s) of the integers -5, 2, and 3 produce a solution of the linear system that consist of only integers. Justify your answer.
x – 3y + 6z = 21
_x + _y + _z = -30
2x – 5y + 2z = -6
Answer:

Question 43.
ABSTRACT REASONING Write a linear system to represent the first three pictures below. Use the system to determine how many tangerines are required to balance the apple in the fourth picture. Note:The first picture shows that one tangerine and one apple balance one grapefruit.
Big Ideas Math Algebra 2 Answer Key Chapter 1 Linear Functions 100
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 43

Maintaining Mathematical Proficiency

Simplify. (Skills Review Handbook)

Question 44.
(x – 2)2

Answer:

Question 45.
(3m + 1)2

Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 45

Question 46.
(2z – 5)2

Answer:

Question 47.
(4 – y)2

Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 47

Write a function g described by the given transformation of f(x) =∣x∣− 5.(Section 1.2)

Question 48.
translation 2 units to the left
Answer:

Question 49.
reflection in the x-axis
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 49

Question 50.
translation 4 units up
Answer:

Question 51.
vertical stretch by a factor of 3
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 51

Linear Functions Performance Task: Secret of the Hanging Baskets

1.3–1.4 What Did You Learn?

Core Vocabulary
Big Ideas Math Algebra 2 Answer Key Chapter 1 Linear Functions 101

Core Concepts
Section 1.3
Writing an Equation of a Line, p. 22
Finding a Line of Fit, p. 24
Section 1.4
Solving a Three-Variable System, p. 31
Solving Real-Life Problems, p. 33

Mathematical Practices

Question 1.
Describe how you can write the equation of the line in Exercise 7 on page 26 using only one of the labeled points.

Question 2.
How did you use the information in the newspaper article in Exercise 30 on page 35 to write a system of three linear equations?

Question 3.
Explain the strategy you used to choose the values for a, b, and c in Exercise 35 part (a) on page 35.

Performance Task

Secret of the Hanging Baskets
A carnival game uses two baskets hanging from springs at different heights. Next to the higher basket is a pile of baseballs. Next to the lower basket is a pile of golf balls. The object of the game is to add the same number of balls to each basket so that the baskets have the same height. But there is a catch—you only get one chance. What is the secret to winning the game?
Big Ideas Math Algebra 2 Solutions Chapter 1 Linear Functions 102
To explore the answers to this question and more, go to BigIdeasMath.com.
Big Ideas Math Algebra 2 Solutions Chapter 1 Linear Functions 103

Linear Functions Chapter Review

Graph the function and its parent function. Then describe the transformation.

Question 1.
f(x) = x + 3

Question 2.
g(x) = | x | – 1

Question 3.
h(x) = \(\frac{1}{2}\)x2

Question 4.
h(x) = 4

Question 5.
f(x) = -| x | – 3

Question 6.
g(x) = -3(x + 3)2

Write a function g whose graph represents the indicated transformations of the graph of f. Use a graphing calculator to check your answer.

Question 7.
f(x) = | x |; reflection in the x-axis followed by a translation 4 units to the left

Question 8.
f(x) = | x | ; vertical shrink by a factor of \(\frac{1}{2}\) followed by a translation 2 units up

Question 9.
f(x) = x; translation 3 units down followed by a reflection in the y-axis

Question 10.
The table shows the total number y (in billions) of U.S. movie admissions each year for x years. Use a graphing calculator to find an equation of the line of best fit for the data.
Big Ideas Math Algebra 2 Solutions Chapter 1 Linear Functions 103.1

Question 11.
You ride your bike and measure how far you travel. After 10 minutes, you travel 3.5 miles. After 30 minutes, you travel 10.5 miles. Write an equation to model your distance. How far can you ride your bike in 45 minutes?

Question 12.
x + y + z = 3
-x + 3y + 2z = -8
x = 4z

Question 13.
2x – 5y – z = 17
x + y + 3z = 19
-4x + 6y + z = -20

Question 14.
x + y + z = 2
2x – 3y + z = 11
-3x + 2y – 2z = -13

Question 15.
x + 4y – 2z = 3
x + 3y + 7z = 1
2x + 9y – 13z = 2

Question 16.
x – y + 3z = 6
x – 2y = 5
2x – 2y + 5z = 9

Question 17.
x + 2y = 4
x + y + z = 6
3x + 3y + 4z = 28

Question 18.
A school band performs a spring concert for a crowd of 600 people. The revenue for the concert is $3150. There are 150 more adults at the concert than students. How many of each type of ticket are sold?
Big Ideas Math Algebra 2 Solutions Chapter 1 Linear Functions 104

Linear Functions Chapter Test

Write an equation of the line and interpret the slope and y-intercept.

Question 1.
Big Ideas Math Algebra 2 Solutions Chapter 1 Linear Functions 105

Question 2.
Big Ideas Math Algebra 2 Solutions Chapter 1 Linear Functions 106

Solve the system. Check your solution, if possible.

Question 3.
-2x + y + 4z = 5
x + 3y – z = 2
4x + y – 6z = 11

Question 4.
y = \(\frac{1}{2}\)z
x + 2y + 5z = 2
3x + 6y – 3z = 9

Question 5.
x – y + 5z = 3
2x + 3y – z = 2
-4x – y – 9z = -8

Graph the function and its parent function. Then describe the transformation.

Question 6.
f(x) = | x – 1 |

Question 7.
f(x) = (3x)2

Question 8.
f(x) = 4

Match the transformation of f(x) = x with its graph. Then write a rule for g.

Question 9.
g(x) = 2f(x) + 3
Big Ideas Math Algebra 2 Solutions Chapter 1 Linear Functions 107

Question 10.
g(x) = 3f(x) – 2
Big Ideas Math Answer Key Algebra 2 Chapter 1 Linear Functions 108

Question 11.
g(x) = -2f(x) – 3
Big Ideas Math Answer Key Algebra 2 Chapter 1 Linear Functions 109

Question 12.
A bakery sells doughnuts, muffins, and bagels. The bakery makes three times as many doughnuts as bagels. The bakery earns a total of $150 when all 130 baked items in stock are sold. How many of each item are in stock? Justify your answer.
Big Ideas Math Answer Key Algebra 2 Chapter 1 Linear Functions 110

Question 13.
A fountain with a depth of 5 feet is drained and then refilled. The water level (in feet) after t minutes can be modeled by f(t) = \(\frac{1}{4}\)|t – 20 |. A second fountain with the same depth is drained and filled twice as quickly as the first fountain. Describe how to transform the graph of f to model the water level in the second fountain after t minutes. Find the depth of each fountain after 4 minutes. Justify your answers.

Linear Functions Cumulative Assessment

Question 1.
Describe the transformation of the graph of f(x) = 2x – 4 represented in each graph.
Big Ideas Math Answer Key Algebra 2 Chapter 1 Linear Functions 111

Question 2.
The table shows the tuition costs for a private school between the years 2010 and 2013.
Big Ideas Math Answer Key Algebra 2 Chapter 1 Linear Functions 112
a. Verify that the data show a linear relationship. Then write an equation of a line of fit.
b. Interpret the slope and y-intercept in this situation.
c. Predict the cost of tuition in 2015.

Question 3.
Your friend claims the line of best fit for the data shown in the scatter plot has a correlation coefficient close to 1. Is your friend correct? Explain your reasoning.
Big Ideas Math Answer Key Algebra 2 Chapter 1 Linear Functions 113

Question 4.
Order the following linear systems from least to greatest according to the number of solutions.
A. 2x + 4y – z = 7
14x + 28y – 7z = 49
-x + 6y + 12z = 13
B. 3x – 3y + 3z = 5
-x + y – z = 5
-x + y – z = 8
14x – 3y + 12z = 108
C. 4x – y + 2z = 18
-x + 2y + z = 11
3x + 3y – 4z = 44

Question 5.
You make a DVD of three types of shows: comedy, drama, and reality-based. An episode of a comedy lasts 30 minutes, while a drama and a reality-based episode each last 60 minutes. The DVDs can hold 360 minutes of programming.
a. You completely fill a DVD with seven episodes and include twice as many episodes of a drama as a comedy. Create a system of equations that models the situation.
b. How many episodes of each type of show are on the DVD in part (a)?
c. You completely fill a second DVD with only six episodes. Do the two DVDs have a different number of comedies? dramas? reality-based episodes? Explain.

Question 6.
The graph shows the height of a hang glider over time. Which equation models the situation?
A. y + 450 = 10x
B. 10y = -x+ 450
C. \(\frac{1}{10}\)y = -x + 450
D. 10x + y = 450
Big Ideas Math Answer Key Algebra 2 Chapter 1 Linear Functions 114

Question 7.
Let f(x) = x and g(x) = -3x – 4. Select the possible transformations (in order) of the graph of f represented by the function g.
A. reflection in the x-axis
B. reflection in the y-axis
C. vertical translation 4 units down
D. horizontal translation 4 units right
E. horizontal shrink by a factor of \(\frac{1}{3}\)
F. vertical stretch by a factor of 3

Question 8.
Choose the correct equality or inequality symbol which completes the statement below about the linear functions f and g. Explain your reasoning.
Big Ideas Math Answer Key Algebra 2 Chapter 1 Linear Functions 115

Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations

Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations available here are free of cost. Students can seek homework help and solve the questions covered in the Ch 1 lessons. Also, these solutions of BIM Algebra 1 Ch 1 Answer Book improve their math skills and become math proficient in the future. Choosing the right track during the preparation can change your education life. So, accessing the Big Ideas Math Book Solutions Algebra 1 Ch 1 solving linear equations is a better & fun-learning way for students of high school. Also, it clarifies all your doubts about the concepts of chapter 2 BIM Textbook Solutions.

Big Ideas Math Book Algebra 1 Answer Key Chapter 1 Solving Linear Equations

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Solving Linear Equations Maintaining Mathematical Proficiency

Add or subtract.

Question 1.
-5 + (-2)
Answer:
-5 + (-2 ) = -7

Explanation:
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
-5 + ( -2 ) = -5 – 2
= – ( 5 + 2 )
= -7
Hence, from the above,
We can conclude that,
-5 + ( -2 ) = -7

Question 2.
0 + (-13)
Answer:
0 + -13 = -13

Explanation:
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
0 + ( -13 ) = 0 – 13
= -13
Hence, from the above,
We can conclude that,
0 + ( -13 ) = -13

Question 3.
-6 + 14
Answer:
-6 + 14 = 8

Explanation:
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
-6 + 14 = +14 – 6
= +8
= 8
Hence, from the above,
We can conclude that
-6 +14 = 8

Question 4.
19 – (-13)
Answer:
19 – ( -13 ) = 32

Explanation:
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
19 – ( -13 ) = 10 + 9 + 10 + 3
= 20 + 12
= 32
Hence, from the above,
We can conclude that,
19 – (-13 ) = 32

Question 5.
-1 – 6
Answer:
-1 -6 = -7

Explanation:
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
-1 – 6 = – ( 1 + 6 )
= -7
Hence, from the above,
We can conclude that
-1 -6 = -7

Question 6.
– 5 – (-7)
Answer:
-5 – ( -7 ) = 2

Explanation:
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
-5 – ( -7 ) = -5 + 7
= 7 – 5
= 2
Hence, from the above,
We can conclude that
-5 – ( -7 ) = 2

Question 7.
17 + 5
Answer:
17 + 5 = 22

Explanation:
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
17 + 5 = 15 + 2 + 5
= 22
Hence, from the above,
We can conclude that
17 + 5 = 22

Question 8.
8 + (-3)
Answer:
8 + ( -3 ) = 5

Explanation:
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
8 + ( -3 ) = 8 – 3
= 5
Hence, from the above,
We can conclude that
8  + ( -3 ) = 5

Question 9.
11 – 15
Answer:
11 – 15 = -4

Explanation:
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
11 – 15 = -15 + 11
= – ( 15 – 11 )
= -4
Hence, from the above,
We can conclude that,
11 – 15 = -4

Multiply or divide.

Question 10.
-3(8)
Answer:
-3(8) = -3 × 8
= -24

Explanation:
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
-3(8) = -3 × +8
= -24
Hence, from the above,
We can conclude that
-3 ( 8 ) = -24

Question 11.
-7 • (-9)
Answer:
-7 . ( -9 ) = -7 × +9
= -63

Explanation:
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
-7 . ( 9 ) = -7 × +9
= -63
Hence, from the above,
We can conclude that
-7 . ( 9 ) = -63

Question 12.
4 • (-7)
Answer:
4 . ( -7 ) = 4 × ( -7 )
= -28

Explanation:
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
4. (-7 ) = +4 × -7
= -28
Hence, from the above,
We can conclude that
4 . ( -7 ) = -28

Question 13.
-24 ÷ (-6)
Answer:
-24 ÷ ( -6 ) = 4

Explanation:
We know that,
A) – ÷ – = +
B) + ÷ – = –
C) + ÷ + = +
D) – ÷ + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
-24 ÷ ( -6 ) = + ( 24 ÷ 6 )
= 4
Hence, from the above,
We can conclude that
-24 ÷ ( -6 ) = 4

Question 14.
-16 ÷ 2
Answer:
-16 ÷ 2 = -8

Explanation:
We know that,
A) – ÷ – = +
B) + ÷ – = –
C) + ÷ + = +
D) – ÷ + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
-16 ÷ 2 = -8
Hence, from the above,
We can conclude that
-16 ÷ 2 = -8

Question 15.
12 ÷ (-3)
Answer:
12 ÷ ( -3 ) = -4

Explanation:
We know that,
A) – ÷ – = +
B) + ÷ – = –
C) + ÷ + = +
D) – ÷ + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
12 ÷ ( -3 )
= 12 ÷ -3
= -4
Hence, from the above,
We can conclude that
12 ÷ ( -3 ) = -4

Question 16.
6 • 8
Answer:
6 . 8 = 6 × 8
= 48

Explanation:
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
6 . 8 = 6 × 8
= 48
Hence, from the above,
We can conclude that
6 . 8 = 48

Question 17.
36 ÷ 6
Answer:
36 ÷ 6 = 6

Explanation:
We know that,
A) – ÷ – = +
B) + ÷ – = –
C) + ÷ + = +
D) – ÷ + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
36 ÷ 6 = ( 30 + 6 ) ÷ 6
= ( 30 ÷ 6 ) + ( 6 ÷ 6 )
= 5 + 1
= 6
Hence, from the above,
We can conclude that
36 ÷ 6 = 6

Question 18.
-3(-4)
Answer:
-3 ( -4 ) = -3 × -4
= -12

Explanation:
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
-3 ( -4 ) = -3 × -4
= +12
Hence, from the above,
We can conclude that
-3 ( – 4 ) = 12

Question 19.
ABSTRACT REASONING
Summarize the rules for
(a) adding integers,
(b) subtracting integers,
(c) multiplying integers, and
(d) dividing integers.
Give an example of each.
Answer:
a) Adding integers:
We know that,
The result of any mathematical operation i.e., Addition or subtraction follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
Example:
Let take the two numbers -2 and 3
So,
The addition of -2 and 3 is:
-2 + 3 = +1
= 1 ( Since the big number has a positive sign )

b) Subtracting integers:
We know that,
The result of any mathematical operation i.e., Addition or subtraction follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
Example:
Let take the two numbers -3 and +8
So,
The subtraction of -3 and +8 is:
-3 – ( +8 ) = -3 – 8
= – ( 3 + 8 )
= -11 ( Since both the numbers have a negative sign )

c) Multiplying integers:
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
Example:
Let take the two numbers +8 and -3
So,
The multiplication of +8 and -3 is:
+8 ( -3 ) = 8 × -3
= -24 ( Since + × – = – )

d) Dividing integers:
We know that,
A) – ÷ – = +
B) + ÷ – = –
C) + ÷ + = +
D) – ÷ + = –
Example:
Let take the two numbers -12 and -2
So,
The division of -12 and -2 is:
-12 ÷ ( -2 ) = -12 ÷ -2
= ( -10 + -2 ) ÷ -2
= ( -10 ÷ -2 ) + ( -2 ÷ -2 )
= 5 + 1
= 6
Hence, from the above,
We can conclude all the rules for the four basic mathematical operations.

Solving Linear Equations Monitoring Progress

Solve the problem and specify the units of measure.

Question 1.
The population of the United States was about 280 million in 2000 and about 310 million in 2010. What was the annual rate of change in population from 2000 to 2010?
Answer:
The annual rate of change in population from 2000 to 2010 is: 30 million

Explanation:
It is given that the population of the United States was about 280 million in 2000 and about 310 million in 2010.
So,
The annual rate of change in population from 2000 to 2010 = ( The population of United States in 2010 ) – ( The population of United States in 2000 )
= 310 – 280
= 30 million
Hence, from the above,
We can conclude that the annual rate of change in population from 2000 to 2010 is: 30 million

Question 2.
You drive 240 miles and use 8 gallons of gasoline. What was your car’s gas mileage (in miles per gallon)?
Answer:
Your car’s gas mileage ( in miles per gallon ) is: 30

Explanation:
It is given that you drive 240 miles and use 8 gallons of gasoline.
So,
The mileage of your car = ( The total number of miles driven by your car ) ÷ ( The number of gallons of gasoline used by your car )
= 240 ÷ 8
= ( 160 + 80 ) ÷ 8
= ( 160 ÷ 8 ) + ( 80 ÷ 8 )
= 20 + 10
= 30 miles
Hence, from the above,
We can conclude that the mileage of your car is: 30 miles per gallon

Question 3.
A bathtub is in the shape of a rectangular prism. Its dimensions are 5 feet by 3 feet by 18 inches. The bathtub is three-fourths full of water and drains at a rate of 1 cubic foot per minute. About how long does it take for all the water to drain?
Answer:
The total time taken for the water to drain is: 2,430 minutes

Explanation:
It is given that a bathtub is in the shape of a rectangular prism. Its dimensions are 5 feet by 3 feet by 18 inches. The bathtub is three-fourths full of water and drains at a rate of 1 cubic foot per minute.
So,
The volume of the rectangular prism = The dimensions of the rectangular prism
= 5 × 3 × 18 × 12
= 3,240 cubic feet
Now,
The volume of the bathtub which is three-fourths full of water = \(\) {3}{4}[\latex] × 3240
= 2,430 cubic feet
It is also given that that the bathtub drains at a rate of 1 cubic foot per minute.
So,
The time taken to drain 2,430 cubic feet of water in minutes = 2,430 × 1
= 2,430 minutes
Hence, from the above,
We can conclude that the time taken for the water to drain from the bathtub at a rate of 1 cubic foot per minute is: 2,430 minutes

Lesson 1.1 Solving Simple Equations

Essential Question
How can you use simple equations to solve real-life problems?

Exploration 1
Measuring Angles

Work with a partner. Use a protractor to measure the angles of each quadrilateral. Copy and complete the table to organize your results. (The notation m∠A denotes the measure of angle A.) How precise are your measurements?

Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 2

EXPLORATION 2
Making a Conjecture

Work with a partner. Use the completed table in Exploration 1 to write a conjecture about the sum of the angle measures of a quadrilateral. Draw three quadrilaterals that are different from those in Exploration 1 and use them to justify your conjecture.
Answer:
The completed table is:

From the above table,
We can say that the sum of all the angle in any quadrilateral is: 360 degrees
So,
From the above table,
The angles in Quadrilateral a is: 60 degrees, 125 degrees, 120 degrees, and 55 degrees
We know that,
The quadrilateral will have n angles based on the shape.
The shape which has more than 3 sides is called a Quadrilateral.
Ex: Rectangle, Square, Pentagon, Hexagon, etc.

From the above quadrilaterals,
We can say that all the sides in each quadrilateral are equal.
So,
The angles in each quadrilateral are also equal.
So,
In a rectangle, there are 4 sides
So,
By measuring, we can observe that each angle of a rectangle is: 90 degrees
Hence,
The sum of all angles in a rectangle = 90 + 90 + 90 + 90 = 360 degrees
In a pentagon, there are 5 sides
By measuring, we can observe that each rectangle of a pentagon
Hence,
The sum of all the angles in a pentagon = = 72+ 72 + 72 + 72 + 72 = 360 degrees
In a Hexagon, there are 6 sides
So,
By measuring, we can observe that each angle of a hexagon is: 60 degrees
Hence,
The sum of all angle is a Hexagon = 60 + 60 + 60 + 60 + 60 + 60 = 360 degrees
Hence, from the above-drawn quadrilaterals,
We can conclude that our conjecture is proven.

EXPLORATION 3
Applying Your Conjecture

Work with a partner. Use the conjecture you wrote in Exploration 2 to write an equation for each quadrilateral. Then solve the equation to find the value of x. Use a protractor to check the reasonableness of your answer.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 3
Answer:
The given figure is:
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 3
From Exploration 2, the proven conjecture is:
The sum of all angles in a quadrilateral is: 360 degrees
Now,
In Quadrilateral a.,
By using the above-proven conjecture,
85 + 80 + 100 + x = 360 degrees
265 + x = 360 degrees
x = 360 – 265
= 95 degrees
So,
The angle x is: 95 degrees
In Quadrilateral b.,
By using the above-proven conjecture,
72 + 78 + 60 + x = 360 degrees
210 + x = 360 degrees
x = 360 – 210
= 150 degrees
So,
The angle of x is: 150 degrees
In Quadrilateral c.,
By using the above-proven conjecture,
90 + 90  +30 + x = 360 degrees
210 + x = 360 degrees
x = 360 – 210
= 150 degrees
So,
The angle of x is: 150 degrees

Communicate Your Answer

Question 4.
How can you use simple equations to solve real-life problems?

Question 5.
Draw your own quadrilateral and cut it out. Tear off the four corners of the quadrilateral and rearrange them to affirm the conjecture you wrote in Exploration 2. Explain how this affirms the conjecture.
Answer:
Your Quadrilateral is:

From the above Quadrilateral,
We can observe that the tear-off corners of the quadrilateral are joined and it becomes the triangle.
So,
In the above Quadrilateral, there are two quadrilaterals.
We know that,
The sum of all angles in a triangle is: 180 degrees
So,
The sum of all angles in the two triangles = 180 + 180 = 360 degrees
These two triangles form a quadrilateral.
So,
The sum of all angles in a quadrilateral is: 360 degrees
Hence, from the above,
We can conclude that we can prove Exploration 2’s conjecture by your own example.

1.1 Lesson

Monitoring Progress

Solve the equation. Justify each step. Check your solution.

Question 1.
n + 3 = -7
Answer:
The value of n is: -10

Explanation:
The given equation is:
n + 3 = -7
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
n + 3 = -7
n = -7 – (+3 )
n = -7 – 3
= -10
Hence from the above,
We can conclude that the value of n is: -10

Question 2.
g – \(\frac{1}{3}\) = –\(\frac{2}{3}\)
Answer:
The value of g is: –\(\frac{1}{3}\)

Explanation:
The given equation is:
g – \(\frac{1}{3}\) = –\(\frac{2}{3}\)
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
g – \(\frac{1}{3}\) = –\(\frac{2}{3}\)
g = –\(\frac{2}{3}\) + \(\frac{1}{3}\)
g = \(\frac{-2 + 1}{3}\)
g = \(\frac{-1}{3}\)
g = –\(\frac{1}{3}\)
Hence, fromthe above,
We can conclude that the value of g is: –\(\frac{1}{3}\)

Question 3.
-6.5 = p + 3.9
Answer:
The value of p is: -10.4

Explanation:
The given equation is:
-6.5 = p + 3.9
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
-6.5 = p + 3.9
p = -6.5 – 3.9
= – ( 6.5 + 3.9 )
= – 10.4
Hence, from the above,
We can conclude that the value of p is: -10.4

Monitoring Progress

Solve the equation. Justify each step. Check your solution.

Question 4.
\(\frac{y}{3}\) = -6
Answer:
The value of y is: -18

Explanation:
The given equation is:
\(\frac{y}{3}\) = -6
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
\(\frac{y}{3}\) = -6
\(\frac{y}{1}\) × \(\frac{1}{3}\) = -6
\(\frac{y}{1}\) = -6 ÷ \(\frac{1}{3}\)
y = -6 × -3
y = -18
Hence, from the above,
We can conclude that the value of y is: -18

Question 5.
9π = πx
Answer:
The value of x is: 9

Explanation:
The given equation is:
9π = πx
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
9π = πx
9 × π = π × x
x = ( 9 × π ) ÷ π
x = 9
Hence, from the above,
We can conclude that the value of x is: 9

Question 6.
0.05w = 1.4
Answer:
The value of w is: 28

Explanation:
The given equation is:
0.05w = 1.4
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
0.05w = 1.4
0.05 × w = 1.4
\(\frac{5}{100}\) × w = \(\frac{14}{10}\)
w = \(\frac{14}{10}\) ÷ \(\frac{5}{100}\)
w = \(\frac{14}{10}\) × \(\frac{100}{5}\)
w = \(\frac{14 × 100}{10 × 5}\)
w = \(\frac{28}{1}\)
w = 28
Hence, from the above,
We can conclude that the value of w is: 28

Monitoring Progress

Question 7.
Suppose Usain Bolt ran 400 meters at the same average speed that he ran the 200 meters. How long would it take him to run 400 meters? Round your answer to the nearest hundredth of a second.
Answer:
The time it took for him to run 400 meters is: 0.50 seconds

Explanation:
It is given that Usain Bolt ran 400 meters at the same average speed that he ran the 200 meters.
We know that,
Speed = Distance ÷ Time
But, it is given that the average speed is the same.
Hence,
Speed = Constant
So,
Since speed is constant, distance is directly proportional to time.
So,
The time taken by Usain Bolt to run 400 meters = 200 ÷ 400
= ( 2 × 100 ) ÷ ( 4 × 100 )
= 10 ÷ 20
= 0.50 seconds ( 0.5 and 0.50 are the same values Only for the representation purpose, we will add ‘0’ after 5 )
Hence from the above,
We can conclude that the time is taken by Usain Bolt to run 400 meters when rounded-off to the nearest hundredth is: 0.50 seconds

Monitoring Progress

Question 8.
You thought the balance in your checking account was $68. When your bank statement arrives, you realize that you forgot to record a check. The bank statement lists your balance as $26. Write and solve an equation to find the amount of the check that you forgot to record.
Answer:
The amount of the check that you forgot to record is: $42

Explanation:
It is given that you thought the balance in your checking account was $68 and when your bank statement arrives, you realize that you forgot to record a check and the bank statement lists your balance as $26.
Now,
Let the amount you forgot to record be: x
So,
The total balance in your checking account = ( The listed balance ) + ( The amount that you forgot to record a check )
68 = 26 + x
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
68 = 26 + x
x = 68 – 26
x = $42
Hence, from the above,
We can conclude that the amount that forgot to record is: $42

Solving Simple Equations 1.1 Exercises

Monitoring Progress and Modeling with Mathematics

In Exercises 5–14, solve the equation. Justify each step. Check your solution.

Question 1.
VOCABULARY Which of the operations +, -, ×, and ÷ are inverses of each other?
Answer:
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
Hence, from the above,
We can conclude that,
+ is inverse of –  and vice-versa
× is inverse of ÷ and vice-versa

Question 2.
VOCABULARY Are the equations -2x = 10 and -5x = 25 equivalent? Explain.
Answer:
The equations -2x = 10 and -5x = 25 are equivalent

Explanation:
The given equations are:
-2x = 10 and -5x = 25
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
We know that,
A) – ÷ – = +
B) + ÷ – = –
C) + ÷ + = +
D) – ÷ + = –
So,
From -2x = 10,
x = 10 ÷ ( -2 )
x = -10 ÷ 2
x = -5
From -5x = 25,
x = 25 ÷ ( -5 )
x = -25 ÷ 5
x = -5
Hence, from the above,
We can conclude that the equations -2x = 10 and -5x = 25 are equivalent

Question 3.
WRITING Which property of equality would you use to solve the equation 14x = 56? Explain.
Answer:
The given equation is:
14x = 56
So,
It can be re-written as
14 × x = 56
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
x = 56 ÷ 14
x = 4
Hence, from the above,
We can conclude that the value of x is: 4

Question 4.
WHICH ONE DOESN’T BELONG? Which expression does not belong with the other three? Explain your reasoning.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 501
Answer:
The equation C) does not belong to the other three

Explanation:
Let the given equations be named as A), B), C), and D)
So,
The given equations are:
A) 8 = x ÷ 2
B) 3 = x ÷ 4
C) x – 6 = 5
D) x ÷ 3 = 9
So,
From the above equations,
The equations A, B), and D) are dividing the numbers whereas equation C) subtracting the numbers
Hence, from the above,
We can conclude that,
The equation C) does not belong to the other three.

Question 5.
x + 5 = 8

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q5

Question 6.
m + 9 = 2
Answer:
The value of m is: -5

Explanation:

Question 7.
y – 4 = 3

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q7

Question 8.
s – 2 = 1
Answer:
The value of s is: 3

Explanation;

Question 9.
w + 3 = -4

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q9

Question 10.
n – 6 = -7
Answer:
The value of n is: -1

Explanation:

Question 11.
-14 = p – 11

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q11

Question 12.
0 = 4 + q
Answer:
The value of q is: -4

Explanation;

Question 13.
r + (-8) = 10

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q13

Question 14.
t – (-5) = 9
Answer:
The value of t is: 4

Explanation;

Question 15.
MODELING WITH MATHEMATICS
A discounted amusement park ticket costs $12.95 less than the original price p. Write and solve an equation to find the original price.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 4

Answer:
The equation for the original price is:
p = x + $12.95

Explanation:
It is given that a discounted amusement park ticket costs $12.95 less than the original price p.
So,
Let the discounted amusement park ticket be: x
The given original price is: p
So,
The discounted amusement park ticket cost = p – $12.95
x = p – 12.95
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
p = x + $12.95
Hence, from the above,
We can conclude that the equation for the original price is:
p = x + $12.95

Question 16.
MODELING WITH MATHEMATICS
You and a friend are playing a board game. Your final score x is 12 points less than your friend’s final score. Write and solve an equation to find your final score.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 5

Answer:
Your final score is:
x = ( The score of your friend ) – 12

Explanation:
It is given that you and a friend are playing a board game. Your final score x is 12 points less than your friend’s final score.
So,
The scores table is shown below:
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 5
From the above table,
The final score of your friend is: 195
Let the final score of yours is: x
So,
x = ( The final score of your friend ) – 12
= 195 – 12
= 183 points
Hence, from the above,
We can conclude that the final score of yours is: 183 points

USING TOOLS
The sum of the angle measures of a quadrilateral is 360°. In Exercises 17–20, write and solve an equation to find the value of x. Use a protractor to check the reasonableness of your answer.

Question 17.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 6

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q17

Question 18.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 7

Answer:
The value of x is: 85 degrees

Explanation:
We know that,
The sum of angles in a quadrilateral is: 360 degrees
So,
150 + 48 + 77 + x = 360
275 + x = 360
x = 360 – 275
x = 85 degrees
Hence, from the above,
We can conclude that the value of x is: 85 degrees

Question 19.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 8

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q19

Question 20.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 9

Answer:
The value of x is: 100 degrees

Explanation:
We know that,
The sum of all angles in a quadrilateral is: 360 degrees
So,
115 + 85 + 60 + x = 360
260 + x = 360
x = 360 – 260
x = 100 degrees
Hence, from the above,
We can conclude that the value of x is: 100 degrees

In Exercises 21–30, solve the equation. Justify each step. Check your solution.

Question 21.
5g = 20

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q21

Question 22.
4q = 52
Answer:
The value of g is: 13

Explanation:
The given equation is:
4g = 52
4 × g = 52
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
g = 52 ÷ 4
= ( 44 + 8 ) ÷ 4
= ( 44 ÷ 4 ) + ( 8 ÷ 4 )
= 11 + 2
= 13
Hence, from the above,
We can conclude that the value of g is: 13

Question 23.
p ÷ 5 = 3

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q23

Question 24.
y ÷ 7 = 1
Answer:
The value of y is: 7

Explanation:
The given equation is:
y ÷ 7 = 1
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
y = 1 × 7
y = 7
Hence, from the above,
We can conclude that the value of y is: 7

Question 25.
-8r = 64

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q25

Question 26.
x ÷(-2) = 8
Answer:
The value of x is: -16

Explanation:
The given equation is:
x ÷ ( -2 ) = 8
We know that,
A) – ÷ – = +
B) + ÷ – = –
C) + ÷ + = +
D) – ÷ + = –
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
x ÷ ( -2 ) = 8
x = 8 × ( -2 )
x = -16
Hence, from the above,
We can conclude that the value of x is: -16

Question 27.
\(\frac{x}{6}\) = 8

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q27

Question 28.
\(\frac{w}{-3}\) = 6
Answer:
The value of w is: -18

Explanation:
The given equation is:
\(\frac{w}{-3}\) = 6
We know that,
A) – ÷ – = +
B) + ÷ – = –
C) + ÷ + = +
D) – ÷ + = –
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
\(\frac{w}{-3}\) = 6
w = 6 × ( -3 )
w = -18
Hence, from the above,
We can conclude that the value of w is: -18

Question 29.
-54 = 9s

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q29

Question 30.
-7 = \(\frac{t}{7}\)
Answer:
The value of t is: -49

Explanation:
The given equation is:
-7 = \(\frac{t}{7}\)
We know that,
A) – ÷ – = +
B) + ÷ – = –
C) + ÷ + = +
D) – ÷ + = –
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
-7 = \(\frac{t}{7}\)
t = -7 × 7
t = -49
Hence, from the above
We can conclude that the value of t is: -49

In Exercises 31– 38, solve the equation. Check your solution.

Question 31.
\(\frac{3}{2}\) + t = \(\frac{1}{2}\)

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q31

Question 32.
b – \(\frac{3}{16}\) = \(\frac{5}{16}\)
Answer:
The value of b is: \(\frac{1}{2}\)

Explanation:
The given equation is:
b – \(\frac{3}{16}\) = \(\frac{5}{16}\)
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
b = \(\frac{5}{16}\) + \(\frac{3}{16}\)
Since the denominators of both the numerators are equal, add the numerators making the denominator common
So,
b = \(\frac{5 + 3}{16}\)
b = \(\frac{8}{16}\)
b = \(\frac{1}{2}\)
Hence, from the above,
We can conclude that the value of b is: \(\frac{1}{2}\)

Question 33.
\(\frac{3}{7}\)m = 6

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q33

Question 34.
–\(\frac{2}{5}\)y = 4
Answer:
The value of y is: 10

Explanation:
The given equation is:
–\(\frac{2}{5}\)y = 4
We know that,
A) – ÷ – = +
B) + ÷ – = –
C) + ÷ + = +
D) – ÷ + = –
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
–\(\frac{2}{5}\)y = 4
–\(\frac{2}{5}\) × y = 4
y = 4 ÷ –\(\frac{2}{5}\)
y = 4 × –\(\frac{5}{2}\)
y = -4 × –\(\frac{5}{2}\)
y = –\(\frac{4}{1}\) × –\(\frac{5}{2}\)
y = –\(\frac{4 × 5}{1 × 2}\)
y = 10
Hence, from the above,
We can conclude that the value of y is: 10

Question 35.
5.2 = a – 0.4

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q35

Question 36.
f + 3π = 7π
Answer:
The value of f is: 4π

Explanation:
The given equation is:
f + 3π = 7π
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
f + 3π = 7π
f = 7π – 3π
f = π ( 7 – 3 )
f = π ( 4 )
f = 4π
Hence, from the above,
We can conclude that the value of f is: 4π

Question 37.
– 108π = 6πj

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q37

Question 38.
x ÷ (-2) = 1.4
Answer:
The value of x is: –\(\frac{14}{5}\)

Explanation:
The given equation is:
x ÷ ( -2 ) = 1.4
We know that,
A) – ÷ – = +
B) + ÷ – = –
C) + ÷ + = +
D) – ÷ + = –
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
x ÷ ( -2 ) = 1.4
x ÷ ( -2 ) = \(\frac{14}{10}\)
x ÷ ( -2 ) = \(\frac{7}{5}\)
x = \(\frac{7}{5}\) × ( -2 )
x = – \(\frac{7}{5}\) × \(\frac{2}{1}\)
x = –\(\frac{14}{5}\)
Hence, from the above,
We can conclude that the value of x is: –\(\frac{14}{5}\)

ERROR ANALYSIS
In Exercises 39 and 40, describe and correct the error in solving the equation.

Question 39.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 10

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q39

Question 40.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 11

Answer:
A negative 3 should have been multiplied to each side.
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
So,
-(\(\frac{m}{3}\) ) =-4
-3  ( \(\frac{m}{3}\) ) = -4 ( -3 )
3 ( \(\frac{m}{3}\) ) = -4 ( -3 )
3 ( \(\frac{m}{3}\) ) = 12
\(\frac{m}{3}\)  × \(\frac{3}{1}\) = 12
m = 12
Hence, from the above,
We can conclude that the value of m is: 12

Question 41.
ANALYZING RELATIONSHIPS
A baker orders 162 eggs. Each carton contains 18 eggs. Which equation can you use to find the number x of cartons? Explain your reasoning and solve the equation.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 12

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q41

MODELING WITH MATHEMATICS
In Exercises 42– 44, write and solve an equation to answer the question.

Question 42.
The temperature at 5 P.M. is 20°F. The temperature at 10 P.M. is -5°F. How many degrees did the temperature fall?
Answer:
The fall in temperature is: 25 degrees Fahrenheit

Explanation:
It is given that the temperature at 5 P.M. is 20°F and the temperature at 10 P.M. is -5°F.
So,
The fall in temperature = ( The temperature at 5 P.M ) – ( The temperature at 10 P.M )
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
So,
The fall in temperature = 20 – ( -5 )
= 20 + 5
= 25 degrees Fahrenheit
Hence, from the above,
We can conclude that the fall in temperature is: 25 degrees Fahrenheit

Question 43.
The length of an American flag is 1.9 times its width. What is the width of the flag?
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 13

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q43

Question 44.
The balance of an investment account is $308 more than the balance 4 years ago. The current balance of the account is $4708. What was the balance 4 years ago?
Answer:
The balance 4 years ago is: $4,400

Explanation:
It is given that the balance of an investment account is $308 more than the balance 4 years ago. The current balance of the account is $4708.
So,
The current balance of the account = ( The balance of an investment account 4 years ago ) + $308
Let the balance of an investment account four years ago be x.
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
$4,708 = x + $308
x = 4,708 – 308
x = $4,400
Hence, from the above,
We can conclude that the balance of an investment account four years ago is: $4,400

Question 45.
REASONING
Identify the property of equality that makes Equation 1 and Equation 2 equivalent.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 14

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q45

PROBLEM-SOLVING
Question 46.

Tatami mats are used as a floor covering in Japan. One possible layout uses four identical rectangular mats and one square mat, as shown. The area of the square mat is half the area of one of the rectangular mats.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 15
a. Write and solve an equation to find the area of one rectangular mat.
Answer:
The area of one rectangular mat is: 18 ft²

Explanation:
It is given that the tatami mats are used as a floor covering in Japan and for that, one layout of tatami mats requires the four identical rectangular mats and the one square mat
So,
The total area of the tatami mats = ( The area of the four rectangular mats ) + ( The area of the one square mat )
The give tatami mat layout is:
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 15
From the layout,
We can observe that
The total area of the layout is: 81 ft²
So,
The total area of tatami mats = 81 ft²
( The area of the four rectangular mats ) + ( The area of the one square mat ) = 81 ft²
It is also given that
The area of a square mat is half of one of the rectangular mats
So,
Area of the square mat = ( Area of the rectangular mat ) ÷ 2
So,
( The area of the four rectangular mats ) + \(\frac{Area of the rectangular mat}{2}\)  = 81 ft²
4 ( The area of the rectangular mat ) + \(\frac{Area of the rectangular mat}{2}\)  = 81 ft²
\(\frac{8}{2}\) ( The area of the rectangular mat ) + \(\frac{1}{2}\) ( The area of the rectangular mat ) = 81 ft²
( The area of the rectangular mat ) [ \(\frac{8}{2}\) + \(\frac{1}{2}\) ] = 81 ft²
\(\frac{9}{2}\) ( The area of the rectangular mat ) = 81 ft²
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
The area of the rectangular mat = 81 ÷ \(\frac{9}{2}\)
= 81 × \(\frac{2}{9}\)
= \(\frac{81}{1}\) × \(\frac{2}{9}\)
= \(\frac{81 × 2}{9 × 1}\)
= 18 ft²
Hence, from the above,
We can conclude that the area of one rectangular mat is: 18 ft²

b. The length of a rectangular mat is twice the width. Use Guess, Check, and Revise to find the dimensions of one rectangular mat.
Answer:
The dimensions of the rectangular mat are:
Length: 6 ft
Width: 3 ft

Explanation:
From the above problem,
The area of the rectangular mat = 18 ft²
It is given that the length of a rectangular mat is twice the width.
We know that the area of the rectangle = ( Length ) × ( Width )
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
It is also given that the length of a rectangular mat is twice the width
So,
Length of a rectangular mat = 2 × Width
Now,
The area of the rectangular mat = Length × Width
18 = 2 × Width × Width
Width × Width = 18 ÷ 2
Width × Width = 9
From guessing,
We can say that
Width of the rectangular mat = 3 ft
Now,
The length of the rectangular mat = 2 × 3
= 6 ft
Hence, from the above,
We can conclude that the dimensions of the rectangular mat are:
Length: 6 ft
Width: 3 ft

Question 47.
PROBLEM-SOLVING
You spend $30.40 on 4 CDs. Each CD costs the same amount and is on sale for 80% of the original price.
a. Write and solve an equation to find how much you spend on each CD.
b. The next day, the CDs are no longer on sale. You have $25. Will you be able to buy 3 more CDs? Explain your reasoning.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 15.1

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q47

Question 48.
ANALYZING RELATIONSHIPS
As c increases, does the value of x increase, decrease, or stay the same for each equation? Assume c is positive.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 16

Answer:
Let assume the values of c be: 0,1,2,3
So,
The completed table by taking the values of c is:

By taking the values of c positive i.e., 0, 1, 2, 3
We can observe that as the value of c increases, the values of x sometimes increasing and sometimes stays the same but not decreasing.

Question 49.
USING STRUCTURE
Use the values -2, 5, 9, and 10 to complete each statement about the equation ax = b – 5.
a. When a = ___ and b = ___, x is a positive integer.
b. When a = ___ and b = ___, x is a negative integer.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q51

Question 50.
HOW DO YOU SEE IT?
The circle graph shows the percents of different animals sold at a local pet store in 1 year.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 17
a. What percent is represented by the entire circle?
Answer:
The percent represented by the entire circle is = 69 % + x %

Explanation:
It is given that the circle represents the percent of different animals sold at a local store for 1 year
Now,
The given circle is:
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 17
We know that,
In terms of percentages, any circle represents 100%
So,
The equation representing the circle is:
The percentage of different animals in the circle = 48 + 5 + 9 + 7 +x
The percentage of different animals in the circle= 69% + x%

b. How does the equation 7 + 9 + 5 + 48 + x = 100 relate to the circle graph? How can you use this equation to find the percent of cats sold?
Answer:
The percent of cats sold is: 31%

Explanation:
We know that,
In terms of percentages, any circle represents 100%
So,
The total percent of animals = The percent of animals that are represented by the circle
100% = 69% + x%
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
x% = 100% – 69%
x% = 31%
Hence, from the above,
We can conclude that the percent of cats is: 31%

Question 51.
REASONING
One-sixth of the girls and two-sevenths of the boys in a school marching band are in the percussion section. The percussion section has 6 girls and 10 boys. How many students are in the marching band? Explain.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q51

Question 52.
THOUGHT-PROVOKING
Write a real-life problem that can be modeled by an equation equivalent to the equation 5x = 30. Then solve the equation and write the answer in the context of your real-life problem.

Answer:
Let suppose there is some number of boys. The number of girls is five times of the boys and the total number of girls is 30. Find the number of boys?
Ans:
Let,
The number of boys is x.
It is given that the number of girls is five times of boys.
So,
The number of girls = 5x
It is also given that
The number of girls = 30
So,
5x = 30
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
5 × x = 30
x = 30 ÷ 5
x = 6
Hence, from the above,
We can conclude that the number of boys is: 6

MATHEMATICAL CONNECTIONS
In Exercises 53–56, find the height h or the area of the base B of the solid.

Question 53.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 18

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q53

Question 54.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 19

Answer:
The height of the cuboid is: 9 cm

Explanation:
The given figure is:
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 19
We know that,
The volume of a cuboid = L × B × H
We know that,
The cuboid is made from a rectangle
We know that,
The area of a rectangle = L × H
So,
The volume of a cuboid = A × B
From the above figure,
The given volume is: 1323 cm³
The given Area is: 147 cm²
So,
1323 = 147 × H
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
H = 1323 ÷ 147
H = 9
Hence, from the above,
We can conclude that the height of the cuboid is: 9 cm

Question 55.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 20

Answer:

Question 56.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 21

Answer:
The height of the prism is: \(\frac{5}{6}\) ft

Explanation:
The given figure is:
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 21
We know that,
The volume of the prism =  Area × Height
From the above figure,
The volume of the prism = 35 ft³
The area of the prism = 30 ft²
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
35 = 30 × H
H = 35 ÷ 30
H = \(\frac{5}{6}\) ft
Hence, from the above,
We can conclude that the height of the prism is: \(\frac{5}{6}\) ft

Question 57.
MAKING AN ARGUMENT
In baseball, a player’s batting average is calculated by dividing the number of hits by the number of at-bats. The table shows Player A’s batting average and the number of at-bats for three regular seasons.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 22
a. How many hits did Player A have in the 2011 regular season? Round your answer to the nearest whole number.
b. Player B had 33 fewer hits in the 2011 season than Player A but had a greater batting average. Your friend concludes that Player B had more at-bats in the 2011 season than Player A. Is your friend correct? Explain. Maintaining

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q57

Maintaining Mathematical Proficiency

Use the Distributive Property to simplify the expression.

Question 58.
8(y + 3)
Answer:
8 ( y + 3 ) = 8y + 24

Explanation:
The given expression is: 8 ( y + 3 )
We know that,
By using the Distributive Property,
a ( b + c ) = ( a × b ) + ( a × c )
So,
By using the above Property,
8 ( y + 3 ) = ( 8 × y ) + ( 8 × 3 )
= 8y + 24
Hence, from the above,
We can conclude that
8 ( y + 3 ) = 8y + 24

Question 59.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 23

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q59

Question 60.
5(m + 3 + n)
Answer:
5 ( m + 3 + n ) = 5m + 5n + 15

Explanation:
The given expression is: 5 ( m + 3 + n )
By using the Distributive Property,
a ( b + c ) = ( a × b ) + ( a × c )
So,
By using the above Property,
5 ( m + 3 + n ) = ( 5 × m ) + ( 5 × 3 ) + ( 5 × n )
= 5m + 15 + 5n
Hence, from the above,
We can conclude that,
5 ( m + 3 + n ) = 5m + 15 + 5n

Question 61.
4(2p + 4q + 6)

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q61

Copy and complete the statement. Round to the nearest hundredth, if necessary.

Question 62.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 24

Answer:
The missing number is: \(\frac{1}{12}\)

Explanation:
Let the missing number be: x
So,
The given equation is:
\(\frac{5L}{min}\) = \(\frac{x L}{h}\)
We know that,
1 hour = 60 minutes
So,
1 min = \(\frac{1}{60}\) hour
So,
\(\frac{5 L}{min}\) = \(\frac{5 L × 1}{60h}\)
\(\frac{5 L}{min}\) = \(\frac{1 L }{12h}\)
So,
x = \(\frac{1}{12}\)
Hence, from the above,
We can conclude that,
The missing number is: \(\frac{1}{12}\)

Question 63.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 25

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q63

Question 64.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 26

Answer:
The missing number is: \(\frac{1}{12}\)

Explanation:
Let the missing number be: x
So,
The given equation is:
\(\frac{7 gal}{min}\) = \(\frac{x qin}{sec}\)
We know that,
1 min = 60 seconds
1 quintal = 100 kg
1 gallon = 3.78 kg = 4 kg
So,
1 gallon = 0.04 quintal
1 sec = \(\frac{1}{60}\) min
So,
\(\frac{7 gal}{min}\) = \(\frac{x qin × 1}{60min}\)
\(\frac{7 gal}{min}\) = \(\frac{1 L }{12h}\)
So,
x = \(\frac{1}{12}\)
Hence, from the above,
We can conclude that,
The missing number is: \(\frac{1}{12}\)

Question 65.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 27

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q65

Lesson 1.2 Solving Multi-step Equations

Essential Question

How can you use multi-step equations to solve real-life problems?

EXPLORATION 1
Solving for the Angle Measures of a Polygon

Work with a partner. The sum S of the angle measures of a polygon with n sides can be found using the formula S = 180(n – 2). Write and solve an equation to find each value of x. Justify the steps in your solution. Then find the angle measures of each polygon. How can you check the reasonableness of your answers?
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 27.1
Answer:
The given polygons are:

It is given that,
The sum S of the angle measures of a polygon with n sides can be found using the formula S = 180(n – 2).
a)
The number of sides (n ) = 3
So,
The sum of angles ( S ) = 180 ( n – 2 )
= 180 ( 3 – 2 )
= 180 ( 1 )
= 180
Now,
The given sides of a polygon are: 30, 9x, (30 + x )
So,
30 + 9x + 30 + x = 180
60 + 10x = 180
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
10x = 180 – 60
10x = 120
x = 120 ÷ 10
x = 12
Hence, from the above,
The angle measures of the given polygon are:
30, 9 × 12, 30 + 12
= 30, 108, 45 degrees
b)
The number of sides (n ) = 3
So,
The sum of angles ( S ) = 180 ( n – 2 )
= 180 ( 3 – 2 )
= 180 ( 1 )
= 180
Now,
The given sides of a polygon are: 30, 9x, (30 + x )
So,
50 + x + 10 + 20 + x = 180
80 + 2x = 180
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
2x = 180 – 80
2x = 100
x = 100 ÷ 2
x = 50
Hence, from the above,
The angle measures of the given polygon are:
50, 50 + 10, 50 + 20
= 50, 60, 70 degrees
c)
The number of sides (n ) = 4
So,
The sum of angles ( S ) = 180 ( n – 2 )
= 180 ( 4 – 2 )
= 180 ( 2 )
= 360
Now,
The given sides of a polygon are: 50, x, ( 2x + 20), ( 2x + 30 )
So,
50 + x + 2x + 20 + 2x + 30 = 360
100 + 5x = 360
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
5x = 360 – 100
5x = 260
x = 260 ÷ 5
x = 52
Hence, from the above,
The angle measures of the given polygon are:
50,52, 2 (52) + 20, 2(52) + 30
= 50, 52, 124, 134 degrees
d)
The number of sides (n ) = 4
So,
The sum of angles ( S ) = 180 ( n – 2 )
= 180 ( 4 – 2 )
= 180 ( 2 )
= 360
Now,
The given sides of a polygon are: x, x + 42, x + 35, x – 17
So,
x + x + 42 + x + 35 + x – 17 = 360
60 + 4x = 360
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
4x = 360 – 60
4x = 300
x = 300 ÷ 4
x = 75
Hence, from the above,
The angle measures of the given polygon are:
75,  75 + 42,  75 + 35 , 75 – 17
= 75, 117, 110, 58 degrees
e)
The number of sides (n ) = 5
So,
The sum of angles ( S ) = 180 ( n – 2 )
= 180 ( 5 – 2 )
= 180 ( 3 )
= 540
Now,
The given sides of a polygon are: (4x + 15), (5x + 10), (8x + 8), (3x + 5), (5x + 2)
So,
(4x + 15)+ (5x + 10)+ (8x + 8)+ (3x + 5)+ (5x + 2) = 540
40 + 25x = 540
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
25x = 540 – 40
25x = 500
x = 500 ÷ 25
x = 20
Hence, from the above,
The angle measures of the given polygon are:
(4. 20 + 15)+ (5. 20 + 10)+ (8.20 + 8)+ (3. 20 + 5)+ (5. 20 + 2)
= 95, 110, 168, 65, 102 degrees
f)
The number of sides (n ) = 5
So,
The sum of angles ( S ) = 180 ( n – 2 )
= 180 ( 5 – 2 )
= 180 ( 3 )
= 540
Now,
The given sides of a polygon are: (2x + 25), (3x + 16), (2x + 8), (4x – 18), (3x – 7)
So,
(2x + 25) + (3x + 16) + (2x + 8) + (4x – 18) + (3x – 7) = 540
24 + 14x = 540
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
14x = 540 – 24
14x = 516
x = 540 ÷ 14
x = 38.5
x = 39
Hence, from the above,
The angle measures of the given polygon are:
(2. 39 + 25), (3.39 + 16), (2.39 + 8), (4.39 – 18), (3.39 – 7)
= 103, 133, 86, 138, 110 degrees

EXPLORATION 2
Work with a partner.

a. Draw an irregular polygon.
Answer:

b. Measure the angles of the polygon. Record the measurements on a separate sheet of paper.
Answer:

c. Choose a value for x. Then, using this value, work backward to assign a variable expression to each angle measure, as in Exploration 1.
d. Trade polygons with your partner.
e. Solve an equation to find the angle measures of the polygon your partner drew. Do your answers seem reasonable? Explain.

Communicate Your Answer

Question 3.
How can you use multi-step equations to solve real-life problems?

Question 4.
In Exploration 1, you were given the formula for the sum S of the angle measures of a polygon with n sides. Explain why this formula works.
Answer:
We know that,
The sum of the angles in a triangle is: 180 degrees
The triangle is also a quadrilateral
So,
A quadrilateral can be formed by the minimum of the three lines
So,
The minimum sum of all the angles in a quadrilateral is: 180 degrees
Now,
Let suppose we form a quadrilateral with 4 sides.
So,
The sum of all the angles in a quadrilateral = 360 degrees = 180 degrees × 2
= 180 degrees ( 4 sides -2 )
Let suppose we form a quadrilateral with 5 sides
So,
The sum of all the angles in a quadrilateral = 540 degrees = 180 degrees × 3
= 180 degrees ( 5 -2 )
Hence, in general,
We can conclude that the sum of all the angles with n sides in a quadrilateral = 180 degrees ( n-2 )

Question 5.
The sum of the angle measures of a polygon is 1080º. How many sides does the polygon have? Explain how you found your answer.
Answer:
The number of sides the polygon with 1080° have: 6

Explanation:
It is given that the sum of all angle measures of a polygon is: 1080°
We know that,
The sum of angle measures with n sides in a polygon = 180° ( n – 2 )
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
1080° =180° ( n – 2 )
n – 2 = 1080 ÷ 180
n – 2 = 6
n = 6 + 2
n = 8
Hence, from the above,
We can conclude that the number of sides of the polygon with sum of the angles 1080° is: 6

1.2 Lesson

Monitoring Progress

Solve the equation. Check your solution.

Question 1.
-2n + 3 = 9
Answer:
The value of n is: -3

Explanation:
The given equation is:
-2n + 3 = 9
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
-2n + 3 = 9
-2n = 9 – (+3 )
n = 6 ÷ ( -2 )
= -3
Hence from the above,
We can conclude that the value of n is: -3

Question 2.
-21 = \(\frac{1}{2}\) – 11

Question 3.
-2x – 10x + 12 = 18
Answer:
The value of x is: –\(\frac{1}{2}\)

Explanation:
The given equation is:
-2x – 10x + 12 = 18
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
-( 2x + 10x ) = 18 – 12
-12x = 6
x = 6 ÷ ( -12 )
x = –\(\frac{1}{2}\)
Hence, from the above,
We can conclude that the value of x is: –\(\frac{1}{2}\)

Monitoring Progress

Solve the equation. Check your solution.

Question 4.
3(x + 1) + 6 = -9
Answer:
The value of x is: -6

Explanation:
The given equation is:
3 ( x + 1 ) + 6 = -9
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
3 ( x + 1 ) = -9 – (+6 )
By using the Distributive property,
3 ( x + 1 ) = 3x + 3
So,
3x + 3 = -15
3x = -15 – ( +3 )
3x = -18
x = -18 ÷ 3
x = -6
Hence, from the above,
We can conclude that the value of x is: -6

Question 5.
15 = 5 + 4(2d – 3)
Answer:
The value of d is:\(\frac{11}{4}\)

Explanation:
The given equation is:
15 = 5 + 4 ( 2d – 3 )
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
4 ( 2d – 3 ) = 15 -5
4 ( 2d – 3 ) = 10
By using the Distributive property,
4 ( 2d – 3 ) = 4 (2d ) -4 (3 )
= 8d – 12
So,
8d – 12 = 10
8d = 10 + 12
8d = 22
d = 22 ÷ 8
d = \(\frac{11}{4}\)
Hence, from the above,
We can conclude that the value of d is: \(\frac{11}{4}\)

Question 6.
13 = -2(y – 4) + 3y
Answer:
The value of y is: 5

Explanation:
The given equation is:
13 = -2 ( y – 4 ) + 3y
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
By using the Distributive Property,
-2 ( y – 4 ) = -2y + 8
So,
13 = -2y + 8 + 3y
13 = y + 8
y = 13 – 8
y = 5
Hence, from the above,
We can conclude that the value of y is: 5

Question 7.
2x(5 – 3) – 3x = 5
Answer:
The value of x is: 5

Explanation:
The given equation is:
2x ( 5 – 3 ) – 3x = 5
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
2x ( 2 ) – 3x = 5
4x – 3x = 5
x = 5
Hence, from the above,
We can conclude that the value of y is: 5

Question 8.
-4(2m + 5) – 3m = 35
Answer:
The value of m is: -5

Explanation:
The given equation is:
-4 ( 2m + 5 ) – 3m = 35
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
Now,
By using the Distributive Property,
-4 ( 2m + 5 ) = -4 (2m ) + 5 ( -4 )
= -8m -20
So,
-8m -20 -3m = 35
-11m – 20 = 35
-11m = 35 + 20
-11m = 55
m = 55 ÷ ( -11 )
m = -5
Hence, from the above,
We can conclude that the value of m is: -5

Question 9.
5(3 – x) + 2(3 – x) = 14
Answer:
The value of x is: 1

Explanation:
The given equation is:
5 ( 3 – x ) + 2 ( 3 – x ) = 14
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
Now,
By using the Distributive property,
5 ( 3 – x ) = 5 (3 ) -5 (x)
= 15 – 5x
2 ( 3 – x ) = 2 (3) – 2 ( x)
= 6 – 2x
So,
15 – 5x + 6 – 2x = 14
21 – 7x = 14
7x = 21 – 14
7x = 7
x = 7 ÷ 7
x = 1
Hence, from the above,
We can conclude that the value of x is: 1

Monitoring Progress

Question 10.
The formula d = \(\frac{1}{2}\)n + 26 relates the nozzle pressure n (in pounds per square inch) of a fire hose and the maximum horizontal distance the water reaches d (in feet). How much pressure is needed to reach a fire 50 feet away?
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 28
Answer:
The pressure needed to reach a fire 50 feet away (n ) is: 48 pounds per square inch

Explanation:
It is given that the formula
d = \(\frac{1}{2}\)n + 26
relates the nozzle pressure n (in pounds per square inch) of a fire hose and the maximum horizontal distance the water reaches d (in feet).
So,
The given equation is:
d = \(\frac{1}{2}\)n + 26
Where,
d is the maximum horizontal distance
n is the pressure
It is also given that the maximum horizontal distance is: 50 feet
So,
50 = \(\frac{1}{2}\)n + 26
\(\frac{1}{2}\)n = 50 – 26
\(\frac{1}{2}\)n = 24
\(\frac{1}{2}\) × n = 24
n = 24 × 2
n = 48 pounds per square inch
Hence, from the above
We can conclude that the pressure needed to reach 50 feet away is: 48 pounds per square inch

Question 11.
Monitoring Progress
You have 96 feet of fencing to enclose a rectangular pen for your dog. To provide sufficient running space for your dog to exercise, the pen should be three times as long as it is wide. Find the dimensions of the pen.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 29
Answer:
The dimensions of the pen are:
length of the pen: 12 feet
Width of the pen: 36 feet

Explanation;
It is given that you  have 96 feet of fencing to enclose a rectangular pen for your dog. To provide sufficient running space for your dog to exercise, the pen should be three times as long as it is wide.
So,
The perimeter of the rectangular pen is: 96 feet
We know that,
The perimeter of the rectangle = 2 (Length + Width )
It is also given that the pen is three times as long as it is wide
So,
Width = 3 × Length
So,
The perimeter of the rectangular pen =2 (  Length + ( 3 × Length ) )
96 = 2 ( 4 × Length )
4 × Length = 96 ÷ 2
4 × Length = 48
Length = 48 ÷ 4
Length = 12 feet
So,
Width = 3 × Length
= 3 × 12 = 36 feet
hence, from the above,
We can conclude that
The dimensions of the rectangular pen are:
Length of the pen is: 12 feet
Width of the pen is: 36 feet

Solving Multi-step Equations 1.2 Exercises

Monitoring Progress and Modeling with Mathematics

In Exercises 3−14, solve the equation. Check your solution.

Vocabulary and Core ConceptCheck

Question 1.
COMPLETE THE SENTENCE To solve the equation 2x + 3x = 20, first combine 2x and 3x because they are _________.
Answer:
The given equation is:
2x + 3x = 20
As 2x and 3x are combined by the symbol “+”, add 2x and 3x
So,
2x + 3x = 5x
So,
5x = 20
x = 20 ÷ 4
x = 5

Question 2.
WRITING Describe two ways to solve the equation 2(4x – 11) = 10.
Answer:
The given equation is:
2 (4x – 11) = 10
Way-1:
2 × (4x – 11) = 10
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
4x – 11 = 10 ÷ 2
4x – 11 = 5
4x = 5 + 11
4x = 16
x = 16 ÷ 4
x = 4
Hence,
The value of x is: 4

Way-2:
By using the Distributive Property,
2 (4x – 11) = 2 (4x) – 2 (11)
= 8x – 22
So,
8x – 22 = 10
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
8x = 10 + 22
8x = 32
x = 32 ÷ 8
x = 4
Hence,
The value of x is: 4

Question 3.
3w + 7 = 19

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q3

Question 4.
2g – 13 = 3
Answer:
The value of g is: 8

Explanation:
The given equation is:
2g – 13 = 3
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
2g = 3 + 13
2g = 16
2 × g = 16
g = 16 ÷ 2
g = 8
Hence, from the above,
We can conclude that the value of g is: 8

Question 5.
11 = 12 – q

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q5

Question 6.
10 = 7 – m
Answer:
The value of m is: -3

Explanation:
The given equation is:
10 = 7 – m
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
-m = 10 – 7
-m = 3
Multiply with “-” on both sides
– (-m ) = -3
m = -3
Hence, from the above,
We can conclude that the value of m is: -3

Question 7.
5 = \(\frac{z}{-4}\) – 3

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q7

Question 8.
\(\frac{a}{3}\) + 4 = 6
Answer:
The value of a is: 6

Explanation:
The given equation is:
\(\frac{a}{3}\) + 4 = 6
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
\(\frac{a}{3}\) = 6 – 4
\(\frac{a}{3}\) = 2
a = 2 × 3
a = 6
Hence, from the above,
We can conclude that the value of a is: 6

Question 9.
\(\frac{h + 6}{5}\) = 2

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q9

Question 10.
\(\frac{d – 8}{-2}\) = 12
Answer:
The value of d is: -16

Explanation:
The given equation is:
\(\frac{d – 8}{-2}\) = 12
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
d – 8 = 12 × (-2)
d – 8 = -24
d = -24 + 8
d = -16
Hence, from the above,
We can conclude that the value of d is: -16

Question 11.
8y + 3y = 44

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q11

Question 12.
36 = 13n – 4n
Answer:
The value of n is: 4

Explanation:
The given equation is:
36 = 13n – 4n
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
36 = 9n
9n = 36
n = 36 ÷ 9
n = 4
Hence, from the above,
We can conclude that the value of n is: 4

Question 13.
12v + 10v + 14 = 80

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q13

Question 14.
6c – 8 – 2c = -16
Answer:
The value of c is: -2

Explanation:
The given equation is:
6c – 8 – 2c = -16
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
4c – 8 = -16
4c = -16 + 8
4 × c = -8
c = -8 ÷ 4
c = -2
Hence, from the above,
We can conclude that the value of c is: -2

Question 15.
MODELING WITH MATHEMATICS
The altitude a (in feet) of a plane in minutes after liftoff is given by a = 3400t + 600. How many minutes after liftoff is the plane at an altitude of 21,000 feet?
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 30

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q15

Question 16.
MODELING WITH MATHEMATICS
A repair bill for your car is $553. The parts cost $265. The labor cost is $48 per hour. Write and solve an equation to find the number of hours of labor spent repairing the car.
Answer:
The number of hours of labor spent repairing the car is: 6 hours

Explanation:
It is given that a repair bill for your car is $553. The parts cost $265. The labor cost is $48 per hour.
Let the number of hours of labor spent repairing the car be: x
So,
The total bill to repair your car = ( The labor cost per hour ) × ( The number of hours of labor spent repairing the car ) +  (The cost of the parts )
553 = 48x + 265
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
48x = 553 – 265
48x = 288
48 × x = 288
x = 288 ÷ 48
x = 6
Hence, from the above,
We can conclude that the number of hours of labor spent repairing the car is: 6 hours

In Exercises 17−24, solve the equation. Check your solution.

Question 17.
4(z + 5) = 32

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q17

Question 18.
-2(4g – 3) = 3018.
Answer:
The value of g is: 378

Explanation:
The given equation is:
-2 (4g – 3) = 3018
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
-2 × ( 4g – 3 ) = 3018
4g – 3 = 3018 ÷ 2
4g – 3 = 1,509
4g = 1,509 +3
4 × g = 1,512
g = 1,512 ÷ 4
g = 378
Hence, from the above,
We can conclude that the value of g is: 378

Question 19.
6 + 5(m + 1) = 26

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q19

Question 20.
5h+ 2(11 – h) = -5
Answer:
The value of h is: -9

Explanation:
The given equation is:
5h + 2 ( 11-h ) = -5
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
By using the Distributive Property of Multiplication,
2 ( 11 – h ) = 2 (11 ) – 2 ( h )
= 22 – 2h
So,
5h + 22 – 2h = -5
3h + 22 = -5
3h = -5 – (+22)
3h = -5 -22
3h = -27
h = -27 ÷ 3
h = -9
Hence, from the above,
We can conclude that the value of h is: -9

Question 21.
27 = 3c – 3(6 – 2c)

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q21

Question 22.
-3 = 12y – 5(2y – 7)
Answer:
The value of y is: -19

Explanation:
The given equation is:
-3 = 12y – 5 (2y – 7)
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
12y – 5 (2y – 7) = -3
By using the Distributive Property of Multiplication,
5 ( 2y – 7 ) = 5 (2y ) – 5 (7 )
= 10y – 35
So,
12y – ( 10y – 35 ) = -3
12y – 10y + 35 = -3
2y + 35 = -3
2y = -3 – (+35 )
2y = -3 – 35
2y = -38
y = -38 ÷ 2
y = -19
Hence, from the above,
We can conclude that the value of y is: -19

Question 23.
-3(3 + x) + 4(x – 6) = -4

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q23

Question 24.
5(r + 9) – 2(1 – r) = 1
Answer:
The value of r is: -6

Explanation:
The given equation is:
5 ( r + 9 ) – 2 ( 1 – r ) = 1
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
Now,
By using the Distributive Property of Multiplication,
5 ( r + 9 ) = 5 ( r ) + 5 ( 9 )
= 5r + 45
2 ( 1 – r ) = 2 ( 1 ) – 2 ( r )
= 2 – 2r
So,
5r + 45 – ( 2 – 2r ) = 1
5r + 45 – 2 + 2r = 1
7r + 43 = 1
7r = 1 – 43
7r = -42
r = -42 ÷ 7
r = -6
Hence, from the above,
We can conclude that the value of r is: -6

USING TOOLS
In Exercises 25−28, find the value of the variable. Then find the angle measures of the polygon. Use a protractor to check the reasonableness of your answer.

Question 25.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 31

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q25

Question 26.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 32
Answer:
The angle measures of the rhombus are:
60°, 60°, 120°, 120°

Explanation:
The given figure is:
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 32
From the above figure,
The angle measures of the rhombus are: a°, 2a°, a°, 2a°
It is also given that the sum of all the angle measures is: 360°
So,
a° + 2a° + a° + 2a° = 360°
6a° = 360°
a = 360° ÷ 6
a = 60°
Hence, from the above,
We can conclude that the angle measures of the rhombus are:
a°, 2a°, a°, 2a° = 60°, 2 ( 60° ), 60°, 2 ( 60° )
= 60°, 60°, 120°, 120°

Question 27.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 33

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q27

Question 28.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 34
Answer:
The angle measures of the hexagon are:
120°, 120°, 100°, 120°, 250°, 260°

Explanation:
The given figure is:
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 34
From the given figure,
The angle measures of the hexagon are:
120°, 120°, 100°, 120°, x°, (x + 10)°
It is also given that the sum of the angle measures of the hexagon is: 720°
So,
120° + 120° + 100° + 120° + x° + (x + 10)° = 720°
470° + x = 720°
x = 720° – 470°
x = 250°
Hence, from the above,
We can conclude that the angle measures of the hexagon are:
120°, 120°, 100°, 120°, x°, (x + 10)° = 120°, 120°, 100°, 120°, 250°, (250 + 10)°
= 120°, 120°, 100°, 120°, 250°, 260°

In Exercises 29−34, write and solve an equation to find the number.

Question 29.
The sum of twice a number and 13 is 75.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q29

Question 30.
The difference of three times a number and 4 is -19.
Answer:
The number is: -5

Explanation:
It is given that the difference of three times of a number and 4 is -19
Now,
Let the number be x
So,
The three times of a number = 3 (x) = 3x
So,
3x – 4 = -19
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
3x = -19 + 4
3x = -15
x = -15 ÷ 3
x = -5
Hence, from the above,
We can conclude that the number is: -5

Question 31.
Eight plus the quotient of a number and 3 is -2.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q31

Question 32.
The sum of twice a number and half the number is 10.
Answer:
The number is: 4

Explanation:
It is given that the sum of twice of a number and half the number is 10.
Let the number be x.
So,
The twice of a number = 2 (x ) = 2x
Half of the number = x ÷ 2 = \(\frac{x}{2}\)
So,
2x + \(\frac{x}{2}\) = 10
2x can be rewritten as: \(\frac{4x}{2}\)
So,
\(\frac{4x}{2}\) + \(\frac{x}{2}\) = 10
\(\frac{4x + x}{2}\) = 10
\(\frac{5x}{2}\) = 10
5x = 10 × 2
5x = 20
x = 20 ÷ 5
x = 4
Hence, from the above,
We can conclude that the numebr is: 4

Question 33.
Six times the sum of a number and 15 is -42.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q33

Question 34.
Four times the difference of a number and 7 is 12.
Answer:
The number is: 4

Explanation:
It is given that the four times the difference of a number and 7 is 12
Let the number be x
So,
Four times of the number = 4 ( x ) = 4x
So,
4x – x = 12
3x = 12
x = 12 ÷ 3
x = 4
Hence, from the above,
We can conclude that the number is: 4

USING EQUATIONS
In Exercises 35−37, write and solve an equation to answer the question. Check that the units on each side of the equation balance.

Question 35.
During the summer, you work 30 hours per week at a gas station and earn $8.75 per hour. You also work as a landscaper for $11 per hour and can work as many hours as you want. You want to earn a total of $400 per week. How many hours must you work as a landscaper?

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q35

Question 36.
The area of the surface of the swimming pool is 210 square feet. What is the length d of the deep end (in feet)?
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 35

Answer:
The length d of the deep end is: 12 feet

Explanation:
The given figure is:
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 35
It is given that the area of the surface of the swimming pool is 210 square feet
From the above figure,
We can observe that the shape of the swimming pool is a rectangle.
So,
Length of the swimming pool = 10 ft
Width of the swimming pool = d + 9 ft
So,
The area of the swimming pool = Length × Width
= 10 × ( d + 9 )
Now,
210 = 10 × ( d + 9 )
d + 9 = 210 ÷ 10
d + 9 = 21
d = 21 – 9
d = 12 feet
Hence, from the above,
We can conclude that the length d of the deep end is: 12 feet

Question 37.
You order two tacos and a salad. The salad costs $2.50. You pay 8% sales tax and leave a $3 tip. You pay a total of $13.80. How much does one taco cost?

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q37

JUSTIFYING STEPS
In Exercises 38 and 39, justify each step of the solution.

Question 38.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 36

Answer:
–\(\frac{1}{2}\) ( 5x – 8 ) – 1 = 6                          Write the equation
–\(\frac{1}{2}\) ( 5x – 8 ) = 6 + 1                         Arrange the similar terms
–\(\frac{1}{2}\) ( 5x – 8 ) = 7                                  Simplify
– ( 5x – 8 ) = 7 × 2                                                              Divide by 2 on both sides
– ( 5x – 8 ) = 14                                                                    Simplify
5x – 8 = -14                                                                       Multiply with “-” on both sides
5x = -14 + 8                                                                         Arrange the similar terms
5x = -6                                                                               Divide by 6 on both sides
x = –\(\frac{6}{5}\)                                              The result
Hence,
The solution is: x = –\(\frac{6}{5}\)

Question 39.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 37

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q39

ERROR ANALYSIS
In Exercises 40 and 41, describe and correct the error in solving the equation.

Question 40.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 38

Answer:
The given equation is:
-2 ( 7 – y ) + 4 = -4
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
-2 ( 7 – y ) = -4 –  (+4 )
-2 ( 7 – y ) = -4 – 4
-2 ( 7 – y ) = -8
Now,
By using the Distributive Property of Multiplication,
2 ( 7 – y ) = 2 ( 7 ) – 2 ( y )
= 14 – 2y
So,
– ( 14 – 2y ) = -8
2y – 14 = -8
2y = -8 + 14
2y = 6
y = 6 ÷ 2
y = 3
Hence,
The value of y is: 3

Question 41.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 39

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q41

MATHEMATICAL CONNECTIONS
In Exercises 42−44, write and solve an equation to answer the question.

Question 42.
The perimeter of the tennis court is 228 feet. What are the dimensions of the court?
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 40

Answer:
The dimensions of the court are:
The Length of the court is: 36 feet
The width of the court is: 78 feet

Explanation:
The given figure is:
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 40
It is given that the perimeter of the tennis court is: 228 feet
From the above figure,
We can observe that the shape of the tennis court is the rectangle
So,
The length of the tennis court = w feet
The width of the tennis court =  (2w + 6 ) feet
We know that,
The perimeter of the rectangle = 2 ( Length + Width )
So,
The perimeter of the tennis court = 2 ( Length + Width )
228 = 2 ( w + 2w + 6 )
By using the Distributive Property of Multiplication,
2 ( w + 2w + 6 ) = 2 ( 3w + 6 )
= 2 ( 3w ) + 2 ( 6 )
= 6w + 12
So,
228 = 6w + 12
6w = 228 – 12
6w = 216
w = 216 ÷ 6
w = 36
Hence, from the above,
We can conclude that
The length of the tennis court is: 36 feet
The width of the tennis court is: 2w + 6  = 2 ( 36 ) + 6 = 78 feet

Question 43.
The perimeter of the Norwegian flag is 190 inches. What are the dimensions of the flag?
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 41

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q43

Question 44.
The perimeter of the school crossing sign is 102 inches. What is the length of each side?
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 42

Answer:
The length of each side is: 15 inches

Explanation:
The given figure is:
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 42
It is given that the perimeter of the crossing sign is 102 inches
We know that,
The perimeter of any polygon is the sum of all the sides of that polygon
So,
The perimeter of the crossing sign = s + ( s + 6 ) + ( s + 6 ) + s + 2s
102 = 6s + 12
102 – 12 = 6s
6s = 90
s = 90 ÷ 6
s = 15 inches
Hence, from the above,
We can conclude that the length of each side is: 15 inches

Question 45.
COMPARING METHODS
Solve the equation 2(4 – 8x) + 6 = -1 using (a) Method 1 from Example 3 and (b) Method 2 from Example 3. Which method do you prefer? Explain.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q45
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q45-i

Question 46.
PROBLEM – SOLVING
An online ticket agency charges the amounts shown for basketball tickets. The total cost for an order is $220.70. How many tickets are purchased?
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 43

Answer:

Question 47.
MAKING AN ARGUMENT
You have quarters and dimes that total $2.80. Your friend says it is possible that the number of quarters is 8 more than the number of dimes. Is your friend correct? Explain.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q47

Question 48.
THOUGHT-PROVOKING
You teach a math class and assign a weight to each component of the class. You determine final grades by totaling the products of the weights and the component scores. Choose values for the remaining weights and find the necessary score on the final exam for a student to earn an A (90%) in the class, if possible. Explain your reasoning.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 44

Answer:
The completed table is:

From the above table,
The weights can be calculated by the difference between the total participation and the class participation and divide the total value by 100.
So,
The weight of homework = [ ( 100 – 95 ) ÷ 100]
= 5 ÷ 100
= 0.50
The weight of midterm exam = ( 100 – 88 ) ÷ 100
= 12 ÷ 100
= 0.12
So,
The necessary score of the final exam = ( 92 + 95 + 88 ) % ÷ 3
= 275 % ÷ 3
= 91.6 %

Question 49.
REASONING
An even integer can be represented by the expression 2n, where n is an integer. Find three consecutive even integers that have a sum of 54. Explain your reasoning.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q49

Question 50.
HOW DO YOU SEE IT?
The scatter plot shows the attendance for each meeting of a gaming club.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 45
a. The mean attendance for the first four meetings is 20. Is the number of students who attended the fourth meeting greater than or less than 20? Explain.
Answer:
The number of students who attended the fourth meeting is greater than 20

Explanation:
The given graph is:
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 45
We know that
The mean = ( Sum of the given numbers ) ÷ (The total number of the numbers)
From the above graph,
The attendance of the 1st meeting = 18
The attendance of the 2nd meeting = 21
The attendance of the 3rd meeting = 17
Let the attendance of the 4th meeting be: x
So,
The mean attendance  of the first four meetings = ( The attendance of the 4 meetings ) ÷  ( The total number of meetings )
= ( 18 + 21 + 17 + x ) ÷ 4
It is given that the mean attendance of the first four meetings is: 20
So,
20 =  ( 18 + 21 + 17 + x ) ÷ 4
( 56 + x ) ÷ 4 = 20
56 + x = 20 × 4
56 + x = 80
x = 80 – 56
x = 24
Hence, from the above,
We can conclude that the attendance of the 4th meeting is greater than 20

b. Estimate the number of students who attended the fourth meeting.
Answer:
The number of students who attended the fourth meeting is: 24

Explanation:
The mean attendance  of the first four meetings = ( The attendance of the 4 meetings ) ÷  ( The total number of meetings )
= ( 18 + 21 + 17 + x ) ÷ 4
It is given that the mean attendance of the first four meetings is: 20
So,
20 =  ( 18 + 21 + 17 + x ) ÷ 4
( 56 + x ) ÷ 4 = 20
56 + x = 20 × 4
56 + x = 80
x = 80 – 56
x = 24
Hence, from the above,
We can conclude that the number of students who attended the 4th meeting is: 24

c. Describe a way you can check your estimate in part (b).
Answer:
The estimate in part (b) can be checked by using the property of the mean
So,
The mean attendance of the four meetings = ( The attendance of the four meetings ) ÷ ( The total number of meetings )
= ( 18 + 21 + 17 + 24 ) ÷ 4
= 80 ÷ 4
= 20
Hence, from the above,
We can conclude that the mean attendance of the four meetings is the same as given above.

REASONING
In Exercises 51−56, the letters a, b, and c represent nonzero constants. Solve the equation for x.

Question 51.
bx = -7

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q51

Question 52.
x + a = \(\frac{3}{4}\)
Answer:
The value of x is: \(\frac{3}{4}\) – a

Explanation:
The given equation is:
x + a = \(\frac{3}{4}\)
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
x =  \(\frac{3}{4}\) – a
Hence, from the above,
We can conclude that the value of a is: \(\frac{3}{4}\) – a

Question 53.
ax – b = 12.5

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q53

Question 54.
ax + b = c
Answer:
The value of x is: \(\frac{c – b}{a}\)

Explanation:
The given equation is:
ax + b = c
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
ax = c – b
x = \(\frac{c – b}{a}\)
Hence, from the above,
We can conclude that the value of x is: \(\frac{c – b}{a}\)

Question 55.
2bx – bx = -8

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q55

Question 56.
cx – 4b = 5b
Answer
The value of x is: \(\frac{9b}{c}\)

Explanation:
The given equation is:
cx – 4b = 5b
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
cx = 5b + 4b
cx = 9b
x = \(\frac{9b}{c}\)
Hence, from the above,
We can conclude that the value of x is: \(\frac{9b}{c}\)

Maintaining Mathematical Proficiency

Simplify the expression.

Question 57.
4m + 5 – 3m

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q57

Question 58.
9 – 8b + 6b
Answer:
9 – 8b + 6b
= 9 – (8b – 6b )
= 9 – 2b

Question 59.
6t + 3(1 – 2t) – 5

Answer:

Determine whether (a) x = −1 or (b) x = 2 is a solution of the equation.

Question 60.
x – 8 = -9
Answer:
x = -1 is a solution to the given equation

Explanation:
The given equation is:
x – 8 = -9
a) Let x = -1
So,
-1 – 8 = -9
-9 = -9
As LHS is equal to RHS
x = -1 is a solution of the given equation
b) Let x = 2
So,
2 – 8 = -9
-6 = -9
As LHS is not equal to RHS,
x = 2 is not a solution of the given equation

Question 61.
x + 1.5 = 3.5

Answer:

Question 62.
2x – 1 = 3
Answer:
x = 2 is a solution to the given equation

Explanation:
The given equation is:
2x – 1 = 3
a) Let x = -1
So,
2 ( -1 ) – 1 = 3
-2 – 1 = 3
-3 = 3
As LHS is not equal to RHS
x = -1 is not a solution to the given equation
b) Let x = 2
So,
2 ( 2 ) -1 = 3
4 – 1 = 3
3 = 3
As LHS is equal to RHS,
x = 2  is a solution to the given equation

Question 63.
3x + 4 = 1

Answer:

Question 64.
x + 4 = 3x
Answer:
x = 2 is a solution to the given equation

Explanation:
The given equation is:
x + 4 = 3x
a) Let x = -1
So,
-1 + 4 = 3 ( -1 )
= 3 = -3
As LHS is not equal to RHS,
x = -1 is not a solution to the given equation
b) Let x = 2
So,
2 + 4 = 3 ( 2 )
6 = 6
As LHS is equal to RHS,
x = 2 is a solution to the given equation.

Question 65.
-2(x – 1) = 1 – 3x

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q65

Lesson 1.3 Solving Equations with Variables on Both Sides

EXPLORATION 1
Perimeter

Work with a partner. The two polygons have the same perimeter. Use this information to write and solve an equation involving x. Explain the process you used to find the solution. Then find the perimeter of each polygon.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 46
Answer:
The perimeter of the hexagon is: 6
The perimeter of the square is: 6

Explanation:
The given figures are:
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 46
Based on the number of sides (n) in the polygon,
We can say the name of that polygon
So,
In the first figure,
The number of sides is: 6
So,
It is Hexagon
In the second figure,
The number of sides is: 4
So,
It is square ( As all the angles are 90° )
It is given that the two polygons have the same perimeter
We know that,
“Perimeter” of a polygon is defined as the sum of all the sides in the polygon
So,
The sum of all sides in the hexagon = 5 + 2 + 5 + 2 + x + x
= 14 + 2x
The sum of all sides in the square = \(\frac{3x}{2}\) + 3 + 4 + 5
= \(\frac{3x}{2}\) + 12
It is given that the perimeter of both the polygons are equal
So,
14 + 2x = \(\frac{3x}{2}\) + 12
14 – 12 = \(\frac{3x}{2}\) – 2x
\(\frac{3x}{2}\) – 2x = 2
We can write 2x as \(\frac{4x}{2}\)
So,
\(\frac{3x}{2}\) – \(\frac{4x}{2}\) = 2
\(\frac{3x – 4x}{2}\) = 2
\(\frac{-x}{2}\) = 2
– \(\frac{x}{2}\) = 2
-x = 2 × 2
-x = 4
x = -4
Hence,
The perimeter of the Hexagon = 14 + 2x = 14 + 2 ( -4 )
= 14 – 8 = 6
The perimeter of the square = \(\frac{3x}{2}\) + 12
\(\frac{3 × -4}{2}\) + 12
= \(\frac{-12}{2}\) + 12
= \(\frac{-12}{2}\) + \(\frac{24}{2}\)
= \(\frac{24 – 12}{2}\)
= \(\frac{12}{2}\)
= 6
Hence, from the above,
We can conclude that
The perimeter of the Hexagon is: 6
The perimeter of the square is: 6

EXPLORATION 2
Perimeter and Area

Work with a partner.

  • Each figure has the unusual property that the value of its perimeter (in feet) is equal to the value of its area (in square feet). Use this information to write an equation for each figure.
  • Solve each equation for x. Explain the process you used to find the solution.
  • Find the perimeter and area of each figure.

Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 47

Question 3.
How can you solve an equation that has variables on both sides?
Answer:
If the variable is the same on both sides in an equation, then rearrange the like terms
So,
Separate the variables and the numbers and simplify the variables and the numbers
In this way,
We can solve an equation with a single variable

Question 4.
Write three equations that have the variable x on both sides. The equations should be different from those you wrote in Explorations 1 and 2. Have your partner solve the equations.
Answer:
Let the three equations that have variable x on both sides and different from Explorations 1 and 2 are:
a) 6x + 2 = 5x-6
b) 16x = 18x – 2
c) 12x = 15x + 63
Now,
a)
The given equation is:
6x + 2 = 5x – 6
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
6x – 5x = -6 – 2
x = -8
Hence,
The value of x is: -8
b) The given equation is:
9x = 18x – 2
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
18x – 16x = 2
2x = 2
x = 2 ÷ 2
x = 1
Hence,
The value of x is: 1
c) The given equation is:
12x = 15x + 63
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
12x -15x = 63
-3x = 63
x = 63 ÷ ( -3 )
x = -63 ÷ 3
x = -21
Hence,
The value of x is: -21

1.3 Lesson

Monitoring Progress

Solve the equation. Check your solution.

Question 1.
-2x = 3x + 10
Answer:
The value of x is: -2

Explanation:
The given equation is:
-2x = 3x + 10
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
-2x – 3x = 10
-5x = 10
x = 10 ÷ (-5)
x = -10 ÷ 5
x = -2
Hence, from the above,
We can conclude that the value of x is: -2

Question 2.
\(\frac{1}{2}\)(6h – 4) = -5h + 1
Answer:
The value of h is: \(\frac{3}{8}\)

Explanation:
The given equation is:
\(\frac{1}{2}\) ( 6h – 4 ) = -5h + 1
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
6h – 4 = 2 ( -5h + 1 )
6h – 4 = 2 ( -5h ) + 2 ( 1 ) [ By using the Distributive Property of Multiplication )
6h – 4 = -10h + 2
6h + 10h = 2 + 4
16h = 6
h = \(\frac{6}{16}\)
h= \(\frac{3}{8}\)
Hence, from the above,
We can conclude that the value of h is: \(\frac{3}{8}\)

Question 3.
–\(\frac{3}{4}\)(8n + 12) = 3(n – 3)
Answer:
The value of n is: 0

Explanation:
The given equation is:
–\(\frac{3}{4}\) ( 8n + 12 ) = 3 ( n – 3 )
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
8n + 12 = –\(\frac{4}{3}\) × 3 ( n – 3 )
8n + 12 = –\(\frac{4}{3}\) \(\frac{3}{1}\) ( n – 3 )
8n + 12 = –\(\frac{3 × 4}{3 × 1}\) ( n – 3 )
8n + 12 = -4 ( n – 3 )
8n + 12 = -4n – 4 ( -3 )
8n + 12 = -4n + 12
8n + 4n =12 – 12
12n = 0
n = 0
Hence, from the above,
We can conclude that the value of n is: 0

Monitoring Progress

Solve the equation.

Question 4.
4(1 – p) = 4p – 4
Answer:
The value of p is: 1

Explanation:
The given equation is:
4 ( 1 -p ) = 4p – 4
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
4 ( 1 ) – 4 ( p ) = 4p – 4
4 – 4p = 4p – 4
4p + 4p = 4 + 4
8p = 8
p = 8 ÷ 8
p = 1
Hence, from the above,
We can conclude that the value of p is: 1

Question 5.
6m – m = –\(\frac{5}{6}\)(6m – 10)
Answer:
The value of m is: \(\frac{5}{6}\)

Explanation:
The given equation is:
6m – m = –\(\frac{5}{6}\) ( 6m – 10 )
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
5m = –\(\frac{5}{6}\) ( 6m – 10 )
5m = –\(\frac{5}{6}\) ( 6m ) – ( –\(\frac{5}{6}\) ( 10 ) )
5m = –\(\frac{5}{6}\) × \(\frac{6m}{1}\) + \(\frac{5}{6}\) \(\frac{10}{1}\)
5m = –\(\frac{5 × 6m}{6 × 1}\) + \(\frac{5 × 10}{6 × 1}\)
5m = -5m + \(\frac{25}{3}\)
5m + 5m = \(\frac{25}{3}\)
10m = \(\frac{25}{3}\)
m = \(\frac{25}{3}\) ÷ \(\frac{10}{1}\)
m = \(\frac{25}{3}\) × \(\frac{1}{10}\)
m = \(\frac{25}{30}\)
m = \(\frac{5}{6}\)
Hence, from the above,
We can conclude that the value of m is: \(\frac{5}{6}\)

Question 6.
10k + 7 = -3 – 10k
Answer:
The value of k is: –\(\frac{1}{2}\)

Explanation:
The given equation is:
10k + 7 = -3 – 10k
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
10k + 10k = -3 – ( +7 )
20k = -3 –
20k = -10
k = -10 ÷ 20
k = –\(\frac{1}{2}\)
Hence, from the above,
We can conclude that the value of k is: –\(\frac{1}{2}\)

Question 7.
3(2a – 2) = -2(3a – 3)
Answer:
The value of a is: 1

Explanation:
The given equation is:
3 ( 2a – 2 ) = -2 ( 3a – 3 )
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
By using the Distributive Property of Multiplication,
3 ( 2a ) – 3 ( 2 ) = -2 ( 3a ) + 2 ( 3 )
6a – 6 = -6a + 6
6a + 6a = 6 + 6
12a = 12
a = 12 ÷ 12
a = 1
Hence, from the above,
We can conclude that the value of a is: 1

Concept Summary

Steps for Solving Linear Equations
Here are several steps you can use to solve a linear equation. Depending on the equation, you may not need to use some steps.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 48
Step 1
Use the Distributive Property to remove any grouping symbols.
Step 2
Simplify the expression on each side of the equation.
Step 3
Collect the variable terms on one side of the equation and the constant terms on the other side.
Step 4
Isolate the variable.
Step 5
Check your solution.

Monitoring Progress

Question 8.
A boat travels upstream on the Mississippi River for 3.5 hours. The return trip only takes 2.5 hours because the boat travels 2 miles per hour faster downstream due to the current. How far does the boat travel upstream? Answer:
The distance the boat travel upstream is: 17.5 miles

Explanation:
It is given that a boat travels upstream on the Mississippi River for 3.5 hours. The return trip only takes 2.5 hours because the boat travels 2 miles per hour faster downstream due to the current.
Now,
Let x be the speed of the boat traveled upstream
We know that,
Speed = Distance ÷ Time
Distance = Speed × Time
It is given that the time taken by the boat traveled upstream is: 3.5 hours
So,
Distance traveled upstream = 3.5 × x = 3.5x
Now,
It is also given that the speed of the boat is 2 miles per hour faster downstream
So,
Distance traveled downstream by boat = 2.5 ( x + 2 )
SO,
As both the distances are the same,
3.5x = 2.5 ( x + 2 )
By using the Distributive Property of Multiplication,
3.5x = 2.5 ( x) + 2.5 ( 2 )
3.5x = 2.5x + 5
3.5x – 2.5x = 5
x = 5
So,
The distance traveled upstream by boat = 3.5x = 3.5 ( 5 )
= 17.5 miles per hour
Hence, from the above,
We can conclude that the distance traveled upstream by boat is: 17.5 miles per hour

Solving Equations with Variables on Both Sides 1.3 Exercises

Monitoring Progress and Modeling with Mathematics

In Exercises 3–16, solve the equation. Check your solution.

Question 1.
VOCABULARY Is the equation -2(4 – x) = 2x + 8 an identity? Explain your reasoning.
Answer:
-2 ( 4 – x ) = 2x + 8 is not an identity

Explanation:
The given equation is:
-2 ( 4 – x ) = 2x + 8
By using the Distributive Property of Multiplication,
-2 ( 4 ) + 2 ( x ) = 2x + 8
-8 + 2x = 2x + 8
2x – 8 = 2x + 8
As
LHS ≠ RHS
-2 ( 4 – x ) = 2x + 8  is not an identity

Question 2.
WRITING Describe the steps in solving the linear equation 3(3x – 8) = 4x + 6
Answer:
The value of x is: 6

Explanation:
The given equation is:
3 ( 3x – 8 ) = 4x + 6
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
Now,
By using the Distributive Property of Multiplication,
3 ( 3x ) – 3 ( 8 ) = 4x + 6
9x – 24 = 4x + 6
9x – 4x = 6 + 24
5x = 30
x = 30 ÷ 5
x = 6
Hence, from the above,
We can conclude that the value of x is: 6

Question 3.
15 – 2x = 3x

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q3

Question 4.
26 – 4s = 9s
Answer:
The value of s is: 2

Explanation:
The given equation is:
26 – 4s = 9s
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
26 = 9s + 4s
13s = 26
s = 26 ÷ 13
s = 2
Hence, from the above,
We can conclude that the value of s is: 2

Question 5.
5p – 9 = 2p + 12

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q5

Question 6.
8g + 10 = 35 + 3g
Answer:
The value of g is: 5

Explanation:
The given equation is:
8g + 10 = 35 + 3g
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
8g – 3g = 35 – 10
5g = 25
g = 25 ÷ 5
g = 5
Hence, from the above,
We can conclude that the value of g is: 5

Question 7.
5t + 16 = 6 – 5t

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q7

Question 8.
-3r + 10 = 15r – 8
Answer:
The value of r is: 1

Explanation:
The given equation is:
-3r + 10 = 15r – 8
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
-3r – 15r = -10 – 8
-18r = -18
r = -18 ÷ ( -18 )
r = 1 [ since – ÷ – = + ]
Hence, from the above,
We can conclude that the value of r is: 1

Question 9.
7 + 3x – 12x = 3x + 1

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q9

Question 10.
w – 2 + 2w = 6 + 5w
Answer:
The value of w is: -4

Explanation:
The given equation is:
w – 2 + 2w = 6 + 5w
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
w + 2w -5w =6 + 2
-2w = 8
w = -8 ÷ 2
w = -4
Hence, from the above,
We can conclude that the value of w is: -4

Question 11.
10(g + 5) = 2(g + 9)

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q11

Question 12.
-9(t – 2) = 4(t – 15)
Answer:
The value of t is: 6

Explanation:
The given equation is:
-9 ( t – 2 ) = 4 ( t – 15 )
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
Now,
By using the Distributive Property of Multiplication,
-9 ( t ) +  9 ( 2 ) = 4 ( t ) – 4 ( 15 )
-9t + 18 = 4t – 60
-9t – 4t = -60 – 18
-13t = -78
t = -78 ÷ ( -13 )
t = 6 [ Since  -÷ – = + ]
Hence, from the above,
We can conclude that the value of t is: 6

Question 13.
\(\frac{2}{3}\)(3x + 9) = -2(2x + 6)

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q13

Question 14.
2(2t + 4) = \(\frac{3}{4}\)(24 – 8t)
Answer:
The value of t is: 1

Explanation:
The given equation is:
2 ( 2t + 4 ) = \(\frac{3}{4}\) ( 24 – 8t )
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
By using the Distributive Property of Multiplication,
2 ( 2t ) + 2 ( 4 ) = \(\frac{3}{4}\) ( 24 ) – 8t  (\(\frac{3}{4}\) )
4t + 8 = \(\frac{3}{4}\) × \(\frac{24}{1}\) – \(\frac{8t}{1}\) × \(\frac{3}{4}\)
4t + 8 = \(\frac{3 × 24}{4 × 1}\) – \(\frac{3 × 8t}{4 × 1}\)
4t + 8 = \(\frac{18}{1}\) – \(\frac{6t}{1}\)
4t + 8 = 18 – 6t
4t + 6t = 18 – 8
10t = 10
t = 10 ÷ 10
t = 1
Hence, from the above,
We can conclude that the value of t is: 1

Question 15.
10(2y + 2) – y = 2(8y – 8)

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q15

Question 16.
2(4x + 2) = 4x – 12(x – 1)
Answer:
The value of x is: \(\frac{1}{2}\)

Explanation:
The given equation is:
2 ( 4x + 2 ) = 4x – 12 ( x – 1 )
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
Now,
By using the Distributive Property of Multiplication,
2 ( 4x ) + 2 ( 2 ) = 4x – 12 ( x ) + 12 ( 1 ) [ Since – × – = + ]
8x + 4 = 4x – 12x + 12
8x + 4 =12 – 8x
8x + 8x = 12 – 4
16x = 8
x = 8 ÷ 16
x = \(\frac{1}{2}\)
Hence, from the above,
We can conclude that the value of x is: \(\frac{1}{2}\)

Question 17.
MODELING WITH MATHEMATICS
You and your friend drive toward each other. The equation 50h = 190 – 45h represents the number h of hours until you and your friend meet. When will you meet?

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q17

Question 18.
MODELING WITH MATHEMATICS
The equation 1.5r + 15 = 2.25r represents the number r of movies you must rent to spend the same amount at each movie store. How many movies must you rent to spend the same amount at each movie store?
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 49

Answer:
The number of movies you rent to spend the same amount at each movie store is: 20

Explanation:
It is given that
The equation 1.5r + 15 = 2.25r represents the number r of movies you must rent to spend the same amount at each movie store.
Now,
We have to find the value of r to find the number f movies you must rent to spend the same amount at each movie store
So,
The given equation is:
1.5r + 15 = 2.25r
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
2.25r – 1.5r = 15
0.75r = 15
\(\frac{75}{100}\)r = 15
r = 15 × \(\frac{100}{75}\)
r = \(\frac{15}{1}\) × \(\frac{100}{75}\)
r = \(\frac{15 × 100}{1 × 75}\)
r = 20
Hence, from the above,
We can conclude that the number of movies you rent to spend the same amount at each movie store is: 20

In Exercises 19–24, solve the equation. Determine whether the equation has one solution, no solution, or infinitely many solutions.

Question 19.
3t + 4 = 12 + 3t

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q19

Question 20.
6d + 8 = 14 + 3d
Answer:
The value of d is: 2

Explanation:
The given equation is:
6d + 8 = 14 + 3d
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
6d – 3d = 14 – 8
3d = 6
d = 6 ÷ 3
d = 2
Hence, from the above,
We can conclude that the value of d is: 2

Question 21.
2(h + 1) = 5h – 7

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q21

Question 22.
12y + 6 = -6(2y + 1)
Answer:
The value of y is: –\(\frac{1}{2}\)

Explanation:
The given equation is:
12y + 6 = -6 ( 2y + 1 )
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
By using the Distributive Property of Multiplication,
12y + 6 = -6 ( 2y ) – 6 ( 1 )
12y + 6 = -12y – 6
12y + 12y = -6 – ( +6 )
24y = -6 – 6
24y = -12
y = -12 ÷ 24
y = –\(\frac{1}{2}\)
Hence, from the above,
We can conclude that the value of y is: –\(\frac{1}{2}\)

Question 23.
3(4g + 6) = 2(6g + 9)

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q23

Question 24.
5(1 + 2m) = \(\frac{1}{2}\)(8 + 20m)
Answer:
m has indefinite solutions

Explanation:
The given equation is:
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
By using the Distributive Property of Multiplication,
5 ( 1 ) + 5 ( 2m ) = \(\frac{1}{2}\) ( 8 ) + \(\frac{1}{2}\) ( 20m )
2 ( 5 + 10m ) = 8 + 20m
2 ( 5 ) + 2 ( 10m ) = 8 + 20m
10 + 20m = 8 + 20m
20m – 20m = 8 – 10
20m – 20m = -2
As  m has the same coefficients and have the opposite signs, m has indefinite solutions
Hence, from the above,
We can conclude that the equation has the indefinite solutions

ERROR ANALYSIS
In Exercises 25 and 26, describe and correct the error in solving the equation.

Question 25.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 50

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q25

Question 26.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 51

Answer:
The given equation is:
6 ( 2y + 6 ) = 4 ( 9 + 3y )
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
Now,
By using the Distributive Property of Multiplication,
6 ( 2y ) + 6 ( 6 ) = 4 ( 9 ) + 4 ( 3y )
12y + 36 = 36 + 12y
12y – 12y = 36 – 36
0 = 0
As the coefficients of y are zero, the equation has no solution
Hence, from the above,
We can conclude that there is no error in the analysis of the equation.

Question 27.
MODELING WITH MATHEMATICS
Write and solve an equation to find the month when you would pay the same total amount for each Internet service.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 52

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q27

Question 28.
PROBLEM-SOLVING
One serving of granola provides 4% of the protein you need daily. You must get the remaining 48 grams of protein from other sources. How many grams of protein do you need daily?
Answer:
The number of grams of protein you need daily is: 50 grams

Explanation:
It is given that one serving of granola provides 4% of the protein you need daily. You must get the remaining 48 grams of protein from other sources.
So,
Let the number of grams of protein you need daily be: x
So,
The number of grams of protein you need daily = 4 % of x + 48
We know that,
100%  = 1
So,
4 % = 0.04
So,
The number of grams of protein you need daily = 0.04x + 48
x = 0.04x + 48
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
x – 0.04x = 48
0.96x = 48
\(\frac{96}{100}\)x = 48
x = 48 × \(\frac{100}{96}\)
x = \(\frac{48}{1}\) × \(\frac{100}{96}\)
x = \(\frac{48 × 100}{1 × 96}\)
x = \(\frac{50}{1}\)
x = 50 grams
Hence, from the above,
We can conclude that the number of proteins you need daily is: 50 grams

USING STRUCTURE
In Exercises 29 and 30, find the value of r.

Question 29.
8(x + 6) – 10 + r = 3(x + 12) + 5x

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q29

Question 30.
4(x – 3) – r + 2x = 5(3x – 7) – 9x
Answer:
The value of r is: 23

Explanation:
The given equation is:
4 ( x – 3 ) – r + 2x = 5 ( 3x – 7 ) – 9x
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
By using the Distributive Property of Multiplication,
4x – 4 ( 3 ) – r + 2x = 5 ( 3x ) – 5 ( 7 ) – 9x
4x – 12 – r + 2x = 15x – 35 – 9x
6x – 12 – r = 6x – 35
r = 6x – 6x – 12 + 35
r = 23
Hence, from the above,
We can conclude that the value of r is: 23

MATHEMATICAL CONNECTIONS
In Exercises 31 and 32, the value of the surface area of the cylinder is equal to the value of the volume of the cylinder. Find the value of x. Then find the surface area and volume of the cylinder.

Question 31.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 53

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q31

Question 32.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 54
Answer:
The Surface Area of the cylinder is: 461.49 cm²
The volume of the cylinder is: 488.58 cm³

Explanation:
The given figure is:
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 54
From the above figure,
The radius of the cylinder is: 7\(\frac{1}{5}\) feet
The height of the cylinder is: x feet
It is given that the Total Surface Area of the cylinder and the volume of the cylinder are equal.
We know that,
The Surface Area of the cylinder = 2πr² + 2πrh
The volume of the cylinder = πr²h
The value of π is: 3.1416
Now,
The representation of 7\(\frac{1}{5}\) in the improper fraction form is: \(\frac{36}{5}\)
So,
2πr² + 2πrh = πr²h
[ 2 × 3.1416 × \(\frac{36}{5}\) × \(\frac{36}{5}\) ] + [ 2 × 3.1416 × \(\frac{36}{5}\) × x ] = [3.1416 × latex]\frac{36}{5}[/latex] × \(\frac{36}{5}\) × x ]
325.72 + 45.23x = 162.86x
162.86x – 45.23x = 325.72
117.63x = 325.72
x = 2.76
x = 3
So,
The Surface Area of the cylinder = 2πr² + 2πrh
= [ 2 × 3.1416 × \(\frac{36}{5}\) × \(\frac{36}{5}\) ] + [ 2 × 3.1416 × \(\frac{36}{5}\) × x ]
= 325.72 + 45.23x
= 325.72 + 45.23 ( 3 )
= 461.49 cm²
The volume of the cylinder = πr²h
= [3.1416 × latex]\frac{36}{5}[/latex] × \(\frac{36}{5}\) × x ]
= 162.86x
= 162.86 ( 3 )
= 488.58 cm³
Hence, from the above,
We can conclude that
The Surface Area of the cylinder is: 461.49 cm²
The volume of the cylinder is: 488.58 cm³

Question 33.
MODELING WITH MATHEMATICS
A cheetah that is running 90 feet per second is 120 feet behind an antelope that is running 60 feet per second. How long will it take the cheetah to catch up to the antelope?

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q33

Question 34.
MAKING AN ARGUMENT
A cheetah can run at top speed for only about 20 seconds. If an antelope is too far away for a cheetah to catch it in 20 seconds, the antelope is probably safe. Your friend claims the antelope in Exercise 33 will not be safe if the cheetah starts running 650 feet behind it. Is your friend correct? Explain.
Answer:
Your friend is not correct

Explanation:
It is given that a cheetah can run at top speed for only about 20 seconds. If an antelope is too far away for a cheetah to catch it in 20 seconds, the antelope is probably safe. Your friend claims the antelope in Exercise 33 will not be safe if the cheetah starts running 650 feet behind it.
Let the distance of running antelope be x.
Let ‘t’ be the time taken
So,
The distance of running Antelope is:
x = 650 + 60t
The cheetah must arrive at the same position to catch the antelope
So,
x = 90t
Now,
90t = 650 + 60t
90t – 60t = 650
30t = 650
t = 650 ÷ 30
t = 21.7 seconds
But it is given that the cheetah has to reach the same position as the antelope in 20 seconds
But according to the calculation, it takes 21.7 seconds
So,
According to your friend, the antelope is not safe if the cheetah is running 650 meters behind it.
Hence, from the above,
We can conclude that your friend is not correct.

REASONING
In Exercises 35 and 36, for what value of a is the equation an identity? Explain your reasoning.

Question 35.
a(2x + 3) = 9x + 15 + x

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q35

Question 36.
8x – 8 + 3ax = 5ax – 2a
Answer:
The given equation becomes an identity at a = 4

Explanation:
The given equation is:
8x – 8 + 3ax = 5ax – 2a
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
5ax – 3ax = 8x – 8 – 2a
2ax = 8 ( x – 1 ) – 2a
2ax + 2a = 8x – 8
Equate the like coefficients of x and the like constants in both LHS and RHS
So,
2ax = 8x                                   2a = -8
a = 8x ÷ 2x                               a = -8 ÷ 2
a = 4                                         a = -4
Now,
At a = 4,
The equation becomes
8x – 8 + 3ax = 5ax – 2a
8x – 8 + 3x ( 4 ) = 5x ( 4 ) -2 ( 4 )
8x – 8 + 12x = 20x – 8
20x – 8 = 20x – 8
Hence,
At a =4,
The given equation is an Identity
At a = -4,
The equation becomes
8x – 8 + 3ax = 5ax – 2a
8x – 8 + 3x ( -4 ) = 5x ( -4 ) -2 ( -4 )
8x – 8 – 12x = -20x + 8
-4x – 8 = -20x + 8
Hence,
At a = -4, the given equation is not an Identity
Hence, from the above,
We can conclude that the given equation is an Identity at a = 4

Question 37.
REASONING
Two times the greater of two consecutive integers is 9 less than three times the lesser integer. What are the integers?

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q37

Question 38.
HOW DO YOU SEE IT?
The table and the graph show information about students enrolled in Spanish and French classes at a high school.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 55
a. Use the graph to determine after how many years there will be equal enrollment in Spanish and French classes.
Answer:
The year where there will be equal enrollment in Spanish and French classes is: 6

Explanation:
The given table and the graph of the students for the Spanish and French classes are:
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 55
So,
From the above table,
To find the year where there is equal enrollment in both Spanish and French, we have to find the point in the graph that the two points of Spanish and French intersect
So,
From the above graph,
The point that is intersecting is at 6th year
Hence, from the above,
We can conclude that there is an equal enrollment of students in the 6th year in both Spanish and French classes

b. How does the equation 355 – 9x = 229 + 12x relate to the table and the graph? How can you use this equation to determine whether your answer in part (a) is reasonable?
Answer:
The given equation
355 – 9x = 229 + 12x
represents the total number of students enrolled in the different years in both Spanish and French classes
Now,
In part (a),
We observed that there is an equal enrollment of the students at 6th year in both Spanish and French classes
So,
Here,
x is: The number of years
So,
In part (a),
x = 6
Now,
Substitute x = 6 in the given equation.
Now,
355 – 9x = 229 + 12x
355 – 9 ( 6 ) = 229 + 12 ( 6 )
355 – 54 = 229 + 72
301 = 301
As
LHS = RHS
We can say that the answer is reasonable in part (a)

Question 39.
WRITING EQUATIONS
Give an example of a linear equation that has (a) no solution and (b) infinitely many solutions. Justify your answers.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q39

Question 40.
THOUGHT-PROVOKING
Draw a different figure that has the same perimeter as the triangle shown. Explain why your figure has the same perimeter.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 56

Answer:

Maintaining Mathematical Proficiency

Order the values from least to greatest.

Question 41.
9, | -4|, -4, 5, | 2 |

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q41

Question 42.
| -32 |, 22, -16, -| 21 |, | -10 |
Answer:
We know that,
| -x | = x
| x | = x
So,
| -32 | = 32
| 21 | = 21
| -10 | = 10
Hence,
The order of the values from the least to the greatest is:
-21, -16, 10, 22, 32

Question 43.
-18, | -24 |, -19, | -18 |, | 22 |

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q43

Question 44.
-| – 3 |, | 0 |, -1, | 2 |, -2
Answer:
We know that,
| -x | = x
| x | = x
So,
| -3 | = 3
| 0 | = 0
| 2 | = 2
Hence,
The order of the numbers from the least to the greatest is:
-3, -2, -1, 0, 2

Solving Linear Equations Study Skills: Completing

1.1-1.3 What Did You Learn

Core Vocabulary

Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 57

Core Concepts

Section 1.1
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 58

Section 1.2

Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 59

Section 1.3

Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 60

Mathematical Practices

Question 1.
How did you make sense of the relationships between the quantities in Exercise 46 on page 9?
Answer:
In Exercise 46 on page 9,
There is a layout of the tatami mat which comprises the four identical rectangular mats and the one square mat.
it is also given that the area of the square mat is half of one of the rectangular mats
Now,
We know that,
The area of the square mat = Area² [ Since all the sides of the square are equal ]
The area of the rectangular mat = Length × Width
So,
According to the given condition,
The relation between the area of the square mat and one of the rectangular mat is:
Area of the square mat = \(\frac{1}{2}\) Area of one of the rectangular mat
Side² = \(\frac{1}{2}\) ( Length × Width )

Question 2.
What is the limitation of the tool you used in Exercises 25–28 on page 16?
Answer:
The limitations of the tool you used in Exercises 25 – 28 on page 16 are:
A) The calculated values and the values measured using the tool will be different
B) We won’t get the exact values of the angle measures using the tool

Question 3.
What definition did you use in your reasoning in Exercises 35 and 36 on page 24?
Answer:
The definition you used in your reasoning in Exercises 35 and 36 on page 24 is:
Make the like coefficients of the same variable in both LHS and RHS equal so that we get the value of the variable.

Study Skills

Completing Homework Efficiently

Before doing homework, review the Core Concepts and examples. Use the tutorials at BigIdeasMath.com for additional help.

Complete homework as though you are also preparing for a quiz. Memorize different types of problems, vocabulary, rules, and so on.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 61

Solving Linear Equations 1.1-1.3 Quiz

Solve the equation. Justify each step. Check your solution. (Section 1.1)

Question 1.
x + 9 = 7
Answer:
The value of x is -2

Explanation:
The given equation is:
x + 9 = 7
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
x = 7 –  ( +9 )
x = 7 – 9
x = -2
Hence, from the above,
We can conclude that the value of x is: -2

Question 2.
8.6 = z – 3.8
Answer:
The value of z is: 12.4

Explanation:
The given equation is:
8.6 = z – 3.8
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
z = 8.6 + 3.8
z = 12.4
Hence, from the above,
We can conclude that the value of z is: 12.4

Question 3.
60 = -12r
Answer:
The value of r is:  -5

Explanation:
The given equation is:
60 = -12r
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
r = 60 ÷ ( -12 )
r = -60 ÷ 12
r = -5
Hence, from the above,
We can conclude that the value of r is: -5

Question 4.
\(\frac{3}{4}\)p = 18
Answer:
The value of p is: 24

Explanation:
The given equation is:
\(\frac{3}{4}\)p = 18
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
p = 18 × \(\frac{4}{3}\)
p = \(\frac{18}{1}\) × \(\frac{4}{3}\)
p = \(\frac{18 × 4}{1 × 3}\)
p = \(\frac{24}{1}\)
p = 24
Hence, from the above,
We can conclude that the value of p is: 24

Solve the equation. Check your solution. (Section 1.2)

Question 5.
2m – 3 = 13
Answer:
The value of m is: 8

Explanation:
The given equation is:
2m – 3 = 13
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
2m = 13 + 3
2m = 16
m = 16 ÷ 2
m = 8
Hence, from the above,
We can conclude that the value of m is: 8

Question 6.
5 = 10 – v
Answer:
The value of v is: 5

Explanation:
The given equation is:
5 = 10 – v
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
v= 10 – 5
v = 5
Hence, from the above,
We can conclude that the value of v is: 5

Question 7.
5 = 7w + 8w + 2
Answer:
The value of w is: \(\frac{1}{5}\)

Explanation:
The given equation is:
5 = 7w + 8w + 2
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
5 – 2 = 7w + 8w
15w = 3
w = 3 ÷ 15
w = \(\frac{1}{5}\)
Hence, from the above,
We can conclude that the value of w is: \(\frac{1}{5}\)

Question 8.
-21a + 28a – 6 = -10.2
Answer:
The value of a is: –\(\frac{3}{5}\)

Explanation:
The given equation is:
-21a + 28a  – 6 = -10.2
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
-21a + 28a = -10.2 + 6
7a = -4.2
a = -4.2 ÷ 7
a= –\(\frac{42}{10}\) ÷ 7
a = –\(\frac{42}{10}\) × \(\frac{1}{7}\)
a = –\(\frac{42 × 1}{10 × 7}\)
a = – \(\frac{6}{10}\)
a = –\(\frac{3}{5}\)
Hence, from the above,
We can conclude that the value of a is: –\(\frac{3}{5}\)

Question 9.
2k – 3(2k – 3) = 45
Answer:
The value of k is: -9

Explanation:
The given equation is:
2k – 3 ( 2k – 3 ) = 45
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
SO,
2k – 3 ( 2k ) + 3 ( 3 ) = 45
2k – 6k + 9 = 45
2k – 6k = 45 – 9
-4k = 36
k = 36 ÷ -4
k = -9
Hence, from the above,
We can conclude that the value of k is: -9

Question 10.
68 = \(\frac{1}{5}\)(20x + 50) + 2
Answer:
The value of x is: 14

Explanation:
The given equation is:
68 = \(\frac{1}{5}\) [ 20x + 50 ] + 2
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
68 – 2 = \(\frac{1}{5}\) [ 20x + 50 ]
66 = \(\frac{1}{5}\) [ 20x + 50 ]
66 × 5 = 20x + 50
330 = 20x + 50
20x = 330 – 50
20x = 280
x = 280 ÷ 20
x = 14
Hence, from the above,
We can conclude that the value of x is: 14

Solve the equation. (Section 1.3)

Question 11.
3c + 1 = c + 1
Answer:
The value of c is: 0

Explanation:
The given equation is:
3c + 1 = c + 1
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
3c – c = 1 – 1
2c = 0
c = 0
Hence, from the above,
We can conclude that the value of c is: 0

Question 12.
-8 – 5n = 64 + 3n
Answer:
The value of n is: -9

Explanation:
The given equation is:
-8 – 5n = 64 + 3n
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
-8 – 64 = 3n + 5n
-72 = 8n
n = -72 ÷ 8
n = -9
Hence, from the above,
We can conclude that the value of n is: -9

Question 13.
2(8q – 5) = 4q
Answer:
The value of q is: \(\frac{5}{6}\)

Explanation:
Te given equation is:
2 ( 8q – 5 ) = 4q
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
2 ( 8q ) – 2 ( 5 ) = 4q
16q – 10 = 4q
16q – 4q = 10
12q = 10
q = 10 ÷ 12
q = \(\frac{5}{6}\)
Hence, from the above,
We can conclude that the value of q is: \(\frac{5}{6}\)

Question 14.
9(y – 4) – 7y = 5(3y – 2)
Answer:
The value of y is: -2

Explanation:
The given equation is:
9 ( y – 4 ) – 7y = 5 ( 3y – 2 )
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
9 ( y ) – 9 ( 4 ) – 7y = 5 ( 3y ) – 5 ( 2 )
9y – 36 – 7y = 15y – 10
2y – 36 = 15y – 10
15y – 2y = 10 – 36
13y = -26
y = -26 ÷ 13
y = -2
Hence, from the above,
We can conclude that the value of y is: -2

Question 15.
4(g + 8) = 7 + 4g
Answer:
The given equation has no solution

Explanation:
The given equation is:
4 ( g + 8 ) = 7 + 4g
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,the given equation has no solution
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
4 ( g ) + 4 ( 8 ) = 7 + 4g
4g + 32 = 7 + 4g
4g – 4g + 7 = 32
7 = 32
Hence, from the above,
We can conclude that the given equation has no solution.

Question 16.
-4(-5h – 4) = 2(10h + 8)
Answer:
The value of h is: 0

Explanation:
The given equation is:
-4 ( 5h – 4 ) = 2 ( 10h + 8 )
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
-4 ( 5h ) + 4 ( 4 ) = 2 ( 10h ) + 2 ( 8 )
-20h + 16 = 20h + 16
-20h – 20h = 1 – 16
-40h =0
h = 0
Hence, from the above,
We can conclude that the value of h is: 0

Question 17.
To estimate how many miles you are from a thunderstorm, count the seconds between when you see lightning and when you hear thunder. Then divide by 5. Write and solve an equation to determine how many seconds you would count for a thunderstorm that is 2 miles away. (Section 1.1)
Answer:

Question 18.
You want to hang three equally-sized travel posters on a wall so that the posters on the ends are each 3 feet from the end of the wall. You want the spacing between posters to be equal. Write and solve an equation to determine how much space you should leave between the posters. (Section 1.2)
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 62

Answer:
The space you should leave between the posters is: \(\frac{3}{2}\) ft

Explanation:
The given figure is:
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 62
It is given that you want to hang three equally-sized travel posters on a wall so that the posters on the ends are each 3 feet from the end of the wall. You want the spacing between posters to be equal.
So,
From the figure,
The total space = 15 ft
The total spacing covered at the ends = 3 + 3 = 6 ft
Let the space between the equally spaced posters be x
So,
The total spacing between the travel posters = 2x + 2x + 2x = 6x ft
SO,
The total space = ( The total spacing covered at the ends ) + ( The total spacing between the travel posters )
15 = 6 + 6x
6x = 15 – 6
6x = 9
x = 9 ÷ 6
x = \(\frac{3}{2}\) ft
Hence, from the above,
We can conclude that the spacing between the travel posters is: \(\frac{3}{2}\) ft

Question 19.
You want to paint a piece of pottery at an art studio. The total cost is the cost of the piece plus an hourly studio fee. There are two studios to choose from. (Section 1.3)
a. After how many hours of the painting are the total costs the same at both studios? Justify your answer.
b. Studio B increases the hourly studio fee by $2. How does this affect your answer in part (a)? Explain.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 63

Answer:
a)
It is given that,
The total cost = Cost of the vase + The hourly studio fee
Let the number of hours be x
So,
The total cost for studio A = 10 + 8x
The total cost of studio B = 16 + 6x
It is given that the total costs are the same
So,
10 + 8x = 16 + 6x
8x – 6x = 16 – 10
2x = 6
x = 6 ÷ 2
x = 3
Hence, from the above,
We can conclude that the total cost will be the same after 3 hours for both the studios

b)
It is given that the studio B increases the hourly studio fee by $2
So,
The total hourly studio fee for studio B = 6 + 2 = $8
So,
Now,
As in part (a), the same process will be repeated but in the studio B’s hourly fee of $6, we have to put $8
So,
10 + 8x = 16 + 8x
8x – 8x = 6 – 10
10 = 16
Hence, from the above,
We can conclude that the value of x has no solutions

Lesson 1.4 Solving Absolute Value Equations

Essential Question

How can you solve an absolute value equation?
EXPLORATION 1
Solving an Absolute Value Equation Algebraically
Work with a partner. Consider the absolute value equation
| x + 2 | = 3.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 64
a. Describe the values of x + 2 that make the equation true. Use your description to write two linear equations that represent the solutions of the absolute value equation.
Answer:
We know that,
| x | = x
-| x | = -x
So,
| x + 2 | = 3
x + 2 = 3
x = 3 – 2
x = 1
Now,
-| x + 2 | = 3
| x + 2  | = -3
-x – 2 = -3
-x = -3 + 2
-x = -1
x = 1
So,
The values of x + 2 that make the equation true is: 3 and 3
The value of x is: 1 and 1

b. Use the linear equations you wrote in part (a) to find the solutions of the absolute value equation.
Answer:
We know that,
| x | = x
-| x | = -x
So,
| x + 2 | = 3
x + 2 = 3
x = 3 – 2
x = 1
Now,
-| x + 2 | = 3
-| x + 2  | = -3
-x – 2 = -3
-x = -3 + 2
-x = -1
x = 1
So,
The solutions of | x + 2 | are: 1 and 1

c. How can you use linear equations to solve an absolute value equation?
Answer:
We use linear equations to solve an absolute value equation by using the following properties. They are:
A) | x | = x
B) -| x | = -x

EXPLORATION 2
Solving an Absolute Value Equation Graphically
Work with a partner.
Consider the absolute value equation
| x + 2 | = 3.
a. On a real number line, locate the point for which x + 2 = 0.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 64.1
Answer:
The given equation is:
x + 2 = 0
x = 0 – 2
x = -2
So,
On a real number line, we have to locate the point x = -2
Hence,
The point we have to locate on the real number line is:

b. Locate the points that are 3 units from the point you found in part (a). What do you notice about these points?
Answer:
From part (a).
We found that x = -2
Now,
To locate the points that are 3 units away or 3 units behind from the point you found in part (a), i.e., x = -2
We know that,
3 units away imply ” Add 3 ”
3 units behind imply ” Subtract 3 ”
Now,
We have to add 3 and subtract 3 to the point we obtained in part (a)
So,
When we add 3 to x = -2,
x = -2 + 3
x =1
When we subtract 3 from x = -2,
x = -2 – 3
x = -5
Hence,
The points we have to locate in the real number line are: 1 and -5
So,
The real number line with the located points is:

c. How can you use a number line to solve an absolute value equation?
Answer:
The given absolute value equation is:
| x + 2 | = 3
We know that,
| x | = x
-| x | = -x
So,
| x + 2 | = 3
x + 2 = 3
x = 3 – 2
x = 1
– | x + 2 | = -3
-x – 2 = -3
-x = -3 + 2
-x = -1
x = 1
So,
The values we have to locate in the number line is:

EXPLORATION 3
Solving an Absolute Value Equation Numerically

Work with a partner. Consider the absolute value equation
| x + 2 | = 3.
a. Use a spreadsheet, as shown, to solve the absolute value equation.
b. Compare the solutions you found using the spreadsheet with those you found in Explorations 1 and 2. What do you notice?
c. How can you use a spreadsheet to solve an absolute value equation?
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 65

Communicate Your Answer

Question 4.
How can you solve an absolute value equation?
Answer:
We can solve the absolute equation by using the following properties. They are:
A) | x| = x
B) – | x | = -x

Question 5.
What do you like or dislike about the algebraic, graphical, and numerical methods for solving an absolute value equation? Give reasons for your answers.
Answer:
The algebraic, numerical, and graphical methods have their own advantages in their own perspective.
The algebraic methods used to solve the linear equations whereas the graphical method used to indicate the linear equations.
The numerical method is applicable for mathematical operations

1.4 Lesson

Monitoring Progress

Solve the equation. Graph the solutions, if possible.

Question 1.
| x | = 10
Answer:
The value of x is: 10

Explanation:
The given absolute value equation is:
| x | = 10
We know that,
| x | = x
So,
x = 10
Hence, from the above,
We can conclude that the value of x is: 10

Question 2.
| x – 1 | = 4
Answer:
The value of x is: 5

Explanation:
The given absolute value equation is:
| x – 1 | = 4
We know that,
| x | = x
So,
x – 1 = 4
x = 4 + 1
x = 5
Hence, from the above,
We can conclude that the value of x is: 5

Question 3.
| 3 + x | = -3
Answer:
The value of x is: -6

Explanation:
The given absolute value equation is:
| 3 + x | = -3
We know that,
| x | = x
So,
x + 3 = -3
x = -3 – 3
x = -6
Hence, from the above,
We can conclude that the value of x is: -6

Solve the equation. Check your solutions.

Question 4.
| x – 2 | + 5 = 9
Answer:
The value of x is: 6

Explanation:
The given absolute value equation is:
| x – 2 | + 5 = 9
| x – 2 | = 9 – 5
| x – 2 | = 4
We know that,
| x | = x
So,
x – 2 = 4
x = 4 + 2
x = 6
Hence, from the above,
We can conclude that the value of x is: 6

Question 5.
4 | 2x + 7 | = 16
Answer:
The value of x is: –\(\frac{3}{2}\)

Explanation:
The given absolute value equation is:
4 | 2x + 7 | = 16
| 2x + 7 | = 16 ÷ 4
| 2x + 7| = 4
We know that,
| x | = x
So,
2x + 7 = 4
2x = 4 – 7
2x = -3
x = –\(\frac{3}{2}\)
Hence, from the above,
We can conclude that the value of x is: –\(\frac{3}{2}\)

Question 6.
-2 | 5x – 1 | – 3 = -11
Answer:
The value of x is: 1

Explanation:
The given absolute value equation is:
-2 | 5x – 1 | – 3 = -11
-2 | 5x – 1 | = -11 + 3
-2 | 5x – 1 | = -8
| 5x – 1 | = -8 ÷ ( -2 )
| 5x – 1 | = 4
We know that,
| x | = x
So,
5x – 1 = 4
5x = 4 + 1
5x = 5
x = 5 ÷ 5
x = 1
Hence, from the above,
We can conclude that the value of x is: 1

Question 7.
For a poetry contest, the minimum length of a poem is 16 lines. The maximum length is 32 lines. Write an absolute value equation that represents the minimum and maximum lengths.
Answer:
The minimum value length is: 16
The maximum length is: 32

Explanation:
It is given that for a poetry contest, the minimum length of a poem is 16 lines. The maximum length is 32 lines.
So,
The absolute value equation that represents the minimum length of a poem = | The minimum length of a poem |
= | 16 |
= 16
The absolute value equation that represents the maximum length of a poem = | The maximum length of a poem |
= | 32 |
= 32
Hence, from the above,
We can conclude that
The minimum value length is: 16
The maximum length is: 32

Solve the equation. Check your solutions.

Question 8.
| x + 8 | = | 2x + 1 |
Answer:
The value of x is: 7

Explanation:
The given absolute value equation is:
| x + 8 | = | 2x + 1 |
We know that,
| x | = x
So,
x + 8 = 2x + 1
2x – x = 8 – 1
x = 7
Hence from the above,
We can conclude that the value of x is: 7

Question 9.
3 | x – 4 | = | 2x + 5 |
Answer:
The value of x is: 17

Explanation:
The given absolute equation is:
3 | x – 4 | = | 2x + 5 |
We know that,
| x |  = x
So,
3 ( x – 4 ) = 2x + 5
3 ( x ) – 3 ( 4 ) = 2x + 5
3x – 12 = 2x + 5
3x – 2x = 5 + 12
x = 17
Hence, from the above,
We can conclude that the value of x is: 17

Solve the equation. Check your solutions.

Question 10.
| x + 6 | = 2x
Answer:
The value of x is: 6

Explanation:
The absolute value equation is:
| x + 6 | = 2x
We know that,
| x | = x
So,
x + 6 = 2x
2x – x = 6
x = 6
Hence, from the above,
We can conclude that the value of x is: 6

Question 11.
| 3x – 2 | = x
Answer:
The value of x is: 1

Explanation:
The given absolute value equation is:
| 3x – 2 | = x
We know that,
| x | = x
So,
3x – 2 = x
Soo,
3x – x = 2
2x = 2
x = 2 ÷ 2
x = 1
Hence, from the above,
We can conclude that the value of x is: 1

Question 12.
| 2 + x | = | x – 8 |
Answer:
The given absolute value equation has no solution

Explanation:
The given absolute value equation is:
| 2 + x | = | x – 8 |
We know that,
| x | = x
So,
2 + x = x – 8
2 = x – x – 8
2 = -8
Hence, from the above,
We can conclude that the given absolute value equation has no solution

Question 13.
| 5x – 2 | = | 5x + 4 |
Answer:
The given absolute value equation has no solution

Explanation:
The given absolute value equation is:
| 5x – 2 | = | 5x + 4 |
We know that,
| x | = x
So,
5x – 2 = 5x + 4
5x – 5x – 2 = 4
-2 = 4
Hence, from the above,
We can conclude that the given absolute value equation has no solution

Solving Absolute Value Equations 1.4 Exercises

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
What is an extraneous solution?
Answer:
Extraneous solutions are values that we get when solving equations that are not really solutions to the equation.
Example for extraneous solution:
| 5x – 2 | = | 5x + 4 |

Question 2.
WRITING
Without calculating, how do you know that the equation | 4x – 7 | = -1 has no solution?
Answer:
The given absolute value equation is:
| 4x – 7 | = -1
We know that,
An absolute value can never equal a negative number.
So,
By the above,
We can say that
| 4x – 7 | must not equal to a negative number.
Hence, from the above,
We can conclude that | 4x – 7 | = -1 has no solution without calculating its solution

Monitoring Progress and Modeling with Mathematics

In Exercises 3−10, simplify the expression.

Question 3.
| -9 |

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q3

Question 4.
– | 15 |
Answer:
The value of -| 15 | is: -15

Explanation:
The given absolute value is: -| 15 |
We know that,
| x | = x
| -x | = x
-| x | = -x
So,
-| 15 | = -15
Hence, from the above,
We can conclude that the value of -| 15 | is: -15

Question 5.
| 14 | – | -14 |

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q5

Question 6.
| -3 | + | 3 |
Answer:

The value of | -3 | + | 3 | is: 6

Explanation:
The given absolute value expression is:
| -3 | + | 3 |
We know that,
| x | = x
| -x | = x
-| x | = -x
So,
| -3 | + | 3 |
= 3 + 3
= 6
Hence, from the above,
We can conclude that the value of | -3 | + | 3 | is: 6

Question 7.
– | -5 • (-7) |

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q7

Question 8.
| -0.8 • 10 |
Answer:
The value of | -0.8 ⋅ 10 | is: 8

Explanation:
The given absolute value expression is:
| -0.8 ⋅ 10 |
We know that,
| x | = x
| -x | = x
-| x | = -x
So,
| -0.8 ⋅ 10 |
= | – ( 8 ⁄ 10 ) ⋅ ( 10 ⁄ 1 ) |
= | – ( 8 × 10 ) ⁄ ( 10 × 1 ) |
= | -8 |
= 8
hence, from the above,
We can conclude that the value of | -0.8 ⋅ 10 | is: 8

Question 9.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 66

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q9

Question 10.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 67
Answer:
The value of | -12 ⁄ 4 | is: 3

Explanation:
The given absolute value expression is: | -12 ⁄ 4 |
We know that,
| x | = x
| -x | = x
– | x | = -x
So,
| -12 ⁄ 4 | = | -3 |
= 3
Hence, from the above,
We can conclude that the value of | -12 ⁄ 4 | is: 3

In Exercises 11−24, solve the equation. Graph the solution(s), if possible.

Question 11.
| w | = 6

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q11

Question 12.
| r | = -2
Answer:
The absolute value of a number must be greater than or equal to 0 and can not be equal to -2.
Hence,
The given absolute eqution has no solution

Question 13.
| y | = -18

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q13

Question 14.
| x | = 13
Answer:
The value of x is: 13 or -13

Explanation:
The given absolute value equation is:
| x | = 13
We know that,
|x | = x
– | x | = -x
So,
| x | = 13 or – 13
Hence, from the above,
We can conclude that the value of x is: 13 or -13

Question 15.
| m + 3 | = 7

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q15

Question 16.
| q – 8 | = 14
Answer:
The value of q is: 22 or -6

Explanation:
The given absolute value equation is:
| q – 8 | = 14
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
q – 8 = 14                                                      q – 8 = -14
q = 14 + 8                                                     q = -14 + 8
q = 22                                                             q = -6
Hence, from the above,
We can conclude that the value of q is: 22 or -6

Question 17.
| -3d | = 15

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q17

Question 18.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 68
Answer:
The value of t is: 12 or -12

Explanation:
The given absolute value equation is:
| t / 2 | = 12
We know that,
| x | = x  for x > 0
| x | = -x for x < 0
So,
t / 2 = 6                                       t / 2 = -6
t = 6 × 2                                      t = 6 × -2
t = 12                                           t = -12
Hence, from the above,
We can conclude that the value of t is: 12 or -12

Question 19.
| 4b – 5 | = 19

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q19

Question 20.
| x – 1 | + 5 = 2
Answer:
The given absolute value equation has no solution

Explanation:
The given absolute value equation is:
| x – 1 | + 5 = 2
| x – 1 | = 2 – 5
| x – 1  | = -3
We know that,
The absolute value of an equation must be greater than or equal to zero
So,
| x – 1 | = -3 has no solution
Hence, from the above,
We can conclude that the given absolute value equation has no solution

Question 21.
-4 | 8 – 5n | = 13

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q21

Question 22.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 69
Answer:
The value of y is: -3 or 6

Explanation:
The given absolute value equation is:
-3 | 1 – ( 2 / 3 ) y | = -9
| 1 – (2 / 3  ) y | = -9 ÷ ( -3 )
| 1 – ( 2 / 3 ) y | = 3
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
1 – ( 2 / 3 ) y = 3                                    1 – ( 2 /3 ) y = -3
2/3 y = 1 – 3                                           2/3 y = 1 + 3
2 / 3 y = -2                                              2 / 3 y = 4
2y = -2 × 3                                               2y = 4 × 3
2y = -6                                                      2y = 12
y = -6 ÷ 2                                                  y = 12 ÷ 2
y = -3                                                         y = 6
Hence, from the above,
We can conclude that the value of y is: -3 or 6

Question 23.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 70

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q23

Question 24.
9 | 4p + 2 | + 8 = 35
Answer:
The value of p is: 1 / 4 or -5 / 4

Explanation:
The given absolute value equation is:
9 | 4p + 2 | + 8 = 35
9 | 4p + 2 | = 35 – 8
9 | 4p + 2 | = 27
| 4p + 2 | = 27 ÷ 9
| 4p + 2 | = 3
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
4p + 2 = 3                    4p + 2 = -3
4p = 3 – 2                     4p = -3 – 2
4p = 1                           4p = -5
p = 1 / 4                         p = -5 / 4
Hence, from the above,
We can conclude that the value of p is: 1 / 4 or -5 / 4

Question 25.
WRITING EQUATIONS
The minimum distance from Earth to the Sun is 91.4 million miles. The maximum distance is 94.5 million miles.
a. Represent these two distances on a number line.
b. Write an absolute value equation that represents the minimum and maximum distances.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q25

Question 26.
WRITING EQUATIONS
The shoulder heights of the shortest and tallest miniature poodles are shown.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 71
a. Represent these two heights on a number line.
b. Write an absolute value equation that represents these heights.

Answer:
a)
The number line that represents the two heights on a number line is:

b)
The minimum shoulder height = ( 15 – 10 ) / 2
= 5 / 2
= 2.5 inches
The maximum shoulder height = 10 + 2.5
= 12.5 inches
Now,
Let the heights between poodles be x.
Hence,
The absolute value equation is:
| x – 12.5 | = 2.5

USING STRUCTURE In Exercises 27−30, match the absolute value equation with its graph without solving the equation.

Question 27.
| x + 2 | = 4

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q27

Question 28.
| x + 4 | = 2
Answer:
The given absolute value equation is:
| x + 4 | = 2
To find the halfway point, made the absolute value equation equal to 0.
So,
| x + 4  | = 0
So,
x = -4
From the given absolute value equation,
We can say that the distance from the halfway point to the minimum and maximum points is: 2

Question 29.
| x – 2 | = 4

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q29

Question 30.
| x + 4 | = 2
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 72

Answer:

In Exercises 31−34, write an absolute value equation that has the given solutions.

Question 31.
x = 8 and x = 18

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q31

Question 32.
x = -6 and x = 10
Answer:
The given absolute value equation is:
| x – 2 | = 8

Explanation:
The given values of x are:
x = -6 and x = 10
Now,
The halfway point between 10 and -6 = [ 10 – ( -6 ) ] / 2
= [ 10 + 6 ] / 2
= 16 / 2
= 8
The minimum distance from the halfway point = 8 – 6 = 2
Hence,
The absolute value equation is:
| x – 2 | = 5

Question 33.
x = 2 and x = 9

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q33

Question 34.
x = -10 and x = -5

Answer:
The given values of x are:
x = -10 and x = -5
Now,
The halfway point between -10 and -5 = [ 10 – ( 5 ) ] / 2
= [ 10 – 5 ] / 2
= 5 / 2
= 2.5
So,
The minimum value from the half-point = 2.5 + ( -10 )
= 2.5 – 10
= -7.5
Hence,
The absolute value equation is:
| x – ( -7.5 ) | = 2.5
| x + 7.5 | = 2.5

In Exercises 35−44, solve the equation. Check your solutions. 

Question 35.
| 4n – 15 | = | n |

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q35

Question 36.
| 2c + 8 | = | 10c |
Answer:
The values of c are: 1 and 2 / 3

Explanation:
The given absolute value equation is:
| 2c + 8 | = | 10c |
We know that,
| x | = x for x > 0
| x | = -x for x < 0
Now,
2c + 8 = 10c                                      2c + 8 = -10c
10c – 2c = 8                                       2c + 10c = 8
8c = 8                                                12c = 8
c = 8 / 8                                              c = 8 / 12
c = 1                                                    c = 2 /3
Hence, from the above,
We can conclude that the values of c are: 1 and 2 / 3

Question 37.
| 2b – 9 | = | b – 6 |

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q37

Question 38.
| 3k – 2 | = 2 | k + 2 |
Answer:
The values of k are: 6 and -2 / 5

Explanation:
The given absolute equation is:
| 3k – 2 | = 2 | k + 2 |
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
2 ( k + 2 ) = 3k – 2                                            2 ( k + 2 ) = – ( 3k – 2 )
2k + 4 = 3k – 2                                                 2k + 4 = -3k + 2
3k – 2k = 4 + 2                                                 2k + 3k = 2 – 4
k = 6                                                                 5k = -2
k = 6                                                                 k = -2 / 5
Hence, from the above,
We can conclude that the values of k are: 6 and -2 / 5

Question 39.
4 | p – 3 | = | 2p + 8 |

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q39

Question 40.
2 | 4w – 1 | = 3 | 4w + 2 |
Answer:
The value of w is: -2

Explanation:
The given absolute value equation is:
2 | 4w – 1 | = 3 | 4w+ 2 |
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
2 ( 4w – 1 ) = 3 ( 4w + 2 )                        -2 ( 4w – 1 ) = -3 ( 4w + 2 )
8w – 2 = 12w + 6                                      -8w + 2 = -12w -6
12w – 8w = -6 – 2                                      -12w + 8w = 6 + 2
4w = -8                                                       -4w = 8
w = -8 / 4                                                      w = 8 / -4
w = -2                                                            w = -2
Hence, from the above,
We can conclude that the value of w is: -2

Question 41.
| 3h + 1 | = 7h

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q41

Question 42.
| 6a – 5 | = 4a
Answer:
The value of a is: 5 / 2 and 1 / 2

Explanation:
The given absolute value equation is:
| 6a – 5 | = 4a
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
6a – 5 = 4a                                                6a – 5 = -4a
6a – 4a = 5                                                 6a + 4a = 5
2a = 5                                                         10a = 5
a = 5 / 2                                                       a = 5 / 10
a = 5 / 2                                                       a = 1 / 2
Hence, from the above,
We can conclude that the values of a are: 5 / 2 and 1 / 2

Question 43.
| f – 6 | = | f + 8 |

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q43

Question 44.
| 3x – 4 | = | 3x – 5 |
Answer:
The given absolute value equation has no solution

Explanation:
The given absolute value equation is:
| 3x – 4 | = | 3x – 5 |
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
3x – 4 = 3x – 5                                – ( 3x – 4 ) = – ( 3x – 5 )
4 = 5                                                 4 = 5
Hence, from the above,
We can conclude that the given absolute value equation has no solution

Question 45.
MODELING WITH MATHEMATICS
Starting from 300 feet away, a car drives toward you. It then passes by you at a speed of 48 feet per second. The distance d (in feet) of the car from you after t seconds is given by the equation d = | 300 – 48t |. At what times is the car 60 feet from you?

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q45

Question 46.
MAKING AN ARGUMENT
Your friend says that the absolute value equation | 3x + 8 | – 9 = -5 has no solution because the constant on the right side of the equation is negative. Is your friend correct? Explain.
Answer:
Yes, your friend is correct

Explanation:
The given absolute value equation is:
| 3x + 8 | – 9 = -5
We know that,
The absolute value equation value must have greater than or equal to 0
But here
The value of the absolute value equation is less than 0
Hence,
The given absolute value equation has no solution.
Hence, from the above,
We can conclude that your friend is correct.

Question 47.
MODELING WITH MATHEMATICS
You randomly survey students about year-round school. The results are shown in the graph.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 73
The error given in the graph means that the actual percentage could be 5% more or 5% less than the percent reported by the survey.
a. Write and solve an absolute value equation to find the least and greatest percents of students who could be in favor of the year-round school.
b. A classmate claims that \(\frac{1}{3}\) of the student body is actually in favor of the year-round school. Does this conflict with the survey data? Explain.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q47

Question 48.
MODELING WITH MATHEMATICS
The recommended weight of a soccer ball is 430 grams. The actual weight is allowed to vary by up to 20 grams.
a. Write and solve an absolute value equation to find the minimum and maximum acceptable soccer ball weights.
Answer:
It is given that the recommended weight of a soccer ball is 430 grams and the actual weight is allowed to vary up to 20 grams
Hence,
The absolute value equation that represents the minimum and maximum acceptable soccer ball weights is:
| x – 430 | = 20
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
x – 430 = 20                               x – 430 = -20
x = 20 + 430                              x = -20 + 430
x = 460 grams                            x = 410 grams
Hence, from the above,
We can conclude that the maximum and minimum acceptable soccer weights respectively are: 460 grams and 410 grams

b. A soccer ball weighs 423 grams. Due to wear and tear, the weight of the ball decreases by 16 grams. Is the weight acceptable? Explain.
Answer:
The weight that caused due to wear and tear is not acceptable

Explanation:
From the above problem,
We get the maximum weight of the soccer ball to be 460 grams with 20 grams increase or decreased to the weight of the ball
Now,
It is given that the weight of the ball is decreased by 16 grams due to wear and tear
So,
The weight of the ball now = 460 – 16 = 444 grams
But it is given that the weight of the ball becomes 423 grams due to wear and tear.
Hence, from the above,
We can conclude that the weight is not acceptable

ERROR ANALYSIS
In Exercises 49 and 50, describe and correct the error in solving the equation.

Question 49.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 74

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q49

Question 50.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 75

Answer:
The values of x are: -2 and -4 / 3

Explanation:
The given absolute value equation is:
| 5x + 8 | = x
We know that,
| x | = x for x > 0
| x | = – x for x < 0
So,
5x + 8 = x                                                      5x + 8 = -x
5x – x = -8                                                      5x + x = -8
4x = -8                                                            6x = -8
x = -8 / 4                                                         x = -8 / 6
x = -2                                                               x = -4 / 3
Hence, from the above,
We can conclude that the values of x are: -2 and -4 / 3

Question 51.
ANALYZING EQUATIONS
Without solving completely, place each equation into one of the three categories.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 76

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q51

Question 52.
USING STRUCTURE
Fill in the equation Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 77 with a, b, c, or d so that the equation is graphed correctly.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 77.1

Answer:

ABSTRACT REASONING
In Exercises 53−56, complete the statement with always, sometimes, or never. Explain your reasoning.

Question 53.
If x2 = a2, then | x | is ________ equal to | a |.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q53

Question 54.
If a and b are real numbers, then | a – b | is _________ equal to | b – a |.
Answer:
If a and b are real numbers, then
| a – b | is equal to | b – a |

Explanation:
Let,
| a | = 5 and | b | = 9
We know that,
| x | =x for  x > 0
| x | = -x for x < 0
So,
| a – b | = | 5 – 9 |
= | -4 | = 4
| b – a | = | 9 – 5 |
=  | 4 |
= 4
Hence, from the above,
We can conclude that value of
| a – b | is equal to | b – a | if a and b are real numbers

Question 55.
For any real number p, the equation | x – 4 | = p will ________ have two solutions.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q55

Question 56.
For any real number p, the equation | x – p | = 4 will ________ have two solutions.
Answer:
For any real number,
| x – p | = 4 will have two solutions

Explanation:
The given absolute value equation is:
| x – p | = 4
Let the value of p be 1
We know that,
| x | = x for x > 0
| x | = – x for x < 0
So,
| x – 1 | = 4
| x – 1 | = 4                                        | x – 1 | = -4
x = 4 + 1                                           x = -4 + 1
x = 5                                                  x = -3
Hence, from the above,
We can conclude that
| x – p | = 4 will have two solutions for any real number p

Question 57.
WRITING
Explain why absolute value equations can have no solution, one solution, or two solutions. Give an example of each case.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q55

Question 58.
THOUGHT-PROVOKING
Describe a real-life situation that can be modeled by an absolute value equation with the solutions x = 62 and x = 72.
Answer:
Suppose in a school, an exam is conducted. In that examination, 67% of the students are passed. If the error of the pass percentage is 5 %, then what are the minimum and the maximum number of students passed in the examination?
Now,
The absolute value equation for the given real-life situation is:
| x – 67 | = 5
We know that,
| x | = x for x> 0
| x | =-x for x < 0
So,
x – 67 = 5                                         x – 67 = -5
x = 5 + 67                                        x = -5 + 67
x = 72                                               x = 62
Hence, from the above,
We can conclude that the minimum and maximum number of students passed in the examination  respectively are: 72 and 67

Question 59.
CRITICAL THINKING
Solve the equation shown. Explain how you found your solution(s).
8 | x + 2 | – 6 = 5 | x + 2 | + 3

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q59
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q59-i

Question 60.
HOW DO YOU SEE IT?
The circle graph shows the results of a survey of registered voters on the day of an election.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 78
The error given in the graph means that the actual percentage could be 2% more or 2% less than the percent reported by the survey.
a. What are the minimum and maximum percents of voters who could vote Republican?
Answer:
The minimum percentage of voters for Republicans is: 40%
The maximum percentage of voters for Republicans is: 44 %

Explanation:
The given graph is:
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 78
From the graph,
The vote percentage for Republicans is: 42 %
The error percentage is: ±2%
So,
The absolute value equation for the maximum and the minimum number of voters is:
| x – 42 | = 2
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
x – 42 = 2                                                                 x – 42 = -2
x = 2 + 42                                                                x = -2 + 42
x = 44                                                                       x = 40
Hence, from the above,
We can conclude that
The minimum percentage of voters for Republicans is: 40%
The maximum percentage of voters for Republicans is: 44 %

b. How can you use absolute value equations to represent your answers in part (a)?
Answer:
From the property of absolute values,
We know that,
| x | = x for x > 0
| x | = -x for x < 0
From the part ( a ),
The absolute value equation is:
| x – 42 | = 2
So,
x – 42 = 2                                                                 x – 42 = -2
x = 2 + 42                                                                x = -2 + 42
x = 44                                                                       x = 40
Hence, from the above,
We can conclude that we can use absolute values in the above way to represent the answers

c. One candidate receives 44% of the vote. Which party does the candidate belong to? Explain.
Answer:
The candidate of the Republican party receives 44 % of the vote.

Explanation:
The given graph is:
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 78
From the above graph,
We can say that,
The percentage of the vote received by the candidate of the Republican party = 42 %
The error percentage = ± 2 %
So,
Now,
The percentage of the vote received by the Republicans = 42 + 2  ( or ) 42 – 2
= 44 ( or ) 40
Hence, from the above,
We can conclude that the candidate of the Republican party received the 44 % of the vote

Question 61.
ABSTRACT REASONING
How many solutions does the equation a | x + b | + c = d have when a > 0 and c = d? when a < 0 and c > d? Explain your reasoning.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q61

Maintaining Mathematical Proficiency

Identify the property of equality that makes Equation 1 and Equation 2 equivalent. (Section 1.1)

Question 62.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 79

Answer:
The given equations are:
Equation 1: 3x + 8 = x – 1
Equation 2: 3x + 9= x
From Equation 1,
3x + 8 = x – 1
3x + 8 + 1 = x
3x + 9 = x
Hence, from the above,
We can conclude that we can get Equation2 by rearranging the Equation 1

Question 63.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 80

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q63

Use a geometric formula to solve the problem.

Question 64.
A square has an area of 81 square meters. Find the side length.
Answer:
The side length of the square is: 9 meters

Explanation:
The given area of the square is: 81 square meters
We know that,
Area of the square = Side × Side
81 = Side × Side
Side² = 81
Apply square root on both sides
√Side² = √81
Side = 9 meters
Hence, from the above,
We can conclude that the side of the square is: 9 meters

Question 65.
A circle has an area of 36π square inches. Find the radius.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q65

Question 66.
A triangle has a height of 8 feet and an area of 48 square feet. Find the base.
Answer:
The base of the triangle is: 12 feet

Explanation:
It is given that a triangle has a height of 8 feet and an area of 48 square feet
We know that,
The area of the triangle = ( 1 /  2 ) × Base × Height
48 = ( 1 / 2 ) × Base × 8
Base × 8 = 48 × 2
Base = ( 48 × 2 ) ÷ 8
Base = 96 ÷ 8
Base = 12 feet
Hence, from the above,
We can conclude that the base of the triangle is: 12 feet

Question 67.
A rectangle has a width of 4 centimeters and a perimeter of 26 centimeters. Find the length.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q67

Lesson 1.5 Rewriting Equations and Formulas

Essential Question
How can you use a formula for one measurement to write a formula for a different measurement?
Answer:
Write the formula for one measurement and then solve the formula for the different measurement you want to find and use this new formula to find that measurement
Hence, in the above way,
We can use a formula for one measurement to write a formula for a different measurement

EXPLORATION 1
Using an Area Formula
Work with a partner.

a. Write a formula for the area A of a parallelogram.
Answer:
We know that,
The area of the parallelogram ( A) = Base × Height

b. Substitute the given values into the formula. Then solve the equation for b. Justify each step.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 81
Answer:
The value of b is: 6 in

Explanation:
The given figure is:
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 81
From the given figure,
Area ( A) = 30 in²
Height ( h ) = 5 in
Base = b
From part ( a),
Area of the parallelogram = Base × Height
30 = 5 × b
b = 30 ÷ 5
b = 6 in
Hence, from the above,
We can conclude that the value of b is: 6 in

c. Solve the formula in part (a) for b without first substituting values into the formula. Justify each step.
Answer:
From part ( a ),
Area of the parallelogram = Base × Height
Base = ( Area of the parallelogram ) ÷ Height of the parallelogram
From the given figure,
Base = b
So,
b = ( Area of the parallelogram ) ÷ Height of the parallelogram

d. Compare how you solved the equations in parts (b) and (c). How are the processes similar? How are they different?
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 82
Answer:
We know that,
The area of the parallelogram = Base × Height
Using the above formula,
We solved parts (b ) and ( c )

EXPLORATION 2
Using Area, Circumference, and Volume Formulas
Work with a partner. Write the indicated formula for each figure. Then write a new formula by solving for the variable whose value is not given. Use the new formula to find the value of the variable.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 83
Answer:
The given geometrical figures are:
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 83
Now,
a)
The given figure is:

We know that,
Area of the trapezoid = h ( a + b ) / 2
Where,
h is the height between the two sides
a + b is the sum of the parallel sides [ Here, b1 and b2 ]
From the given figure,
Area of the trapezoid = 63 cm²
1st parallel side ( b1 ) = 8 cm
2nd parallel side ( b2 ) = 10 cm
So,
Area of the trapezoid = h ( 8 + 10 ) / 2
63 = h ( 18 ) / 2
63 × 2 = h × 18
h = ( 63 × 2 ) ÷ ( 18 × 1 )
h = 7 cm
Hence, from the above,
We can conclude that the value of h is: 7 cm
b)
The given figure is:

We know that,
Circumference of a circle = 2πr
Where
r is the radius of the circle
From the above figure,
Circumference of the circle ( C ) = 24π ft
So,
24π = 2πr
r = ( 24π ) ÷ ( 2π )
r = 12 ft
Hence, from the above,
We can conclude that the value of r is: 12 ft
c)
The given figure is:

We know that,
The volume of the rectangular prism ( V ) = Length × Width × Height
The area of the rectangle ( B) = Length × Width
So,
The volume of the rectangular prism (V ) = B × Height
From the above figure,
The volume of the rectangular prism ( V ) = 75 yd³
The area of the rectangle ( B ) = 15 yd²
So,
75 = 15 × Height
Height = 75 ÷ 15
Height = 15 yd
Hence, from the above,
We can conclude that the value of h is: 15 yd
d)
The given figure is:

We know that,
The volume of cone ( V ) = πr²h / 3
The area of the circle ( B ) = πr²
Where,
r is the radius of the circle
h is the height of the cone
So,
The volume of the cone ( V ) = Bh / 3
From the above figure,
V = 24π m³
B = 12π m³
So,
24π = 12π × h / 3
h / 3 = 24π ÷ 12π
h / 3 = 2
h = 2 × 3
h = 6 m
Hence, from the above,
We can conclude that the height of the cone is: 6 m

Communicate Your Answer

Question 3.
How can you use a formula for one measurement to write a formula for a different measurement? Give an example that is different from those given in Explorations 1 and 2.
Answer:
Write the formula for one measurement and then solve the formula for the different measurement you want to find and use this new formula to find that measurement
Example:
The given figure is:

We know that,
The area of the rectangle ( A ) = Length × Width
From the above figure,
A = 20 cm²
L = 10 cm
Let,
W be the width of the rectangle
So,
20 = 10 × W
W = 20 ÷ 10
W = 2 cm
Hence, from the above,
We can conclude that the value of W is: 2 cm

1.5 Lesson

Monitoring Progress

Solve the literal equation for y.

Question 1.
3y – x = 9
Answer:
The value of y is: ( x + 9 ) / 3

Explanation:
The given equation is:
3y – x = 9
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
3y = 9 + x
y = ( x + 9 ) / 3
Hence, from the above,
We can conclude that the value of y is: ( x + 9 ) / 3

Question 2.
2x – 2y = 5
Answer:
The value of y is: ( 2x – 5 ) / 2

Explanation:
The given equation is:
2x – 2y = 5
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
2y = 2x – 5
y = ( 2x – 5 ) / 2
Hence, from the above,
We can conclude that the value of y is: ( 2x – 5 ) / 2

Question 3.
20 = 8x + 4y
Answer:
The value of y is: 5 – 2x

Explanation:
The given equation is:
20 = 8x + 4y
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
4y = 20 – 8x
y = ( 20 – 8x ) / 4
y = ( 20 ÷ 4 ) – ( 8x ÷ 4 )
y = 5 – 2x
Hence, from the above,
We can conclude that the value of y is: 5 – 2x

Solve the literal equation for x.

Question 4.
y = 5x – 4x
Answer:
The value of x is: y

Explanation:
The given equation is:
y = 5x – 4x
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
x = y
Hence, from the above,
We can conclude that the value of x is: y

Question 5.
2x + kx = m
Answer:
The value of x is: m / ( k + 2  )

Explanation:
The given equation is:
2x + kx = m
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
x ( k + 2 ) = m
x = m / ( k + 2 )
Hence, from the above,
We can conclude that the value of x is: m / ( k + 2 )

Question 6.
3 + 5x – kx = y
Answer:
The value of x is: ( y – 3 ) / ( 5 – k )

Explanation:
The given equation is:
3 + 5x – kx = y
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
5x – kx = y – 3
x ( 5 – k ) = y – 3
x = ( y – 3 ) / ( 5 – k )
Hence, from the above,
We can conclude that the value of x is: ( y – 3 ) / ( 5 – k )

Solve the formula for the indicated variable. 

Question 7.
Area of a triangle: A = \(\frac{1}{2}\)bh; Solve for h.
Answer:
The value of h is: \(\frac{2A}{b}\)

Explanation:
The given area of a triangle is:
A = \(\frac{1}{2}\) bh
bh = 2A
h = \(\frac{2A}{b}\)
Hence, from the above,
We can onclude that the value of h is: \(\frac{2A}{b}\)

Question 8.
The surface area of a cone: S = πr2 + πrℓ; Solve for ℓ.
Answer:
The value of l is: \(\frac{S}{πr}\) – r

Explanation:
The given surface area of a cone is:
S = πr² + πrl
S= πr ( r + l )
r + l = \(\frac{S}{πr}\)
l = \(\frac{S}{πr}\) – r
Hence, from the above,
We can conclude that the value of l is: \(\frac{S}{πr}\) – r

Monitoring Progress

Question 9.
A fever is generally considered to be a body temperature greater than 100°F. Your friend has a temperature of 37°C. Does your friend have a fever?
Answer:
Your friend does not have a fever

Explanation:
It is given that a fever is generally considered to be a body temperature greater than 100°F.
We know that,
To convert Fahrenheit into Celsius,
°C = ( °F – 32 ) × \(\frac{5}{9}\)
°C = ( 100 – 32 ) × \(\frac{5}{9}\)
°C = 68 × \(\frac{5}{9}\)
°C = 37.7°
But it is given that your friend has a temperature of 37°C
So, for fever, the temperature has to be 37.7°C
Hence, from the above,
We can conclude that your friend does not have a fever

Question 10.
How much money must you deposit in a simple interest account to earn $500 in interest in 5 years at 4% annual interest?
Answer:
The money you deposit in simple interest is: $2,500

Explanation:
It is given that you earned $500 in a simple interest to earn in 5 years at 4% annual interest
Let,
The money you deposited be: $x
We know that,
Simple interest = ( Principle × Time × Rate ) / 100
The principle is the money you deposited
So,
500 = ( x × 5 × 4 ) / 100
( x × 5 × 4 ) = 500 × 100
x × 20 = 500 × 100
x = ( 500 × 100 ) ÷ 20
x = $2,500
Hence, from the above,
We can conclude that the money you deposited is: $2,500

Question 11.
A truck driver averages 60 miles per hour while delivering freight and 45 miles per hour on the return trip. The total driving time is 7 hours. How long does each trip take?
Answer:
The time taken for each trip is: 3 hours and 4 hours respectively

Explanation:
It is given that a truck driver averages 60 miles per hour while delivering freight and 45 miles per hour on the return trip. The total driving time is 7 hours.
We know that,
Speed = \(\frac{Distance}{Time}\)
Time = \(\frac{Distance}{Speed}\)
Let the distance be D
It is given that the total driving time is: 7 hours
So,
7 = \(\frac{D}{60}\) + \(\frac{D}{45}\)
7 / D = \(\frac{60 + 45}{60 × 45}\)
7 / D = \(\frac{105}{2,700}\)
D = 7 / \(\frac{105}{2,700}\)
D = 7 × \(\frac{2,700}{105}\)
D = \(\frac{7}{1}\) × \(\frac{2,700}{105}\)
D = \(\frac{7 × 2,700}{1 × 105}\)
D = 180 miles
So,
The time taken to deliver = \(\frac{180}{60}\) = 3 hours
The time taken to return = \(\frac{180}{45}\) = 4 hours
Hence, from the above,
We can conclude that the time taken for each trip is: 3 hours and 4 hours respectively

Rewriting Equations and Formulas 1.5 Exercices

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
Is Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 84 a literal equation? Explain.
Answer:
The ” Literal equation” is an equation that contains only letters
Now,
The given equation is:
9r + 16 = π / 5
From the above definition,
We can say that the given equation is a ” Literal equation ”

Question 2.
DIFFERENT WORDS, SAME QUESTION?
Which is different? Find “both” answers.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 85
Answer:
The given problems are:
a) Solve 3x + 6y = 24 for x
b) Solve 24 – 3x = 6y for x
c) Solve 6y = 24 – 3x in terms of x
d) Solve 24 – 6y = 3x for x in terms of y
So,
From the above-given problems,
We can observe that d) is different as we have to find x in terms of y whereas in the remaining three problems, we have to find x

Monitoring Progress and Modeling with Mathematics

In Exercises 3–12, solve the literal equation for y.

Question 3.
y – 3x = 13

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q3

Question 4.
2x + y = 7
Answer:
The value of y is: 7 – 2x

Explanation:
The given literal equation is:
2x + y = 7
Now,
y = 7 – 2x
Hence, from the above,
We can conclude that the value of y is: 7 – 2x

Question 5.
2y – 18x = -26

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q5

Question 6.
20x + 5y = 15
Answer:
The value of y is: 3 – 4x

Explanation:
The given literal equation is:
20x + 5y = 15
Now,
5y = 15 – 20x
y = ( 15 – 20x ) / 5
y = ( 15  / 5 ) – ( 20x / 5 )
y = 3 – 4x
Hence, from the above,
We can conclude that the value of y is: 3 – 4x

Question 7.
9x – y = 45

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q7

Question 8.
6 – 3y = -6
Answer:
The value of y is: 4

Explanation:
The given literal equation is:
6 – 3y = -6
-3y = -6 – ( +6 )
-3y = -6 -6
-3y = -12
y = -12 ÷ ( -3 )
y = 12 ÷ 3
y = 4
Hence, from the above,
We can conclude that the value of y is: 4

Question 9.
4x – 5 = 7 + 4y

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q9

Question 10.
16x + 9 = 9y – 2x
Answer:
The value of y is: 18x + 9

Explanation:
The given literal equation is:
16x + 9 = y – 2x
So,
16x + 2x + 9 = y
18x + 9 = y
y = 18x + 9
Hence, from the above,
We can conclude that the value of y is: 18x + 9

Question 11.
2 +\(\frac{1}{6}\)y = 3x + 4

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q11

Question 12.
11 – \(\frac{1}{2}\)y = 3 + 6x
Answer:
The value of y is: 16 – 12x

Explanation:
The given literal equation is:
11 – \(\frac{1}{2}\)y = 3 + 6x
So,
–\(\frac{1}{2}\)y = 3 + 6x – 11
-y = 2 ( 3 + 6x – 11 )
y = -2 ( 3 + 6x – 11 )
y = -2 ( 3 ) -2 ( 6x ) + 2 ( 11 )
y = -6 – 12x + 22
y = 16 – 12x
Hence, from the above,
We can conclude that the value of y is: 16 – 12x

In Exercises 13–22, solve the literal equation for x.

Question 13.
y = 4x + 8x

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q13

Question 14.
m = 10x – x
Answer:
The value of x is: m / 9

Explanation:
The given literal equation is:
m = 10x – x
m = 9x
x = m / 9
Hence, from the above,
We can conclude that the value of x is: m / 9

Question 15.
a = 2x + 6xz

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q15

Question 16.
y = 3bx – 7x
Answer:
The value of x is: y / ( 3b – 7 )

Explanation:
The given literal equation is:
y = 3bx – 7x
So,
y = x ( 3b – 7 )
x = y / ( 3b – 7 )
Hence, from the above,
We can conclude that the value of x is: y / ( 3b – 7 )

Question 17.
y = 4x + rx + 6

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q17

Question 18.
z = 8 + 6x – px
Answer:
The value of x is: ( z – 8 ) / ( 6 – p )

Explanation:
The given literal equation is:
z = 8 + 6x – px
So,
z – 8 = 6x – px
z – 8 = x ( 6 – p )
x = ( z – 8 ) / ( 6 – p )
Hence, from the above,
We can conclude that the value of x is: ( z – 8 ) / ( 6 – p )

Question 19.
sx + tx = r

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q19

Question 20.
a = bx + cx + d
Answer:
The value of x is: ( a – d ) / ( b + c )

Explanation:
The given literal equation is:
a = bx + cx + d
a – d = bx + cx
a – d = x ( b + c )
x = ( a – d ) / ( b + c )
Hence, from the above,
We can conclude that the value of x is: ( a – d ) / ( b + c )

Question 21.
12 – 5x – 4kx = y

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q21

Question 22.
x – 9 + 2wx = y
Answer:
The value of x is: ( y – 9 ) / ( 1 – 2w )

Explanation:
The given literal equation is:
x – 9 + 2wx = y
x – 2wx = y + 9
x ( 1 – 2w ) = y + 9
x = ( y – 9 ) / ( 1 – 2w )
Hence, from the above,
We can conclude that the value of x is: ( y – 9 ) / ( 1 – 2w )

Question 23.
MODELING WITH MATHEMATICS
The total cost C (in dollars) to participate in a ski club is given by the literal equation C = 85x + 60, where x is the number of ski trips you take.
a. Solve the equation for x.
b. How many ski trips do you take if you spend a total of $315? $485?
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 86

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q23

Question 24.
MODELING WITH MATHEMATICS
The penny size of a nail indicates the length of the nail. The penny size d is given by the literal equation d = 4n – 2, where n is the length (in inches) of the nail.
a. Solve the equation for n.
b. Use the equation from part (a) to find the lengths of nails with the following penny sizes: 3, 6, and 10.
Answer:
a)
The given literal equation is:
d = 4n – 2
Where,
n is the length ( in inches ) of the nail
So,
4n = d + 2
n = ( d + 2 ) / 4
b)
It is given that,
The penny sizes ( d ) are: 3, 6, and 10
From part ( a ),
The literal equation is:
n = ( d + 2  ) / 4
Put, d= 3, 6 and 10
So,
n = ( 3 + 2 ) /4 = 5 / 4 inches
n = ( 6 + 2 ) / 4 = 2 inches
n = ( 10 + 2 ) / 4 = 3 inches

ERROR ANALYSIS
In Exercises 25 and 26, describe and correct the error in solving the equation for x.

Question 25.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 87

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q25

Question 26.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 72.1

Answer:
The given literal equation is:
10 = ax – 3b
So,
ax = 10 + 3b
x = ( 10 + 3b ) / a

In Exercises 27–30, solve the formula for the indicated variable.

Question 27.
Profit: P = R – C; Solve for C.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q27

Question 28.
Surface area of a cylinder: S = 2πr2 + 2πrh; Solve for h.
Answer:
The given Surface area of a cylinder is:
S = 2πr² + 2πrh
So,
S = 2πr ( r + h )
S / 2πr = r + h
h = S / 2πr – r
Hence, from the above,
We can conclude that the value of h is: S / ( 2π

Question 29.
Area of a trapezoid: A = \(\frac{1}{2}\)h(b1 + b2); Solve for b2.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q29

Question 30.
The average acceleration of an object: Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 88; Solve for v1.
Answer:
The given average acceleration of an object is:
a = ( v1 – v0 ) / t
So,
at = v1 – v0
v1 = at + v0
Hence, from the above,
We can conclude that the value of v1 is: at + v0

Question 31.
REWRITING A FORMULA
A common statistic used in professional football is the quarterback rating. This rating is made up of four major factors. One factor is the completion rating given by the formula
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 88.1
where C is the number of completed passes and A is the number of attempted passes. Solve the formula for C.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q31

Question 32.
REWRITING A FORMULA
Newton’s law of gravitation is given by the formula
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 89
where F is the force between two objects of masses m1 and m2, G is the gravitational constant, and d is the distance between the two objects. Solve the formula for m1.
Answer:
The value of m1 is: Fd² / Gm2

Explanation:
The given Newton’s law of gravitation is given by:
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 89
where
F is the force between two objects of masses m1 and m2
G is the gravitational constant
d is the distance between the two objects.
Now,
( m1m2 ) = Fd² / G
m1 = Fd² / Gm2
Hence, from the above,
We can conclude that the value of m1 is: Fd² / Gm2

Question 33.
MODELING WITH MATHEMATICS
The sale price S (in dollars) of an item is given by the formula S = L – rL, where L is the list price (in dollars) and r is the discount rate (in decimal form).
a. Solve the formula for r.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 90
b. The list price of the shirt is $30. What is the discount rate?

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q33

Question 34.
MODELING WITH MATHEMATICS
The density d of a substance is given by the formula d = \(\frac{m}{V}\), where m is its mass and V is its volume.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 91
a. Solve the formula for m?
Answer:
The given density d of a substance is given by:
d = \(\frac{m}{V}\)
So,
d × V = m
Hence, from the above,
We can conclude that the value of m is: d × V

b. Find the mass of the pyrite sample.
Answer:
The mass of the pyrite sample is: 6.012 gm

Explanation:
The given figure is:
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 91
From the above figure,
The density of Pyrite = 5.01 g/cm³
The volume of Pyrite = 1.2 cm³
From part (a),
The mass of Pyrite = Density × Volume
So,
The mass of Pyrite ( m ) = 5.01 × 1.2
= 6.012 gm
Hence, from the above,
We can conclude that the mass of Pyrite is: 6.012 gm

Question 35.
PROBLEM-SOLVING
You deposit $2000 in an account that earns simple interest at an annual rate of 4%. How long must you leave the money in the account to earn $500 in interest?

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q35

Question 36.
PROBLEM-SOLVING
A flight averages 460 miles per hour. The return flight averages 500 miles per hour due to a tailwind. The total flying time is 4.8 hours. How long is each flight? Explain.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 91.1

Answer:
The time taken for flight is: 2.5 hours
The time taken for return is: 2.3 hours

Explanation:
It is given that a flight averages 460 miles per hour. The return flight averages 500 miles per hour due to a tailwind. The total flying time is 4.8 hours.
We know that,
Speed = Distance / Time
Time = Distance / Speed
It is also given that the total flying time is 4.8 hours
Let the distance be D
So,
\(\frac{D}{460}\) + \(\frac{D}{500}\) = 4.8
\(\frac{460 + 500}{230,000}\) = 4.8 / D
\(\frac{960}{230,000}\) = 4.8 / D
D = 4.8 × \(\frac{230,000}{960}\)
D = 1,150 miles
Hence,
The time taken for flight = 1,150 ÷ 460 = 2.5 hpurs
The time taken for return = 1,150 ÷ 500 = 2.3 hours

Question 37.
USING STRUCTURE
An athletic facility is building an indoor track. The track is composed of a rectangle and two semicircles, as shown.
Big Ideas Math Algebra 1 Solutions Chapter 1 Solving Linear Equations 92
a. Write a formula for the perimeter of the indoor track.
b. Solve the formula for x.
c. The perimeter of the track is 660 feet, and r is 50 feet. Find x. Round your answer to the nearest foot.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q37

Question 38.
MODELING WITH MATHEMATICS
The distance d (in miles) you travel in a car is given by the two equations shown, where t is the time (in hours) and g is the number of gallons of gasoline the car uses.
Big Ideas Math Algebra 1 Solutions Chapter 1 Solving Linear Equations 93
a. Write an equation that relates g and t.
Answer:
The given equations are:
d = 55t ——————— (1)
d = 20g ——————— (2)
By the law of Equality,
55t = 20g [ As the LHS for both the equations are equal, make the RHS equal ]
t / g = 20 / 55
t / g = 4 / 11

b. Solve the equation for g.
Answer:
From the given figure,
d = 20g
d = 55t
From part (a),
t / g = 4 / 11
11t = 4g
g = 11t / 4
Hence, from the above,
We can conclude that the value of g is: 11t / 4

c. You travel for 6 hours. How many gallons of gasoline does the car use? How far do you travel? Explain.
Answer:
From part (b),
g = 11t / 4
Where,
g is the number of gallons of gasoline
It is given that you travel for 6 hours
So,
t = 6 hours
Now,
g = ( 11 × 6 ) / 4
g = 66/4 gallons
Hence, from the above,
We can conclude that the number of gallons of gasoline is: 66 / 4 gallons

Question 39.
MODELING WITH MATHEMATICS
One type of stone formation found in Carlsbad Caverns in New Mexico is called a column. This cylindrical stone formation connects to the ceiling and the floor of a cave.
Big Ideas Math Algebra 1 Solutions Chapter 1 Solving Linear Equations 94
a. Rewrite the formula for the circumference of a circle, so that you can easily calculate the radius of a column given its circumference.
b. What is the radius (to the nearest tenth of a foot) of a column that has a circumference of 7 feet? 8 feet? 9 feet?
c. Explain how you can find the area of a cross-section of a column when you know its circumference.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q39

Question 40.
HOW DO YOU SEE IT?
The rectangular prism shown has bases’ with equal side lengths.
Big Ideas Math Algebra 1 Solutions Chapter 1 Solving Linear Equations 95
a. Use the figure to write a formula for the surface area S of the rectangular prism.
Answer:
The given figure is:
Big Ideas Math Algebra 1 Solutions Chapter 1 Solving Linear Equations 95
From the above figure,
The surface area of the rectangular prism ( S) = 2 ( lb + bh + lh )
Where,
l is the length of the rectangular prism
b is the Width of the rectangular prism
h is the height of the rectangular prism

b. Your teacher asks you to rewrite the formula by solving for one of the side lengths, b or ℓ. Which side length would you choose? Explain your reasoning.
Answer:
From part (a),
The surface area of the rectangular prism ( S ) = 2 ( lb + bh + lh )
S / 2 = lb + bh + lh
S / 2 = b ( l + h ) + bh
S / 2 = b ( l + b + h )
b = S / 2 ( l + b + h )
Hence, from the above,
We can conclude that the value of b is: S / 2 ( l + b + h )

Question 41.
MAKING AN ARGUMENT
Your friend claims that Thermometer A displays a greater temperature than Thermometer B. Is your friend correct? Explain your reasoning.
Big Ideas Math Algebra 1 Solutions Chapter 1 Solving Linear Equations 96

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q41

Question 42.
THOUGHT-PROVOKING
Give a possible value for h. Justify your answer. Draw and label the figure using your chosen value of h.
Big Ideas Math Algebra 1 Solutions Chapter 1 Solving Linear Equations 97

Answer:
The completed figure is:

The value of h is: 5 cm

Explanation:
The given figure is:
Big Ideas Math Algebra 1 Solutions Chapter 1 Solving Linear Equations 97
From the given figure,
We can say that the geometrical figure is the parallelogram
We know that,
Area of the parallelogram = Base × Height
From the given figure,
Area of the parallelogram = 40 cm²
The base of the parallelogram = 8 cm
So,
40 = 8 × Height
Height = 40 ÷ 8
Height = 5 cm
Hence, from the above,
We can conclude that
The completed figure is:

The value of h is: 5 cm

MATHEMATICAL CONNECTIONS
In Exercises 43 and 44, write a formula for the area of the regular polygon. Solve the formula for the height h.

Question 43.
Big Ideas Math Algebra 1 Solutions Chapter 1 Solving Linear Equations 98

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q43

Question 44.
Big Ideas Math Algebra 1 Solutions Chapter 1 Solving Linear Equations 99

Answer:
The value of h is: A / 3b

Explanation:
The given figure is:
Big Ideas Math Algebra 1 Solutions Chapter 1 Solving Linear Equations 99
From the above figure,
We can say that the figure is Hexagon
Now,
The Hexagon with the six triangles is:

So,
From the figure,
There are 6 triangles
We know that,
The area of a triangle = \(\frac{1}{2}\) × Base × Height
So,
The area of the Hexagon = The area of the 6 triangles
= 6 ( \(\frac{1}{2}\) ) × Base × Height
Let,
The area of the Hexagon be A
The height of the hexagon be h
The Base of the hexagon be b
So,
A = 6 ( \(\frac{1}{2}\) ) × Base × Height
A = 3 × Base × Height
Base × Height = A / 3
Height = A / ( 3 × Base )
So,
h = A / 3b
Hence, from the above,
We can conclude that the value of h is: A / 3b

REASONING
In Exercises 45 and 46, solve the literal equation for a.

Question 45.
Big Ideas Math Algebra 1 Solutions Chapter 1 Solving Linear Equations 100

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q45

Question 46.
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 101
Answer:
The value of a is: \(\frac{by}{y – bx}\)

Explanation:
The given literal equation is:
y = x [ \(\frac{ab}{a – b}\)
\(\frac{ab}{a – b}\) = y / x
x ( ab ) = y ( a – b )
abx = ay – by
by = ay – abx
by = a ( y – bx )
a = \(\frac{by}{y – bx}\)
Hence, from the above,
We can conclude that the value of a is: \(\frac{by}{y – bx}\)

Maintaining Mathematical Proficiency

Evaluate the expression.

Question 47.
15 – 5 + 52

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q47

Question 48.
18 • 2 – 42 ÷ 8
Answer:
The given expression is:
18 ⋅ 2 – 4² ÷ 8
We have to remember that,
When there is an expression to solve with multiple mathematical symbols, we have to follow the BODMAS rule
BODMAS indicates the hierarchy we have to follow when we will solve mathematical symbols
In BODMAS,
B – Brackets
O – Of
D – Division
M – Multiplication
A – Addition
S – Subtraction
So,
18 ⋅ 2 – 4² ÷ 8 = 18 ⋅ 2 – ( 4 × 4 ) ÷ 8
= 18 ⋅ 2 – 2
= 36 – 2
= 34

Question 49.
33 + 12 ÷ 3 • 5

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q49

Question 50.
25(5 – 6) + 9 ÷ 3
Answer:
The given expression is:
25(5 – 6) + 9 ÷ 3
We have to remember that,
When there is an expression to solve with multiple mathematical symbols, we have to follow the BODMAS rule
BODMAS indicates the hierarchy we have to follow when we will solve mathematical symbols
In BODMAS,
B – Brackets
O – Of
D – Division
M – Multiplication
A – Addition
S – Subtraction
So,
25(5 – 6) + 9 ÷ 3 = ( 2 × 2 × 2 × 2 × 2 ) ( 5 – 6 ) + ( 9 ÷ 3 )
= ( 2 × 2 × 2 × 2 × 2 ) ( 5 – 6 ) + 3
= ( 2 × 2 × 2 × 2 × 2 ) ( -1 ) + 3
= -( 2 × 2 × 2 × 2 × 2 )  + 3
= -32 + 3
= -29

Solve the equation. Graph the solutions, if possible. (Section 1.4)

Question 51.
| x – 3 | + 4 = 9

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q51

Question 52.
| 3y – 12 | – 7 = 2
Answer:
The values of y are: 7 and 1

Explanation:
The given absolute value equation is:
| 3y – 12 | – 7 = 2
| 3y – 12 | = 2 + 7
| 3y – 12 | = 9
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
3y – 12 = 9                      3y – 12 = -9
3y = 9 + 12                     3y = -9 + 12
3y = 21                            3y = 3
y = 21 / 3                         y = 3 / 3
y = 7                                y = 1
Hence, from the above,
We can conclude that the values of y are: 7 and 1

Question 53.
2 | 2r + 4 | = -16

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q53

Question 54.
-4 | s + 9 | = -24
Answer:
The value of s is: -3 and -15

Explanation:
The given absolute value equation is:
-4 | s + 9 | = -24
| s + 9 | = -24 ÷ ( -4 )
| s + 9 | = 6 [ Since – ÷ – = + ]
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
s + 9 = 6                                       s + 9 = -6
s = 6 – 9                                         s = -6 – 9
s = -3                                             s = -15
Hence, from the above,
We can conclude that the values of s are: -3 and -15

Solving Linear Equations Performance Task: Magic of Mathematics

1.4–1.5 What Did You Learn?

Core Vocabulary
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 102

Core Concepts
Section 1.4
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 103
Section 1.5
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 104

Mathematical Practices

Question 1.
How did you decide whether your friend’s argument in Exercise 46 on page 33 made sense?
Answer:
On page 33 in Exercise 46,
The given absolute equation is:
| 3x + 8 | – 9 = -5
| 3x + 8 | = -5 + 9
| 3x + 8 | = 4
So, from the absolute equation,
We can say that the given absolute value equation has a solution
But, according to your friend,
The argument is that the absolute value equation has no solution

Question 2.
How did you use the structure of the equation in Exercise 59 on page 34 to rewrite the equation?
Answer:
The given absolute value equation in Exercise 59 on page 34 is:
8 | x + 2 | – 6 = 5 | x + 2 | + 3
The above equation can be re-written as:
8 | x + 2  | – 5  | x + 2 | = 3 + 6
3 | x + 2  | = 9
Hence, from the above,
We can conclude that the re-written form of the given absolute value equation is:
3 | x + 2 | = 9

Question 3.
What entry points did you use to answer Exercises 43 and 44 on page 42?
Answer:
In Exercises 43 and 44 on page 42,
We used the triangles as an entry point
In Exercise 43,
The given figure is a pentagon
Using the above entry point,
We divided the pentagon into 5 triangles
In Exercise 44,
The given figure is a Hexagon
Using the above entry point,
We divided the hexagon into 6 triangles.

Performance Task

Magic of Mathematics

Have you ever watched a magician perform a number trick? You can use algebra to explain how these types of tricks work.
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 105
To explore the answers to these questions and more, go to Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 106

Solving Linear Equations Chapter Review

1.1 Solving Simple Equations (pp. 3–10)

a. Solve x − 5 = −9. Justify each step.
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 106.1

b. Solve 4x = 12. Justify each step.
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 107

Solve the equation. Justify each step. Check your solution.

Question 1.
z + 3 = -6
Answer:
The value of z is: -9

Explanation:
The given equation is:
z + 3 = -6
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
z = -6 – ( +3 )
z = -6 – 3
z = -9
Hence, from the above,
We can conclude that the value of z is: -9

Question 2.
2.6 = -0.2t
Answer:
The value of t is: -13

Explanation:
The given equation is:
2.6 = -0.2t
\(\frac{26}{10}\) = –\(\frac{2}{10}\)t
t = \(\frac{26}{10}\) ÷ ( –\(\frac{2}{10}\) )
t = – \(\frac{26}{10}\) × \(\frac{10}{2}\)
t = -13
Hence, from the above,
We can conclude that the value of t is: -13

Question 3.
– \(\frac{n}{5}\) = -2
Answer:
The value of n is: 10

Explanation:
The given equation is:
–\(\frac{n}{5}\) = -2
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
-n = -2 × 5
-n = -10
n = 10
Hence, from the above,
We can conclude that the value of n is: 10

1.2 Solving Multi-Step Equations (pp. 11–18)

Solve −6x + 23 + 2x = 15.
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 108

Solve the equation. Check your Solution.

Question 4.
3y + 11 = -16
Answer:
The value of y is: -9

Explanation:
The given equation is:
3y + 11 = -16
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
3y = -16 – 11
3y = -27
y = -27 ÷ 3
y = -9
Hence, from the above,
We can conclude that the value of y is: -9

Question 5.
6 = 1 – b
Answer:
The value of b is: -5

Explanation:
The given equation is:
6 = 1 – b
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
b = 1 – 6
b = -5
Hence, from the above,
We can conclude that the value of b is: -5

Question 6.
n + 5n + 7 = 43
Answer:
The value of n is: 6

Explanation:
The given equation is:
n + 5n + 7 = 43
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
6n + 7 = 43
6n = 43 – 7
6n = 36
n = 36 ÷ 6
n = 6
Hence, from the above,
We can conclude that the value of n is: 6

Question 7.
-4(2z + 6) – 12 = 4
Answer:
The value of z is: -5

Explanation:
The given equation is:
-4 ( 2z + 6 ) – 12 = 4
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
-4 ( 2z + 6 ) = 4 + 12
-4 ( 2z + 6 ) = 16
-4 ( 2z ) – 4 ( 6 ) = 16
-8z – 24 = 16
-8z = 16 + 24
-8z = 40
z = 40 ÷ ( -8 )
z = -5
Hence, from the above,
We can conclude that the value of z is: -5

Question 8.
\(\frac{3}{2}\)(x – 2) – 5 = 19
Answer:
The value of x is: 18

Explanation:
The given equation is:
\(\frac{3}{2}\) ( x – 2 ) – 5 = 19
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
\(\frac{3}{2}\) ( x – 2 ) = 19 + 5
\(\frac{3}{2}\) ( x – 2 ) = 24
x – 2 = 24 × \(\frac{2}{3}\)
x – 2 = \(\frac{24}{1}\) × \(\frac{2}{3}\)
x – 2 = 16
x = 16 + 2
x = 18
Hence, from the above,
We can conclude that the value of x is: 18

Question 9.
6 = \(\frac{1}{5}\)w + \(\frac{7}{5}\)w – 4
Answer:
The value of w is: \(\frac{25}{4}\)

Explanation:
The given equation is:
6 = \(\frac{1}{5}\)w + \(\frac{7}{5}\)w – 4
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
6 + 4 = \(\frac{1}{5}\)w + \(\frac{7}{5}\)w
10 = w [ \(\frac{1 + 7}{5}\) ]
10 = \(\frac{8}{5}\)w
w = 10 × \(\frac{5}{8}\)
w = \(\frac{10}{1}\) × \(\frac{5}{8}\)
w = \(\frac{25}{4}\)
Hence, from the above,
We can conclude that the value of w is: \(\frac{25}{4}\)

Find the value of x. Then find the angle measures of the polygon.

Question 10.
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 109

Answer:
The angle measures of the given polygon are: 110°, 50°, 20°

Explanation:
The given figure is:
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 109
From the above figure,
The angle measures are: 110°, 5x°, 2x°
It is also given that
The sum of the angle measures = 180°
So,
110 + 5x + 2x = 180°
7x = 180 – 110
7x = 70°
x = 70 / 7
x = 10°
Hence, from the above,
We can conclude that the angle measures of the given polygon are: 110°, 50°, 20°

Question 11.
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 110

Answer:
The angle measures of the given polygon are: 126°, 126°, 96°, 96°, 96°

Explanation:
The given figure is:
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 110
From the above figure,
The angle measures of the given polygon are: ( x – 30 )°, x°, x°, ( x – 30 )°, ( x – 30 )°
It is also given that,
The sum of the angle measures of the given polygon = 540°
So,
( x – 30 )° + x° + x° + ( x – 30 )° + ( x – 30 )° = 540°
5x – 90° = 540°
5x = 540° + 90°
5x = 630°
x = 630 / 5
x = 126°
Hence,f rom the above,
We can conclude that the angle measures of the given polygon are: 126°, 126°, 96°, 96°, 96°

1.3 Solving Equations with Variables on Both Sides (pp. 19–24)

Solve 2( y − 4) = −4( y + 8).
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 111

Solve the equation.

Question 12.
3n – 3 = 4n + 1
Answer:
The value of n is: -4

Explanation:
The given equation is:
3n – 3 = 4n + 1
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
4n – 3n = -1 – 3
n = -4
Hence, from the above,
We can conclude that the value of n is: -4

Question 13.
5(1 + x) = 5x + 5
Answer:
The given equation has no solution

Explanation:
The given equation is:
5 ( 1 + x ) = 5x + 5
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
5 ( 1 ) + 5 ( x ) = 5x + 5
5 + 5x = 5x + 5
5 = 5x – 5x + 5
5 = 5
Hence, from the above,
We can conclude that the given equation has no solution

Question 14.
3(n + 4) = \(\frac{1}{2}\)(6n + 4)
Answer:
The given equation has no solution

Explanation:
The given equation is:
3 ( n + 4 ) = \(\frac{1}{2}\) ( 6n + 4 )
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
3 ( n ) + 3 ( 4 ) = \(\frac{1}{2}\) ( 6n + 4 )
3n + 12 = \(\frac{1}{2}\) ( 6n + 4 )
2 ( 3n + 12 ) = 6n + 4
2 ( 3n ) + 2 ( 12 ) = 6n + 4
6n + 24 = 6n + 4
24 = 6n – 6n + 4
24 = 4
Hence, from the above,
We can conclude that the given equation has no solution

1.4 Solving Absolute Value Equations (pp. 27–34)

a. Solve | x − 5 | = 3.
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 112

b. Solve | 2x + 6 | = 4x. Check your solutions.
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 113

Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 114

Check the apparent solutions to see if either is extraneous.
The solution is x = 3. Reject x = -1 because it is extraneous.

Solve the equation. Check your solutions.

Question 15.
| y + 3 | = 17
Answer:
The value of y is: 14 or -20

Explaantion:
The given absolute value equation is:
| y + 3 | = 17
We know that,
| x | = x for x  0
| x | = -x for x < 0
So,
y + 3 = 17                               y + 3 = -17
y = 17 – 3                                y = -17 – 3
y = 14                                      y = -20
Hence, from the above,
We can conclude that the value of y is: 14 or -20

Question 16.
-2 | 5w – 7 | + 9 = -7
Answer:
The value of w is: 3 or –\(\frac{1}{5}\)

Explanation:
The given absolute value equation is:
-2 | 5w – 7 | + 9 = -7
-2 | 5w – 7 | = -7 – 9
-2 | 5w – 7 | = -16
| 5w – 7 | = -16 / ( -2 )
| 5w – 7 | = 8
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
5w – 7 = 8                                     5w – 7 = -8
5w = 8 + 7                                    5w = -8 + 7
5w = 15                                         5w = -1
w = 15 ÷ 5                                     w = –\(\frac{1}{5}\)
w = 3                                              w =-\(\frac{1}{5}\)
Hence, from the above,
We can conclude that the value of w is: 3 or –\(\frac{1}{5}\)

Question 17.
| x – 2 | = | 4 + x |
Answer:
The given absolute equation has no solution

Explanation:
The given absolute value equation is:
| x – 2 | = | 4 + x |
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
x – 2 = 4 + x                                                   – ( x – 2 ) = – ( 4 + x )
-2 = 4                                                                2 = -4
Hence, from the above,
We can conclude that the given absolute equation has no solution

Question 18.
The minimum sustained wind speed of a Category 1 hurricane is 74 miles per hour. The maximum sustained wind speed is 95 miles per hour. Write an absolute value equation that represents the minimum and maximum speeds.
Answer:
The absolute value equation that represents the minimum and maximum speeds is:
| x – 84.5 | = 9.5

Explanation:
It is given that the minimum sustained wind speed of a Category 1 hurricane is 74 miles per hour. The maximum sustained wind speed is 95 miles per hour.
So,
The average wind speed sustained = ( 74 + 95 ) /2
= 169 / 2
= 84.5 miles per hour
Now,
The minimum wind speed from the average speed point = 84.5 – 74
= 9.5 miles per hour
So,
The absolute value equation that represents the minimum and maximum wind speed is:
| x – 84.5 | = 9.5

1.5 Rewriting Equations and Formulas (pp. 35–42)
a. The slope-intercept form of a linear equation is y = mx + b. Solve the equation for m.
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 115

b. The formula for the surface area S of a cylinder is S = 2πr2 + 2πrh. Solve the formula for the height h.
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 116

Solve the literal equation for y.

Question 19.
2x – 4y = 20
Answer:
The value of y is: ( x / 2 ) – 5

Explanation:
The given literal equation is:
2x – 4y = 20
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
4y = 2x – 20
y = ( 2x – 20 ) / 4
y = ( 2x / 4 ) – ( 20 / 4 )
y = ( x / 2 ) – 5
Hence, from the above,
We can conclude that the value of y is: ( x / 2 ) – 5

Question 20.
8x – 3 = 5 + 4y
Answer:
The value of y is: 2x – 2

Explanation:
The given literal equation is:
8x – 3 = 5 + 4y
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
4y = 8x – 3 – 5
4y = 8x – 8
y = ( 8x – 8 ) / 4
y = ( 8x / 4 ) – ( 8 – 4 )
y = 2x – 2
Hence, from the above,
We can conclude that the value of y is: 2x – 2

Question 21.
a a = 9y + 3yx
Answer:
The value of y is: a² / ( 3x + 9 )

Explanation:
The given literal equation is:
a² = 9y + 3yx
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
a² = y ( 3x + 9 )
y = a² / ( 3x + 9 )
Hence, from the above,
We can conclude that the value of y is: a² / ( 3x + 9 )

Question 22.
The volume V of a pyramid is given by the formula V = \(\frac{1}{3}\)Bh, where B is the area of the base and h is the height.
a. Solve the formula for h.
b. Find the height h of the pyramid.
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 117

Answer:
a)
The value of h is: \(\frac{3V}{B}\)

Explanation:
The given formula is:
V = \(\frac{1}{3}\)Bh
Where,
B is the area of the base
h is the height
Now,
3V = Bh
h = \(\frac{3V}{B}\)
Hence, from the above,
We can conclude that the value of h is: \(\frac{3V}{B}\)

b)
The value of h is: 18 cm

Explanation:
From the given figure,
Area of the base ( B ) = 36 cm²
Volume of the base ( V ) = 216 cm³
From part (a),
h = \(\frac{3V}{B}\)
h = \(\frac{3 × 216}{36}\)
h = \(\frac{3 × 216}{36 × 1}\)
h = 18 cm
Hence, from the above,
We can conclud ethat the value of h is: 18 cm

Question 23.
The formula F = \(\frac{9}{5}\)(K – 273.15) + 32 converts a temperature from kelvin K to degrees Fahrenheit F.
a. Solve the formula for K.
b. Convert 180°F to kelvin K. Round your answer to the nearest hundredth.
Answer:
a)
The formula for K is:
K = \(\frac{5}{9}\) ( F – 32 ) + 273.15

Explanation:
The given formula for F is:
F = \(\frac{9}{5}\) ( K – 273.15 ) + 32
Now,
F – 32 = \(\frac{9}{5}\) ( K – 273.15 )
\(\frac{5}{9}\) ( F – 32 ) = K – 273.15
K = \(\frac{5}{9}\) ( F – 32 ) + 273.15
Hence, from the above,
We can conclude that the value of K is: \(\frac{5}{9}\) ( F – 32 ) + 273.15

Solving Linear Equations Chapter Test

Solve the equation. Justify each step. Check your solution.

Question 1.
x – 7 = 15
Answer:
The value of x is: 22

Explanation:
The given equation is:
x – 7 = 15
Now,
x = 15 + 7
x = 22
Hence, from the above,
We can conclude that the value of x is: 22

Question 2.
\(\frac{2}{3}\)x = 5
Answer:
The value of x is: \(\frac{15}{2}\)

Explanation:
The given equation is:
\(\frac{2}{3}\) x = 5
Now,
x = 5 × \(\frac{3}{2}\)
x = \(\frac{5}{1}\) × \(\frac{3}{2}\)
x = \(\frac{15}{2}\)
Hence, from the above,
We can conclude that the value of x is: \(\frac{15}{2}\)

Question 3
11x + 1 = -1 + x
Answer:
The value of x is: –\(\frac{1}{5}\)

Explanation:
The given equation is:
11x + 1 = -1 + x
Now,
11x – x = -1 – 1
10x = -2
x = –\(\frac{2}{10}\)
x = –\(\frac{1}{5}\)
Hence, from the above,
We can conclude that the value of x is: –\(\frac{1}{2}\)

Solve the equation.

Question 4.
2 | x – 3 | – 5 = 7
Answer:
The value of x is: 9 or -3

Explanation:
The given absolute value equation is:
2 | x – 3 | – 5 = 7
2 | x – 3 | = 7 + 5
2 | x – 3 | = 12
| x – 3 | = \(\frac{12}{2}\)
| x – 3 | = 6
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
x – 3 = 6                             x – 3 = -6
x = 6 + 3                            x = -6 + 3
x = 9                                   x = -3
Hence, from the above,
We can conclude that the value of x is: 9 or -3

Question 5.
| 2x – 19 | = 4x + 1
Answer:
The value of x is: -10 or 3

Explanation:
The given absolute value equation is:
| 2x – 19 | = 4x + 1
We know that,
| x | = x for x > 0
| x | = -x for x < 0
4x + 1 = 2x – 19                         4x + 1 = – ( 2x – 19 )
4x – 2x = -19 – 1                         4x + 2x = 19 – 1
2x = -20                                      6x = 18
x = \(\frac{-20}{2}\)      x = \(\frac{18}{6}\)
x = -10                                         x = 3
Hence, from the above,
We can conclude that the value of x is: -10 or 3

Question 6.
-2 + 5x – 7 = 3x – 9 + 2x
Answer:
The given absolute equation has no solution

Explanation:
The given equation is:
-2 + 5x – 7 = 3x – 9 + 2x
5x – 9 = 5x – 9
Hence, from the above,
We can conclude that the given absolute value equation has no solution

Question 7.
3(x + 4) – 1 = -7
Answer:
The value of x is: -6

Explanation:
The given equation is:
3 ( x + 4 ) – 1 = -7
So,
3 ( x ) + 3 ( 4 ) = -7 + 1
3x + 12 = -6
3x = -6 – 12
3x = -18
x = –\(\frac{18}{3}\)
x = -6
Hence, from the above,
We can conclude that the value of x is: -6

Question 8.
| 20 + 2x | = | 4x + 4 |
Answer:
The value of x is: 8

Explanation:
The given absolute value equation is:
| 20 + 2x | = | 4x + 4 |
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
20 + 2x = 4x + 4
4x – 2x = 20 – 4
2x = 16
x = \(\frac{16}{2}\)
x = 8
Hence, from the above,
We can conclude that the value of x is: 8

Question 9.
\(\frac{1}{3}\)(6x + 12) – 2(x – 7) = 19
Answer:
The given equation has no solution

Explanation:
The given equation is:
\(\frac{1}{3}\) ( 6x + 12 ) – 2 ( x – 7 ) = 19
Now,
\(\frac{1}{3}\) ( 6x – 12 ) = 19 + 2 ( x – 7 )
\(\frac{1}{3}\) ( 6x – 12 ) = 19 + 2x – 14
\(\frac{1}{3}\) ( 6x – 12 ) = 2x + 5
1 ( 6x – 12 ) = 3 ( 2x + 5 )
6x – 12 = 6x + 15
6x – 6x = 15 + 12
15 = -12
Hence, from the above,
We can conclude that the given equation has no solution

Describe the values of c for which the equation has no solution. Explain your reasoning.

Question 10.
3x – 5 = 3x – c
Answer:
The value of c is: 5

Explanation:
The given equation is:
3x – 5 = 3x – c
It is given that the equation has no solution
So,
3x – 3x – 5 =-c
-c = -5
c = 5
Hence, from the above,
We can conclude that the value of c is: 5

Question 11.
| x – 7 | = c
Answer:
The value of c is: -7

Explanation:
The given absolute value equation is:
| x – 7 | = c
It is given that the equation has no solution i.e., x = 0
So,
0 – 7 = c
c = -7
Hence, from the above,
We can conclude that the value of c is: -7

Question 12.
A safety regulation states that the minimum height of a handrail is 30 inches. The maximum height is 38 inches. Write an absolute value equation that represents the minimum and maximum heights.
Answer:
The absolute value expression that represents the minimum and maximum heights is:
| x – 64 | = 34

Explanation:
It is given that a safety regulation states that the minimum height of a handrail is 30 inches. The maximum height is 38 inches.
So,
The average height of a handrail = ( 30 + 38 ) / 2
= 68 / 2
= 34 inches
Now,
The minimum height from the average height of a handrail = 34 + 30
= 64 inches
Hence,
The absolute value equation that represents the minimum and maximum height of a handrail is:
| x – 64 | = 34

Question 13.
The perimeter P (in yards) of a soccer field is represented by the formula P = 2ℓ + 2w, where ℓ is the length (in yards) and w is the width (in yards).
a. Solve the formula for w.
Answer:
The given formula is:
P = 2l + 2w
Where,
P is perimeter ( in yards )
l is the length ( in yards )
w is the width ( in yards )
So,
2w = P – 2l
w = ( P – 2l ) / 2
Hence, from the above,
We can conclude that the formula for w is:
w = ( P – 2l ) / 2

b. Find the width of the field.
Answer:
The given figure is:
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 117.1
From the above figure,
Perimeter ( P ) = 330 yd
Length ( l) = 100 yd
From part (a),
w = ( P – 2l ) /2
w = ( 330 – 100 ) / 2
w = 230 / 2
w = 115 yd
Hence, from the above,
We can conclude that the width of the field is: 115 yd

c. About what percent of the field is inside the circle?
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 117.1
Answer:

Question 14.
Your car needs new brakes. You call a dealership and a local mechanic for prices.
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 118
a. After how many hours are the total costs the same at both places? Justify your answer.
Answer:
From the given table,
Let the total labor hours be x
So,
The total cost at the Dealership = ( Cost of parts ) + ( Labor cost per hour ) × ( Total labor hours )
= 24 + 99x
The total cost at the local mechanic = ( Cost of parts ) + ( Labor cost per hour ) × ( Total labor hours )
= 45 + 89x
It is given that the total cost is the same in both places
So,
24 + 99x = 45 + 89x
99x – 89x = 45 – 24
10x = 21
x = 21 / 10
x = 2.1 hours
Hence, from the above,
We can conclude that after 2.1 hours, the total cost will be the same in both places

b. When do the repairs cost less at the dealership? at the local mechanic? Explain.
Answer:
The given table is:
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 118
From the above table,
Compare the labor cost per hour
By comparison,
We can say that the labor cost per hour is less at the local mechanic
Hence, from the above,
We can conclude that the repair cost less at the local mechanic

Question 15.
Consider the equation | 4x + 20 | = 6x. Without calculating, how do you know that x = -2 is an extraneous solution?
Answer:
We know that,
The absolute value equations only accept the values greater than or equal to 0
Hence,
For the given absolute value equation,
| 4x + 20 | = 6x
x = -2 is an extraneous solution

Question 16.
Your friend was solving the equation shown and was confused by the result “-8 = -8.” Explain what this result means.
4(y – 2) – 2y = 6y – 8 – 4y
4y – 8 – 2y = 6y – 8 – 4y
2y – 8 = 2y – 8
-8 = -8
Answer:
The result ” -8 = -8 ” means that the solved equation has no solution

Solving Linear Equations Cumulative Assessment

Question 1.
A mountain biking park has 48 trails, 37.5% of which are beginner trails. The rest are divided evenly between intermediate and expert trails. How many of each kind of trail are there?
A. 12 beginner, 18 intermediate, 18 expert
B. 18 beginner, 15 intermediate, 15 expert
C. 18 beginner, 12 intermediate, 18 expert
D. 30 beginner, 9 intermediate, 9 expert

Answer:
The correct option is: B
The number of beginner trials is: 18
The number of intermediate trials is: 15
The number of expert trials is: 15

Explanation:
It is given that a mountain biking park has 48 trails, 37.5% of which are beginner trails. The rest are divided evenly between intermediate and expert trials.
So,
The number of beginner trials is 3.5 % of the total number of trials
It is given that the total number of trials is: 48
We know that,
The value of 37.5 % is: \(\frac{3}{8}\) [ 37.5 % = 50 % – 12.5 % ]
So,
The number of beginner trials = \(\frac{3}{8}\) × 48
= \(\frac{3}{8}\) × \(\frac{48}{1}\)
= \(\frac{3 × 48}{8 × 1}\)
= 18
So,
The number  of intermediate and expert trials = ( The total number of trials ) – ( The number of beginner trials )
= 48 – 18
= 30 trials
It is also given that the intermediate trials and expert trails are divided evenly
So,
30 ÷ 2 = 15 trials each
Hence, from the above,
We can conclude that
The number of beginner trials is: 18
The number of intermediate trials is: 15
The number of expert trials is: 15

Question 2.
Which of the equations are equivalent to cx – a = b?
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 119

Answer:
The given equations are:
a) cx – a + b = 2b
b) 0 = cx – a + b
c) 2cx – 2a = b / 2
d) x – a = b / 2
e) x = ( a + b ) / c
f) b + a = cx
Now,
We have to find the equations from above that is equivalent to the given equation cx – a = b
Now,
a)
The given equation is:
cx – a + b = 2b
So,
cx – a = b – b
cx – a = b
b)
The given equation is:
0 = cx – a + b
So,
cx – a = -b
c)
The given equation is:
2cx – 2a = b / 2
So,
cx – a = b / 4
d )
The given equation is:
cx – a = b / 2
So,
2 ( cx – a ) = b
e)
The given equation is:
x = ( a + b ) / c
So,
cx = a + b
cx – a = b
f)
The given equation is:
b + a = cx
So,
cx – a = b
Hence, from the above,
We can conclude that the equations that are equivalent to cx – a = b is: a, e, f

Question 3.
Let N represent the number of solutions of the equation 3(x – a) = 3x – 6. Complete each statement with the symbol <, >, or =.
a. When a = 3, N ____ 1.
b. When a = -3, N ____ 1.
c. When a = 2, N ____ 1.
d. When a = -2, N ____ 1.
e. When a = x, N ____ 1.
f. When a = -x, N ____ 1.

Answer:
The given equation is:
3 ( x – a ) = 3x – 6
So,
3x – 3a = 3x – 6
Now,
a) When a = 3,
3x – 3 ( 3 ) = 3x – 6
3x – 9 = 3x – 6
9 = 6
Hence,
When a = 3 there is no solution
Hence,
N < 1
b) When a = -3
3x + 3 ( 3 ) =3 x – 6
9 = -6
Hence,
When a = -3, there is no solution
Hence,
N < 1
c) When a = 2
3x – 3 ( 2 ) = 3x – 6
3x – 6 =3x –
6 = 6
Hence,
When a= 2, there is no solution
Hence,
N < 1
d) When a = -2
3x + 3 ( 2 ) = 3x – 6
3x + 6 = 3x – 6
6 = -6
Hence,
When a = -2, thereis no solution
Hence,
N < 1
e) When a = x
3x – 3 ( x ) = 3x – 6
3x = 6
x = 6 / 3
x = 2
Hence,
When a  x, theer is 1 solution
Hence,
N = 1
f) When a = -x
3x + 3 ( x ) = 3x – 6
6x – 3x = -6
3x = -6
x = -6 / 3
x = -2
Hence,
When a = -x, there is 1 solution
Hence,
N = 1

Question 4.
You are painting your dining room white and your living room blue. You spend $132 on 5 cans of paint. The white paint costs $24 per can, and the blue paint costs $28 per can.
a. Use the numbers and symbols to write an equation that represents how many cans of each color you bought.
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 120
Answer:
The number of cans of white paint is: 2
The number of cans of blue paint is: 3

Explanation:
It is given that you spend $132 on 5 cans of paint
It is also given that the white paint costs $24 per can and the blue paint costs $28 per can
Now,
Let
The number of white cans is: x
The number of blue cans be: 5 – x
So,
The total cost of paint = ( The number of white cans ) × ( The cost of white paint per can ) + ( The number of blue cans ) × ( The cost of blue paint per can )
132 = 24x + 28 ( 5 – x )
24x + 28 ( 5 ) – 28x = 132
140 – 4x = 132
4x = 140 – 132
4x = 8
x = 8 ÷ 4
x = 2
Hence, from the above,
We can conclude that
The number of cans of white paint is: 2
The number of cans of blue paint is: 3

b. How much would you have saved by switching the colors of the dining room and living room? Explain.
Answer:
The money you have saved by switching the colors of the dining room and living room is: $0

Explanation:
From part (a),
The number of white cans is: 2
The number of blue cans is: 3
It is given that white color is used in the dining room and the blue color is used in the living room
So,
The cost of white paint after interchanging the color in the living room = 24 × 2
= $48
The cost of blue paint after interchanging the color in the dining room = 28 × 3
= $84
So,
The total cost of paint after interchanging the colors = 48 + 84
= $132
Hence,
The amount of money saved = ( The money you paid for the paint before interchanging ) – ( The money you paid for the paint after interchanging )
= 132 – 132
=$0

Question 5.
Which of the equations are equivalent?
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 121

Answer:
The given equations are:
a ) 6x + 6 = -14
b ) 8x + 6 = -2x – 14
c ) 5x + 3 = -7
d ) 7x + 3 = 2x – 13
Now,
To find the equivalent equations, find the value of x
So,
a)
The given equation is:
6x + 6 = -14
6x = -14 – 6
6x = -20
x = -20 / 6
x = -10 / 3
b)
The given equation is:
8x + 6 = -2x – 14
8x + 2x = -14 – 6
10x = -20
x = -20 / 10
x = -2
c)
The given equation is:
5x + 3 = -7
5x = -7 -3
5x = -10
x= -10 / 5
x = -2
d)
The given equation is:
7x + 3 = 2x – 13
7x – 2x = -13 – 3
5x = -16
x = -16 / 5
Hence, from the above,
We can conclude that the equations c) and d) are equivalent

Question 6.
The perimeter of the triangle is 13 inches. What is the length of the shortest side?
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 122
Answer:
The length of the shortest side is: 3 inches

Explanation:
We know that,
The perimeter is the sum of all the sides of the given figure
It is given that the perimeter of the triangle is: 13 inches
So,
The perimeter of the triangle = ( x – 5 ) + ( x / 2 ) + 6
13 = x + 1 + ( x / 2 )
13 = ( 2x / 2 ) + ( x / 2 ) + 1
3x / 2 = 13 – 1
3x / 2 = 12
3x = 12 × 2
3x = 24
x = 24 / 3
x = 8
So,
The lengths of all sides are: ( 8 – 5 ), 6, ( 8 / 2 ) = 3 inches, 6 inches, 4 inches
Hence, from the above,
We can conclude that the length of the shortest side is: 3 inches

Question 7.
You pay $45 per month for cable TV. Your friend buys a satellite TV receiver for $99 and pays $36 per month for satellite TV. Your friend claims that the expenses for a year of satellite TV are less than the expenses for a year of cable.
a. Write and solve an equation to determine when you and your friend will have paid the same amount for TV services.
Answer:
It is given that you pay $45 per month for cable TV. Your friend buys a satellite TV receiver for $99 and pays $36 per month for satellite TV
So,
Let the number of months be x
Now,
The time they paid the same amount for TV services is:
45x = 99 + 36x
45x – 36x = 99
9x= 99
x = 99 / 9
x = 11
Hence, from the above,
We can conclude that after 11 months, you and your friend will pay the same amount for TV services

b. Is your friend correct? Explain.
Answer:
Your friend is correct

Explanation:
We know that,
1 year = 12 months
So,
The expenses paid by you for TV services = 45x = 45 × 12 = $540
The expenses paid by your friend for TV services = 99 + 36x
= 99 + 36 ( 12 )
= 99 + 432
= $531
By comparing the expenses of you and your friend,
We can conclude that your friend is correct

Question 8.
Place each equation into one of the four categories.
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 123

Question 9.
A car travels 1000 feet in 12.5 seconds. Which of the expressions do not represent the average speed of the car?
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 124

Answer:
We know that,
Average speed = ( Distance ) ÷ ( Time )
It is given that a car travels 1000 feet in 12.5 seconds
So,
Average speed = 1000 / 12.5
= 80\(\frac{feet}{second}\)
Now,
The given options are:
A) 80\(\frac{second}{feet}\) B) 80\(\frac{feet}{second}\) C) \(\frac{80 feet}{second}\)
D) \(\frac{second}{ 80 feet}\)
Hence, from the above,
We can conclude that option B) represents the average speed

Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers

Big Ideas Math Answers Grade 4 Chapter 2

The basics must be strong to grow with perfect knowledge. Students need to learn the math basics strongly to become a master in maths. Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers is the best reference to make your student’s basics strong. The Big Ideas Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers Math Answers will help your students to achieve their dreams of learning maths with simple methods. From the fundamentals of Add and Subtract Multi-Digit Numbers to difficult problems, all are given on this page.

Big Ideas 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers Math Book Answer Key

Check out the Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers links provided in the below section to begin your preparation. We have given a detailed explanation for each and every problem for your better understanding. So, make use of every given link and try to score the best marks in the exams. The basics mentioned here will not only help you in the exam but also in real life. Be the first to learn the best of math topics by referring to bigideasmathanswers.com

Lesson: 1 Estimate Sum and Differences

Lesson 2.1 Estimate Sum and Differences
Estimate Sum and Differences Homework & Practice 2.1

Lesson: 2 Add Multi-Digit Numbers

Lesson 2.2 Add Multi-Digit Numbers
Add Multi-Digit Numbers Homework & Practice 2.2

Lesson: 3 Subtract Multi-Digit Numbers

Lesson 2.3 Subtract Multi-Digit Numbers
Subtract Multi-Digit Numbers Homework & Practice 2.3

Lesson: 4 Use Strategies to Add and Subtract

Lesson 2.4 Use Strategies to Add and Subtract
Use Strategies to Add and Subtract Homework & Practice 2.4

Lesson: 5 Problem Solving: Addition and Subtraction

Lesson 2.5 Problem Solving: Addition and Subtraction
Problem Solving: Addition and Subtraction Homework & Practice 2.5

Performance Task

Add and Subtract Multi-Digit Numbers Performance Task
Add and Subtract Multi-Digit Numbers Activity
Add and Subtract Multi-Digit Numbers Chapter Practice

Lesson 2.1 Estimate Sum and Differences

Explore and Grow
Estimate to find each sum by rounding to the nearest thousand, hundred, or ten. Explain why you chose to round to that place value.
A football team had 917 spectators at their first game and 872 at their second game. About how many spectators did the team have at both games?

Answer: The answer is 1789.

Explanation:
Number of spectators in First game: 917
Number of spectators in First game: 872
To get the number of spectators in both the games, we have to add the number of spectators in first game and second game.
That means, 917 + 872 = 1789.
A company budgets $1,800 for a company picnic. They spend $917 on the location and $872 on food. Did they stay within their budget?

Answer: Yes, they can stay within their budget.

Explanation:
Total budget: $1,800
Spend for location: $917
Spend for food: $872
Total spend = spend for location + spend for food.
That means, total spend = $917 + $872 = $1789.
Balance budget = Total budget – Total spend
That means, Balance budget: $1800 – $1789 = $11
As, the total spend for the company picnic is lesser than the company budget, they can stay within the budget.
Reasoning
Explain why you may choose to round to different place values in different situations.

Answer: “Round to” are the approximate values we give that are the nearest possible values to the exact values. Sometimes, we may be sufficient with the estimate or nearest values. We may not require the exact values. In such situations, we can go ahead with the round to values. For example, you organized a lunch at your home. The caterer asks you for the number of guests for whom the lunch has to be served. You cannot give him the exact number of guests. You give him the round of value of the number of guests attend based on your past experience and present situation. In such cases, round to is very helpful.
Think and Grow: Estimate Sum and Differences
An estimate is a number that is close to an exact number. You can use rounding to estimate sums and differences.
Example
Estimate 8,675 + 3,214.
One Way: Round each addend to the nearest hundred. Then find the sum.

So, 8,675 + 3,214 is about rounding the two numbers to the nearest hundreds. 8,700 is the nearest hundreds value for 8,675. 3200 is the nearest hundreds value for 3214.
Then, we will add 8,700 and 3,200.
8,700 + 3,200 = 11, 900
Another Way: Round each addend to the nearest thousand. Then find the sum.

So, 8,675 + 3,214 is about rounding the two numbers to the nearest thousands. 9,000 is the nearest thousands value for 8,675. 3,000 is the nearest thousands value for 3214.
Then, we will add 9,000 and 3,000.
9,000 + 3,000 = 12, 000
.
Example
Estimate 827,615 – 54,3006.
One Way: Round each addend to the nearest thousand. Then find the difference.

So, 827,615 – 54,306 is about rounding the two numbers to the nearest thousands. 8,28,000 is the nearest thousands value for 8,27,615. 54,000 is the nearest thousands value for 54,306.
Then, we will subtract 54,000 from 8,28,000.
8,28,000 – 54,000 = 7,74,000.
Another Way: Round each addend to the nearest ten thousand. Then find the difference.

So, 827,615 – 54,306 is about rounding the two numbers to the nearest ten thousands. 8,30,000 is the nearest thousands value for 8,27,615. 50,000 is the nearest thousands value for 54,306.
Then, we will subtract 50,000 from 8,30,000.
8,28,000 – 54,000 = 7,80,000.
Show and Grow
Estimate the sum or difference.

Question 1.
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 5

Answer: We will here round of the values to the nearest hundreds. 63,900 is the nearest hundred for 63,851. For, 19,375 the nearest hundred value is 19,400.

Now we will add 63,900 to 19,400.
63,900 + 19,400 = 83,300.

Question 2.
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 6

Answer: We will here round of the values to the nearest hundreds. 4,900 is the nearest hundred for 4,874. For, 2,530 the nearest hundred value is 2,500.

Now we will subtract 2,500 from 4,900.
4,900 – 2,500 = 2,400.
Apply and Grow: Practice
Estimate the sum or difference.

Question 3.
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 7

Answer:
We here round the values to the nearest hundreds value.
For 27,369 the nearest hundreds value is 27,400. For 14,608 the nearest hundred value is 14,600.

27,400 + 14,600 = 42,000.
So, the estimate of the sum of 27,369 and 14,608 by rounding to the nearest hundreds value is 42,000.

Question 4.
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 8

Answer:
We here round the values to the nearest hundreds value.
For 53,744 the nearest hundreds value is 53,700. For 41,086 the nearest hundred value is 41,100.

53,700 – 41,100 = 12,600.
So, the estimate of the difference of 53,744 and 41,086 by rounding to the nearest hundreds value is 12,600.

Question 5.
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 9

Answer:
We here round the values to the nearest thousands value.
For 68,451 the nearest thousands value is 68,000. For 40,695 the nearest thousands value is 41,000.
68,000 – 41,000 = 27,000.
So, the estimate of the difference of 68,451 and 40,695 by rounding to the nearest thousands value is 27,000.

Question 6.
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 10

Answer:
We here round the values to the nearest thousands value.
For 34,685 the nearest thousands value is 35,000. For 27,043 the nearest thousands value is 27,000.
35,000 + 27,000 = 62,000.
So, the estimate of the sum of 34,685 and 27,043 by rounding to the nearest thousands value is 62,000.

Question 7.
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 11

Answer:
We here round the values to the nearest ten thousands value.
For 908,465 the nearest ten thousands value is 910,000. For 653,629 the nearest ten thousands value is 650,000.
910,000 – 650,000 = 260,000.
So, the estimate of the difference of 908,465 and 653,299 by rounding to the nearest ten thousands value is 260,000.

Question 8.
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 12

Answer:
We here round the values to the nearest ten thousands value.
For 478,633 the nearest ten thousands value is 480,000. For 200,081 the nearest ten thousands value is 200,000.
480,000 + 200,000 = 680,000.
So, the estimate of the sum of 478,633 and 200,081 by rounding to the nearest ten thousands value is 680,000.

Question 9.
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 13

Answer:
We here round the values to the nearest thousands value.
For 395,408 the nearest thousands value is 395,000. For 102,677 the nearest thousands value is 103,000.
395,000 – 103,000 = 292,000.
So, the estimate of the difference of 395,408 and 102,677 by rounding to the nearest thousands value is 292,000.
Question 10.
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 14

Answer:
For 563,427 the nearest ten thousands value is 560,000. For 178,023 the nearest ten thousands value is 180,000.
560,000 + 180,000 = 740,000.
So, the estimate of the sum of 563,427 and 178,023 by rounding to the nearest ten thousands value is 740,000.

Question 11.
888,056 – 423,985 = _______

Answer:
We here round the values to the nearest ten thousands value.
For 888,056 the nearest ten thousands value is 890,000. For 423,985 the nearest ten thousands value is 420,000.
890,000 – 420,000 = 472,000.
So, the estimate of the difference of 888,056 and 423,985 by rounding to the nearest ten thousands value is 470,000.
Question 12.
713,642 + 49,018 = ______

Answer:
We here round the values to the nearest thousands value.
For 713,642 the nearest thousands value is 714,000. For 49,018 the nearest thousands value is 49,000.
714,000 + 49,000 = 763,000.
So, the estimate of the sum of 713,642 and 49,018 by rounding to the nearest thousands value is 763,000.

Question 13.
DIG DEEPER!
Is 20,549 + 9,562 greater than or less than 30,000? Explain how you know without finding the exact sum.

Answer:
We will solve this using round to estimates.
For 20,549 the nearest hundreds value is 20,500. For 9,562 the nearby smaller hundreds value is 9,500.
20,500 + 9,500 = 30,000.
The estimate sum is 30,000. We still have the values of 49 and 62 in both the values. So, the value of the sum of 20,549 and 9,562 is greater than 30,000.

Question 14.
Writing
Describe a real-life situation in which estimation would not be appropriate to use.

Answer:
Think and Grow: Modeling Real Life
Example
About how many more pounds does the whale shark weigh than the orca?
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 15
Round the weight of each animal to the nearest thousand because you do not need a precise answer.
Orca: 8,000 Whale shark: 40,000.
Subtract the estimated weight of the orca from the estimated weight of the whale shark.

The whale shark weighs about 32,000 more pounds than the orca.
Show and Grow

Question 15.
About how many more votes did Candidate A receive than Candidate B?
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 17

Answer:
Round the number of votes to the nearest thousands value.
Candidate A: 250,000. Candidate B: 85,000

Candidate A scored about 165,000 votes more than Candidate B.

Question 16.
Mount Saint Helens is a volcano that is 8,363 feet tall. Mount Fuji is a volcano that is 4,025 feet taller than Mount Saint Helens. About howtall is Mount Fuji?
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 18

Answer:
Round the heights of the volcanoes to the nearest hundreds value.
Mount Saint Helens: 8,400. Mount Fuji: 4,000.

So, Mount Helens is about 4,400 feet taller than Mount Fuji.

Question 17.
An educational video has 6,129 fewer views than a gaming video. The educational video has 483,056 views. About how many views does the gaming video have?

Answer:
Here, it is given that the educational video has 6,129 fewer views than the gaming video.
That means, Gaming Video views – 6,129 = Educational video views.
Or Gaming video views = Educational video views + 6,129.
Given educational video views = 483,056
Therefore, Gaming video views = 483,056 + 6,129 = 489,185.
So, the number of gaming video views = 489,185.

Estimate Sum and Differences Homework & Practice 2.1

Estimate the sum or difference.

Question 1.
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 19

Answer:
Round the values to the nearest hundreds values.
For 7,910 the nearest hundreds value is 7,900. For 1,358, the nearest hundreds value is 1,400.

7,900 + 1,400 = 9,300.

Question 2.
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 20

Answer:
Round the values to the nearest hundreds values.
For 5,608 the nearest hundreds value is 5,600. For 3,217, the nearest hundreds value is 3,200.


5,600 – 3,200 = 2,400

Question 3.
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 21

Answer:
Round the values to the nearest thousands values.
For 73,406 the nearest thousands value is 73,000. For 45,699 the nearest thousands value is 46,000.

73,000 – 46,000 = 27,000

Question 4.
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 22

Answer:
Round the values to the nearest thousands values.
For 82,908 the nearest thousands value is 83,000. For 28,643 the nearest thousands value is 29,000.

83,000 + 29,000 = 1,12,000

Question 5.
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 23

Answer:
Round the values to the nearest thousands values.
For 96,420 the nearest thousands value is 96,000. For 63,877 the nearest thousands value is 64,000.

96,000 – 64,000 = 32,000

Question 6.
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 24

Answer:
Round the values to the nearest ten thousands values.
For 517,605 the nearest ten thousands value is 520,000. For 359,421 the nearest ten thousands value is 360,000.

520,000 + 360,000 = 880,000

Question 7.
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 25

Answer:
Round the values to the nearest ten thousands values.
For 688,203 the nearest ten thousands value is 690,000. For 444,387 the nearest ten thousands value is 440,000.

690,000 – 440,000 = 250,000

Question 8.
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 26

Answer:
Round the values to the nearest ten thousands values.
For 261,586 the nearest ten thousands value is 260,000. For 116,934 the nearest ten thousands value is 120,000.

260,000 + 120,000 = 380,000
Estimate the sum or difference.

Question 9.
864,733 – 399,608 = ______

Answer:
Round the values to the nearest ten thousands values.
864,733: 860,000
399,608: 400,000
860,000 – 400,000 = 460,000

Question 10.
134,034 + 26,987 = ______

Answer:
Round the values to the nearest thousands values.
134,034: 134,000
26,987: 27,000
134,000 + 27,000 = 161,000

Question 11.
Number Sense
Descartes estimates a difference by rounding each number to the nearest ten thousand. His estimate is 620,000. Which problems could he have estimated?
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 27

Answer:
Round off all the given values to their nearest ten thousand value.
690,000 – 70,000 = 620,000
890,000 – 270,000 = 620,000
680,000 – 50,000 = 630,000
700,000 – 80,000 = 620,000.
So, Descartes could have estimated problems 1,2 and 4 and all the three give the estimated difference of 620,000.

Question 12.
Reasoning
When might you estimate the difference of 603,476 and 335,291 to the nearest hundred? to the nearest hundred thousand?

Answer:
When the precision of the difference value should be in hundreds, then round the values to hundred and calculate the difference.
603,476: 603,500 & 335,291: 335,300
603,500 – 335,300 = 268,200.
When the precision of the difference value should be in thousands, then round the values to thousands and calculate the difference.
603,476: 603,000 & 335,291: 335,000
603,000 – 335,000 = 268,000

Question 13.
Modeling Real Life
A storm causes causes 23,890 homes to be without power on the east side of a city and 18,370 homes to be without power on the west side. About how many homes altogether are without power?
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 28

Answer:
Homes without power on east side: 23,890
Homes without power on west side: 18,370
Round the values to the nearest thousands value.
So, homes without power on east side = 24,000 and homes without power on west side = 18,000.
Total homes without power = 24,000 + 18,000 = 42,000
Therefore, total estimate of homes without power due to storm are 42,000.

Question 14.
Modeling Real Life
You walk 5,682 steps. Your teacher walks 4,219 steps more than you. About how many steps does your teacher walk?
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.1 29

Answer:
Round the number of steps to the nearest hundreds value.
Number of steps I walked, 5,682 rounding it to the nearest hundreds value = 5,700
Number of steps teacher walked additional more than me, 4,219 rounding it to the nearest hundreds value = 4,200
Total number of steps teacher walked = Number of steps I walked + additional number of steps teacher walked.
Therefore total number of steps teacher walked = 5,700 + 4,200 = 9,900.
Review & Refresh
Find the product.

Question 15.
3 × 3 × 2 = ______

Answer:
Break the multiplication into steps as follows:
(3 X 3) x 2
= 9 x 2
=18

Question 16.
2 × 4 × 7 = ______

Answer:
Break the multiplication into steps as follows:
2 X (4 X 7)
= 2 X 28
= 56

Question 17.
3 × 3 × 5 = ______

Answer:
Break the multiplication into steps as follows:
(3 X 3) X 5
= 9 X 5
= 45.

Question 18.
6 × 2 × 3 = ______

Answer:
Break the multiplication into steps as follows:
6 X (2 X 3)
=6 X 6
= 36.

Question 19.
4 × 9 × 2 = ______

Answer:
Break the multiplication into steps as follows:
4 X (9 X 2)
= 4 X 18
= 72

Question 20.
4 × 10 × 2 = ______

Answer:
Break the multiplication into steps as follows:
4 X (10 X 2)
= 4 X 20
= 80.

Lesson 2.2 Add Multi-Digit Numbers

Explore and Grow
Which addition problem shows a correct way to find 38 +7? Why?
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 1

Answer:
The second option shows the correct way.
Explanation: As per place value rule, 7 is the units position value. A units digit always should be added to the units digit of the other number. In the first option, 7 is under 3, which is actually in tens digit position. The second option displays it correctly pointing 7 under the units digit value, 8.
Which addition problem shows a correct way to find 403 + 1,248? Why?
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 2

Answer:
The first option shows the correct way.
Explanation: As per place value rule, a units digit always should be added to the units digit of the other number. Similarly, tens digit value to the other number’s tens digit value and the hundreds digit value. First option maintains the rule correctly pointing 8 under 3, both which are units values. Similarly, 4 under 0 and 2 under 8. The second option started the sum from the left, which means, adding thousands digit value of 1,248 to the hundreds digit value of 403.
Reasoning
Why do you need to use place value when adding? Explain.

Answer:
While adding, we use place value i.e., we go from right hand side to the left hand side. In other words, we add the digits as per the place value. The digit of every number holds the place value. That means, a digit in tens place has the value of multiple of ten. Similarly, a digit in hundreds place has the value of multiple of hundred.
For example, In the number 23, 2 digit holds the value of multiple of ten. That means, 2 X 10.
To continue the parity of the sum, we follow the place value rule by adding digits as per the place values.
Think and Grow: Add Multi-Digit Numbers
Example
Add: 307,478 + 95,061.
Estimate: 307,000 + 95,000 = _______
Use place value to line up the addends.
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 3
Add the ones, then the tens, and then the hundreds. Regroup if necessary.
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 4
Add the thousands, then the ten thousands, and then the hundred thousands. Regroup if necessary.
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 5
Show and Grow
Find the sum. Check whether your answer is reasonable.

Question 1.
Estimate: _______
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 6

Answer:
Estimate: Round the values to the nearest hundreds.
17,700 + 53,000 = 70,700.
Add the tens, then hundreds, thousands and then the ten thousands. Regroup if necessary.

Check: Because 70,714 is close to the estimate, 70,700, the answer is reasonable.

Question 2.
Estimate: ______
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 7

Answer:
Round the values to the nearest hundreds.
297,900 + 6,100 = 304,000
Add the ones, tens, then hundreds and thousands. Regroup if necessary.

Check: Because 303,982 is close to the estimate, 304,000, the answer is reasonable.
Apply and Grow: Practice
Find the sum. Check whether your answer is reasonable.

Question 3.
Estimate: ______
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 8

Answer:
Round the values to the nearest hundreds.
Estimate: 6,400 + 3,300 = 9,700
Add the ones, tens, then hundreds and thousands. Regroup if necessary.

Check: Because 9,719 is close to the estimate, 9,700, the answer is reasonable.

Question 4.
Estimate: ______
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 9

Answer:
Round the values to the nearest hundreds.
Estimate: 61,100 + 8,800 = 69,900
Add the ones, tens, then hundreds, then thousands and ten thousands. Regroup if necessary.

Check: Because 69,861 is close to the estimate, 69,900, the answer is reasonable.

Question 5.
Estimate: ______
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 10

Answer:
Round the values to the nearest hundreds.
Estimate: 82,200 + 4,700 = 86,900
Add the ones, tens, then hundreds, then thousands and ten thousands. Regroup if necessary.

Check: Because 86,935 is close to the estimate, 86,900, the answer is reasonable.

Question 6.
Estimate: ______
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 11

Answer:
Round the values to the nearest hundreds.
Estimate: 46,800 + 38,500 = 85,300
Add the ones, tens, then hundreds, then thousands and ten thousands. Regroup if necessary.

Check: Because 85,308 is close to the estimate, 85,300, the answer is reasonable.

Question 7.
Estimate: ______
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 12

Answer:
Round the values to the nearest hundreds.
Estimate: 686,400 + 75,300 = 761,700
Add the ones, tens, then hundreds, then thousands and ten thousands. Regroup if necessary.

Check: Because 761,739 is close to the estimate, 761,700, the answer is reasonable.

Question 8.
Estimate: ______
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 13

Answer:
Round the values to the nearest thousands.
Estimate: 594,000 + 308,000 = 902,000
Add the ones, tens, then hundreds, then thousands, ten thousands and then hundred thousands. Regroup if necessary.

Check: Because 902,240 is close to the estimate, 902,000, the answer is reasonable.

Question 9.
Estimate: ______
246,890 + 13,579 = ______

Answer:
Round the values to the nearest thousands.
Estimate: 247,000 + 14,000 = 261,000
Add the ones, tens, then hundreds, then thousands and then ten thousands. Regroup if necessary.

Check: Because 260,469 is close to the estimate, 261,000, the answer is reasonable.

Question 10.
Estimate: ______
822,450 + 8,651 = _______

Answer:
Round the values to the nearest thousands.
Estimate: 822,000 + 9,000 = 831,000
Add the ones, tens, then hundreds, then thousands and then ten thousands. Regroup if necessary.

Check: Because 831,101 is close to the estimate, 831,000, the answer is reasonable.

Question 11.
A video receives 10,678 views the first day. It receives 25,932 views the second day. How many views does the video receive in two days?
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 14

Answer:
Number of views on first day = 10,678
Number of views on second day = 25,932.
Total number of views = number of views on first day + number of views on second day.
Total number of views = 10,678 + 25,932

Therefore, total number of views = 36,610 views.

Question 12.
YOU BE THE TEACHER
Is Newton correct? Explain.
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 15

Answer:
No, Newton is not correct.
Explanation: Newton is doing the sum without following the place values. He is adding the units place digit of the second number to the hundreds value of the first number. Similarly, he is also adding tens, hundreds and thousands place value digits of the first number to the thousands, ten thousands and hundred thousands place value digits of the first number respectively.
The correct method is as follows:

Question 13.
DIG DEEPER!
Find the missing digits.
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 16

Answer:
The digit missing in the first number is 5.
The digit missing in the second number is 4.
Explanation: Add the tens value digits 7 and 9. 7+9 = 16. Regroup 1 onto the hundreds place numbers.
1+ missing digit + 1 = 7. Therefore, the missing digit in first number becomes 5, as 1+5+1=7.
In the thousands place, 9+missing digit gives the value of 3 in the sum. the highest possible sum of 9 with a missing digit is 9+9=18. To get 3 in the units place of the sum, 9 has to be added with 4 because 9+4 gives a value of 13. Therefore the missing digit of second number is 4.
Think and Grow: Modeling Real Life
Example
A family is traveling in a car from Seattle to Atlanta. They travel 1,099 miles the first two days and 1,082 miles the next two days. Has the family arrived in Atlanta?
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 17
Add the distances traveled.
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 18
Compare the distance traveled to the distance from Seattle to Atlanta.
The family ______ arrived in Atlanta.
Show and Grow

Question 14.
One World Trade Center has 2,226 steps. A visitor enters the building and climbs 1,387 steps, takes a break, and climbs839 more steps. Did the visitor reach the top of the building?
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 19

Answer:
Add the total number of steps:

Number of steps climbed: 2,226 and the number of steps of the world trade center, 2,226 are equal.
Therefore, the visitor has reached the top of the World Trade Center.

Question 15.
There were 51,787 more people who rode the city buses on Saturday than on Sunday. On Sunday, 174,057 people rode the buses. How many people rode the buses on Saturday?

Answer:
Given: People rode on city bus on Saturday = 51,787 + People rode on city bus on Sunday.
Therefore, People rode on city bus on Saturday = 51,787 + 174,057

Therefore, People rode on city bus on Saturday = 225,844.

Question 16.
The deepest part of the Atlantic Ocean is 8,577 feet shallower than the deepest part of the Pacific Ocean. The deepest part of the Atlantic Ocean is 27,493 feet deep. How deep is the deepest part of the Pacific Ocean?

Answer:
Given: Depth of Atlantic Ocean = Depth of Pacific Ocean – 8,577.
So, Depth of Atlantic Ocean = 27,493 – 8,577

Therefore, depth of Atlantic ocean = 18,916 feet.

Add Multi-Digit Numbers Homework & Practice 2.2

Find the sum. Check whether your answer is reasonable

Question 1.
Estimate: ______
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 20

Answer:
Round the values to the nearest hundreds.
Estimate: 8,500 + 4,700 = 13,200
Add the ones, tens, then hundreds and then thousands. Regroup if necessary.

Check: Because 13,141 is close to the estimate, 13,200, the answer is reasonable.

Question 2.
Estimate: ______
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 21

Answer:
Round the values to the nearest hundreds.
Estimate: 75,400 + 8,600 = 84,000
Add the ones, tens, then hundreds and then thousands. Regroup if necessary.

Check: Because 84,016 is close to the estimate, 86,000, the answer is reasonable.

Question 3.
Estimate: ______
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 22

Answer:
Round the values to the nearest thousands.
Estimate: 90,000 + 20,000 = 110,000
Add the ones, tens, then hundreds and then thousands. Regroup if necessary.

Check: Because 110,340 is close to the estimate, 110,000, the answer is reasonable.

Question 4.
Estimate: ______
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 23

Answer:
Round the values to the nearest thousands.
Estimate: 48,000 + 24,000 = 72,000
Add the ones, tens, then hundreds and then thousands. Regroup if necessary.

Check: Because 71,585 is close to the estimate, 72,000, the answer is reasonable.

Question 5.
Estimate: ______
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 24

Answer:
Round the values to the nearest thousands.
Estimate: 505,000 + 65,000 = 570,000
Add the ones, tens, then hundreds and then thousands. Regroup if necessary.

Check: Because 569,821 is close to the estimate, 570,000, the answer is reasonable.

Question 6.
Estimate: ______
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 25

Answer:
Round the values to the nearest thousands.
Estimate: 505,000 + 65,000 = 570,000
Add the ones, tens, then hundreds and then thousands. Regroup if necessary.

Check: Because 569,821 is close to the estimate, 570,000, the answer is reasonable.
Find the sum. Check whether your answer is reasonable.

Question 7.
Estimate: ______
547,795 + 9,418 = ______

Answer:
Round the values to the nearest thousands.
Estimate: 548,000 + 9,000 = 557,000
Add the ones, tens, then hundreds and then thousands. Regroup if necessary.

Check: Because 557,213 is close to the estimate, 557,000, the answer is reasonable.

Question 8.
Estimate: ______
401,269 + 58,135 = ______

Answer:
Round the values to the nearest thousands.
Estimate: 401,000 + 58,000 = 459,000
Add the ones, tens, then hundreds and then thousands. Regroup if necessary.

Check: Because 459,404 is close to the estimate, 459,000, the answer is reasonable.

Question 9.
Number Sense
Find and explain the error. What is the correct sum?
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 26

Answer:
Round the values to the nearest thousands.
Estimate: 175,000 + 29,000 = 204,000
Add the ones, tens, then hundreds and then thousands. Regroup if necessary.

Check: Because 459,404 is close to the estimate, 459,000, the answer is reasonable.

Question 10.
Writing
Write a word problem that can be solved by finding the sum of 75,629 and 23,548.

Answer:
A purchases a house worth $23,548 more than the house purchased by B. House purchased by B is worth $75,629. Find the worth of the house purchased by A.
Worth of house purchased by A = Worth of house purchased by B + $23,548
Worth of house purchased by A = $75,629 + $23,548
Therefore, worth of house purchased by A = $ 99,177.

Question 11.
Modeling Real Life
Two chefs need to make 2,500 food samples for the opening day of their restaurant. Chef A makes 1,346 samples. Chef B makes 1,084 samples. Did the chefs make enough samples?
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 27

Answer:
Food samples made by Chef A = 1,346.
Food samples made by Chef B = 1,084.
Total food samples made = Food samples made by Chef A + Chef B.
So, total food samples made = 1,346+1,084 = 2,430.
Total food samples required = 2,500.
The total food samples made are lesser than the required number of food samples. So, the chefs A & B didn’t make enough food samples.

Question 12.
Modeling Real Life
House A costs $24,450 less than House B. How much does House B cost?
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 28

Answer:
Given, Cost of House A = Cost of House B – $24,450.
So, Cost of House B = Cost of House A + $24,450.
Therefore, Cost of House B = $175,500 + $ 24,450 = $ 199,950.
Review & Refresh
Find the quotient.

Question 13.
6 ÷ 2 = ______

Answer:

Therefore, the quotient is 3.

Question 14.
24 ÷ 4 = ______

Answer:

Quotient = 6.

Question 15.
32 ÷ 8 = ______

Answer:

Quotient = 4.

Question 16.
14 ÷ 7 = ______

Answer:

Quotient = 2.

Question 17.
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 29

Answer:

Quotient = 1

Question 18.
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 30

Answer:

Quotient = 7.

Question 19.
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 31

Answer:

Quotient = 10

Question 20.
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 2.2 32

Answer:

Quotient = 9.

Lesson 2.3 Subtract Multi-Digit Numbers

Explore and Grow
Which subtraction problem shows a correct way to find 94 – 8? Why?
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 1

Answer:
First one shows the correct way to find the subtraction because it follows the place value rule by subtracting the units digit value of second number from the units digit value of the first number.
Which subtraction problem shows a correct way to find 3,710 – 251? Why?
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 2

Answer:
Second one shows the correct way to find the subtraction because it follows the place value rule by subtracting the units, tens and hundreds digit values of second number from the units, tens and hundreds digit values of the first number respectively.
Reasoning
Why do you need to use place value when subtracting? Explain.

Answer:
While subtracting, we use place value rule i.e., we go from right hand side to the left hand side. In other words, we add the digits as per the place value. The digit of every number holds the place value. That means, a digit in tens place has the value of multiple of ten. Similarly, a digit in hundreds place has the value of multiple of hundred.
For example, In the number 23, 2 digit holds the value of multiple of ten. That means, 2 X 10.
To continue the parity of the sum, we follow the place value rule by subtracting digits as per the place values.
Think and Grow: Subtract Multi-Digit Numbers
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 3
Show and Grow
Find the difference. Then check your answer.

Question 1.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 4

Answer: 2073

Check: Use addition to check your answer. 

Question 2.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 5

Answer: 9,821

Check: Use addition to check your answer.

Question 3.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 6

Answer: 452,107

Check: Use addition to check the solution.

Question 4.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 7

Answer: 183099

Check: Use addition to check the solution.
Apply and Grow: Practice
Find the difference. Then check your answer.

Question 5.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 8

Answer: 90,960

Check: Use addition to check the solution.

Question 6.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 9

Answer: 33,479

Check: Use addition to check the solution.

Question 7.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 10

Answer: 53,122

Check: Use addition to check the solution.

Question 8.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 11

Answer: 25,528.

Check: Use addition to check the solution.

Question 9.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 12

Answer: 338,681

Check: Use addition to check the solution.

Question 10.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 13

Answer: 293,189

Check: Use addition to check the solution.

Question 11.
281,660 – 44,521 = ______

Answer: 237,139

Check: Use addition to check the solution.

Question 12.
798,400 – 5,603 = ______

Answer: 792,379.

Check: Use addition to check the solution.

Question 13.
103,219 people attended a championship football game last year. 71,088 people attend the game this year. How many more people attended the game last year than this year?
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 14

Answer: 32,131
People attended last year = 103,219
People attended this year = 71,088
People attended more last year than this year = People attended last year – People attended this year
People attended more last year than this year = 103,219 – 71,088

Therefore, People attended more last year than this year = 32,131.

Question 14.
Number Sense
Find and explain the error. What is the correct difference?
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 15

Answer: The error in the solution is the regrouping is not applied correctly. The correct regrouping process is as follows:

Therefore, the correct difference = 363,505

Question 15.
Number Sense
Which statements describe the difference of 32,064 and 14,950?
The difference is about 17,000.
The difference is less than 17,000.
The difference is greater than 17,000.
The difference is 17,000.

Answer: The first and third statements describe the difference.
Explanation: Find the difference of 32,064 and 14,950 by rounding them to the nearest hundreds values and then to the nearest thousands values.
               
When you round the values to the nearest hundreds values, the statement 3 is satisfied. While rounding the values to the nearest thousands values, statement 1 is satisfied.
Think and Grow: Modeling Real Life
Example
The shoreline of Lake Michigan is 1,090 miles shorter than the shoreline of Lake Superior. How long is the shoreline of Lake Michigan?
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 16
Subtract 1,090 from the length of the shoreline of Lake Superior.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 17
The shoreline of Lake Michigan is 1640 miles long.
Show and Grow

Question 16.
The new car is $15,760 less than the new truck. How much does the new car cost?
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 18

Answer: $ 16,755
Explanation:
Cost of truck = $ 32,535.
Cost of car = Cost of truck – $ 15,760
Therefore Cost of car = $ 32,535 – $ 15,760

Therefore, Cost of car = $ 16,775.

Question 17.
Last year, an amusement park had 770,495 more guests than a water park. The attendance at the amusement park was 875,562 guests. What was the attendance at the water park?

Answer: 105,067
Number of guests in at Amusement park = 875,562
Number of guests at water park = number of guests at amusement park – 770,495 guests.
Number of guests at water park = 875,562 – 770,495

Therefore, the number of guests at water park = 105,067.

Question 18.
How many fewer miles did the pilot fly in Years 1 and 2 combined than in Year 3?
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 19

Answer: 5,406
Number of miles the pilot fly in years 1 and 2 combined = number of miles fly in year 1 + number of miles fly in year 2.
So, total number of miles fly in years 1 and 2 = 5,396 + 10,821 = 16,217.
Fewer miles pilot flew in years 1 and 2 combined than year 3 = Number of miles flew in year 3  Number of miles pilot flew in years 1 & 2 combined.

Therefore, the pilot flew a combined of 5,406 miles fewer in years 1 and 2 compared to year 3.

Subtract Multi-Digit Numbers Homework & Practice 2.3

Find the difference. Then check your answer.

Question 1.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 21

Answer: 2,374
Explanation:

Question 2.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 22

Answer: 25259.
Explanation:

Question 3.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 23

Answer: 21,284
Explanation:

Question 4.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 24

Answer: 13,162
Explanation:

Question 5.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 25

Answer: 140,938
Explanation:

Question 6.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 26

Answer: 292,694
Explanation:

Find the difference. Then check your answer.

Question 7.
319,120 – 278,188 = _______

Answer: 40,932
Explanation:

Check: Add 278,188 to 40,932 to verify.

Question 8.
312,396 – 23,891 = ______

Answer: 288,505
Explanation:

Check: Add 23,891 to 288,505 to verify.

Question 9.
Writing
Write and solve a subtraction word problem using the numbers 34,508 and 8,529.

Answer:
The bike racing 8,529 more curators than the Formula 1 racing. The attendance at the bike racing park was 34,508 guests. What was the attendance at the Formula 1 racing?
Attendance at Bike racing = 34,508.
Attendance at Formula 1 racing = Attendance at Bike racing – 8,529.
So, Attendance at Formula 1 racing = 34,508 – 8,529

Therefore, attendance at Formula 1 racing = 25,979.
Use the time line to answer the questions.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 27

Question 10.
Modeling Real Life
How many years after the first cell phone was invented was the smartphone invented?

Answer: 29 years.
Year of first cell phone invented = 1973.
Year of first smart phone invented = 2002.
Number of years after smart phone invented = 2002 – 1973

Therefore, the years after smart phone invented from first cell phone is 29 years.

Question 11.
Modeling Real Life
How many years passed from the invention of the first telephone to the release of the iPhone?
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 28

Answer: 131 years.
Explanation:
Year when first telephone invented = 1876.
Year of the launch of first iPhone = 2007.
Years passed from invention of first telephone to the invention of first iPhone = Launch year of iPhone – invention year of first telephone.
So, years passed = 2007 – 1876

Therefore, years passed after the invention of first telephone to the launch of first iPhone are 131 years.
Review & Refresh
Find the area of the shape.

Question 12.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 29

Answer: 15 square meter.
Explanation:
Given, one block = 1 square meter.
Total number of blocks highlighted = 15.
Therefore, total area = 15 square meters.

Question 13.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.3 30

Answer: 11 square inch
Explanation:
Given, one block = 1 square inch.
Total number of blocks highlighted = 11.
Therefore, total area = 11 square inch.

Lesson 2.4 Use Strategies to Add and Subtract

Explore and Grow
Choose any strategy to find 8,005 + 1,350.
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 1
For the given numbers, 8,005 and 1,350 the normal addition can be done using grouping using the place value rule. There is no need for the regrouping as well. However, we here will use the partial sums strategy to find the sum.
8000 + 005 = 8005
1000 + 350 = 1350
9000 + 355 = 9355
Choose any strategy to find 54,000 – 10,996.

Answer: 64996
For the given numbers, 54,000 and 10,996 we have to apply compensation strategy.
Add 4 to 10,996 to round it to the nearest thousands value.
10,996 + 4 = 11,000.
54,000 – 11,000 = 43,000.
Now subtract 4 from the result.
43,4000 – 4 = 42,996.
Therefore,54,000 – 10,996 = 42,996.
Reasoning
Explain why you chose your strategies. Compare your strategies to your partner’s strategies. How are they the same or different?

Answer:
Think and Grow: Use Strategies to Add or Subtract
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 2
Show and Grow
Find the sum or difference. Then check your answer.

Question 1.
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 3

Answer: 22,719.
We will use partial sums to add.

Therefore, 10,500 + 12,219 = 22,7199.

Question 2.
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 4

Answer:
We will use compensation strategy to find the difference.
7,008 – 8 = 7,000.
9,318 – 7,000 = 2,318.
2,318 – 8 = 2,310.
Therefore, 9,318 – 7,008 = 2,310.
Apply and Grow: Practice
Find the sum or difference. Then check your answer.

Question 3.
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 5

Answer:  16,457.
We here use regrouping strategy to solve this.

Question 4.
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 6

Answer: 36,081
Subtract the given numbers using regrouping model.

Therefore, the difference of 44,561 and 8480is 36,081

Question 5.
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 7

Answer: 11,063
Use compensation strategy to solve this:
Add 40 to 91,803.
91,803 + 40 = 91,843
91,843 – 80,740 = 11,103.
Now subtract 60 from the result.
11,103 – 40 = 11,063.

Question 6.
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 8

Answer: 100,973
Use regrouping to add the numbers.

Question 7.
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 9

Answer: 281,712
Use partial sum method to calculate this.

Question 8.
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 10

Answer: 822,594
Use regrouping method to find the difference:

Question 9.
780,649 – 13,754 = ______

Answer: 766,895
Use regrouping method:

Question 10.
417,890 + 90,284 = ______

Answer:  508,172
Use regrouping method to sum.

Question 11.
614,008 + 283,192 = ______

Answer: 897,200
Use regrouping method to find the sum.

Question 12.
801,640 – 206,427 = ______

Answer: 595,213
Use regrouping:

Question 13.
Structure
Write an equation shown by the number line.
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 11

Answer:
23,977 + 1000 = 24,977 + 20 = 24,997 + 3 = 25,000.
Think and Grow: Modeling Real Life
Example
Inner Core Earth’s mantle is 1,802 miles thick. Earth’s outer core is 1,367 miles thick. How many miles thinner is Earth’s outer core than its mantle?
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 12
Subtract the thickness of the outer core from the thickness of the mantle. Use compensation to subtract.
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 13
Earth’s outer core is 435 miles thinner than its mantle.
Show and Grow

Question 14.
Your friend’s heart beats 144,000 times in one day. Your heart beats 115,200 times in one day. How many fewer times does your heart beat than your friend’s?
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 14

Answer: 28,800
Number of times heart beats of my friend = 144,000
Number of times heart beats of me = 115,200
The fewer times my heart beats than my friend = 144,000 – 115,200

Question 15.
You, your friend, and your cousin are playing a video game. You score 2,118 more points than your friend. Your friend scores 1,503 fewer points than your cousin. What is each player’s score? Who wins?PlayerScoreYou?Friend6,010Cousin?
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 15

Answer:
Given, friend’s score = 6,010.
My score = My friend’s score + 2,118 = 6,010 + 2,118 = 8,128.
My cousin’s score = My friend’s score – 1,503 = 6,010 – 1,503 = 4,507.

Question 16.
Students at a school want tore cycle a total of 50,000 cans and bottles. So far, the students recycled 40,118 cans and 9,863 bottles. Did the students reach their goal? If not, how many more cans and bottles need recycled?
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 16

Answer: No, the students fell short by 19.
Number of cans recycled = 40,118.
Number of bottles recycled = 9,863
Total recycled = 40,118 + 9,863 = 49,981.
Students planned number for recycling = 50,000.
Shortage = 50,000 – 49,981 = 19.
The students fell short by 19 cans or bottles to recycle.

Use Strategies to Add and Subtract Homework & Practice 2.4

Find the sum or difference. Then check your answer.

Question 1.
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 17

Answer: 13,754
Use regrouping method to add the numbers.

Question 2.
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 18

Answer: 3,030
Use compensation method to subtract.
Subtract 30 from the first number to make it’s hundreds tens and units digits same those of the second number..
7,683 – 30 = 7,653.
Now subtract the second number from the result.
7,653 – 4,683 = 3,000
Now add 30 which was subtracted in the beginning to the result again to compensate.
3,000 + 20 = 3,030.

Question 3.
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 19

Answer: 25,801
Use compensation method to add the numbers.
We, here can see both the numbers are near to their nearest hundreds. The first number is 12 greater than its nearest hundreds value and the second is 11 lesser than its nearest hundreds value.
Subtract 11 from the first number and add 11 to the second number. Adding and subtracting compensates the net value.
18,212 – 11 = 18,201.
7,589 + 11 = 7,600.
Now add both the results.
18,201 + 7,600 = 25,801.

Question 4.
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 20

Answer: 7,625
Use compensation method to do the subtraction.
Subtract 25 from the first number to make the tens and units digits same those of the second number.
9,800 – 25 = 9,775.
Now subtract the second number from the result.
9,775 – 2,175 = 7,600.
Now add 25 to the result which was subtracted initially to compensate.
7,600 + 25 = 7,625.

Question 5.
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 21

Answer: 8,515
Use regrouping to subtract.

Question 6.
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 22

Answer: 161,935
Use regrouping to sum.

Question 7.
814,327 – 32,475 = ______

Answer: 781,852.
Use regrouping method to subtract.

Question 8.
294,801 + 46,030 = ______

Answer:
Use regrouping to sum.

Question 9.
512,006 + 318,071 = ______

Answer: 830,085.
Use compensation method.
Subtract 6 from the first number.
512,006 – 6 = 512,000.
Add the second number to the result.
512,000 + 318,071 = 830,071.
Now add 6  again to the result to compensate.
830,071 + 6 = 830,085.

Question 10.
746,620 – 529,706 = _______

Answer: 216,914
Use grouping to subtract.

Question 11.
YOU BE THE TEACHER
Your friend uses compensation to add. Is your friend correct? Explain.
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 23

Answer: No, the addition is incorrect.
Explanation:
In step 1, 17 was subtracted to round it to the nearest hundreds. To compensate the subtraction, 17 should be added to the result in the final step. But, 17 was subtracted instead.
Correct answer = 75,220 + 17 = 75,237.

Question 12.
Writing
Which strategy would you use to subtract 9,618 from 58,007? Explain.

Answer: 48,389
We here use regrouping strategy.

Question 13.
Modeling Real Life
There are about 500,000 detectable earthquakes each year. About 99,500 of the detectable earthquakes are felt. How many earthquakes are detectable, but are not felt each year?
Big Ideas Math Solutions Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.4 24

Answer:
Given, number of detectable earthquakes = 500,000.
Number of detectable earthquakes that are felt = 99,500.
Number of earthquakes that cannot be felt = 500,000 – 99,500 = 405,000.

Question 14.
DIG DEEPER!
A committee wants to purchase a playground for $52,499. They put a donation of $8,025 towards the purchase of the playground. Then they make 2 payments of $4,275. How much money does the committee have left to pay for the playground?

Answer:
Cost of playground = $52,499.
Donation towards purchase = $8,025.
Next payments made = $4,275.
Amount left to pay = Total cost pf playground – donation – payment.
So, amount left = $52,499 – $8,025 – $4,275 = ($52,499 – $8,025) – $4,275
= $44,474 – $4,275 = $40,199.
So, the money left to pay by the committee to purchase the playground is $40,199
Review & Refresh
Write an equation to solve. Use a letter to represent the unknown number. Check whether your answer is reasonable.

Question 15.
There are 4 boxes. Each box has 6 granola bars. Your soccer team eats 18 granola bars. How many granola bars are left?

Answer: 6.
Let the total number of granola bars = X.
Given X = 4 Boxes of 6 granola bars each.
So, X = 6 * 4 = 24.
Number of Granola bars ate by Soccer team = 18.
Number of granola bars left = total granola bars – granola bars ate = 24 – 18 = 6.
Therefore, number of granola bars left = 6.

Question 16.
Newton has 35 blocks. Descartes has 28 blocks. Newton divides his blocks into7 equal groups and gives Descartes one group. How many blocks does Descartes have now?

Answer: 33.
Total blocks Newton has = 35.
Dividing 35 into 7 equal groups = 35/7.
So, number of blocks in each group = 35/7 = 5.
Number of blocks with Descartes = 28.
Number of blocks with Descartes after Newton giving one group of blocks = 28 + 5 = 33.

Lesson 2.5 Problem Solving: Addition and Subtraction

Explore and Grow
Use addition or subtraction to make a conclusion about the table.
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.5 1

Answer:
We use addition and also subtraction.
Let’s solve this problem. How much is the combined area of Arizona and California is bigger than the combined area of Nevada and Utah.
Combined area (1) of Arizona and California = 113,594 + 155,779 = 269,373.
Combined area (2) of Nevada and Utah = 109,781 + 82,170 = 191,951.
Difference in combined areas = Combined area 1 – Combined area 2 = 269,373 – 191,951.

Therefore, the combined area of Arizona and California is 77,422 square miles bigger than the combined area of Nevada and Utah.
Think and Grow: Problem Solving: Addition and Subtraction
Example
You have3,914 songs in your music library. You download 1,326 more songs. Then you delete 587 songs. How many songs do you have now?
Understand the Problem
What do you know?
• You have 3,914 songs.
• You download 1,326 more
• You delete 587 songs.
What do you need to find?
• You need to find how many songs you have now.
Make a Plan
How will you solve?
• Add 3,914 and 1,326 to find how many songs you have after downloading some songs.
• Then subtract 587 from the sum to find how many songs you have now.
Solve
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.5 2
k = 4,240
n = k – 587 = 5,240 – 587 = 3,653
You have 4,653 songs now.
Show and Grow

Question 1.
Explain how you can check whether your answer above is reasonable.

Answer:
You can check the answer by doing the subtraction first and the addition later.
First, subtract 587 songs that were deleted.
Therefore, 3,914 – 587 = 3,327.
Now add the downloaded songs, 1,326 to the result.
So, 3,327 + 1,326 = 4,653.
Apply and Grow: Practice
Understand the problem. What do you know? What do you need to find? Explain.

Question 2.
There are about 12,762 known ant species. There are about 10,997 known grasshopper species. The total number of known ant, grasshopper, and spider species is 67,437. How many known spider species are there?
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.5 3

Answer: 43,678.
Given, number of ant species: 12,762.
Number of grasshopper species: 10,997.
Total number of ant, grasshopper and spider species: 67,437.
So, Number of (ant species + grasshopper species + spider species) = 67,437.
That means, 12,762 + 10,997 + number of spider species = 67,437.
Now, initially add the number of ant and grasshopper species using compensation strategy.
Add 3 to the number of grasshopper species: 10,997 + 3 = 11,000.
Now add the result to the number of ant species: 12,762 + 11,000 = 23,762.
Now subtract 3 to compensate: 23,762 – 3 = 23,759.
So, Number of spider species = 67,437 – 23,759.
Do the subtraction using regrouping.

Therefore, the known number of spider species = 43,678.

Question 3.
A quarterback threw for 66,111 yards between 2001 and 2016. His all-time high was 5,476 yards in 1 year. In his second highest year, he threw for 5,208 yards. How many passing yards did he throw in the remaining years?

Answer: 55,427
Total number of yards = 66,111 yards.
His first and second all-time high were 5,476 yards and 5,208 yards respectively.
Total yards in first and second year all-time high = 5,476 + 5,208 = 10,684.
The number of yards he threw in the remaining years = 66,111 – 10684
Find the difference using regrouping.

Therefore, number of yards he threw in the remaining years = 55,427 yards
Understand the problem. Then make a plan. How will you solve? Explain.

Question 4.
There are 86,400 seconds in 1 day. On most days, a student spends 28,800 seconds sleeping and 28,500 seconds in school. How many seconds are students not awake, but in school?

Answer:
Given, Number of seconds student sleeps = 28,800.
Number of seconds student spends at school = 28,500.
The
Question asked was the number of seconds the student spend not awake (sleeping), but in school.
In the given data, it’s not mentioned that the student spends time sleeping at school separately. So, with the given information, we cannot calculate the number of seconds student was not awake, but at school.

Question 5.
A pair of rhinoceroses weigh 14,860 pounds together. The female weighs 7,206 pounds. How much more does the male weigh than the female?
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.5 4

Answer: 258 pounds.
Given,
Weight of female rhino =7,206 pounds.
Weight of the rhino pair = 14,860.
Weight of male rhino = 14,860 – 7,206
Solve this using compensation strategy.
Add 6 to the first number = 14,860 + 6 = 14,866.
Now subtract the second number from the result.
That means, 14,866 – 7,206 = 7,460.
Now, subtract 6 from the result to compensate = 7,460 – 6 = 7,454.
So, weight of male rhino = 7,464.
Weight of male rhino more than female rhino = male rhino weight – female rhino weight. = 7,464 – 7,206 = 258 pounds.

Question 6.
Earth is 24,873 miles around. If a person’s blood vessels were laid out in a line, they would be able to circle Earth two times, plus 10,254 more miles. How many miles long are a person’s blood vessels when laid out in a line?

Answer: 60,000 miles.
Given circumference of earth = 24,873 miles.
So, length of person’s blood vessels = 2 times the earth’s circumference + 10,254 miles.
That means, length of person’s blood vessels = (2 X 24,873) + 10,254 = 49,746 + 10,254.
Add the two numbers using grouping technique.

Therefore,, the length of human blood vessels = 60,000 miles.

Question 7.
Alaska has 22,041 more miles of shoreline than Florida and California combined. Alaska has 33,904 miles of shoreline. Florida has 8,436 miles of shoreline. How many miles of shoreline does California have?

Answer: 3,427 miles
Give, Shoreline of Alaska = 33,904.
Shoreline of Florida = 8,436 miles.
Shore lines of Florida + California = Shoreline of Alaska – 22,041 miles.
That means, 8,436 miles + Shoreline of California = 33,904 miles – 22,041 miles.
Subtract the right hand side using regrouping.

So, Shoreline of California + 8436 miles = 11,863 miles.
That means, Shoreline of California = 11,863 miles – 8436 miles

Therefore, shoreline of California = 3,427 miles.
Think and Grow: Modeling Real Life
Example
The attendance on the second day of a music festival is 10,013 fewer people than on the third day.How many total people attend the three-day music festival?
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.5 5
Think: What do you know? What do you need to find? How will you solve?3?
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.5 6
_______ total people attend the music festival.
Show and Grow

Question 8.
A construction company uses 3,239 more bricks to construct Building 1 than Building 2. How many bricks does the company use to construct all three buildings?
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.5 7

Answer:
Given,
Number of bricks used for Building 1 = 11,415.
Number of bricks used for Building 3 = 16,352.
Number of bricks used for building 2 = Number of bricks used for Building 1 – 3,239.
So, Number of bricks used for Building 2 = 11,415 – 3,239.
Use regrouping to subtract.

So, number of bricks used for constructing Building 2 = 8,176.
So, bricks used to construct all three buildings = Bricks used for (Building 1 + Building 2 + Building 3)
So, total bricks used in construction = 11,415 + 8,176 + 16,352.
Use grouping to add the three numbers.

Therefore, the number of bricks used to construct all the three buildings = 35,943.

Problem Solving: Addition and Subtraction Homework & Practice 2.5

Understand the problem. Then make a plan. How will you solve? Explain.

Question 1.
A cargo plane weighs 400,000 pounds. After a load of cargo is removed, the plane weighs 336,985 pounds. Then a 12,395-pound load is removed. How many pounds of cargo are removed in all?

Answer: 75,410
Given,
Weight of plane with cargo = 400,000 pounds.
Weight of plane after removing first load of cargo = 336,985.
So, Weight of first cargo load = 400,000 – 336,985.
Use Count on to strategy to find the difference.

So, the difference is 60,000 + 3,000 + 15 = 63,015.
So, the weight of the first cargo = 63,015.
Given, weight of the second cargo = 12,395.
Total weight of the cargo = 63,015 + 12,395

So, the total weight of the cargo unloaded = 75,410 pounds.

Question 2.
A ski resort uses 5,200 gallons of water per minute to make snow. A family uses 361 gallons of water each day. How many more gallons of water does the ski resort use to make snow in 2 minutes than a family uses in 1 day?

Answer: 10,039 Gallons
Given,
Water required to make snow for 1 minute = 5,200 gallons.
Water required for household usage per day = 361 gallons.
Water required for ski resort to prepare snow per two minutes = 2 X 5,200 = 10,400 gallons.
The more water required than the water consumed by the household one day usage = 10,400 – 361.
Subtract using compensation.
Add 39 to the second number.
361 + 39 = 400.
Now subtract the result from the second number.
10,400 – 400 = 10,000.
Now, add 39 to the result to compensate.
10,000 + 39 = 10,039.
So, the additional water used for ski resort to make snow for 2 minutes than the water used by household per one day is 10,039 gallons.

Question 3.
In July, a website receives 379,162 fewer orders than in May and June combined. The website receives 542,369 orders in May and 453,708 orders in June. How many orders does the website receive in July?

Answer: 542,369
Given,
Orders received in May = 542,369.
Orders received in June = 453,708.
Total orders in May and June = 542,369 + 453,708.

So, orders received in July = 996,077 – 379,162.
Subtract using regrouping.

So, the number of orders received in the month of July = 542,369.

Question 4.
Writing
Write and solve a two-step word problem that can be solved using addition or subtraction.

Answer:
A father wants to buy a laptop to his son worth $3,495. He makes 2 payments of $1,500. How much money does the he have left to pay for the laptop?
Cost of laptop = $3,495.
Number of payments of $1,500 done = 2.
That means $1,500 X 2 = 3,000
Money left to pay by the father for the laptop = Cost of laptop – Amount paid
= $3,495 – $3,000.
$3,495 – $3,000 = $495.
So, money left to buy laptop = $495

Question 5.
Modeling Real Life
World War I lasted from 1914 to1918. World War II lasted from 1939 to 1945. How much longer did World War II last than World War I?

Answer: 2 years
Duration of World war 1 = 1918 – 1914 = 4 years.
Duration of World war 2 = 1945 – 1939 = 6 years.
The longer did World war 2 last = Duration of World war 2 – Duration of World war 1 = 6 years – 4 years
Therefore, World war 2 lasted 2 years longer than World war 1.

Question 6.
Modeling Real Life
Twenty people each donate $9 to a charity. Sixty people each donate $8. The charity organizer wants to raise a total of $1,500. How much more money does the organizer need to raise?

Answer:
Twenty people each donate $9 to a charity = 20 X $9 = $180
Sixty people each donate $8 = 60 X $8 = $480.
Total fund collected = $180 + $480 = $660
The charity organizer wants to raise a total of $1,500.
So, money left by the charity organizer to collect = $1500 – $660 = $740.

Question 7.
DIG DEEPER!
The blackpoll warbler migrates 2,376 miles, stops, and then flies another 3,289 miles to reach its destination. The arctic tern migrates 11,013 miles, stops, and then flies another 10,997 miles to reach its destination. How much farther is the arctic tern’s migration than the blackpoll warbler’s migration?
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.5 8

Answer: 16,345 miles
Total migration of Blackpoll Warbler = 2,376 + 3,289
Total migration of arctic tern = 11,013 + 10,997
Total migration of Blackpoll:      Total migration of arctic tern: 
Farther the arctic tern’s migration than the blackpoll warbler’s migration = 22,010 – 5,665.
Use grouping to subtract.

So, farther the arctic tern’s migration than the blackpoll warbler’s migration is 16,345 miles.
Review & Refresh
Write the time. Write another way to say the time.

Question 8.
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.5 9

Answer: 5 hours 56 minutes (05:56)
Other way: 4 minutes to 6’O Clock.

Question 9.
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.5 10

Answer: 3 hours 43 minutes (03:43)
Other way: 17 minutes to 4’O Clock.

Question 10.
Big Ideas Math Answer Key Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers 2.5 11

Answer: 12 hours 37 minutes (12:37)

Add and Subtract Multi-Digit Numbers Performance Task

Question 1.
The time line shows the population of Austin, Texas from 1995 to 2015.
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 1
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 3
a. Which place would you round to when estimating population? Explain.
b. Estimate Austin’s population each year on the time line.
c. Use your estimates to complete the bar graph.
d. Between the years 1995 and 2015, did Austin’s population increase, decrease, or stay the same? Explain.
e. During which five-year period did the population increase the most? Explain.
f. About how many more people lived in Austin in the year 2015 than in the year 1995?
g. Do you think the population will be more than 1,000,000 in the year 2020? Explain.

Answer:
a) Round the values to the nearest ten thousands because the values are in lakhs.
b)
c)

d) Between the years, 1995 and 2015, Austin’s population was increased. The population was constantly increasing as per the given table and the chart.
e) During the five year period between 1995 and 2000, the population increased more. The graph was more steeper than the other time lines. That indicates more population growth during that period.
f) Population in 2015 = 930,052
Population in 1995 = 555,092.
Round the values to the nearest ten thousands values.

So, the population increase between 1995 and 2015 was estimated to be 370,000.
g) Yes, the population grown more than 100,000 in 5 years time except between 2000 and 2005. So, if that is the case, the population will be more than 1,000,000 by 2020.

Add and Subtract Multi-Digit Numbers Activity

Race to the Moon
Directions:
1. Players take turns.
2. On your turn, flip a Race for the Moon Card and find the sum or difference.
3. Move your piece to the next number on the board that is highlighted in your answer.
Big Ideas Math Answers 4th Grade Chapter 2 Add and Subtract Multi-Digit Numbers 4

Add and Subtract Multi-Digit Numbers Chapter Practice

2.1 Estimate Sum and Differences
Estimate the sum or difference.

Question 1.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers chp 1

Answer: 6,600
Round the numbers to the nearest hundreds values.

Question 2.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers chp 2

Answer: 72,000
Round the numbers to their nearest thousands

Question 3.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers chp 3

Answer: 34,000.
Round the numbers to the nearest thousand values.

Question 4.
742,086 – 486,629 = ______

Answer: 250,000
Round the numbers to the nearest ten thousands value.

Question 5.
216,987 + 72,429 = ______

Answer: 289,000

2.2 Add Multi-Digit Numbers
Find the sum. Check whether your answer is reasonable.

Question 6.
Estimate: ______
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers chp 6

Answer: 53,300
Round the numbers to their nearest hundreds value:

Check: Use grouping to find the exact Sum:

As, 53,276 is nearest to the estimate value 53,300, the estimate is correct.

Question 7.
Estimate: ______
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers chp 7

Answer: 80,000
Round the numbers to the nearest thousands value:

Check: Use grouping to find the exact Sum:

As, 79,960 is nearest to the estimate value 80,000, the estimate is correct.

Question 8.
Estimate: ______
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers chp 8

Answer: 610,000
Round the numbers to the nearest ten thousands values.

Check: Use grouping to find the exact Sum:

As, 609,334 is nearest to the estimate value 610,000, the estimate is correct.

Question 9.
Logic
Find the missing digits
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers chp 9

Answer:

In this given problem, look at the units place. The result is 1 and as the units digit of the first number is 6, then the should be nothing else but 11. So, 11 – 6 = 5. So, the missing units digit of the second number is 5. The tens position also gets satisfied with the result being 11 as the regrouping gives the tens digit of the result 11 again.
If that’s the case, the regrouping again takes another addition of 1 to the hundreds place. It gives the result as follows:
1 + __ + 9 = Units digit of 5.
1+9 = 10. So, the 1 has to go for the thousands place as per regrouping. So, 0 + __ = 5.
So, the hundreds place of the first number is 5.

In this given problem, look at the tens place. The result is 2 and as the tens digit of the second number is 8, There is also a carrying of 1 from the units place to add in the sum.
So, 1 + __ + 8 = 12 (before regrouping of tens place). So, the missing units digit of the first number is 3. The hundreds position also gets satisfied with the result being 12 as the regrouping gives the hundreds digit of the result 15.
The regrouping again takes another addition of 1 to the ten thousands place. It gives the result as follows:
1 + 6 + __ = Units digit of 5. It clearly shows that there will be another regrouping.
7 + __ = 15.
7 + 8 = 15.
So, the ten thousands place of the second number is 8.
2.3 Subtract Multi-Digit Numbers
Find the difference. Then check your answer.

Question 10.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers chp 10

Answer: 4757.
Use grouping to subtract.

Question 11.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers chp 11

Answer: 2,289.
Use compensation method.
Add 11 to the first number.
32,700 + 11 = 32,711
Now subtract the second number from result.
So, 32,711 – 30,411 = 2,300.
Now subtract 11 from the result to compensate.
2,300 – 11 = 2,289.

Question 12.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers chp 12

Answer: 75,575
Use grouping to subtract.

Question 13.
973,287 – 8,345 = ______

Answer: 964,942
Use grouping to subtract.

Question 14.
762,179 – 21,280 = ______

Answer: 740,899
Use grouping to subtract.

2.4 Use Strategies to Add or Subtract
Find the sum or difference. Then check your answer.

Question 15.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers chp 15

Answer: 7,205
Use compensation method.
Subtract 5 from the first number.
13,500 – 5 = 13,495.
Now subtract second number from the result.
13,495 – 6,295 = 7,200.
Now add 5 to the result to compensate.
7,200 + 5 = 7,205.

Question 16.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers chp 16

Answer: 825,528.
Use regrouping.

Check: Add the result to the second number to check.

Question 17.
Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers chp 17

Answer: 911,527.
Use regrouping to solve this.

2.5 Problem Solving: Addition and Subtraction

Question 18.
Modeling Real Life
A library has 12,850 books. There are 1,932 poetry books, 5,047 nonfiction books, and 2,891 graphic novels. The rest are fiction. How many fiction books does the library have?

Answer:
Number of books in the library = 12,850.
Given, Number of (poetry books + notification books + graphic novels + fiction books) = Total books.
So, 1,932 + 5,047 + 2,891 + fiction books = 12,850.

So, 9,870 + Fiction books = 12,850.
So, Fiction books = 12,850 – 9,870.
Use compensation to do this subtraction.
Add 20 to the first number.
12,850 + 20 = 12,870.
12,870 – 9,870 = 3,000.
Now subtract 20 from the result to compensate.
3,000 – 20 = 2,980.
Therefore, the number of fiction books in the library are 2,980.

Question 19.
Modeling Real Life
A technology teacher wants to buy a 3-D printer for $3,495. He makes 3 payments of $999. How much money does the teacher have left to pay for the 3-D printer?

Answer: $498.
Cost of 3-D printer = $3,495.
Number of payments of $999 done = 3.
That means $999 + $999 + $999.
Use compensation technique to add the three numbers.
Add $1 to each payment of $999.
$999 + $1 = $1,000.
So, the addition has to be done for $1,000 + $1,000 + $1,000 = $3,000.
Now, subtract $3 from the result to compensate.
$3,000 – $3 = $2,997.
Money left to pay by the technology teacher for the 3-D printer = Cost of 3-D printer – Amount paid
= $3,495 – $2,997.
Use compensation method to do the subtraction. Add $3 to the second number.
$2,997 + $3 = $3,000.
Now subtract the result from the first number.
$3,495 – $3,000 = $495.
Add $3 to compensate.
$495 + $3 = $498.
So, the money left to pay by the technology teacher for the 3-D printer = $498.

Conclusion:

Come and fall in love with Maths by utilizing the Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers. Make use of the Big Ideas Math Answers Grade 4 Chapter 2 Add and Subtract Multi-Digit Numbers as a reference for all your queries. Keep in touch with our site to avail updates on Class Specific Big Ideas Math Book 4th Grade Answer Key Chapter 2 Add and Subtract Multi-Digit Numbers at your fingertips.

Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series

Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series

Students who feel difficult to understand the concepts of sequence and series can utilize this Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series. This BIM Algebra 2 Ch 8 Answer Key was designed by subject experts to assist high school students during their exam preparation. After solving all the questions given in the below pdf links of Big Ideas math book algebra 2 ch 8 Sequences and Seriessolution key, you will gain more subject knowledge and become pro in math calculations.

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

Learn the fundamental concepts of sequence and series from our Big Ideas Math Book Algebra 2 Solution Key Chapter 8 Sequences and Series and understand the topics covered in this chapter quickly. Furthermore, BIM Math Answers Algebra 2 Chapter 8 Pdf is very helpful for high school strugglers who are practicing the concepts discussed in it. Every question in the Big Ideas math textbook solution key of algebra 2 ch 8 sequence and series are solved in a detailed & simple manner for easy understanding to students. Without any issues, you can learn the topics of sequence and series and solve all basic level to complex level questions asked in the exams.

Sequences and Series Maintaining Mathematical Proficiency

Copy and complete the table to evaluate the function.
Question 1.
y = 3 − 2x
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 1.1
Answer:

Question 2.
y = 5x2 + 1
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 1.2
Answer:

Question 3.
y = −4x + 24
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 1.3
Answer:

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

Question 5.
\(\frac{1}{16}\) = 4 (\(\frac{1}{2}\)x
Answer:

Question 6.
216 = 3(x + 6)
Answer:

Question 7.
2x + 16 = 144
ans;

Question 8.
\(\frac{1}{4}\)x − 8 = 17
Answer:

Question 9.
8(\(\frac{3}{4}\))x = \(\frac{27}{8}\)
Answer:

Question 10.
ABSTRACT REASONING
The graph of the exponential decay function f(x) = bx has an asymptote y = 0. How is the graph of f different from a scatter plot consisting of the points (1, b1), (2, b21 + b2), (3, b1 + b2 + b3), . . .? How is the graph of f similar?
Answer:

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:

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.
a1, a2, a3, a4, . . . , an, . . .Terms of a sequence
Here is an example. 1, 4, 7, 10, . . . , 3n-2, . . .

EXPLORATION 1

Writing Rules for Sequences
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 1
Work with a partner. Match each sequence with its graph. The horizontal axes represent n, the position of each term in the sequence. Then write a rule for the nth term of the sequence, and use the rule to find a10.
a. 1, 2.5, 4, 5.5, 7, . . .
b. 8, 6.5, 5, 3.5, 2, . . .
c. \(\frac{1}{4}, \frac{4}{4}, \frac{9}{4}, \frac{16}{4}, \frac{25}{4}, \ldots\)
d. \(\frac{25}{4}, \frac{16}{4}, \frac{9}{4}, \frac{4}{4}, \frac{1}{4}, \ldots\)
e. \(\frac{1}{2}\), 1, 2, 4, 8, . . .
f. 8, 4, 2, 1, \(\frac{1}{2}\), . . .
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 2

Communicate Your Answer

Question 2.
How can you write a rule for the nth term of a sequence?
Answer:

Question 3.
What do you notice about the relationship between the terms in (a) an arithmetic sequence and (b) a geometric sequence? Justify your answers.
Answer:

Monitoring Progress

Write the first six terms of the sequence.
Question 1.
an = n + 4
Answer:

Question 2.
f(n) = (−2)n-1
Answer:

Question 3.
an = \(\frac{n}{n+1}\)
Answer:

Describe the pattern, write the next term, graph the first five terms, and write a rule for the nth term of the sequence.
Question 4.
3, 5, 7, 9, . . .
Answer:

Question 5.
3, 8, 15, 24, . . .
Answer:

Question 6.
1, −2, 4, −8, . . .
Answer:

Question 7.
2, 5, 10, 17, . . .
Answer:

Question 8.
WHAT IF?
In Example 3, suppose there are nine layers of apples. How many apples are in the ninth layer?
Answer:

Write the series using summation notation.
Question 9.
5 + 10 + 15 +. . .+ 100
Answer:

Question 10.
\(\frac{1}{2}+\frac{4}{5}+\frac{9}{10}+\frac{16}{17}+\cdots\)
Answer:

Question 11.
6 + 36 + 216 + 1296 + . . .
Answer:

Question 12.
5 + 6 + 7 +. . .+ 12
Answer:

Find the sum.
Question 13.
\(\sum_{i=1}^{5}\) 8i
Answer:

Question 14.
\(\sum_{k=3}^{7}\)(k2 − 1)
Answer:

Question 15.
\(\sum_{i=1}^{34}\)1
Answer:

Question 16.
\(\sum_{k=1}^{6}\)k
Answer:

Question 17.
WHAT IF?
Suppose there are nine layers in the apple stack in Example 3. How many apples are in the stack?
Answer:

Defining and Using Sequences and Series 8.1 Exercises

Vocabulary and Core Concept Check
Question 1.
VOCABULARY
What is another name for summation notation?
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 a 1

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

Question 3.
WRITING
Compare sequences and series.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 a 3

Question 4.
WHICH ONE DOESN’T BELONG?
Which does not belong with the other three? Explain your reasoning.
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 3
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 5–14, write the first six terms of the sequence.
Question 5.
an = n + 2
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 a 5

Question 6.
an = 6 − n
Answer:

Question 7.
an = n2
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 a 7

Question 8.
f(n) = n3 + 2
Answer:

Question 9.
f(n) = 4n-1
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 a 9

Question 10.
an = −n2
Answer:

Question 11.
an = n2 − 5
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 a 11

Question 12.
an = (n + 3)2
Answer:

Question 13.
f(n) = \(\frac{2n}{n+2}\)
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 a 13

Question 14.
f(n) = \(\frac{n}{2n-1}\)
Answer:

In Exercises 15–26, describe the pattern, write the next term, and write a rule for the nth term of the sequence.
Question 15.
1, 6, 11, 16, . . .
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 a 15

Question 16.
1, 2, 4, 8, . . .
Answer:

Question 17.
3.1, 3.8, 4.5, 5.2, . . .
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 a 17

Question 18.
9, 16.8, 24.6, 32.4, . . .
Answer:

Question 19.
5.8, 4.2, 2.6, 1, −0.6 . . .
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 a 19

Question 20.
−4, 8, −12, 16, . . .
Answer:

Question 21.
\(\frac{1}{4}, \frac{2}{4}, \frac{3}{4}, \frac{4}{4}, \ldots\)
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 a 21

Question 22.
\(\frac{1}{10}, \frac{3}{20}, \frac{5}{30}, \frac{7}{40}, \ldots\)
Answer:

Question 23.
\(\frac{2}{3}, \frac{2}{6}, \frac{2}{9}, \frac{2}{12}, \ldots\)
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 a 23

Question 24.
\(\frac{2}{3}, \frac{4}{4}, \frac{6}{5}, \frac{8}{6}, \ldots\)
Answer:

Question 25.
2, 9, 28, 65, . . .
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 a 25

Question 26.
1.2, 4.2, 9.2, 16.2, . . .
Answer:

Question 27.
FINDING A PATTERN
Which rule gives the total number of squares in the nth figure of the pattern shown? Justify your answer.
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 4
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 a 27

Question 28.
FINDING A PATTERN
Which rule gives the total number of green squares in the nth figure of the pattern shown? Justify your answer.
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 5
Answer:

Question 29.
MODELING WITH MATHEMATICS
Rectangular tables are placed together along their short edges, as shown in the diagram. Write a rule for the number of people that can be seated around n tables arranged in this manner. Then graph the sequence.
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 6
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 a 29

Question 30.
MODELING WITH MATHEMATICS
An employee at a construction company earns $33,000 for the first year of employment. Employees at the company receive raises of $2400 each year. Write a rule for the salary of the employee each year. Then graph the sequence.
Answer:

In Exercises 31–38, write the series using summation notation.
Question 31.
7 + 10 + 13 + 16 + 19
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 a 31

Question 32.
5 + 11 + 17 + 23 + 29
Answer:

Question 33.
4 + 7 + 12 + 19 + . . .
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 a 33

Question 34.
-1 + 2 + 7 + 14 + …..
Answer:

Question 35.
\(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\cdots\)
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 a 35

Question 36.
\(\frac{1}{4}+\frac{2}{5}+\frac{3}{6}+\frac{4}{7}+\cdots\)
Answer:

Question 37.
−3 + 4 − 5 + 6 − 7
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 a 37

Question 38.
−2 + 4 − 8 + 16 − 32
Answer:

In Exercises 39–50, find the sum.
Question 39.
\(\sum_{i=1}^{6}\)2i
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 a 39

Question 40.
\(\sum_{i=1}^{5}\)7i
Answer:

Question 41.
\(\sum_{n=0}^{4}\)n3
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 a 41

Question 42.
\(\sum_{k=1}^{4}\)3k2
Answer:

Question 43.
\(\sum_{k=3}^{6}\)(5k − 2)
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 a 43

Question 44.
\(\sum_{n=1}^{5}\)(n2 − 1)
Answer:

Question 45.
\(\sum_{i=2}^{8} \frac{2}{i}\)
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 a 45

Question 46.
\(\sum_{k=4}^{6} \frac{k}{k+1}\)
Answer:

Question 47.
\(\sum_{i=1}^{35}\)1
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 a 47

Question 48.
\(\sum_{n=1}^{16}\)n
Answer:

Question 49.
\(\sum_{i=10}^{25}\)i
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 a 49

Question 50.
\(\sum_{n=1}^{18}\)n2
Answer:

ERROR ANALYSIS In Exercises 51 and 52, describe and correct the error in finding the sum of the series.
Question 51.
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 7
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 a 51

Question 52.
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 8
Answer:

Question 53.
PROBLEM SOLVING
You want to save $500 for a school trip. You begin by saving a penny on the first day. You save an additional penny each day after that. For example, you will save two pennies on the second day, three pennies on the third day, and so on.
a. How much money will you have saved after 100 days?
b. Use a series to determine how many days it takes you to save $500.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 a 53

Question 54.
MODELING WITH MATHEMATICS
You begin an exercise program. The first week you do 25 push-ups. Each week you do 10 more push-ups than the previous week. How many push-ups will you do in the ninth week? Justify your answer.
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 9
Answer:

Question 55.
MODELING WITH MATHEMATICS
For a display at a sports store, you are stacking soccer balls in a pyramid whose base is an equilateral triangle with five layers. Write a rule for the number of soccer balls in each layer. Then graph the sequence.
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 10
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 a 55

Question 56.
HOW DO YOU SEE IT?
Use the diagram to determine the sum of the series. Explain your reasoning.
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 11
Answer:

Question 57.
MAKING AN ARGUMENT
You use a calculator to evaluate \(\sum_{i=3}^{1659}\)i because the lower limit of summation is 3, not 1. Your friend claims there is a way to use the formula for the sum of the first n positive integers. Is your friend correct? Explain.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 a 57

Question 58.
MATHEMATICAL CONNECTIONS
A regular polygon has equal angle measures and equal side lengths. For a regular n-sided polygon (n ≥ 3), the measure an of an interior angle is given by an = \(\frac{180(n-2)}{n}\)
a. Write the first five terms of the sequence.
b. Write a rule for the sequence giving the sum Tn of the measures of the interior angles in each regular n-sided polygon.
c. Use your rule in part (b) to find the sum of the interior angle measures in the Guggenheim Museum skylight, which is a regular dodecagon.
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 12
Answer:

Question 59.
USING STRUCTURE
Determine whether each statement is true. If so, provide a proof. If not, provide a counterexample.
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 13
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 a 59.1
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 a 59.2

Question 60.
THOUGHT PROVOKING
In this section, you learned the following formulas.
\(\sum_{i=1}^{n}\)1 = n
\(\sum_{i=1}^{n}\)i = \(\frac{n(n+1)}{2}\)
\(\sum_{i=1}^{n}\)i2 = \(\frac{n(n+1)(2 n+1)}{6}\)
Write a formula for the sum of the cubes of the first n positive integers.
Answer:

Question 61.
MODELING WITH MATHEMATICS
In the puzzle called the Tower of Hanoi, the object is to use a series of moves to take the rings from one peg and stack them in order on another peg. A move consists of moving exactly one ring, and no ring may be placed on top of a smaller ring. The minimum number an of moves required to move n rings is 1 for 1 ring, 3 for 2 rings, 7 for 3 rings, 15 for 4 rings, and 31 for 5 rings.
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 14
a. Write a rule for the sequence.
b. What is the minimum number of moves required to move 6 rings? 7 rings? 8 rings?
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 a 61

Maintaining Mathematical Proficiency

Solve the system. Check your solution.
Question 62.
2x −y − 3z = 6
x + y + 4z =−1
3x − 2z = 8
Answer:

Question 63.
2x − 2y + z = 5
−2x + 3y + 2z = −1
x − 4y + 5z = 4
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 a 63.1
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series 8.1 a 63.2

Question 64.
2x − 3y + z = 4
x − 2z = 1
y + z = 2
Answer:

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 1

Recognizing Graphs of Arithmetic Sequences
Work with a partner. Determine whether each graph shows an arithmetic sequence. If it does, then write a rule for the nth term of the sequence, and use a spreadsheet to fond the sum of the first 20 terms. What do you notice about the graph of an arithmetic sequence?
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 1

EXPLORATION 2

Finding the Sum of an Arithmetic Sequence
Work with a partner. A teacher of German mathematician Carl Friedrich Gauss (1777–1855) asked him to find the sum of all the whole numbers from 1 through 100. To the astonishment of his teacher, Gauss came up with the answer after only a few moments. Here is what Gauss did:
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 2
Explain Gauss’s thought process. Then write a formula for the sum Sn of the first n terms of an arithmetic sequence. Verify your formula by finding the sums of the first 20 terms of the arithmetic sequences in Exploration 1. Compare your answers to those you obtained using a spreadsheet.

Communicate Your Answer

Question 3.
How can you recognize an arithmetic sequence from its graph?
Answer:

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:

Monitoring Progress

Tell whether the sequence is arithmetic. Explain your reasoning.
Question 1.
2, 5, 8, 11, 14, . . .
Answer:

Question 2.
15, 9, 3, −3, −9, . . .
Answer:

Question 3.
8, 4, 2, 1, \(\frac{1}{2}\), . . .
Answer:

Question 4.
Write a rule for the nth term of the sequence 7, 11, 15, 19, . . .. Then find a15.
Answer:

Write a rule for the nth term of the sequence. Then graph the first six terms of the sequence.
Question 5.
a11 = 50, d = 7
Answer:

Question 6.
a7 = 71, a16 = 26
Answer:

Find the sum.
Question 7.
\(\sum_{i=1}^{10}\)9i
Answer:

Question 8.
\(\sum_{k=1}^{12}\)(7k + 2)
Answer:

Question 9.
\(\sum_{n=1}^{20}\)(−4n + 6)
Answer:

Question 10.
WHAT IF?
In Example 6, how many cards do you need to make a house of cards with eight rows?
Answer:

Analyzing Arithmetic Sequences and Series 8.2 Exercises

Vocabulary and Core Concept Check
Question 1.
COMPLETE THE SENTENCE
The constant difference between consecutive terms of an arithmetic sequence is called the _______________.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 a 1

Question 2.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 3
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–10, tell whether the sequence is arithmetic. Explain your reasoning.
Question 3.
1, −1, −3, −5, −7, . . .
Answer:
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 a 3

Question 4.
12, 6, 0, −6, −12, . . .
Answer:

Question 5.
5, 8, 13, 20, 29, . . .
Answer:
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 a 5

Question 6.
3, 5, 9, 15, 23, . . .
Answer:

Question 7.
36, 18, 9, \(\frac{9}{2}\), \(\frac{9}{4}\), . . .
Answer:
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 a 7

Question 8.
81, 27, 9, 3, 1, . . .
Answer:

Question 9.
\(\frac{1}{2}, \frac{3}{4}, 1, \frac{5}{4}, \frac{3}{2}, \ldots\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 a 9

Question 10.
\(\frac{1}{6}, \frac{1}{2}, \frac{5}{6}, \frac{7}{6}, \frac{3}{2}, \ldots\)
Answer:

Question 11.
WRITING EQUATIONS
Write a rule for the arithmetic sequence with the given description.
a. The first term is −3 and each term is 6 less than the previous term.
b. The first term is 7 and each term is 5 more than the previous term.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 a 11

Question 12.
WRITING
Compare the terms of an arithmetic sequence when d > 0 to when d < 0.
Answer:

In Exercises 13–20, write a rule for the nth term of the sequence. Then find a20.
Question 13.
12, 20, 28, 36, . . .
Answer:
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 a 13

Question 14.
7, 12, 17, 22, . . .
Answer:

Question 15.
51, 48, 45, 42, . . .
Answer:
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 a 15

Question 16.
86, 79, 72, 65, . . .
Answer:

Question 17.
−1, −\(\frac{1}{3}\), \(\frac{1}{3}\), 1, . . .
Answer:
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 a 17

Question 18.
−2, −\(\frac{5}{4}\), −\(\frac{1}{2}\), \(\frac{1}{4}\), . . .
Answer:

Question 19.
2.3, 1.5, 0.7, −0.1, . . .
Answer:
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 a 19

Question 20.
11.7, 10.8, 9.9, 9, . . .
Answer:

ERROR ANALYSIS In Exercises 21 and 22, describe and correct the error in writing a rule for the nth term of the arithmetic sequence 22, 9, -4, -17, -30, . . ..
Question 21.
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 4
Answer:
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 a 21

Question 22.
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 5
Answer:

In Exercises 23–28, write a rule for the nth term of the sequence. Then graph the first six terms of the sequence.
Question 23.
a11 = 43, d = 5
Answer:
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 a 23

Question 24.
a13 = 42, d = 4
Answer:

Question 25.
a20 = −27, d =−2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 a 25

Question 26.
a15 = −35, d =−3
Answer:

Question 27.
a17 = −5, d = −\(\frac{1}{2}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 a 27

Question 28.
a21 = −25, d = −\(\frac{3}{2}\)
Answer:

Question 29.
USING EQUATIONS
One term of an arithmetic sequence is a8 = −13. The common difference is −8. What is a rule for the nth term of the sequence?
A. an = 51 + 8n
B. an = 35 + 8n
C. an = 51 − 8n
D. an = 35 − 8n
Answer:
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 a 29

Question 30.
FINDING A PATTERN
One term of an arithmetic sequence is a12 = 43. The common difference is 6. What is another term of the sequence?
A. a3 = −11
B. a4 = −53
C. a5 = 13
D. a6 = −47
Answer:

In Exercises 31–38, write a rule for the nth term of the arithmetic sequence.
Question 31.
a5 = 41, a10 = 96
Answer:
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 a 31

Question 32.
a7 = 58, a11 = 94
Answer:

Question 33.
a6 =−8, a15 = −62
Answer:
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 a 33

Question 34.
a8 = −15, a17 = −78
Answer:

Question 35.
a18 = −59, a21 = −71
Answer:
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 a 35

Question 36.
a12 = −38, a19 = −73
Answer:

Question 37.
a8 = 12, a16 = 22
Answer:
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 a 37

Question 38.
a12 = 9, a27 = 15
Answer:

WRITING EQUATIONS In Exercises 39–44, write a rule for the sequence with the given terms.
Question 39.
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 6
Answer:
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 a 39

Question 40.
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 7
Answer:

Question 41.
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 8
Answer:
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 a 41

Question 42.
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 9
Answer:

Question 43.
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 10
Answer:
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 a 43

Question 44.
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 11
Answer:

Question 45.
WRITING
Compare the graph of an = 3n + 1, where n is a positive integer, with the graph of f(x) = 3x+ 1, where x is a real number.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 a 45

Question 46.
DRAWING CONCLUSIONS
Describe how doubling each term in an arithmetic sequence changes the common difference of the sequence. Justify your answer.
Answer:

In Exercises 47–52, find the sum.
Question 47.
\(\sum_{i=1}^{20}\)(2i − 3)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 a 47

Question 48.
\(\sum_{i=1}^{26}\)(4i + 7)
Answer:

Question 49.
\(\sum_{i=1}^{33}\)(6 − 2i )
Answer:
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 a 49

Question 50.
\(\sum_{i=1}^{31}\)(−3 − 4i )
Answer:

Question 51.
\(\sum_{i=1}^{41}\)(−2.3 + 0.1i )
Answer:
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 a 51

Question 52.
\(\sum_{i=1}^{39}\)(−4.1 + 0.4i )
Answer:

NUMBER SENSE In Exercises 53 and 54, find the sum of the arithmetic sequence.
Question 53.
The first 19 terms of the sequence 9, 2, −5, −12, . . ..
Answer:
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 a 53

Question 54.
The first 22 terms of the sequence 17, 9, 1, −7, . . ..
Answer:

Question 55.
MODELING WITH MATHEMATICS
A marching band is arranged in rows. The first row has three band members, and each row after the first has two more band members than the row before it.
a. Write a rule for the number of band members in the nth row.
b. How many band members are in a formation with seven rows?
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 12
Answer:
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 a 55

Question 56.
MODELING WITH MATHEMATICS
Domestic bees make their honeycomb by starting with a single hexagonal cell, then forming ring after ring of hexagonal cells around the initial cell, as shown. The number of cells in successive rings forms an arithmetic sequence.
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 13
a. Write a rule for the number of cells in the nth ring.
b. How many cells are in the honeycomb after the ninth ring is formed?
Answer:

Question 57.
MATHEMATICAL CONNECTIONS
A quilt is made up of strips of cloth, starting with an inner square surrounded by rectangles to form successively larger squares. The inner square and all rectangles have a width of 1 foot. Write an expression using summation notation that gives the sum of the areas of all the strips of cloth used to make the quilt shown. Then evaluate the expression.
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 14
Answer:
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 a 57

Question 58.
HOW DO YOU SEE IT?
Which graph(s) represents an arithmetic sequence? Explain your reasoning.
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 15
Answer:

Question 59.
MAKING AN ARGUMENT
Your friend believes the sum of a series doubles when the common difference of an arithmetic series is doubled and the first term and number of terms in the series remain unchanged. Is your friend correct? Explain your reasoning.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 a 59

Question 60.
THOUGHT PROVOKING
In number theory, the Dirichlet Prime Number Theorem states that if a and bare relatively prime, then the arithmetic sequence
a, a + b, a + 2b, a + 3b, . . . contains infinitely many prime numbers. Find the first 10 primes in the sequence when a = 3 and b = 4.
Answer:

Question 61.
REASONING
Find the sum of the positive odd integers less than 300. Explain your reasoning.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 a 61

Question 62.
USING EQUATIONS
Find the value of n.
a. \(\sum_{i=1}^{n}\)(3i + 5) = 544
b. \(\sum_{i=1}^{n}\)(−4i − 1) = −1127
c. \(\sum_{i=5}^{n}\)(7 + 12i) = 455
d. \(\sum_{i=3}^{n}\)(−3 − 4i) = −507
Answer:

Question 63.
ABSTRACT REASONING
A theater has n rows of seats, and each row has d more seats than the row in front of it. There are x seats in the last (nth) row and a total of y seats in the entire theater. How many seats are in the front row of the theater? Write your answer in terms of n, x, and y.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 a 63

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:

Question 65.
CRITICAL THINKING
One of the major sources of our knowledge of Egyptian mathematics is the Ahmes papyrus, which is a scroll copied in 1650 B.C. by an Egyptian scribe. The following problem is from the Ahmes papyrus.
Divide 10 hekats of barley among 10 men so that the common difference is \(\frac{1}{8}\) of a hekat of barley.
Use what you know about arithmetic sequences and series to determine what portion of a hekat each man should receive.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 a 65

Maintaining Mathematical Proficiency

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

Question 67.
\(\frac{3^{-2}}{3^{-4}}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 a 67

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

Question 69.
(51/2 • 51/4)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 a 69

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

Question 71.
y = e-3x
Answer:
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 a 71

Question 72.
y = 3e-x
Answer:

Question 73.
y = e0.25x
Answer:
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series 8.2 a 73

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.

EXPLORATION 1

Recognizing Graphs of Geometric Sequences
Work with a partner. Determine whether each graph shows a geometric sequence. If it does, then write a rule for the nth term of the sequence and use a spreadsheet to find the sum of the first 20 terms. What do you notice about the graph of a geometric sequence?
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 1

EXPLORATION 2

Finding the Sum of a Geometric Sequence
Work with a partner. You can write the nth term of a geometric sequence with first term a1 and common ratio r as
an = a1rn-1.
So, you can write the sum Sn of the first n terms of a geometric sequence as
Sn = a1 + a1r + a1r2 + a1r3 + . . . +a1rn-1.
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 2
Rewrite this formula by finding the difference Sn − rSn and solve for Sn. Then verify your rewritten formula by funding the sums of the first 20 terms of the geometric sequences inExploration 1. Compare your answers to those you obtained using a spreadsheet.

Communicate Your Answer

Question 3.
How can you recognize a geometric sequence from its graph?
Answer:

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:

Monitoring Progress

Tell whether the sequence is geometric. Explain your reasoning.
Question 1.
27, 9, 3, 1, \(\frac{1}{3}\), . . .
Answer:

Question 2.
2, 6, 24, 120, 720, . . .
Answer:

Question 3.
−1, 2, −4, 8, −16, . . .
Answer:

Question 4.
Write a rule for the nth term of the sequence 3, 15, 75, 375, . . .. Then find a9.
Answer:

Write a rule for the nth term of the sequence. Then graph the first six terms of the sequence.
Question 5.
a6 = −96, r = −2
Answer:

Question 6.
a2 = 12, a4 = 3
Answer:

Find the sum.
Question 7.
\(\sum_{k=1}^{8}\)5k−1
Answer:

Question 8.
\(\sum_{i=1}^{12}\)6(−2)i−1
Answer:

Question 9.
\(\sum_{i=1}^{7}\)−16(0.5)t−1
Answer:

Question 10.
WHAT IF?
In Example 6, how does the monthly payment change when the annual interest rate is 5%?
Answer:

Analyzing Geometric Sequences and Series 8.3 Exercises

Vocabulary and Core Concept Check
Question 1.
COMPLETE THE SENTENCE
The constant ratio of consecutive terms in a geometric sequence is called the __________.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 a 1

Question 2.
WRITING
How can you determine whether a sequence is geometric from its graph?
Answer:

Question 3.
COMPLETE THE SENTENCE
The nth term of a geometric sequence has the form an = ___________.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 a 3

Question 4.
VOCABULARY
State the rule for the sum of the first n terms of a geometric series.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 5–12, tell whether the sequence is geometric. Explain your reasoning.
Question 5.
96, 48, 24, 12, 6, . . .
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 a 5

Question 6.
729, 243, 81, 27, 9, . . .
Answer:

Question 7.
2, 4, 6, 8, 10, . . .
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 a 7

Question 8.
5, 20, 35, 50, 65, . . .
Answer:

Question 9.
0.2, 3.2, −12.8, 51.2, −204.8, . . .
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 a 9

Question 10.
0.3, −1.5, 7.5, −37.5, 187.5, . . .
Answer:

Question 11.
\(\frac{1}{2}, \frac{1}{6}, \frac{1}{18}, \frac{1}{54}, \frac{1}{162}, \ldots\)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 a 11

Question 12.
\(\frac{1}{4}, \frac{1}{16}, \frac{1}{64}, \frac{1}{256}, \frac{1}{1024}, \ldots\)
Answer:

Question 13.
WRITING EQUATIONS
Write a rule for the geometric sequence with the given description.
a. The first term is −3, and each term is 5 times the previous term.
b. The first term is 72, and each term is \(\frac{1}{3}\) times the previous term.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 a 13

Question 14.
WRITING
Compare the terms of a geometric sequence when r > 1 to when 0 < r < 1.
Answer:

In Exercises 15–22, write a rule for the nth term of the sequence. Then find a7.
Question 15.
4, 20, 100, 500, . . .
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 a 15

Question 16.
6, 24, 96, 384, . . .
Answer:

Question 17.
112, 56, 28, 14, . . .
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 a 17

Question 18.
375, 75, 15, 3, . . .
Answer:

Question 19.
4, 6, 9, \(\frac{27}{2}\), . . .
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 a 19

Question 20.
2, \(\frac{3}{2}\), \(\frac{9}{8}\), \(\frac{27}{32}\), . . .
Answer:

Question 21.
1.3, −3.9, 11.7, −35.1, . . .
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 a 21

Question 22.
1.5, −7.5, 37.5, −187.5, . . .
Answer:

In Exercises 23–30, write a rule for the nth term of the sequence. Then graph the first six terms of the sequence.
Question 23.
a3 = 4, r = 2
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 a 23

Question 24.
a3 = 27, r = 3
Answer:

Question 25.
a2 = 30, r = \(\frac{1}{2}\)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 a 25

Question 26.
a2 = 64, r = \(\frac{1}{4}\)
Answer:

Question 27.
a4 = −192, r = 4
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 a 27

Question 28.
a4 = −500, r = 5
Answer:

Question 29.
a5 = 3, r =− \(\frac{1}{3}\)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 a 29

Question 30.
a5 = 1, r = −\(\frac{1}{5}\)
Answer:

ERROR ANALYSIS In Exercises 31 and 32, describe and correct the error in writing a rule for the nth term of the geometric sequence for which a2 = 48 and r = 6.
Question 31.
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 3
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 a 31

Question 32.
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 4
Answer:

In Exercises 33–40, write a rule for the nth term of the geometric sequence.
Question 33.
a2 = 28, a5 = 1792
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 a 33

Question 34.
a1 = 11, a4 = 88
Answer:

Question 35.
a1 =−6, a5 = −486
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 a 35

Question 36.
a2 = −10, a6 =−6250
Answer:

Question 37.
a2 = 64, a4 = 1
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 a 37

Question 38.
a1 = 1, a2 = 49
Answer:

Question 39.
a2 =−72, a6 = −\(\frac{1}{18}\)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 a 39

Question 40.
a2 =−48, a5 = \(\frac{3}{4}\)
Answer:

WRITING EQUATIONS In Exercises 41–46, write a rule for the sequence with the given terms.
Question 41.
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 5
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 a 41

Question 42.
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 6
Answer:

Question 43.
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 7
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 a 43

Question 44.
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 8
Answer:

Question 45.
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 9
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 a 45

Question 46.
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 10
Answer:

In Exercises 47–52, find the sum.
Question 47.
\(\sum_{i=1}^{9}\)6(7)i−1
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 a 47

Question 48.
\(\sum_{i=1}^{10}\)7(4)i−1
Answer:

Question 49.
\(\sum_{i=1}^{10}\)4(\(\frac{3}{4}\))i−1
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 a 49

Question 50.
\(\sum_{i=1}^{8}\)5(\(\frac{1}{3}\))i−1
Answer:

Question 51.
\(\sum_{i=0}^{8}\)8(−\(\frac{2}{3}\))i
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 a 51

Question 52.
\(\sum_{i=0}^{0}\)9(−\(\frac{3}{4}\))i
Answer:

NUMBER SENSE In Exercises 53 and 54, find the sum.
Question 53.
The first 8 terms of the geometric sequence −12, −48, −192, −768, . . ..
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 a 53

Question 54.
The first 9 terms of the geometric sequence −14, −42, −126, −378, . . ..
Answer:

Question 55.
WRITING
Compare the graph of an = 5(3)n−1, where n is a positive integer, to the graph of f(x) = 5 • 3x−1, where x is a real number.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 a 55

Question 56.
ABSTRACT REASONING
Use the rule for the sum of a finite geometric series to write each polynomial as a rational expression.
a. 1 + x + x2 + x3 + x4
b. 3x + 6x3 + 12x5 + 24x7
Answer:

MODELING WITH MATHEMATICS In Exercises 57 and 58, use the monthly payment formula given in Example 6.
Question 57.
You are buying a new car. You take out a 5-year loan for $15,000. The annual interest rate of the loan is 4%. Calculate the monthly payment.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 a 57

Question 58.
You are buying a new house. You take out a 30-year mortgage for $200,000. The annual interest rate of the loan is 4.5%. Calculate the monthly payment.
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 11
Answer:

Question 59.
MODELING WITH MATHEMATICS
A regional soccer tournament has 64 participating teams. In the first round of the tournament, 32 games are played. In each successive round, the number of games decreases by a factor of \(\frac{1}{2}\).
a. Write a rule for the number of games played in the nth round. For what values of n does the rule make sense? Explain.
b. Find the total number of games played in the regional soccer tournament.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 a 59

Question 60.
MODELING WITH MATHEMATICS
In a skydiving formation with R rings, each ring after the first has twice as many skydivers as the preceding ring. The formation for R = 2 is shown.
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 12
a. Let an be the number of skydivers in the nth ring. Write a rule for an.
b. Find the total number of skydivers when there are four rings.
Answer:

Question 61.
PROBLEM SOLVING
The Sierpinski carpet is a fractal created using squares. The process involves removing smaller squares from larger squares. First, divide a large square into nine congruent squares. Then remove the center square. Repeat these steps for each smaller square, as shown below. Assume that each side of the initial square is 1 unit long.
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 13
a. Let an be the total number of squares removed at the nth stage. Write a rule for an. Then find the total number of squares removed through Stage 8.
b. Let bn be the remaining area of the original square after the nth stage. Write a rule for bn. Then find the remaining area of the original square after Stage 12.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 a 61

Question 62.
HOW DO YOU SEE IT?
Match each sequence with its graph. Explain your reasoning.
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 14
Answer:

Question 63.
CRITICAL THINKING
On January 1, you deposit $2000 in a retirement account that pays 5% annual interest. You make this deposit each January 1 for the next 30 years. How much money do you have in your account immediately after you make your last deposit?
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 a 63

Question 64.
THOUGHT PROVOKING
The first four iterations of the fractal called the Koch snowflake are shown below. Find the perimeter and area of each iteration. Do the perimeters and areas form geometric sequences? Explain your reasoning.
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 15
Answer:

Question 65.
MAKING AN ARGUMENT
You and your friend are comparing two loan options for a $165,000 house. Loan 1 is a 15-year loan with an annual interest rate of 3%. Loan 2 is a 30-year loan with an annual interest rate of 4%. Your friend claims the total amount repaid over the loan will be less for Loan 2. Is your friend correct? Justify your answer.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 a 65

Question 66.
CRITICAL THINKING
Let L be the amount of a loan (in dollars), i be the monthly interest rate (in decimal form), t be the term (in months), and M be the monthly payment (in dollars).
a. When making monthly payments, you are paying the loan amount plus the interest the loan gathers each month. For a 1-month loan, t= 1, the equation for repayment is L(1 +i) −M= 0. For a 2-month loan, t= 2, the equation is [L(1 + i) −M](1 + i) −M = 0. Solve both of these repayment equations for L.
b. Use the pattern in the equations you solved in part (a) to write a repayment equation for a t-month loan. (Hint: L is equal to M times a geometric series.) Then solve the equation for M.
c. Use the rule for the sum of a finite geometric series to show that the formula in part (b) is equivalent to
M = L\(\left(\frac{i}{1-(1+i)^{-t}}\right)\).
Use this formula to check your answers in Exercises 57 and 58.
Answer:

Maintaining Mathematical Proficiency

Graph the function. State the domain and range.
Question 67.
f(x) = \(\frac{1}{x-3}\)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 a 67

Question 68.
g(x) = \(\frac{2}{x}\) + 3
Answer:

Question 69.
h(x) = \(\frac{1}{x-2}\) + 1
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 a 69

Question 70.
p(x) = \(\frac{3}{x+1}\) − 2
Answer:

Sequences and Series Study Skills: Keeping Your Mind Focused

8.1–8.3 What Did You Learn?

Core Vocabulary
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 16

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

• Before doing homework, review the concept boxes and examples. Talk through the examples out loud.
• Complete homework as though you were also preparing for a quiz. Memorize the different types of problems, formulas, rules, and so on.
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series 8.3 17

Sequences and Series 8.1 – 8.3 Quiz

Describe the pattern, write the next term, and write a rule for the nth term of the sequence.
Question 1.
1, 7, 13, 19, . . .
Answer:

Question 2.
−5, 10, −15, 20, . . .
Answer:

Question 3.
\(\frac{1}{20}, \frac{2}{30}, \frac{3}{40}, \frac{4}{50}, \ldots\)
Answer:

Write the series using summation notation. Then find the sum of the series.
Question 4.
1 + 2 + 3 + 4 +. . .+ 15
Answer:

Question 5.
\(0+\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\cdots+\frac{7}{8}\)
Answer:

Question 6.
9 + 16 + 25 + . . .+ 100
Answer:

Write a rule for the nth term of the sequence.
Question 7.
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series q 1
Answer:

Question 8.
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series q 2
Answer:

Question 9.
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series q 3
Answer:

Tell whether the sequence is arithmetic, geometric, or neither. Write a rule for the nth term of the sequence. Then find a9.
Question 10.
13, 6, −1, −8, . . .
Answer:

Question 11.
\(\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \ldots\)
Answer:

Question 12.
1, −3, 9, −27, . . .
Answer:

Question 13.
One term of an arithmetic sequence is a12 = 19. The common difference is d = 7. Write a rule for the nth term. Then graph the first six terms of the sequence.
Answer:

Question 14.
Two terms of a geometric sequence are a6 = −50 and a9 = −6250. Write a rule for the nth term.
Answer:

Find the sum.
Question 15.
\(\sum_{n=1}^{9}\)(3n + 5)
Answer:

Question 16.
\(\sum_{k=1}^{5}\)11(−3)k−2
Answer:

Question 17.
\(\sum_{i=1}^{12}\)−4 (\(\frac{1}{2}\))i+3
Answer:

Question 18.
Pieces of chalk are stacked in a pile. Part of the pile is shown. The bottom row has 15 pieces of chalk, and the top row has 6 pieces of chalk. Each row has one less piece of chalk than the row below it. How many pieces of chalk are in the pile?
Big Ideas Math Algebra 2 Solutions Chapter 8 Sequences and Series q 4
Answer:

Question 19.
You accept a job as an environmental engineer that pays a salary of $45,000 in the first year. After the first year, your salary increases by 3.5% per year.
a. Write a rule giving your salary an for your nth year of employment.
b. What will your salary be during your fifth year of employment?
c. You work 10 years for the company. What are your total earnings? Justify your answer.
Answer:

Lesson 8.4 Finding Sums of Infinite Geometric Series

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

EXPLORATION 1

Finding Sums of Infinite Geometric Series
Big Ideas Math Answer Key Algebra 2 Chapter 8 Sequences and Series 8.4 1
Work with a partner. Enter each geometric series in a spreadsheet. Then use the spreadsheet to determine whether the infinite geometric series has a finite sum. If it does, find the sum. Explain your reasoning. (The figure shows a partially completed spreadsheet for part (a).)
Big Ideas Math Answer Key Algebra 2 Chapter 8 Sequences and Series 8.4 2

EXPLORATION 2

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
a1 + a1r + a1r2 + a1r3 +. . .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 a1 and common ratio r≠ 1 is
Sn = a1\(\left(\frac{1-r^{n}}{1-r}\right)\)
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.

Communicate Your Answer

Question 4.
How can you find the sum of an infinite geometric series?
Answer:

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:

Monitoring Progress

Question 1.
Consider the infinite geometric series
\(\frac{2}{5}+\frac{4}{25}+\frac{8}{125}+\frac{16}{1625}+\frac{32}{3125}+\cdots\)
Find and graph the partial sums Sn for n = 1, 2, 3, 4, and 5. Then describe what happens to Sn as n increases.
Answer:

Find the sum of the infinite geometric series, if it exists.
Question 2.
\(\sum_{n=1}^{\infty}\left(-\frac{1}{2}\right)^{n-1}\)
Answer:

Question 3.
\(\sum_{n=1}^{\infty} 3\left(\frac{5}{4}\right)^{n-1}\)
Answer:

Question 4.
\(3+\frac{3}{4}+\frac{3}{16}+\frac{3}{64}+\cdots\)
Answer:

Question 5.
WHAT IF?
In Example 3, suppose the pendulum travels 10 inches on its first swing. What is the total distance the pendulum swings?
Answer:

Write the repeating decimal as a fraction in simplest form.
Question 6.
0.555 . . . .
Answer:

Question 7.
0.727272 . . . .
Answer:

Question 8.
0.131313 . . . .
Answer:

Finding Sums of Infinite Geometric Series 8.4 Exercises

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

Big Ideas Math Answer Key Algebra 2 Chapter 8 Sequences and Series 8.4 a 1

Question 2.
WRITING
Explain how to tell whether the series \(\sum_{i=1}^{\infty}\)a1ri−1 has a sum.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–6, consider the infinite geometric series. Find and graph the partial sums Sn for n= 1, 2, 3, 4, and 5. Then describe what happens to Sn as n increases.
Question 3.
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{18}+\frac{1}{54}+\frac{1}{162}+\cdots\)
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 8 Sequences and Series 8.4 a 3

Question 4.
\(\frac{2}{3}+\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\cdots\)
Answer:

Question 5.
4 + \(\frac{12}{5}+\frac{36}{25}+\frac{108}{125}+\frac{324}{625}+\cdots\)
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 8 Sequences and Series 8.4 a 5

Question 6.
2 + \(\frac{2}{6}+\frac{2}{36}+\frac{2}{216}+\frac{2}{1296}+\cdots\)
Answer:

In Exercises 7–14, find the sum of the infinite geometric series, if it exists.
Question 7.
\(\sum_{n=1}^{\infty} 8\left(\frac{1}{5}\right)^{n-1}\)
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 8 Sequences and Series 8.4 a 7

Question 8.
\(\sum_{k=1}^{\infty}-6\left(\frac{3}{2}\right)^{k-1}\)
Answer:

Question 9.
\(\sum_{k=1}^{\infty} \frac{11}{3}\left(\frac{3}{8}\right)^{k-1}\)
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 8 Sequences and Series 8.4 a 9

Question 10.
\(\sum_{i=1}^{\infty} \frac{2}{5}\left(\frac{5}{3}\right)^{i-1}\)
Answer:

Question 11.
2 + \(\frac{6}{4}+\frac{18}{16}+\frac{54}{64}+\cdots\)
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 8 Sequences and Series 8.4 a 11

Question 12.
-5 – 2 – \(\frac{4}{5}-\frac{8}{25}-\cdots\)
Answer:

Question 13.
3 + \(\frac{5}{2}+\frac{25}{12}+\frac{125}{72}+\cdots\)
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 8 Sequences and Series 8.4 a 13

Question 14.
\(\frac{1}{2}-\frac{5}{3}+\frac{50}{9}-\frac{500}{27}+\cdots\)
Answer:

ERROR ANALYSIS In Exercises 15 and 16, describe and correct the error in finding the sum of the infinite geometric series.
Question 15.
Big Ideas Math Answer Key Algebra 2 Chapter 8 Sequences and Series 8.4 3
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 8 Sequences and Series 8.4 a 15

Question 16.
Big Ideas Math Answer Key Algebra 2 Chapter 8 Sequences and Series 8.4 4
Answer:

Question 17.
MODELING WITH MATHEMATICS
You push your younger cousin on a tire swing one time and then allow your cousin to swing freely. On the first swing, your cousin travels a distance of 14 feet. On each successive swing, your cousin travels 75% of the distance of the previous swing. What is the total distance your cousin swings?
Big Ideas Math Answer Key Algebra 2 Chapter 8 Sequences and Series 8.4 5
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 8 Sequences and Series 8.4 a 17

Question 18.
MODELING WITH MATHEMATICS
A company had a profit of $350,000 in its first year. Since then, the company’s profit has decreased by 12% per year. Assuming this trend continues, what is the total profit the company can make over the course of its lifetime? Justify your answer.
Answer:

In Exercises 19–24, write the repeating decimal as a fraction in simplest form.
Question 19.
0.222 . . .
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 8 Sequences and Series 8.4 a 19

Question 20.
0.444 . . .
Answer:

Question 21.
0.161616 . . .
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 8 Sequences and Series 8.4 a 21

Question 22.
0.625625625 . . .
Answer:

Question 23.
32.323232 . . .
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 8 Sequences and Series 8.4 a 23

Question 24.
130.130130130 . . .
Answer:

Question 25.
PROBLEM SOLVING
Find two infinite geometric series whose sums are each 6. Justify your answers.
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 8 Sequences and Series 8.4 a 25

Question 26.
HOW DO YOU SEE IT?
The graph shows the partial sums of the geometric series a1 + a2 + a3 + a4+. . . .What is the value of \(\sum_{n=1}^{\infty}\)an ? Explain.
Big Ideas Math Answer Key Algebra 2 Chapter 8 Sequences and Series 8.4 6
Answer:

Question 27.
MODELING WITH MATHEMATICS
A radio station has a daily contest in which a random listener is asked a trivia question. On the first day, the station gives $500 to the first listener who answers correctly. On each successive day, the winner receives 90% of the winnings from the previous day. What is the total amount of prize money the radio station gives away during the contest?
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 8 Sequences and Series 8.4 a 27

Question 28.
THOUGHT PROVOKING
Archimedes used the sum of a geometric series to compute the area enclosed by a parabola and a straight line. In “Quadrature of the Parabola,” he proved that the area of the region is \(\frac{4}{3}\) the area of the inscribed triangle. The first term of the series for the parabola below is represented by the area of the blue triangle and the second term is represented by the area of the red triangles. Use Archimedes’ result to find the area of the region. Then write the area as the sum of an infinite geometric series.
Big Ideas Math Answer Key Algebra 2 Chapter 8 Sequences and Series 8.4 7
Answer:

Question 29.
DRAWING CONCLUSIONS
Can a person running at 20 feet per second ever catch up to a tortoise that runs 10 feet per second when the tortoise has a 20-foot head start? The Greek mathematician Zeno said no. He reasoned as follows:
Big Ideas Math Answer Key Algebra 2 Chapter 8 Sequences and Series 8.4 8
Looking at the race as Zeno did, the distances and the times it takes the person to run those distances both form infinite geometric series. Using the table, show that both series have finite sums. Does the person catch up to the tortoise? Justify your answer.
Big Ideas Math Answer Key Algebra 2 Chapter 8 Sequences and Series 8.4 9
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 8 Sequences and Series 8.4 a 29

Question 30.
MAKING AN ARGUMENT
Your friend claims that 0.999 . . . is equal to 1. Is your friend correct? Justify your answer.
Answer:

Question 31.
CRITICAL THINKING
The Sierpinski triangle is a fractal created using equilateral triangles. The process involves removing smaller triangles from larger triangles by joining the midpoints of the sides of the larger triangles as shown. Assume that the initial triangle has an area of 1 square foot.
Big Ideas Math Answer Key Algebra 2 Chapter 8 Sequences and Series 8.4 10
a. Let an be the total area of all the triangles that are removed at Stage n. Write a rule for an.
b. Find \(\sum_{n=1}^{\infty}\)an. Interpret your answer in the context of this situation.
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 8 Sequences and Series 8.4 a 31

Maintaining Mathematical Proficiency

Determine the type of function represented by the table.
Question 32.
Big Ideas Math Answer Key Algebra 2 Chapter 8 Sequences and Series 8.4 11
Answer:

Question 33.
Big Ideas Math Answer Key Algebra 2 Chapter 8 Sequences and Series 8.4 12
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 8 Sequences and Series 8.4 a 33

Determine whether the sequence is arithmetic, geometric, or neither.
Question 34.
−7, −1, 5, 11, 17, . . .
Answer:

Question 35.
0, −1, −3, −7, −15, . . .
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 8 Sequences and Series 8.4 a 35

Question 36.
13.5, 40.5, 121.5, 364.5, . .
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 an is related to one or more preceding terms.

EXPLORATION 1

Evaluating a Recursive Rule
Work with a partner. Use each recursive rule and a spreadsheet to write the first six terms of the sequence. Classify the sequence as arithmetic, geometric, or neither. Explain your reasoning. (The figure shows a partially completed spreadsheet for part (a).)
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 1

EXPLORATION 2

Writing a Recursive Rule
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 2
Work with a partner. Write a recursive rule for the sequence. Explain your reasoning.
a. 3, 6, 9, 12, 15, 18, . . .
b. 18, 14, 10, 6, 2, −2, . . .
c. 3, 6, 12, 24, 48, 96, . . .
d. 128, 64, 32, 16, 8, 4, . . .
e. 5, 5, 5, 5, 5, 5, . . .
f. 1, 1, 2, 3, 5, 8, . . .

EXPLORATION 3

Writing a Recursive RuleWork with a partner. Write a recursive rule for the sequence whose graph is shown.
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 3

Communicate Your Answer

Question 4.
How can you define a sequence recursively?
Answer:

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:

Monitoring Progress

Write the first six terms of the sequence.
Question 1.
a1 = 3, an = an-1 − 7
Answer:

Question 2.
a0 = 162, an = 0.5an-1
Answer:

Question 3.
f(0) = 1, f(n) = f(n− 1) + n
Answer:

Question 4.
a1 = 4, an = 2an-1 − 1
Answer:

Write a recursive rule for the sequence.
Question 5.
2, 14, 98, 686, 4802, . . .
Answer:

Question 6.
19, 13, 7, 1, −5, . . .
Answer:

Question 7.
11, 22, 33, 44, 55, . . .
Answer:

Question 8.
1, 2, 2, 4, 8, 32, . . .
Answer:

Write a recursive rule for the sequence.
Question 9.
an = 17 − 4n
Answer:

Question 10.
an = 16(3)n-1
Answer:

Write an explicit rule for the sequence.
Question 11.
a1 = −12, an = an-1 + 16
Answer:

Question 12.
a1 = 2, an = −6an-1
Answer:

Question 13.
WHAT IF?
In Example 6, suppose 75% of the fish remain each year. What happens to the population of fish over time?
Answer:

Question 14.
WHAT IF?
How do the answers in Example 7 change when the annual interest rate is 7.5% and the monthly payment is $1048.82?
Answer:

Using Recursive Rules with Sequences 8.5 Exercises

Vocabulary and Core Concept Check
Question 1.
COMPLETE THE SENTENCE
A recursive _________ tells how the nth term of a sequence is related to one or more preceding terms.
Answer:
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 a 1

Question 2.
WRITING
Explain the difference between an explicit rule and a recursive rule for a sequence.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–10, write the first six terms of the sequence.
Question 3.
a1 = 1
an = an-1 + 3
Answer:
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 a 3

Question 4.
a1 = 1
an = an-1 − 5
Answer:

Question 5.
f(0) = 4
f(n) = 2f (n− 1)
Answer:
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 a 5

Question 6.
f(0) = 10
f(n) = \(\frac{1}{2}\)f(n− 1)
Answer:

Question 7.
a1 = 2
an = (an-1)2 + 1
Answer:
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 a 7

Question 8.
a1 = 1
an = (an-1)2 − 10
Answer:

Question 9.
f(0) = 2, f (1) = 4
f(n) = f(n − 1) − f(n − 2)
Answer:
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 a 9

Question 10.
f(1) = 2, f(2) = 3
f(n) = f(n − 1) • f(n − 2)
Answer:

In Exercises 11–22, write a recursive rule for the sequence.
Question 11.
21, 14, 7, 0, −7, . . .
Answer:
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 a 11

Question 12.
54, 43, 32, 21, 10, . . .
Answer:

Question 13.
3, 12, 48, 192, 768, . . .
Answer:
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 a 13

Question 14.
4, −12, 36, −108, . . .
Answer:

Question 15.
44, 11, \(\frac{11}{4}\), \(\frac{11}{16}\), \(\frac{11}{64}\), . . .
Answer:
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 a 15

Question 16.
1, 8, 15, 22, 29, . . .
Answer:

Question 17.
2, 5, 10, 50, 500, . . .
Answer:
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 a 17

Question 18.
3, 5, 15, 75, 1125, . . .
Answer:

Question 19.
1, 4, 5, 9, 14, . . .
Answer:
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 a 19

Question 20.
16, 9, 7, 2, 5, . . .
Answer:

Question 21.
6, 12, 36, 144, 720, . . .
Answer:
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 a 21

Question 22.
−3, −1, 2, 6, 11, . . .
Answer:

In Exercises 23–26, write a recursive rule for the sequence shown in the graph.
Question 23.
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 4
Answer:
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 a 23

Question 24.
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 5
Answer:

Question 25.
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 6
Answer:
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 a 25

Question 26.
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 7
Answer:

ERROR ANALYSIS In Exercises 27 and 28, describe and correct the error in writing a recursive rule for the sequence 5, 2, 3, -1, 4, . . ..
Question 27.
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 8
Answer:
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 a 27

Question 28.
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 9
Answer:

In Exercises 29–38, write a recursive rule for the sequence.
Question 29.
an = 3 + 4n
Answer:
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 a 29

Question 30.
an =−2 − 8n
Answer:

Question 31.
an = 12 − 10n
Answer:
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 a 31

Question 32.
an = 9 − 5n
Answer:

Question 33.
an = 12(11)n-1
Answer:
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 a 33

Question 34.
an = −7(6)n-1
Answer:

Question 35.
an = 2.5 − 0.6n
Answer:
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 a 35

Question 36.
an = −1.4 + 0.5n
Answer:

Question 37.
an= −\(\frac{1}{2}\left(\frac{1}{4}\right)^{n-1}\)
Answer:
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 a 37

Question 38.
an = \(\frac{1}{4}\)(5)n-1
Answer:

Question 39.
REWRITING A FORMULA
You have saved $82 to buy a bicycle. You save an additional $30 each month. The explicit rule an= 30n+ 82 gives the amount saved after n months. Write are cursive rule for the amount you have saved n months from now.
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 10
Answer:
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 a 39

Question 40.
REWRITING A FORMULA
Your salary is given by the explicit rule an = 35,000(1.04)n-1, where n is the number of years you have worked. Write a recursive rule for your salary.
Answer:

In Exercises 41–48, write an explicit rule for the sequence.
Question 41.
a1 = 3, an = an-1 − 6
Answer:
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 a 41

Question 42.
a1 = 16, an = an-1 + 7
Answer:

Question 43.
a1 = −2, an = 3an-1
Answer:
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 a 43

Question 44.
a1 = 13, an = 4an-1
Answer:

Question 45.
a1 = −12, an = an-1 + 9.1
Answer:
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 a 45

Question 46.
a1 = −4, an = 0.65an-1
Answer:

Question 47.
an = 5, an = an-1 − \(\frac{1}{3}\)
Answer:
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 a 47

Question 48.
a1 = −5, an = \(\frac{1}{4}\)an-1
Answer:

Question 49.
REWRITING A FORMULA
A grocery store arranges cans in a pyramid-shaped display with 20 cans in the bottom row and two fewer cans in each subsequent row going up. The number of cans in each row is represented by the recursive rule a1 = 20, an = an-1 − 2. Write an explicit rule for the number of cans in row n.
Answer:
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 a 49

Question 50.
REWRITING A FORMULA
The value of a car is given by the recursive rule a1 = 25,600, an = 0.86an-1, where n is the number of years since the car was new. Write an explicit rule for the value of the car after n years.
Answer:

Question 51.
USING STRUCTURE
What is the 1000th term of the sequence whose first term is a1 = 4 and whose nth term is an = an-1 + 6? Justify your answer.
A. 4006
B. 5998
C. 1010
D. 10,000
Answer:
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 a 51

Question 52.
USING STRUCTURE
What is the 873rd term of the sequence whose first term is a1 = 0.01 and whose nth term is an = 1.01an-1? Justify your answer.
A. 58.65
B. 8.73
C. 1.08
D. 586,459.38
Answer:

Question 53.
PROBLEM SOLVING
An online music service initially has 50,000 members. Each year, the company loses 20% of its current members and gains 5000 new members.
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 11
a. Write a recursive rule for the number an of members at the start of the nth year.
b. Find the number of members at the start of the fifth year.
c. Describe what happens to the number of members over time.
Answer:
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 a 53

Question 54.
PROBLEM SOLVING
You add chlorine to a swimming pool. You add 34 ounces of chlorine the first week and 16 ounces every week thereafter. Each week, 40% of the chlorine in the pool evaporates.
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 12
a. Write a recursive rule for the amount of chlorine in the pool at the start of the nth week.
b. Find the amount of chlorine in the pool at the start of the third week.
c. Describe what happens to the amount of chlorine in the pool over time.
Answer:

Question 55.
OPEN-ENDED
Give an example of a real-life situation which you can represent with a recursive rule that does not approach a limit. Write a recursive rule that represents the situation.
Answer:
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 a 55

Question 56.
OPEN-ENDED
Give an example of a sequence in which each term after the third term is a function of the three terms preceding it. Write a recursive rule for the sequence and find its first eight terms.
Answer:

Question 57.
MODELING WITH MATHEMATICS
You borrow $2000 at 9% annual interest compounded monthly for 2 years. The monthly payment is $91.37.
a. Find the balance after the fifth payment.
b. Find the amount of the last payment.
Answer:
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 a 57.1
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 a 57.2

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:

Question 59.
COMPARING METHODS
In 1202, the mathematician Leonardo Fibonacci wrote Liber Abaci, in which he proposed the following rabbit problem:
Begin with a pair of newborn rabbits. When a pair of rabbits is two months old, the rabbits begin producing a new pair of rabbits each month. Assume none of the rabbits die.
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 13
This problem produces a sequence called the Fibonacci sequence, which has both a recursive formula and an explicit formula as follows.
Recursive: a1 = 1, a2 = 1, an = an-2 + an-1
Explicit: fn = \(\frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^{n}-\frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^{n}\), n ≥ 1
Use each formula to determine how many rabbits there will be after one year. Justify your answers.
Answer:
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 a 59

Question 60.
USING TOOLS
A town library initially has 54,000 books in its collection. Each year, 2% of the books are lost or discarded. The library can afford to purchase 1150 new books each year.
a. Write a recursive rule for the number an of books in the library at the beginning of the nth year.
b. Use the sequence mode and the dot mode of a graphing calculator to graph the sequence. What happens to the number of books in the library over time? Explain.
Answer:

Question 61.
DRAWING CONCLUSIONS
A tree farm initially has 9000 trees. Each year, 10% of the trees are harvested and 800 seedlings are planted.
a. Write a recursive rule for the number of trees on the tree farm at the beginning of the nth year.
b. What happens to the number of trees after an extended period of time?
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 14
Answer:
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 a 61

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.
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?
c. How does doubling the dosage affect the maintenance level of the drug? Justify your answer.
Answer:

Question 63.
FINDING A PATTERN
A fractal tree starts with a single branch (the trunk). At each stage, each new branch from the previous stage grows two more branches, as shown.
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 15
a. List the number of new branches in each of the first seven stages. What type of sequence do these numbers form?
b. Write an explicit rule and a recursive rule for the sequence in part (a).
Answer:
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 a 63

Question 64.
THOUGHT PROVOKING
Let a1 = 34. Then write the terms of the sequence until you discover a pattern.
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 16
Do the same for a1 = 25. What can you conclude?
Answer:

Question 65.
MODELING WITH MATHEMATICS
You make a $500 down payment on a $3500 diamond ring. You borrow the remaining balance at 10% annual interest compounded monthly. The monthly payment is $213.59. How long does it take to pay back the loan? What is the amount of the last payment? Justify your answers.
Answer:
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 a 65

Question 66.
HOW DO YOU SEE IT?
The graph shows the first six terms of the sequence a1 = p, an = ran-1.
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 17
a. Describe what happens to the values in the sequence as n increases.
b. Describe the set of possible values for r. Explain your reasoning.
Answer:

Question 67.
REASONING
The rule for a recursive sequence is as follows.
f(1) = 3, f(2) = 10
f(n) = 4 + 2f(n − 1) −f (n − 2)
a. Write the first five terms of the sequence.
b. Use finite differences to find a pattern. What type of relationship do the terms of the sequence show?
c. Write an explicit rule for the sequence.
Answer:
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 a 67

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 69.
CRITICAL THINKING
The first four triangular numbers Tn and the first four square numbers Sn are represented by the points in each diagram.
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 18
a. Write an explicit rule for each sequence.
b. Write a recursive rule for each sequence.
c. Write a rule for the square numbers in terms of the triangular numbers. Draw diagrams to explain why this rule is true.
Answer:
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 a 69.1
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 a 69.2

Question 70.
CRITICAL THINKING
You are saving money for retirement. You plan to withdraw $30,000 at the beginning of each year for 20 years after you retire. Based on the type of investment you are making, you can expect to earn an annual return of 8% on your savings after you retire.
a. Let an be your balance n years after retiring. Write a recursive equation that shows how an is related to an-1.
b. Solve the equation from part (a) for an-1. Find a0, the minimum amount of money you should have in your account when you retire. (Hint: Let a20 = 0.)
Answer:

Maintaining Mathematical Proficiency

Solve the equation. Check your solution.
Question 71.
\(\sqrt{x}\) + 2 = 7
Answer:
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 a 71

Question 72.
2\(\sqrt{52}\) − 5 = 15
Answer:

Question 73.
\(\sqrt [ 3 ]{ x }\) + 16 = 19
Answer:
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 a 73

Question 74.
2\(\sqrt [ 3 ]{ x }\) − 13 = −5
ans;

The variables x and y vary inversely. Use the given values to write an equation relating x and y. Then find y when x = 4.
Question 75.
x = 2, y = 9
Answer:
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 a 75

Question 76.
x =−4, y = 3
Answer:

Question 77.
x = 10, y = 32
Answer:
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 a 77

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

In April of 1965, an engineer named Gordon Moore noticed how quickly the size of electronics was shrinking. He predicted how the number of transistors that could fit on a 1-inch diameter circuit would increase over time. In 1965, only 50 transistors fit on the circuit. A decade later, about 65,000 transistors could fit on the circuit. Moore’s prediction was accurate and is now known as Moore’s Law. What was his prediction? How many transistors will be able to fit on a 1-inch circuit when you graduate from high school?
To explore the answers to this question and more, go to BigIdeasMath.com.
Big Ideas Math Answers Algebra 2 Chapter 8 Sequences and Series 8.5 19

Sequences and Series Chapter Review

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

Question 1.
Describe the pattern shown in the figure. Then write a rule for the nth layer of the figure, where n = 1 represents the top layer.
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series cr 1
Answer:

Write the series using summation notation.
Question 2.
7 + 10 + 13 +. . .+ 40
Answer:

Question 3.
0 + 2 + 6 + 12 +. . . .
Answer:

Find the sum
Question 4.
\(\sum_{i=2}^{7}\)(9 – i3)
Answer:

Question 5.
\(\sum_{i=1}^{46}\)i
Answer:

Question 6.
\(\sum_{i=1}^{12}\)i2
Answer:

Question 7.
\(\sum_{i=1}^{5} \frac{3+i}{2}\)
Answer:

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

Question 8.
Tell whether the sequence 12, 4, −4, −12, −20, . . . is arithmetic. Explain your reasoning.
Answer:

Write a rule for the nth term of the arithmetic sequence. Then graph the first six terms of the sequence.
Question 9.
2, 8, 14, 20, . . .
Answer:

Question 10.
a14 = 42, d = 3
Answer:

Question 11.
a6 = −12, a12 = −36
Answer:

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

Question 13.
You take a job with a starting salary of $37,000. Your employer offers you an annual raise of $1500 for the next 6 years. Write a rule for your salary in the nth year. What are your total earnings in 6 years?
Answer:

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

Question 14.
Tell whether the sequence 7, 14, 28, 56, 112, . . . is geometric. Explain your reasoning.
Answer:

Write a rule for the nth term of the geometric sequence. Then graph the first six terms of the sequence.
Question 15.
25, 10, 4, \(\frac{8}{5}\) , . . .
Answer:

Question 16.
a5 = 162, r =−3
Answer:

Question 17.
a3 = 16, a5 = 256
Answer:

Question 18.
Find the sum \(\sum_{i=1}^{9}\)5(−2)i−1 .
Answer:

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

Question 19.
Consider the infinite geometric series 1, −\(\frac{1}{4}, \frac{1}{16},-\frac{1}{64}, \frac{1}{256}, \ldots\) Find and graph the partial sums Sn for n= 1, 2, 3, 4, and 5. Then describe what happens to Sn as n increases.
Answer:

Question 20.
Find the sum of the infinite geometric series −2 + \(\frac{1}{2}-\frac{1}{8}+\frac{1}{32}+\cdots\), if it exists.
Answer:

Question 21.
Write the repeating decimal 0.1212 . . . as a fraction in simplest form.
Answer:

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

Write the first six terms of the sequence.
Question 22.
a1 = 7, an = an-1 + 11
Answer:

Question 23.
a1 = 6, an = 4an-1
Answer:

Question 24.
f(0) = 4, f(n) = f(n − 1) + 2n
Answer:

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:

Question 27.
7, 3, 4, −1, 5, . . .
Answer:

Question 28.
Write a recursive rule for an = 105 (\(\frac{3}{5}\))n−1 .
Answer:

Write an explicit rule for the sequence.
Question 29.
a1 = −4, an = an-1 + 26
Answer:

Question 30.
a1 = 8, an = −5an-1
Answer:

Question 31.
a1 = 26, an = \(\frac{2}{5}\)an-1.
Answer:

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 Pn of the town in year n. Let n = 1 represent 2010.
Answer:

Question 33.
The numbers 1, 6, 15, 28, . . . are called hexagonal numbers because they represent the number of dots used to make hexagons, as shown. Write a recursive rule for the nth hexagonal number.
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series cr 33
Answer:

Sequences and Series Chapter Test

Find the sum.
Question 1.
\(\sum_{i=1}^{24}\)(6i− 13)
Answer:

Question 2.
\(\sum_{n=1}^{16}\)n2
Answer:

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

Question 4.
\(\sum_{i=1}^{6}\)4(−3)i−1
Answer:

Determine whether the graph represents an arithmetic sequence, geometric sequence, or neither. Explain your reasoning. Then write a rule for the nth term.
Question 5.
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series ct 5
Answer:

Question 6.
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series ct 6
Answer:

Question 7.
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series ct 7
Answer:

Write a recursive rule for the sequence. Then find a9.
Question 8.
a1 = 32, r = \(\frac{1}{2}\)
Answer:

Question 9.
an = 2 + 7n
Answer:

Question 10.
2, 0, −3, −7, −12, . . .
Answer:

Question 11.
Write a recursive rule for the sequence 5, −20, 80, −320, 1280, . . .. Then write an explicit rule for the sequence using your recursive rule.
Answer:

Question 12.
The numbers a, b, and c are the first three terms of an arithmetic sequence. Is b half of the sum of a and c? Explain your reasoning.
Answer:

Question 13.
Use the pattern of checkerboard quilts shown.
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series ct 8
a. What does n represent for each quilt? What does an represent?
b. Make a table that shows n and an for n= 1, 2, 3, 4, 5, 6, 7, and 8.
c. Use the rule an = \(\frac{n^{2}}{2}+\frac{1}{4}\)[1 − (−1)n] to find an for n = 1, 2, 3, 4, 5, 6, 7, and 8.
Compare these values to those in your table in part (b). What can you conclude? Explain.
Answer:

Question 14.
During a baseball season, a company pledges to donate $5000 to a charity plus $100 for each home run hit by the local team. Does this situation represent a sequence or a series? Explain your reasoning.
Answer:

Question 15.
The lengthℓ1 of the first loop of a spring is 16 inches. The lengthℓ2 of the second loop is 0.9 times the length of the first loop. The lengthℓ3 of the third loop is 0.9 times the length of the second loop, and so on. Suppose the spring has infinitely many loops, would its length be finite or infinite? Explain. Find the length of the spring, if possible.
Big Ideas Math Algebra 2 Answer Key Chapter 8 Sequences and Series ct 15
Answer:

Sequences and Series Cumulative Assessment

Question 1.
The frequencies (in hertz) of the notes on a piano form a geometric sequence. The frequencies of G (labeled 8) and A (labeled 10) are shown in the diagram. What is the approximate frequency of E at (labeled 4)?
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series ca 1
Answer:

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 3.
The table shows that the force F (in pounds) needed to loosen a certain bolt with a wrench depends on the length ℓ (in inches) of the wrench’s handle. Write an equation that relates ℓ and F. Describe the relationship.
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series ca 3
Answer:

Question 4.
Order the functions from the least average rate of change to the greatest average rate of change on the interval 1 ≤ x ≤ 4. Justify your answers.
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series ca 4
Answer:

Question 5.
A running track is shaped like a rectangle with two semicircular ends, as shown. The track has 8 lanes that are each 1.22 meters wide. The lanes are numbered from 1 to 8 starting from the inside lane. The distance from the center of a semicircle to the inside of a lane is called the curve radius of that lane. The curve radius of lane 1 is 36.5 meters, as shown in the figure.
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series ca 5
a. Is the sequence formed by the curve radii arithmetic, geometric, or neither? Explain.
b. Write a rule for the sequence formed by the curve radii.
c. World records must be set on tracks that have a curve radius of at most 50 meters in the outside lane. Does the track shown meet the requirement? Explain.
Answer:

Question 6.
The diagram shows the bounce heights of a basketball and a baseball dropped from a height of 10 feet. On each bounce, the basketball bounces to 36% of its previous height, and the baseball bounces to 30% of its previous height. About how much greater is the total distance traveled by the basketball than the total distance traveled by the baseball?
Big Ideas Math Algebra 2 Answers Chapter 8 Sequences and Series ca 6
A. 1.34 feet
B. 2.00 feet
C. 2.68 feet
D. 5.63 feet
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. 3x2 – 14 = -20
d. x2 + 2x = -3
e. x2 = 16
f. x2 – 5x – 8 = 0
Answer:

Big Ideas Math Answers Grade 7 Chapter 8 Statistics

Big Ideas Math Answers Grade 7 Chapter 8 Statistics

To enhance your knowledge you have to practice all the problems of Big Ideas Math Answers Grade 7 Chapter 8 Statistics. You can get the pdf’s of Big Ideas Math Book 7th Grade Answer Key Chapter 8 Statistics from bigideasmathanswers.com for free of cost. Refer to the Big Ideas Math Answers Grade 7 Chapter 8 Statistics book and score the maximum marks in the exam. Follow the reference problems and attend the exam with good preparation.

It is necessary to know the important questions according to the latest syllabus. Follow the complete article and you will come to know the various important points regarding Statistics. You can find the complete syllabus in one pdf of Big Ideas Math Book 7th Grade Solutions Chapter 8 Statistics. Maximize your scores with the help of the given information in the below sections.

Big Ideas Math Book 7th Grade Answer Key Chapter 8 Statistics

We provide the weightage key of each topic by which you can figure out the important topics and prepare a schedule according to it. The schedule helps you to prepare each topic within the time period. Know the various methods of solving each problem. Become perfect in the concept of statistics with the given solved examples, guide, exploration, and also the formulae, etc.

There are various topics involved in Statistics like Samples and Populations, Using Random Samples to Describe Populations, Comparing Populations, Using Random Samples to Compare Populations, Statistics, etc. Big Ideas Math Book 7th Grade Answer Key Chapter 8 Statistics pdf download link is available in the next sections. You can also get the video links to understand the concept more clearly.

Performance Task

Lesson: 1 Samples and Populations

Lesson: 2 Using Random Samples to Describe Populations

Lesson: 3 Comparing Populations

Lesson: 4 Using Random Samples to Compare Populations

Chapter: 8 – Statistics 

Statistics STEAM Video/Performance Task

STEAM Video

Comparing Dogs
Although dogs and wolves are the same species, they can have very different characteristics. How are dogs and wolves similar?
Watch the STEAM Video “Comparing Dogs.” Then answer the following questions.
Big Ideas Math Answer Key Grade 7 Chapter 8 Statistics 1
1. In the video, the dogs Devo and Etta are walking in a population park. Describe the of the dogs shown in the sample video. Then describe a of the dogs shown in the video. Explain your reasoning.
2. Dogs, wolves, and dingos are all the same species. This species is called Canis lupus
a. Describe one possible sample of the species. Explain your reasoning.
b. You want to know the average height of an animal in the Canis lupus species. Would you use the entire population of the species or would you use a sample to gather data? Explain.
c. The entire Canis lupus species is a sample of what population? Explain.

Answer:
1. In the video, the dogs Devo and Etta are walking in a population park. The population of the dogs is very big. They may differ from color, size, hair, size and skin yet they all are the same species. Here they have shown Canis Lupus bread dogs. They all are the same bone body structured dogs yet  are differently living for their specification of work performance.

2. Dogs, wolves, and Dingo’s are all the same species. This species is called Canis lupus.
a. A wolf with a grey coat living in forested northern regions of North America is one possible sample of the species for Canis Lupus because they resemble like wolfs yet they match to dogs bone structure and near to dogs D.N.A.

b. If you want to know the average height of an animal in the Canis lupus species you should  use a sample to gather data because they are grand children of extinct dogs species having same D.N.A yet different in color, shapes, hair because they having been changing from one generation to another.

c. The entire Canis lupus species is a sample of what population of mammals and under this genus he listed the dog-like carnivores including domestic dogs, wolves, and jackals.

Performance Task

Estimating Animal Populations
After completing this chapter, you will be able to use the concepts you learned to answer the questions in the STEAM Video Performance Task. You will be given a double box-and-whisker plot that represents the weights of male and female gray wolves.
Big Ideas Math Answer Key Grade 7 Chapter 8 Statistics 2
You will be asked to compare the weights of male and female gray wolves. Why might a researcher want to compare data from two different groups of wildlife?

Statistics Getting Ready for Chapter 8

Chapter Exploration
A population is an entire group of people or objects. A sample is a part of the population. You can use a sample to make an inference, or conclusion about a population.
Big Ideas Math Answer Key Grade 7 Chapter 8 Statistics 3
1. Work with a partner. Identify the population and the sample in each pair.
Big Ideas Math Answer Key Grade 7 Chapter 8 Statistics 4
Answer:
A population is an entire group of people or objects. A sample is a part of the population.

a) In the given picture, the school bus students is the population and the class room students  is sample because individual class strength is used for collecting total school population.

b) In the given picture, the grizzly bear with GPS collars in a park is used as sample to collect the population of grizzly bear in a park.

c) In the given picture, 150 quarters coins are used as sample for the calculating all the total  collection in all quarters circulation .

d) In the given picture, 10 fiction books in the library are used as sample to get the total count of all fiction books present in the library.

2. Work with a partner. When a sample is random, each member of the population is equally likely to be selected. You want to know the favorite activity of students at your school. Tell whether each sample is random. Explain your reasoning.
a. members of the school band
b. students in your math class
c. students who enter your school in a morning
d. school newspaper readers

Answer:
c) Students who enter your school in a morning is a random sample because every student data entering into the school is collected and can help to know the favorite activity of them.

Explanation:
2. a) Members of the school band is a not random sample because it gives only the data of the students in the school band not others.

b) Students in your math class is a not random sample because it gives the data of the students who like math.

c) Students who enter your school in a morning is a random sample because every student data entering into the school is collected and can help to know the favorite activity of them.

d) School newspaper readers is not a random sample because here it only collects the data of the students who are in that respective activity of newspaper readers not other students.

Vocabulary
The following vocabulary terms are defined in this chapter. Think about what each term might mean and record your thoughts.
population
unbiased sample
sample
biased sample

Lesson 8.1 Samples and Populations

A population is an entire group of people or objects. A sample is a part of a population. You can gain information about a population by examining samples of the population.

EXPLORATION 1

Using Samples of Populations
Work with a partner. You want to make conclusions about the favorite extracurricular activities of students at your school.
Big Ideas Math Answer Key Grade 7 Chapter 8 Statistics 8.1 1
a. Identify the population. Then identify five samples of the population.
b. When a sample is selected at random, each member of the population is equally likely to be selected. Are any of the samples in part (a) selected at random? Explain your reasoning.
c. How are the samples below different? Is each conclusion valid? Explain your reasoning.
Big Ideas Math Answer Key Grade 7 Chapter 8 Statistics 8.1 2
d. Write a survey question about a topic that interests you. How can you choose people to survey so that you can use the results to make a valid conclusion?

Answer:
Option B is correct to make conclusions about the favorite extracurricular activities of students at your school.

Explanation:
Option B is correct because the information is collected randomly and gives the accurate result conclusions about the favorite extracurricular activities of students at your school.

An unbiased sample  is representative of a population. It is selected at random unbiased sample and is large enough to provide accurate data. A biased sample is not representative of a population. One or more parts of the population are favored over others
Big Ideas Math Answer Key Grade 7 Chapter 8 Statistics 8.1 3

Try It

Question 1.
WHAT IF?
You want to estimate the number of twelfth-grade students in a high school who ride a bus to school. Which sample is unbiased? Explain.
Answer:
Option A is correct.

Explanation:
Option A is correct to collect the twelfth- grade population. Then identity five samples of the population n make a conclusion of the students who come in bus to school.

Question 2.
You want to estimate the number of eighth-grade students in your school who find it relaxing to listen to music. You consider two samples.

  • fifteen randomly selected members of the band
  • every fifth student whose name appears on an alphabetical list of eighth-grade students
    Which sample is unbiased? Explain.

Answer:
Every fifth student whose name appears on an alphabetical list of eighth-grade students.

Explanation:
Every fifth student whose name appears on an alphabetical list of eighth-grade students is a unbiased sample because here it covers large data n gives us the correct data about the students who are relaxed while listening to music.

Question 3.
Four out of five randomly chosen teenagers support the new land fill. So, you conclude that 80% of the residents of your town support the new land fill. Is the conclusion valid? Explain.
Answer:
Yes, its a valid conclusion.

Explanation:
Yes, its a valid conclusion because 4 out of 5 people in the town means 80% of the people are supporting the new land bill mathematically.

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 4.
WRITING
You want to estimate the number of students in your school who play a school sport. You ask 40 honors students at random whether they play a school sport. Is this sample biased or unbiased? Explain.
Answer:
Yea, its a unbiased sample.

Explanation:
Yes, its a unbiased sample because it gives the information about the number of students who play a school sport in the school randomly and accurate data.

Question 5.
ANALYZING A CONCLUSION
You survey 50 randomly chosen audience members at a theater about whether the theater should produce a new musical. The diagram shows the results. You conclude that 80% of the audience members support production of a new musical. Is your conclusion valid? Explain.
Big Ideas Math Answer Key Grade 7 Chapter 8 Statistics 8.1 4
Answer:
Yes, its a valid conclusion.

Explanation:
Yes, its a valid conclusion because 80% of 50 people in the audience members at a theater is equal to 40 people, they support production of a new musical. Out of which 20% of 50 people is equal to 10 people, who are against the production of a new musical.

Self-Assessment for Problem Solving
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 6.
You want to estimate the mean photo size on your cell phone. You choose 30 photos at random from your phone. The total size of the sample is 186 megabytes. Explain whether you can use the sample to estimate the mean size of photos on your cell phone. If so, what is your estimate?
Big Ideas Math Answer Key Grade 7 Chapter 8 Statistics 8.1 5
Answer:
Yes, we can use it for sample to estimate the mean size of photos on your cell phone.

Explanation:
Yes, we can use it for sample to estimate the mean size of photos on your cell phone because it allows to estimate the size of the cell phone.
Mean size of photos = total size of the sample / photos chose
= 186 × 30
= 6.2 bytes.

Question 7.
DIG DEEPER!
You ask 50 randomly chosen employees of a company how many books they read each month. The diagram shows the results. There are 600 people employed by the company. Estimate the number of employees who read at least one book each month.
Big Ideas Math Answer Key Grade 7 Chapter 8 Statistics 8.1 6
Answer:
Number of employees who read at least one book each month = 12

Explanation:
Total number of chosen employee = 50
Total number of people employed = 600
Number of employees who read at least one book each month = Total number of chosen employee / Total number of people employed
= 600 × 50
= 12.

Samples and Populations Homework & Practice 8.1

Review & Refresh

Design a simulation that you can use to model the situation. Then use your simulation to find the experimental probability.
Big Ideas Math Answer Key Grade 7 Chapter 8 Statistics 8.1 7
Question 1.
The probability that a meal at a restaurant is overcooked is 10%. Estimate the probability that exactly 1 of the next 2 meals is overcooked.

Answer:
The probability that exactly 1 of the next 2 meals is overcooked = 3.34%

Explanation:
The probability that a meal at a restaurant is overcooked is 10%
The probability that exactly 1 of the next 2 meals is overcooked = out of 3  meals 1 meal is overcooked of 10%
= 1 × 3 × 10%
= 10 × 3 %
= 3.34 %

Question 2.
The probability that you see a butterfly during a nature center tour is 80%. The probability that you see a turtle is 40%. What is the probability of seeing both?

Answer:
The probability of seeing both = 60%

Explanation:
The probability that you see a butterfly during a nature center tour = 80%.
The probability that you see a turtle = 40%
The probability of seeing both = 80 % + 40%/2
= 120 × 2%
= 60%

Solve the inequality. Graph the solution.
Question 3.
2x – 5 < 9
Answer:
Graph:
Big-Ideas-Math-Book-7th-Grade-Answer-Key-Chapter-8-Statistics-Solve the inequality.-Graph the solution3

Explanation:
2x – 5 < 9
add 5 on both sides
=> 2x – 5 + 5 < 9 + 5
=>2x < 14
=> Divide both sides by 2
=> 2x ÷ 2 < 14 ÷ 2
=> x < 7.

Question 4.
5q + 2 ≥ – 13
Answer:
Graph:
Big-Ideas-Math-Book-7th-Grade-Answer-Key-Chapter-8-Statistics-Solve the inequality.-Graph the solution4

Explanation:
5q + 2 ≥ – 13
Subtract 2 from both sides
=> 5q + 2 – 2 ≥ – 13 – 2
=> 5q ≥ –  15
Divide both sides by 5
=> 5q ÷  5 ≥ – 15 ÷ 5
=> q ≥ – 3

Question 5.
2 > 6 – 3r
Answer:
Graph :
Big-Ideas-Math-Book-7th-Grade-Answer-Key-Chapter-8-Statistics-Solve the inequality.-Graph the solution5

Explanation:
2 > 6 – 3r
subtract 6 from both sides.
2 – 6 > 6 – 3r – 6
=> -4 > -3r
=> -4 ÷ -3 > r
=> 1. 333 > r

Concepts, Skills, & Problem Solving
USING SAMPLES OF POPULATIONS You ask 50 randomly chosen artists in your town about their favorite art form. Determine whether your conclusion is valid. Justify your answer. Favorite Art Form(See Exploration 1, p. 325.)
Big Ideas Math Answer Key Grade 7 Chapter 8 Statistics 8.1 8
Question 6.
You conclude that drawing is the favorite art form of 60% of artists in your town.
Answer:
The conclusion made is not valid that drawing is the favorite art form of 60% of artists in your town.

Explanation:
Total strength of town people = 50
Number of people interested in drawing = 20
Percentage of people interested in drawing = Total strength of town people \times Number of people interested in drawing
= 20 × 50 ×100
= 40%
Hence, the conclusion made is not valid that drawing is the favorite art form of 60% of artists in your town.

Question 7.
You conclude that ceramics is the favorite art form of 10% of people in your town.

Answer:
The conclusion made is  valid that ceramics is the favorite art form of 10% of people in your town.

Explanation:
Total strength of town people = 50
Number of people interested in ceramics = 5
Percentage of people interested in drawing = Total strength of town people / Number of people interested in ceramics
= 5 × 50 × 100
= 10%
Hence, the conclusion made is  valid that ceramics is the favorite art form of 10% of people in your town.

IDENTIFYING POPULATIONS AND SAMPLES Identify the population and the sample.
Question 8.
Big Ideas Math Answer Key Grade 7 Chapter 8 Statistics 8.1 9
Answer:
Residents of New Jersey is the population.
Residents of Ocean Country is the sample.

Question 9.
Big Ideas Math Answer Key Grade 7 Chapter 8 Statistics 8.1 10
Answer:
4 cards is the sample.
All cards in a desk is the population.

IDENTIFYING BIASED AND UNBIASED SAMPLES Determine whether the sample is biased or unbiased. Explain.
Question 10.
You want to estimate the number of books students in your school read over the summer. You survey every fourth student who enters the school.

Answer:
Its a unbiased sample.

Explanation:
You can estimate the number of students who are reading books over the summer by the survey every fourth student who enters the school  because its a unbiased sample which gives you the accurate rate of information.

Question 11.
You want to estimate the number of people in a town who think that a park needs to be remodeled. You survey every 10th person who enters the park.

Answer:
Its a biased sample.

Explanation:
You cannot estimate the number of people in a town who think that a park needs to be remodeled by the survey asking every 10th person who enters the park because its a biased sample as every person entering into the park would like it to be remodeled and its not going to be accurate.

Question 12.
MODELING REAL LIFE
You want to determine the number of students in your school who have visited a science museum. You survey 50 students at random. Twenty have visited a science museum, and thirty have not. So, you conclude that40% of the students in your school have visited a science museum. Is your conclusion valid? Explain.
Big Ideas Math Answer Key Grade 7 Chapter 8 Statistics 8.1 11
Answer:
Its a valid Conclusion that 40% of the students in your school have visited a science museum.

Explanation:
Total number of students the survey taken = 50
Number of students in visited science museum = 20
Number of students who did not visit science museum = 30
Percentage of students visited a science museum = Total number of students the survey taken – Number of students in visited science museum
= 20 ×50 × 100
= 40%

Question 13.
USING A SAMPLE
Which sample is better for making an estimate? Explain.
Big Ideas Math Answer Key Grade 7 Chapter 8 Statistics 8.1 12

Answer:
Sample B -A random sample of 500 pencils from 1 machines  is better than Sample A – A random sample of 500 pencils from 20 machines

Explanation:
Sample A = A random sample of 500 pencils from 20 machines.
Sample B = A random sample of 500 pencils from 1 machines.

Sample B -A random sample of 500 pencils from 1 machines  is better than Sample A – A random sample of 500 pencils from 20 machines  because its gives us the accurate and sufficient data of pencils produced and how many are defective among them. Its a biased sample with correct data.

CONDUCTING SURVEYS Determine whether you should survey the population or a sample. Explain.

Question 14.
You want to know the average height of seventh graders in the United States.
Answer:
Survey on sample of the seventh graders should be taken.

Explanation:
Survey on sample of  the seventh graders should be taken because the total population in United States is very large and its very difficult to survey on the population. Sample survey will be easy and we can get the information  regarding the average height of seventh graders in the United States accurately.

Question 15.
You want to know the favorite types of music of students in your homeroom.
Answer:
Survey on Population should be taken.

Explanation:
Survey on the population in your homeroom should be taken because its a very small area of some limited people and its better to take than sample survey to know the favorite types of music of students.

Question 16.
CRITICAL THINKING

Does increasing the size of a sample necessarily make the sample more representative of a population? Give an example to support your explanation.

Answer:
Yes, increasing the size of the sample necessarily makes the sample more representative of a population. A representative sample is a group or set chosen from a larger statistical population according to specified characteristics. A random sample is a group or set chosen in a random manner from a larger population.

Explanation:
Yes, increasing the size of the sample necessarily makes the sample more representative of a population lead to more accurate or representative results at the same time when it comes to surveying large populations, bigger isn’t always better.
For Example:
You want to know the average height of students in the UK .

Question 17.
LOGIC
A person surveys residents of a town to determine whether a skateboarding ban should be overturned. Describe how the person can conduct the survey so that the sample is biased toward overturning the ban.
Answer:
Yes, the sample is biased toward overturning the ban.

Explanation:
If the person takes the survey on a skateboarding ban should be overturned from every tenth person living in a town then the person surveys is  going to be a biased  sample toward overturning the ban because every tenth person wishes to overturning on the ban.

Question 18.
MODELING REAL LIFE
You ask 20 randomly chosen environmental scientists from your state to name their favorite way to eliminate waste. There are 200 environmental scientists in your state. Estimate the number of environmental scientists in your state whose favorite way to eliminate waste is recycling.
Big Ideas Math Answer Key Grade 7 Chapter 8 Statistics 8.1 13
Answer:
Number of environmental scientists in your state whose favorite way to eliminate waste is recycling = 20

Explanation:
Total number of scientists in your state = 200
Number of scientists randomly ask in your state = 20
Out of 20 randomly asked scientists whose favorite way to eliminate waste is recycling = 2
=> 2 × 20 = 10%
Number of environmental scientists in your state whose favorite way to eliminate waste is recycling = Total number of scientists in your state x Out of 20 randomly asked scientists whose favorite way to eliminate waste is recycling \times100
= 200  × 10 × 100 = 20.

Question 19.
MODELING REAL LIFE
To predict the result of a mayoral election, you survey 50 likely voters at random. The diagram shows the results. Describe whether the sample can be used to predict the outcome of the election. If so, what is your prediction for the number of votes received by the winner assuming that500 people vote?
Big Ideas Math Answer Key Grade 7 Chapter 8 Statistics 8.1 14
Answer:
Yes,  the sample can be used to predict the outcome of the election because  it lets us know how voters are going to vote and whose going to be the winner.
Number of votes candidate A gets  = 330

Explanation:
Number of voters randomly taken in the survey = 50
Total number of people voting = 500
Yes,  the sample can be used to predict the outcome of the election because  it lets us know how voters are going to vote and whose going to be the winner.
According to survey candidate A gets 33 out of 50.
=> 33 ÷ 50 ×100 = 66 %
According to survey candidate B gets 12 out of 50.
=> 12 ÷ 50 ×100 = 24%
According to survey others gets 5 out of 50.
=> 5 ÷ 50 ×100 = 10%
So, the winner according to survey prediction is candidate A.
Number of votes candidate A gets  = 66% of 500
=> 330 votes.

Question 20.
DIG DEEPER!
You ask 100 randomly chosen dog owners in your town how many dogs they own. The results are shown in the table. There are 500 dog owners in your town.
Big Ideas Math Answer Key Grade 7 Chapter 8 Statistics 8.1 15
a. Estimate the median number of dogs per dog owner in your town. Justify your answer.
b. Estimate the mean number of dogs per dog owner in your town. Justify your answer.

Answer:
a) The median number of dogs per dog owner in your town = 4.
b) The mean number of dogs per dog owner in your town = 20.

Explanation:
a) The median is the middle number in a sorted, ascending or descending, list of numbers and can be more descriptive of that data set than the average.
The sorted order of the frequency = 1 3 4 38 54
The median number of dogs per dog owner in your town = 4

b) Explanation:
The mean is the average  of the numbers.
Number of  listed dogs = 5
Total sum of the frequency of the dogs listed = 54 + 38 + 3 + 1 + 4 = 100
The mean number of dogs per dog owner in your town = 100 ÷ 5 = 20

Lesson 8.2 Using Random Samples to Describe Populations

EXPLORATION 1

Exploring Variability in Samples
Work with a partner. Sixty percent of all seventh graders have visited a planetarium.
a. Design a simulation using packing peanuts. Mark 60% of the packing peanuts and put them in a paper bag. What does choosing a marked peanut represent?
Big Ideas Math Answers 7th Grade Chapter 8 Statistics 8.2 1
b. Simulate a sample of 25 students by choosing peanuts from the bag, replacing the peanut each time. Record the results.
c. Find the percent of students in the sample who have visited a planetarium. Compare this value to the actual percent of all seventh graders who have visited a planetarium.
d. Record the percent in part(c) from each pair in the class. Use a dot plot to display the data. Describe the variation in the data.
Big Ideas Math Answers 7th Grade Chapter 8 Statistics 8.2 2

Try It

Question 1.
Use each sample to make an estimate for the number of students in your school who prefer rap music. Describe the center and the variation of the estimates.

Answer:
The Variance is defined as the average of the squared differences from the Mean.
The center of the estimates = 2
The variation of the estimates = 0.64
The Standard Deviation of the estimates = 0.8

Explanation:
Survey randomly took in the school = 20
Total number of students in the school = 840
Number of students who choose rap music by me =  1
Number of students who choose rap music by friend A = 2
Number of students who choose rap music by friend B = 1
Number of students who choose rap music by friend C = 1
Number of students who choose rap music by friend D = 3
The center of the estimates =  1 2 3 = 2
Mean of the estimates =  1+ 2 + 1 + 1 + 3 ÷5= 8 ÷ 5 = 1.6
The variation of the estimates =  (1- 1.6)^2 + (2 – 1.6)^2 +(1- 1.6)^2 + (1- 1.6)^2 + (3 – 1.6)^2 ÷ 5
= (-0.6)^2 + (0.4)^2 + (-0.6)^2 + (-0.6)^2 + (1.4)^2 ÷ 5
= 0.36 + 0.16 + 0.36 + 0.36 + 1.96  ÷  5
= 3.20 ÷ 5
= 0.64
The Standard Deviation of the estimates = \sqrt{0.64} = 0.8

Question 2.
Repeat Example 2, but estimate the medians instead of the means.
Answer:
Median of the students who choose pop music = 5

Explanation:
The six estimates are that students with part time jobs work = 5 7 7 3 7 7
8 and 9 hours per week. The estimates have an average  range from 9 – 5 = 4 hours.
Sorted estimates that students with part time jobs work =  3 5 7
Median of the students who choose pop music = 5

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 3.
USING MULTIPLE RANDOM SAMPLES
Use each sample in Example 1 to make an estimate for the number of students in your school who prefer rock music. Describe the variation of the estimates.

Answer:
The variation of the estimates = 1.84

Explanation:
Estimate for the number of students in your school who prefer rock music = 4 7 5 4 3
Sorted estimate for the number of students in your school who prefer rock music = 3 4 5  7
Mean of the estimate for the number of students in your school who prefer rock music = 4 +7 +5 +4 +3 ÷  5
= 4.6
The variation of the estimates = (4-4.6)^2 + ( 7-4.6)^2 + ( 5 – 4.6)^2 + ( 4-4.6)^2 + ( 3 – 4.6)^2 ÷ 5
= (-0.6)^2 + ( 2.4)^2 + ( 0.4)^2 + (-0.6)^2  + (- 1.6)^2 ÷ 5
= 0.36 + 5.76 + 0.16 + 0.36 + 2.56 ÷ 5
= 9.2 ÷ 5
= 1.84

Question 4.
ESTIMATING AN AVERAGE OF A POPULATION
You want to know the mean number of hours music students at your school practice each week. At each of three music classes you randomly survey 10 students. Your results are shown. Use all three samples to make one estimate for the mean number of hours music students practice each week.
Big Ideas Math Answers 7th Grade Chapter 8 Statistics 8.2 3
Answer:
Mean value of hours for number of students practiced for each week for all three classes A,B,C
= 4.93

Explanation:
Mean value  of hours for number of students practiced for each week of class A
= 6 + 5 + 5 + 6 + 4 + 6 + 8 + 5 + 5 + 2+ 6 \div 10
= 58 ÷ 10
= 5.8
Mean value of hours for number of students practiced for each week of class B
= 0 + 6 + 6 + 5 + 4 + 5 + 6 + 3 + 4 + 9 ÷ 10
= 48 ÷ 10
=4.8
Mean value of hours for number of students practiced for each week of class C
= 4 + 5 + 6 + 4 + 3 + 2 + 2 + 3 + 12 + 1 ÷ 10
=  42 ÷10
= 4.2
Mean value of hours for number of students practiced for each week for all three classes A,B,C
= 5.8 + 4.8 + 4.2 ÷3
= 14.8 ÷ 3
= 4.93

Self-Assessment for Problem Solving
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 5.
Repeat Example 3 with the assumption that 50% of all seventh graders have visited a planetarium.

Answer:
So, most of the samples are within 10% of the actual population that of all seventh graders have visited a planetarium.

Explanation:
The actual percentage of 60% , the number of samples is 200 and the sample size is 50.
The estimates are clustered around 50% , most are between 45 and 60.
So, most of the samples are within 10% of the actual population.

Question 6.
Forty percent of all seventh graders have visited a state park. How closely do 200 random samples of 50 students estimate the percent of seventh graders who have visited a state park? Use a simulation to support your answer.
Big Ideas Math Answers 7th Grade Chapter 8 Statistics 8.2 4
Answer:
Its a biased sample of 200 random samples of 50 students for  estimating  the percent of seventh graders who have visited a state park.

Explanation:
40% of the all seventh graders have visited a state park.
Total samples taken = 200
Number of students randomly taken = 50
Its a biased sample of 200 random samples of 50 students for estimating the percent of seventh graders who have visited a state park because it does not gives the correct information of the seventh graders and they  may or may not be the visitors of the park.

Using Random Samples to Describe Populations Homework & Practice 8.2

Review & Refresh

You ask 100 randomly chosen high school students whether they support a new college in your town. Determine whether your conclusion is valid.
Big Ideas Math Answers 7th Grade Chapter 8 Statistics 8.2 5
Question 1.
You conclude that 85% of high school students in your town support the new college.
Answer:
No, its not a valid conclusion.

Explanation:
Because it is randomly chosen for only 100 people for one set.
It can be 85  do not support for another set of 100 people who are randomly chosen.
You support the statement if total 85 % of the people in the town support the new college .

 

Question 2.
You conclude that 15% of residents in your town do not support the new college.
Answer:
No, its not a valid conclusion.

Explanation:
No, its not a valid conclusion because this percentage which is taken is only for respected chosen students not on the total population of the town who does not prefer new colleges in the town.

Write and solve a proportion to answer the question.
Question 3.
What percent of 30 is 12?
Answer:
3.6 is the percent of 30 is 12.

Explanation:
30 = 100 %
X = 12
Cross multiply the equation
=> X × 100 = 30 ×  12
=> X ×  100 = 360
=> X = 360 × 100
=> X = 3.6

Question 4.
17 is what percent of 68?
Answer:
25 percent of 68 is 17.

Explanation:
68 = 100%
17 = X%
Cross multiply the equation
X × 68 = 17 × 100
=> X × 68 = 1700
=> X = 1700 ÷ 68
=> X = 25

Concepts, Skills, & Problem Solving
EXPLORING VARIABILITY IN SAMPLES Thirty percent of all seventh graders own a bracelet. Explain whether the sample closely estimates the percentage of seventh graders who own a bracelet. (See Exploration 1, p. 331.)
Big Ideas Math Answers 7th Grade Chapter 8 Statistics 8.2 6
Question 5.
50 seventh graders, 14 own a bracelet
Answer:
Its  not a valid conclusion that 50 seventh graders 14 own a bracelet.

Explanation:
Thirty percent of all seventh graders own a bracelet.
30 percent of 50 seventh graders.
=> 30 × 50 ÷ 100
=> 1500 ÷ 100
=> 15.
Its  not a valid conclusion that 50 seventh graders 14 own a bracelet because 15 out of 30 own a bracelet.

Question 6.
30 seventh graders, 3 own a bracelet
Answer:
Its  not a valid conclusion that 30 seventh graders 3 own a bracelet.

Explanation:
30 percent of 30 seventh graders.
30 × 30 ÷ 100
=> 900 ÷ 100
=> 9.

Question 7.
USING MULTIPLE RANDOM SAMPLES
A store owner wants to know how many of her 600 regular customers prefer canned vegetables. Each of her three cashiers randomly surveys 20 regular customers. The table shows the results.
Big Ideas Math Answers 7th Grade Chapter 8 Statistics 8.2 7
a. Use each sample to make an estimate for the number of regular customers of the store who prefer fresh vegetables.
b. Describe the variation of the estimates.

Answer:
a) The Estimation for the number of regular customers of the store who prefer fresh
vegetables = 47.
b) The variation of the estimates = 12.6889.

Explanation:
a) Total number of regular customers to the store = 600
Total number of cashier who takes the survey = 3
Number of customers they have taken survey = 3 × 20 = 60
According to the three survey, number of customers who willing to take fresh vegetables
= (11 + 14 + 12) = 47

b. Describe the variation of the estimates.
Mean of estimates = (11 + 14 + 12) ÷ 3 = 47 ÷ 3 = 15.67
Variation of the estimates = (11 – 15.67)^2 + (14 – 15.67)^2  + (12 – 15.67)^2 ÷ 3
= ( -4.67)^2 + (-1.67)^2 + (-3.67)^2 ÷ 3
= 21.8089 + 2.7889 + 13.4689 ÷ 3
= 38.0667 ÷ 3
= 12.6889

Question 8.
USING MULTIPLE RANDOM SAMPLES
An arcade manager wants to know how many of his 750 regular customers prefer to visit in the winter. Each of five state members randomly surveys 25 regular customers. The table shows the results.
Big Ideas Math Answers 7th Grade Chapter 8 Statistics 8.2 8
a. Use each sample to make an estimate for the number of regular customers who prefer to visit in the winter.
b. Describe the variation of the estimates.
Answer:
a) An estimate for the number of regular customers who prefer to visit in the winter according to survey sample = 52.

b) Variation of the estimates = 1.04

Explanation:
a) Total number of regular customers who prefer to visit in the winter taken in the five survey sample = 25 × 5 = 125
An estimate for the number of regular customers who prefer to visit in the winter according to survey sample = 11 + 10 +9 + 10 +12
= 52

b) Mean of estimates = 11 + 10 +9 + 10 +12 ÷ 5 = 52 ÷ 5 = 10.4
Variation of the estimates = (11 – 10.4)^2 + ( 10 -10.4)^2 + (9  – 10.4)^2 + ( 10 – 10.4)^2 + ( 12 -10.4)^ 2 ÷5
= ( 0.6)^2 + ( -0.4)^2 + ( -1.4)^2 + ( -0.4)^2 + (1.6)^2 ÷5
= 0.36 + 0.16 + 1.96 + 0.16 + 2.56 ÷5
= 5.2 ÷5
= 1.04

Question 9.
ESTIMATING A MEAN OF A POPULATION
A park ranger wants to know the mean number of nights students in your school plan to camp next summer. The park ranger randomly surveys 10 students from each class. The results are shown.
Big Ideas Math Answers 7th Grade Chapter 8 Statistics 8.2 9
a. Use each sample to make an estimate for the mean number of nights students in your school plan to camp next summer. Describe the variation of the estimates.
b. Use all four samples to make one estimate for the mean number of nights students plan to camp next summer.

Answer:
a) The variation of the estimates is quite increasing from one to other sample.

b) One estimate for the mean number of nights students plan to camp next summer = 3.3.

Explanation:
a)  Mean value of  Sample A= 0+5+2+3+0+6+0+10+3+0 ÷ 10 =  29 ÷ 10 = 2.9
Variation of Sample A= (0-2.9)^2 + (5-2.9)^2 + (2-2.9)^2 + (3-2.9)^2 + (0-2.9)^2 + (6-2.9)^2 + (0-2.9)^2 + (10-2.9)^2 + (3-2.9)^2 + (0-2.9)^2  ÷ 10
= (-2.9)^2 + (2.1)^2 + (-0.9)^2 + (0.1)^2 + (-2.9)^2 + (3.1)^2 + (-2.9)^2 + (3.1)^2 + (0.1)^2 + (-2.9)^2  ÷ 10
= 8.41 + 4.41 + 0.81 + 0.01 + 8.41 + 9.61 + 8.41 + 9.61 +0.01 + 8.41 ÷ 10
= 58.1 ÷ 10
=5.81

Mean value of Sample B= 14+0+0+6+5+0+1+2+2+5 ÷ 10 = 35 ÷ 10 = 3.5
Variation of Sample B= (14-3.5)^2 + (0-3.5)^2 + (0-3.5)^2 + (6-3.5)^2 + (5-3.5)^2 + (0-3.5)^2 + (1-3.5)^2 + (2-3.5)^2 + (2-3.5)^2 + (5-3.5)^2 ÷ 10
= (10.5)^2 + (-3.5)^2 + (-3.5)^2 + (2.5)^2 + (1.5)^2 + (-3.5)^2 + (-2.5)^2 + (-1.5)^2 + (-1.5)^2 + (1.5)^2 ÷ 10
= 110.25 + 12.25 + 12.25 + 6.25 + 2.25 + 12.25 + 6.25 +2.25 + 2.25 +2.25 ÷ 10
= 168.5 ÷ 10
= 16.85

Mean value of Sample C = 8+8+2+3+4+1+0+0+0+6 ÷ 10 = 32 ÷ 10 = 3.2
Variation of Sample C= (8-3.2)^2 + (8-3.2)^2 + (2-3.2)^2 + (3-3.2)^2 + (4-3.2)^2 + (1-3.2)^2 + (0-3.2)^2 + (0-3.2)^2 + (0-3.2)^2 + (6-3.2)^2 ÷ 10
= (4.8)^2 + (4.8)^2 + (-1.2)^2 + (-0.2)^2 + (0.8)^2 + (-2.2)^2 + (-2.2)^2 + (-2.2)^2 + (-3.2)^2 + (2.8)^2 ÷ 10
= 23.04 + 23.04 + 1.44 + 0.04 + 0.64 + 4.84 + 4.84 + 4.84 + 10.24 + 7.84 ÷ 10
= 80.8 ÷ 10
= 8.08

Mean value = 10+10+5+6+1+0+0+0+4+0 ÷ 10 = 36 ÷ 10 = 3.6
Variation of Sample D= (10-3.6)^2 + (10-3.6)^2 + (5-3.6)^2 + (6-3.6)^2 + (1-3.6)^2 + (0-3.6)^2 + (0-3.6)^2 + (0-3.6)^2 + (4-3.6)^2 + (0-3.6)^2 ÷ 10
= (6.4)^2 + (6.4)^2 + (1.4)^2 + (2.4)^2 + (-2.6)^2 + (-3.6)^2 + (-3.6)^2 + (-3.6)^2 + (0.4)^2 + (-3.6)^2 ÷ 10
= 40.96 + 40.96 + 1.96 + 5.76 + 6.76 + 12.96 + 12.96 + 12.96 + 0.16 +12.96 ÷ 10
= 148.4 ÷ 10
= 14.84

b) One estimate for the mean number of nights students plan to camp next summer
= 2.9 +3.5+3.2+3.6 ÷ 4
= 13.2 ÷ 4
= 3.3

Question 10.
ESTIMATING A MEDIAN OF A POPULATION
Repeat Exercise 9, but estimate the medians instead of the means.
Answer:
Median value of Sample A = 3 + 5 ÷2 = 8 ÷ 2 = 4.
Median value of Sample B = 2 + 5 ÷2 = 7 ÷ 2 = 3.5.
Median value of Sample C = 3 + 4 ÷ 2 = 7 ÷ 2 = 3.5.
Median value  of Sample C = 4 + 5 ÷ 2 = 9 ÷ 2 = 4.5.

Explanation:
Sample A= 0+5+2+3+0+6+0+10+3+0
Ascending order of  Sample A= 0 2 3 5 6 10
Median value of Sample A = 3 + 5 ÷2 = 8 ÷ 2 = 4

Sample B = 14+0+0+6+5+0+1+2+2+5
Ascending order of Sample B = 0 1 2 5 6 14
Median value of Sample B = 2 + 5 ÷2 = 7 ÷ 2 = 3.5

Sample C = 8+8+2+3+4+1+0+0+0+6
Ascending order of Sample C = 0 1 2 3 4 6 8
Median value of Sample C = 3 + 4 ÷ 2 = 7 ÷ 2 = 3.5

Sample D = 10+10+5+6+1+0+0+0+4+0
Ascending order of Sample D = 0 1 4 5 6 10
Median value  of Sample C = 4 + 5 ÷ 2 = 9 ÷ 2 = 4.5

Question 11.
DESCRIBING SAMPLE VARIATION
Fifty-five percent of doctors at a hospital prescribe a particular medication. A simulation with 200 random samples of 50 doctors each is shown. Describe how the sample percentages vary.
Big Ideas Math Answers 7th Grade Chapter 8 Statistics 8.2 10
Answer:
There is a lot of change in the range of the sample percentages with respective to the frequency. The graph increases from 0 range and reaches to the highest range and later it declines to zero in the last.

Explanation:
The range of sample variation  of frequency from 0.30 to 0.50 is from 0 to 18, here the graph range constantly increases from low to some extent.
The range of the sample variation of frequency from 0.50 to 0.60 is from 18 to 22, here the graph range continuously increased and reached to the highest range and declines to middle range.
The range of the sample variation of frequency from 0.60 to 0.80 is from  18 to 0 , here the graph declines and in between its increases and in the last it declines to zero completely.

Question 12.
MODELING REAL LIFE
Sixty percent of vacationers enjoy water parks. Use technology to generate 20 samples of size 100. How closely do the samples estimate the percent of all vacationers who enjoy water parks?
Answer:
Samples vary slightly in terms of their estimation of  the percent of all vacationers who enjoy water parks compared to the people who does not enjoy the water parks. Out of 2000 total vacationers,  50.65% of  people loves the water parks that compared to 49.35% of  the people who does not like water parks.

Explanation:
Big-Ideas-Math-Book-7th-Grade-Answer-Key-Chapter-8-Statistics-MODELING-REAL-LIFE-12
Total number of vacationers = 20 × 100 = 2000
Total number of vacationers who enjoys the water parks = 48 + 53+ 65 + 79+ 87 +11 + 54 + 25 + 36 + 68 + 51 + 45 + 85 + 73 + 44 + 66 + 33 + 28 + 2 + 60 = 1013
Percent of Total number of vacationers who enjoys the water parks = 1013 ÷2000 ×100
= 50. 65 %
Total number of vacationers who enjoys the water parks  = 52 + 47 + 35 + 21 + 13 + 89 + 46 + 75 + 64 + 32 + 49 + 55 + 15 + 27 + 56 + 34 + 67 + 72 + 98 + 40 = 987.
Percent of Total number of vacationers who enjoys the water parks  = 987 ÷2000 ×100
= 49.35%

Question 13.
MODELING REAL LIFE
Thirty percent of all new wooden benches have a patch of chipped paint. Use technology to simulate 100 random samples of 10 wooden benches. How closely do the samples estimate the percent of all wooden benches with a patch of chipped paint?
Big Ideas Math Answers 7th Grade Chapter 8 Statistics 8.2 11
Answer:
The samples estimate the percent of all wooden benches with a patch of chipped paint are completely differ because almost 79 % are having chipped paint on the wooden bench and 21 % does not have chipped paint on the wooden bench. Majority goes with yes.

Explanation:
Big-Ideas-Math-Book-7th-Grade-Answer-Key-Chapter-8-Statistics-MODELING-REAL-LIFE-13
This means 79% of the students are having patch of chipped paint and 21% of the students are not having patch of chipped paint.

Question 14.
DIG DEEPER!
You want to predict whether a proposal will be accepted by likely voters. You randomly sample 3 different groups of 100 likely voters. The results are shown. Do you expect the proposal to be accepted? Justify your answer.
Big Ideas Math Answers 7th Grade Chapter 8 Statistics 8.2 12
Answer:
No, I think proposal is not to be accepted because out of 300 people opposing people strength is more than supporting people the proposal.

Explanation:
Number of people in the each group = 100
Total number of people in the three groups = 3 × 100 = 300
Number of people who are supporting the proposal in three samples A, B and C
= 48 + 52 +47 = 147
Number of people who are opposing the proposal in three samples A, B and C
= 52 + 48 + 53 = 153
No, I think proposal is not to be accepted because out of 300 people opposing people strength is more than supporting people the proposal.

Question 15.
CRITICAL THINKING
Explain why public opinion polls use sample sizes of more than 1000 people instead of using a smaller sample size.
Answer:
Public opinion polls use sample sizes of more than 1000 people instead of using a smaller sample size because sampling more than 1000 people normally wont add much to the accuracy  given the extra time an money it would cost.

Explanation:
Opinion polls are usually designed to represent the opinions of a population by conducting a series of questions and then extrapolating generalities in ratio or within confidence intervals.
Public opinion polls use sample sizes of more than 1000 people instead of using a smaller sample size because sampling more than 1000 people normally wont add much to the accuracy  given the extra time an money it would cost.
For example, in a population of 5000, 10% would be 500. In a population of 200,000, 10% would be 20,000. This exceeds 1000, so in this case the maximum would be 1000. Even in a population of 200,000, sampling 1000 people will normally give a fairly accurate result.

Lesson 8.3 Comparing Populations

EXPLORATION 1

Comparing Two Data Distributions
Work with a partner.
a. Does each data display show? Explain.
Big Ideas Math Answers Grade 7 Chapter 8 Statistics 8.3 1
b. How can you describe the overlap of two data distributions using words? How can you describe the overlap numerically?
c. In which pair of data sets is the difference in the measures of center the most significant? Explain your reasoning.

Answer:
a)  Yes, the female students and male students data is overlapping as they are having same common sleeping hours in them from 8.5 to 10.5 hours and other data distribution gets overlapping from 59 to 63 inches.

b) The data distribution hours for sleeping for both female and male students is commonly same and in the heights both of them are completely different as they is no common height point in them as same. The female students and male students data is overlapping as they are having same common sleeping hours in them  from 8.5 to 10.5 hours. Here, the data distribution gets overlapping from 59 to 63 inches.

c) There is a lot of difference in the graph of age chart data distribution that of  the first and second data distribution because in the center of 4, there is a big difference of 25 in this set.

Explanation:
a) Overlap means to have something in common with. Female students sleeping hours starts from 5 to 10.5 hours, whereas male students sleeping hours starts from 6 to 10.5 hours. The female students and male students data is overlapping as they are having same common sleeping hours in them  from 8.5 to 10.5 hours.

b) In the first data distribution, Female students and male students maximum sleep for 4 hours. Female students sleeping hours are maximum from 5 to 9 hours and male students sleeping hours are maximum from 6 to 8.5 hours. In the second distribution, Female students height starts from 57 to 63 inches. Male students height starts from 59 to 65 inches. Here, the data distribution gets overlapping from 59 to 63 inches. Maximum height of male students is 62 inches whereas female students is 60 inches.

c) The data distribution in the center is more significant because it shows almost same difference. the center difference in data set 1 of  9.5 – 8.5 = 1hour  for female students and male students 9 – 7 = 2hours. And maximum height of the both female and male students and who attended the class more. In the chart, the center of difference  of exercises at  the center of 4, class of 10am has total 9+7+3+2+2+2= 25 and the 8pm class strength is zero, which is a big difference in this set.


Try It
Question 1.
Which data set is more likely to contain a value of 70?
Answer:
No data sets has more likely to contain a value of 70 in them.

Explanation:
About 25% in the data set A data values are between 80 to 130.
About 50%  in the data set B data values are between  80 to 100.

Question 2.
Which data set is more likely to contain a value that differs from the center by no more than 3?
Answer:
Data set B has more likely to contain a value that differs from the center by no more than 3.

Explanation:
Data set A center is 60.
Data set B center is 90.
Difference in the intervals from center of set A = 60 to 30 = 3
Difference in the intervals from center of set B = 90 to 80=1

Question 3.
WHAT IF?
Each value in the dot plot for Data set A increases by 30. How does this affect your answers? Explain.
Answer:
If each value in the dot plot for Data set A increases by 30 there is going to be a difference of 2 increased value in the mean value and MAD Value of both the Data Sets.

Explanation:
Data Set A                                                            Data Set B
Mean = 810 ÷ 15 = 54                                         Mean = 420 ÷ 15 = 28
MAD = 244 ÷ 15 = 16                                        MAD = 236 ÷ 15 = 16
Difference in Mean ÷ MAD = 26 ÷ 16 = 1.6
If 30 increased in Data Set A:
=> Mean = 810+30 ÷ 15 = 840 ÷ 15 = 56
=> MAD = 244 + 30 ÷ 15 = 274 ÷ 15 = 18.267


Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 4.
COMPARING POPULATIONS
The double dot plot shows two data sets. Compare the data sets using measures of center and variation. Then express the difference in the measures of center as a multiple of the measure of variation.
Big Ideas Math Answers Grade 7 Chapter 8 Statistics 8.3 2
Answer:
So, the difference in the means is about 1.25 times the MAD.

Explanation:
Total value of Data Set A = 1 +2+2+3+3+3+3+4+4+5 = 30
Mean of Data Set A = 30 ÷ 10 = 3
MAD of Data Set A = |∣−3∣ + |2−3∣ +|2−3∣ +|3−3∣ +|3−3∣ + |3−3∣ + |3−3∣ + |4−3∣ + |4−3∣ + |5−3∣ ÷ 10
= 2 + 1 + 1 + 0 + 0 + 0 + 0 + 0+ 1 + 1 +2 ÷ 10
= 8 ÷ 10
= 0.8
Variance of Data Set A :
= (1 – 3 )^2 + (2 – 3 )^2 + (2 – 3)^2 + (3 – 3)^2 + (3 – 3)^2 + (3 – 3)^2 + (3 – 3)^2 + (4 – 3)^2 + (4 – 3)^2 +
(5 -3)^2  ÷10
= (-2)^2 + (-1)^2 + (-1)^2 + (0)^2 + (0)^2 + (0)^2 + (0)^2 + (1)^2 + (1)^2 + (2)^2  ÷10
= 4 + 1 + 1 + 0 + 0 + 0 + 0 + 1 +1 + 4 ÷10
= 12 ÷10
= 1.2

Total value of Data Set B = 1+1+1+1+2+2+2+2+3+3+3+3 = 24
Mean of Data Set B = 24 ÷ 12 = 2
MAD of  Data Set B = |∣−2∣ + |1−2∣ +|1−2∣ +|1−2∣ +|2−2∣ + |2−2∣ + |2−2∣ + |2−2∣ + |3−2∣ + |3−2∣ +  |3−2∣ + |3−2∣ ÷ 12
= 1 + 1 + 1 + 1 + 0 + 0 + 0 + 0 + 1 + 1 + 1 + 1
= 8
= 0.8
Variance of Data Set B :
(1 -2)^2 + (1 -2)^2 + (1 -2)^2 + (1 -2)^2 + (2 – 2)^2 + (2 – 2)^2 + (2 – 2)^2 + (2 – 2)^2 + (3 -2)^2 +
(3 -2)^2 + (3 -2)^2 + (3 -2)^2  ÷12
= (-1)^2 + (-1)^2 + (-1)^2 + (-1)^2 + (0)^2 + (0)^2 + (0)^2 + (0)^2 + (1)^2 + (1)^2 + (1)^2 + (1)^2 ÷12
= 1 + 1 + 1 + 1 + 0 + 0 + 0 + 0 + 1 + 1 + 1 + 1 ÷ 12
= 8 ÷ 12
= 0.67
Difference in mean ÷ MAD = 3 – 2 ÷ 0.8 = 1 ÷ 0.8 = 1.25
So, the difference in the means is about 1.25 times the MAD.

Question 5.
WHICH ONE DOESN’T BELONG?
You want to compare two populations represented by skewed distributions. Which measure not does belong with the other three? Explain your reasoning.
Big Ideas Math Answers Grade 7 Chapter 8 Statistics 8.3 3
Answer:
IQR of first data set does not belong to the other three measurements because its the difference between the first and third quartiles, whereas median is related to the center value of the data and MAD of a data set is the average distance between each data value and the mean.

Explanation:
Data of first set = 1 2 4
Median of first data =  2
Data of second set = 3
Median of second data = 3
MAD of second data set = 0.8
IQR of first data set = 4 – 1 = 3


Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 6.
The double box-and-whisker plot represents the weights of cats at two shelters. Are the cats significantly heavier at one shelter than at the other? Explain.
Big Ideas Math Answers Grade 7 Chapter 8 Statistics 8.3 4
Answer:
The cats are  not significantly heavier at one shelter than at the other.

Explanation:
Data of cats weights in shelter A =  4 5 7 9 12 pounds
Data of cats weights in shelter B =  6  7 8 11 12 pounds
Median of shelter A cats = 7                                                      Median of shelter B cats = 8
IQR of shelter A cats = 9 – 5 = 4                                                IQR of shelter B cats =11 – 7 = 4
Because the variables are same, you can describe the vision overlap by the expression of  the difference in the medians as a multiple of the IQR .
Difference in Median ÷ IQR = 8 – 7 ÷ 4 = 1 ÷ 4 = 0.25
Because the quotient is less than 2, the difference in the medians is not significant.
The cats are  not significantly heavier at one shelter than at the other.

Question 7.
DIG DEEPER!
Tornadoes in Region A travel significantly farther than tornadoes in Region B. The tornadoes in Region A travel a median of 10 miles. Create a double box-and-whisker plot that can represent the distances traveled by the tornadoes in the two regions.
Answer:
Double box-and-whisker plot that can represent the distances traveled by the tornadoes in the two regions.

Big-Ideas-Math-Book-7th-Grade-Answer-Key-Chapter-8-Statistics-8.3-DIG DEEPER7

Explanation:
Data of Tornadoes in Region A = 4 6 10 14 20
Median of Tornadoes in Region A = 10 miles
Data of Tornadoes in Region B = 8 10 12 18 20
Median of Tornadoes in Region B =12

Comparing Populations Homework & Practice 8.3

Review & Refresh

Twenty percent of all seventh graders have watched a horse race. Explain whether the sample closely estimates the percentage of seventh graders who have watched a horse race.
Big Ideas Math Answers Grade 7 Chapter 8 Statistics 8.3 5
Question 1.
In a sample of 15 seventh graders, 4 have watched a horse race.
Answer:
No, the sample  estimation of  the percentage of 15 seventh graders  who have watched a horse race is not valid.

Explanation:
Number of seventh graders = 15
Twenty percent of all seventh graders have watched a horse race.
=> 20 % of the number of seventh graders = 20 % × 15 = 3
Hence, 3 only watched the horse race not 4.

Question 2.
In a sample of 10 seventh graders, 6 have watched a horse race.
Answer:
No, the sample  estimation of  the percentage of 10 seventh graders  who have watched a horse race is not valid.

Explanation:
Number of seventh graders = 10
Twenty percent of all seventh graders have watched a horse race.
=> 20 % of the number of seventh graders = 20 % × 10 = 2
Hence, 2 only watched the horse race not 6.

 

Find the unit rate.
Question 3.
60 kilometers in 2 hours
Answer:
60 kilometers in 2 hours = 30 kilometer per hour.

Explanation:
Time taken to cover 60 kilometers = 2 hours
Time taken to cover 1 kilometer = 60kilometers ÷ 2 hours
= 30 kilometer per hour

Question 4.
$11.40 for 5 cans
Answer:
Cost for 1 can = $2.28.

Explanation:
Cost for 5 cans = $11.40
Cost for 1 can = $11.40 ÷ 5
= $2.28

 

Concepts, Skills, & Problem Solving

COMPARING TWO DATA DISTRIBUTIONS The double box-and-whisker plot represents the values in two data sets. (See Exploration 1, p. 337.)
Big Ideas Math Answers Grade 7 Chapter 8 Statistics 8.3 6
Question 5.
Does the data display show overlap? Explain.
Answer:18.2
Yes, the data display show overlap.

Explanation:
Data of set A = 1 3 5 7 9
Mean  of set A = 1 + 3 +  5 + 7 + 9  ÷ 5
= 25 ÷ 5 = 5
MAD of set A = |∣−5∣ + |3−5∣ +|5−5∣ +|7−5∣ + |9−5∣  ÷ 5
= 4 + 2 + 0 + 2 + 4 = 20 ÷ 5
= 20 ÷ 5 =4

Data of set B = 10 11 14 15 16
Mean  of set B = 8.2 + 6.7 + 4.2 + 3.2 + 2.2 ÷ 5
=  24.5 ÷ 5 = 4.9
MAD of set B = |∣0−18.2∣ + |11.5−18.2∣ +|14−18.2∣ +|15−18.2∣ + |16−18.2∣  ÷ 5
= 8.2 + 6.7 + 4.2 + 3.2 + 2.2 ÷ 5
= 24.5 ÷ 5 = 4.9
Differences in means ÷ MAD = 18.2 – 4.9 ÷ 5
= 13.3 ÷ 5 = 2.66
So, the differences in means is about  2.66 times the MAD.

Question 6.
Is there a significant difference in the measures of center for the pair of data sets? Explain.
Answer:
No, there is no significant difference in the measures of center for the pair of data sets because the quotient is less than 2, the difference in the medians is not significant.

Explanation:
Data of set A = 1 3 5 7 9                                     Data of set B = 10 11 14 15 16
Median of set A = 5                                            Median of set B = 14
IQR of set A = 7 – 3 = 4                                      IQR of set B = 15 – 11.5 = 3.5
Differences in Median ÷ IQR of set A = 5 – 4 ÷ 4 = 1 ÷ 4 =  0.25
Differences in Median ÷ IQR of set B = 5 – 4 ÷ 3.5 = 0.286

 

COMPARING POPULATIONS Two data sets contain an equal number of values. The double box-and-whisker plot represents the values in the data sets.
Big Ideas Math Answers Grade 7 Chapter 8 Statistics 8.3 7
Question 7.
Compare the data sets using measures of center and variation.
Answer:
Comparison:
Variation of set A  and Variation of set B
=> 0.5432  is greater than 0.1576.

Explanation:
Data of set A = 0.8 1.0 1.1 1.2 1.5
Mean of set A = 0.8 + 1.0 + 1.1 + 1.2 + 1.5 ÷ 5 = 5.6 ÷ 5 = 1.12
Variation of set A = (0.8 -1.12)^2 + (1.0-1.12)^2 + (1.1 -1.12)^2 + (1.2 – 1.12)^2 + (1.5 – 1.12)^2 ÷ 5
= (-0.32)^2 + (-0.12)^2  + (-0.02)^2  + (0.08)^2  + (0.38)^2  ÷ 5
= 0.1024 + 0.0144 + 0.004 + 0.0064 + 0.1444 ÷ 5
= 0.2716 ÷ 5
= 0.5432

Data of set B = 0.3 0.4 0.6 0.9 1.4
Mean of set B = 0.3 + 0.4 + 0.6 + 0.9 + 1.4 ÷ 5 = 3.6 ÷ 5 = 0.72
Variation of set B = (0.3 -0.72)^2 + (0.4 – 0.72)^2 + (0.6 -0.72)^2 + (0.9 – 0.72)^2 + (1.4 – 0.72)^2 ÷ 5
= (-0.42)^2 + (-0.32)^2  + (-0.12)^2  + (0.18)^2  + (0.68)^2  ÷ 5
= 0.1764 + 0.1024 + 0.0144 + 0.0324 + 0.4624 ÷ 5
= 0.788  ÷ 5
= 0.1576

Comparison:
Variation of set A  and Variation of set B
=> 0.5432  is greater than 0.1576.

Question 8.
Which data set is more likely to contain a value of 1.1?
Answer:
Data of set A is more likely to contain a value of 1.1.

Explanation:
Data of set A = 0.8 1.0 1.1 1.2 1.5                                  Data of set B = 0.3 0.4 0.6 0.9 1.4
Median of set A = 1.1                                                     Median of set B = 0.6

Question 9.
Which data set is more likely to contain a value that differs from the center by 0.3?
Answer:
No, data set is more likely to contain a value that differs from the center by 0.3.

Explanation:
Data of set A = 0.8 1.0 1.1 1.2 1.5                                      Data of set B = 0.3 0.4 0.6 0.9 1.4
Median of set A = 1.1                                                         Median of set B = 0.6

Question 10.
DESCRIBING VISUAL OVERLAP
The double dot plot shows the values in two data sets. Express the difference in the measures of center as a multiple of the measure of variation.
Big Ideas Math Answers Grade 7 Chapter 8 Statistics 8.3 8
Answer:
Because the quotient is less than 2, the difference in the median is not significant of data set A.
Because the quotient is more than 2, the difference in the median is significant of data set B.

Explanation:
Data of set A = 30 35 40 45 50 55                                            Data of set B = 10 15 20 25 30
Median of set A = 40 +45 ÷ 2 = 85 ÷ 2 = 42.5                        Median of set B = 20
IQR of set A = 50 – 35 = 15                                                      IQR of set B = 30 – 10 = 20
Differences in medians ÷ IQR of set A = 42.5 – 20 ÷ 15 = 22.5 ÷ 15 = 1.5
Because the quotient is less than 2, the difference in the median is not significant of data set A.
Differences in medians ÷ IQR of set B = 42.5 – 20 ÷ 10 = 22.5 ÷ 10 = 2.25
Because the quotient is more than 2, the difference in the median is significant of data set B.

Question 11.
The distributions of attendance at basketball games and volleyball games at your school are symmetric. Your friend makes a conclusion based on the calculations shown below. Is your friend correct? Explain your reasoning.
Big Ideas Math Answers Grade 7 Chapter 8 Statistics 8.3 9
Answer:
No, my friend is not correct.

Explanation:
Volleyball Game attendance:
Mean = 80
MAD = 20
Basketball Game attendance:
Mean = 160
MAD = 20
The difference in means is 4 times the MAD, so attendance at basketball games is significantly greater than the attendance at volleyball games. Since the MAD is “large”, it implies that the mean of 160 is not a reliable indicator of the other values within the data set.

Question 12.
MODELING REAL LIFE
The double box-and-whisker plot represents the goals scored per game by two hockey teams during a 20-game season. Is the number of goals scored per game significantly greater for one team than the other? Explain.
Big Ideas Math Answers Grade 7 Chapter 8 Statistics 8.3 10
Answer:
Yes, the number of goals scored per game significantly greater for one team than the other.

Explanation:
Data of Team A = 0 2 3 4 6                                              Data of Team B = 0 6 7 8 10
Median of data Team A = 3                                             Median of data Team B = 7
IQR of Team A =  4 – 2 = 2                                               IQR of Team B = 8 – 6 = 2
Differences in medians ÷ IQR = 7 – 3 ÷ 2 = 4  ÷ 2 = 2
So, the quotient is equal to 2 the difference in the medians is significant.

Question 13.
MODELING REAL LIFE
The dot plots show the test scores for two classes taught by the same teacher. Are the test scores significantly greater for one class than the other? Explain.
Big Ideas Math Answers Grade 7 Chapter 8 Statistics 8.3 11
Answer:
No, the test scores  are not significantly greater for one class than the other.

Explanation:
Data of Class A = 70 75 80 85 90 95 100                              Data of Class B = 70 75 80 85 90 95 100
Median of data Class A = 85                                                 Median of data Class B = 85
IQR of Class A =  95 – 85 = 10                                               IQR of Class B = 80 – 70 = 10
Both the variables and the center are similar. So, it is not significant.

Question 14.
PROBLEM SOLVING
A scientist experiments with mold colonies of equal area. She adds a treatment to half of the colonies. After a week, she measures the area of each colony. If the areas are significantly different, the scientist will repeat the experiment. The results are shown. Should the scientist repeat the experiment? Justify your answer.
Big Ideas Math Answers Grade 7 Chapter 8 Statistics 8.3 12
Answer:
Yes,  the scientist should repeat the experiment because areas are significantly different from one to other.

Explanation:
Order of Data of untreated Mold = 2 3 4 5 6 7                               Order of Data of treated Mold = 0 1 2 3 6
Median of untreated Mold = 4 + 5 ÷ 2 = 9 ÷ 2 = 4.5                    Median of treated Mold = 2
IQR of untreated Mold = 4 – 2 = 2                                                  IQR of treated Mold = 2 – 0 = 2
Because the variables are same, you can describe the overlap in areas by expressing the difference in the medians as a multiple of the IQR.
Difference in Medians ÷ IQR = 4.5 – 2 ÷ 2
= 2.5 ÷ 2 = 1.25
Because the quotient is  less than 2, the differences in the medians is not significant.
The areas are not significantly greater than one and other colony.

Lesson 8.4 Using Random Samples to Compare Populations

EXPLORATION 1

Using Random Samples
Work with a partner. You want to compare the numbers of hours spent on homework each week by male and female students in your state. You take a random sample of 15 male students and 15 female students throughout the state.
Big Ideas Math Solutions Grade 7 Chapter 8 Statistics 8.4 1
a. Compare the data in each sample.
b. Are the samples likely to be representative of all male and female students in your state? Explain.
c. You take 100 random samples of 15 male students in your state and 100 random samples of 15 female students in your state and record the median of each sample. The double box-and-whisker plot shows the distributions of the sample medians. Compare the distributions in the double box-and-whisker plot with the distributions of the data in the tables.
Big Ideas Math Solutions Grade 7 Chapter 8 Statistics 8.4 2
d. What can you conclude from the double box-and-whisker plot? Explain. d. How can you use random samples to make accurate comparisons of two populations?
Answer:
a) According to the given data in the table, the center of the Female students is greater than the center of the Male students.

b) Yes, the samples likely to be representative of all male and female students in your state because they are randomly selected students allover the state.

c) The data given in the table and the data given in the double box-and-whisker plot are both completely different and there is a lot of variation in the values. The double box-and-whisker plot has greater values than in the table.

d) According to the data in double box-and-whisker plot, the medians of male students is less than female students. So, the hours of male students spent are less than female students.

Explanation:
a)Data of Male students = 1.5  3  0  2.5  1  8  2.5  1  3  0  6.5  1  5  0  5
Arrange the data in order : 0  1  1.5  2.5  3  5  6.5  8
Median of Male students =  2.5 + 3 ÷ 2 = 5.5 ÷ 2 = 2.75

Data of Female students = 4   0  3  1  1  5  1  3  5.5  10  2  0  6  3.5  2
Arrange the data in order : 0  1  2  3  3.5  4  5.5  6
Median of Female students = 3 + 3.5 ÷ 2 = 6.5 ÷  2 = 3.25

b) Data of Male students = 1.5  3  0  2.5  1  8  2.5  1  3  0  6.5  1  5  0  5
Data of Female students = 4   0  3  1  1  5  1  3  5.5  10  2  0  6  3.5  2
Yes, the samples likely to be representative of all male and female students in your state because they are randomly selected students allover the state.

c)                                                                         Data in the Table:
Data of Male students = 1.5  3  0  2.5  1  8  2.5  1  3  0  6.5  1  5  0  5
Median of Male students =  2.5 + 3 ÷ 2 = 5.5 ÷ 2 = 2.75
Mean of Male students = 1.5 + 3 + 0 + 2.5 + 1 + 8 + 2.5 + 1 + 3 + 0 + 6.5 + 1 + 5 + 0 + 5 ÷ 15
= 40  ÷ 15
=  2.67
Data of Female students = 4   0  3  1  1  5  1  3  5.5  10  2  0  6  3.5  2
Median of Female students = 3 + 3.5 ÷ 2 = 6.5 ÷  2 = 3.25
Mean of Female students = 4  + 0  +  3 + 1 + 1 + 5  + 1 + 3 + 5.5 + 10 + 2 + 0 + 6 + 3.5 + 2 ÷ 15
= 47  ÷ 15
= 3.13
Data in the double box-and-whisker plot:
Data of Male students = 1.5  2.5  3  3.5   6                          Data of Female students = 2  3.5  4  5  7
Mean of Male students = 1.5 + 2.5 + 3 + 3.5 + 6 ÷ 5        Mean of Female students = 2 + 3.5 + 4 + 5 + 7 ÷ 5
=  16.5 ÷ 5 = 3.3                                                                  = 21.5 ÷ 5 = 4.3

d)                                                                  Data in the double box-and-whisker plot:
Data of Male students = 1.5  2.5  3  3.5   6                          Data of Female students = 2  3.5  4  5  7
Median of Male students = 3                                              Median of Female students = 4

 

Try It

Question 1.
The double dot plot shows the weekly reading habits of a random sample of 10 students in each of two schools. Compare the samples using measures of center and variation. Can you determine which school’s students read less? Explain.
Big Ideas Math Solutions Grade 7 Chapter 8 Statistics 8.4 3
Answer:
According to the data of distribution, school A students are reading less compared to that of school B students.

Explanation:
Data of school A = 0  0  1  2  2  3  3  4  5  5
Mean of school A = 0 + 0 + 1 + 2 + 2 +3 + 3 + 4 + 5 + 5 ÷ 10
= 25 ÷ 10 = 2.5
Variation of school A = (0-2.5)^2 + (0-2.5)^2 + (1-2.5)^2 + (2-2.5)^2 + (2-2.5)^2 + (3-2.5)^2 + (3-2.5)^2 + (4-2.5)^2 + (5-2.5)^2 + (5-2.5)^2 ÷ 10
= (-2.5)^2 + (-2.5)^2  + (-1.5)^2 + (-0.5)^2 + (-0.5)^2 + (0.5)^2 + (0.5)^2  + (-1.5)^2  + (2.5)^2 + (2.5)^2 ÷ 10
=  6.25 + 6.25 + 2.25 + 0.25 + 0.25 +  0.25 +  0.25 + 2.25 + 6.25 + 6.25 ÷ 10
= 30.5 ÷ 10
= 3.05

Data of school B = 0  2  3  3  3   4  4  4 5  7
Mean of school B = 0 + 2 + 3 + 3 + 3 + 4 + 4 + 4+ 5 +7 ÷ 10
= 60 ÷ 10 = 6
Variation of school A = (0-6)^2 + (2-6)^2 + (3-6)^2 + (3-6)^2 + (3-6)^2 + (4-6)^2 + (4-6)^2  + (4-6)^2  + (5-6)^2 + (7-6)^2 ÷ 10
= (-6)^2 + (-4)^2 + (-3)^2 + (-3)^2 + (-3)^2 + (-2)^2 + (-2)^2 + (-2)^2 + (-1)^2 + (1)^2 ÷ 10
= 36 + 16 + 9 + 9+ 9 + 4 + 4 + 4 + 1 + 1÷ 10
= 93 ÷ 10
= 9.3

Question 2.
WHAT IF?
Each value in the box-and-whisker plot of the sample medians for Bag A decreases by2. Does this change your answer?
Answer:
Yes, this changes the answer because which makes the medians of Bag A and Bag B same.

Explanation:
Data of bag A = 4  5  5.5  6  6.5                                                        Data of bag B = 2.5  3  3.5  4  5
Median of bag A = 5.5                                                                      Median of bag B = 3.5
IQR of bag A = 6 – 5 = 1                                                                    IQR of bag B = 4 – 3 = 1
If the sample medians for Bag A decreases by 2:
Median of bag A = 5.5 – 2 = 3.5


Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 3.
COMPARING RANDOM SAMPLES
Two boxes each contain 600 numbered tiles. The double dot plot shows a random sample of 8 numbers from each box. Compare the samples using measures of center and variation. Can you determine which box contains tiles with greater numbers? Explain.
Big Ideas Math Solutions Grade 7 Chapter 8 Statistics 8.4 4
Answer:
Box A contains greater numbers more than Box B because its has greater median and also in variation.

Explanation:
Data of Box A = 3  4  5  5  6  6  7  8
Order of Box A =3 4 5 6 7 8
Median of Box A = 5 + 6 ÷ 2 = 11 ÷ 2 =5.5
Mean of Box A = 3 + 4 + 5 + 5 + 6 + 6 + 7 + 8 ÷ 8
= 44 ÷ 8
= 5.5
Variation of Box A = (3 -5.5)^2 + (4-5.5)^2 + (5-5.5)^2 + (5-5.5)^2 + (6-5.5)^2 + (6-5.5)^2 + (7-5.5)^2 + (8-5.5)^2 ÷ 8
=  (-2.5)^2 + (-1.5)^2 + (-0.5)^2 + (-0.5)^2 + (0.5)^2 + (1.5)^2 + (1.5)^2 + (2.5)^2   ÷ 8
=  6.25 + 2.25 + 0.25 + 0.25 + 0.25 + 2.25 + 2.25 + 6.25 ÷ 8
= 20 ÷ 8
= 2.5

Data of Box B = 0  0  0 1  2  3  3  3
Order of Box B = 0 1 2 3
Median of Box B =  1 +2 ÷ 2 = 3 ÷ 2 =1.5
Mean of Box B = 0 + 0 + 0 + 1 + 2 + 3 + 3 + 3 ÷ 8
= 12 ÷ 8
= 1.5
Variation of Box B = (0-1.5)^2 + (0-1.5)^2 + (0-1.5)^2 + (1-1.5)^2 + (2-1.5)^2 + (3-1.5)^2 + (3-1.5)^2 + (3-1.5)^2  ÷ 8
=  (-1.5)^2 + (-1.5)^2 + (-1.5)^2 + (-0.5)^2 + (0.5)^2 + (1.5)^2 + (1.5)^2 + (1.5)^2  ÷ 8
=  2.25 + 2.25 +2.25 + 0.25 +0.25 + 2.25  + 2.25 + 2.25  ÷ 8
=  14 ÷ 8
= 1.75

Question 4.
USING MULTIPLE RANDOM SAMPLES
Two crates each contain 750 objects. The double box-and-whisker plot shows the median weights of 50 random samples of 10 objects from each crate. Can you determine which crate weighs more? Explain.
Big Ideas Math Solutions Grade 7 Chapter 8 Statistics 8.4 5
Answer:
Crate B weighs more than Crate A.

Explanation:
Data of crate A = 6   6.5   7.5   9   9.5
Median of crate A = 7.5
Mean of crate A = 6 + 6.5 + 7.5 + 9 + 9.5 ÷ 5 = 38.5 ÷5 = 7.7
IQR = 9 – 6.5 = 2.5
Data of crate B =  9   10   11   12.5   13.5
Median of crate B = 11
IQR = 12.5 – 10 = 2.5
Mean of crate B = 9 + 10  + 11  + 12.5  + 13.5 ÷ 5 = 56 ÷ 5 = 11.2
The variation in the center of crate B is greater than the measure of crate A. The variation of crate A is similar to the variation of crate B.


Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 5.
The double box-and-whisker plot represents the medians of 100 random samples of 20 battery lives for two cell phone brands. Compare the battery lives of the two brands.
Big Ideas Math Solutions Grade 7 Chapter 8 Statistics 8.4 6
Answer:
The variation in the center of Brand B batteries is greater than the measure of Brand A batteries and the variation of Brand A batteries is similar to the variation of Brand B batteries.

Explanation:
Data of Brand A = 16.5  20  26  28   30                                 Data of Brand B = 24  26  30  34  36
Median of Brand A = 26                                                        Median of Brand B =  30
IQR of Brand A = 28 – 20 = 8                                                IQR of Brand B = 34 – 28 = 6

Question 6.
DIG DEEPER!
The double box-and-whisker plot represents the medians of 50 random samples of 10 wait times at two patient care facilities. Which facility should you choose? Explain your reasoning.
Big Ideas Math Solutions Grade 7 Chapter 8 Statistics 8.4 7
Answer:
Facility B should be chosen than the Facility A.

Explanation:
Data of Facility A =  10  15  20  25  30                      Data of Facility B =  5   15   20   35  45
Median of Facility A = 20                                           Median of Facility B = 20
IQR of Facility A = 25 – 15 = 10                                  IQR of Facility B = 35 – 15 = 20
The variation in the center of  Facility A  is similar to  the measure of Facility B and the variation of Facility B  is greater than  the variation of Facility A .

Using Random Samples to Compare Populations Homework & Practice 8.4

Review & Refresh

The double dot plot shows the values in two data sets.
Big Ideas Math Solutions Grade 7 Chapter 8 Statistics 8.4 8
Question 1.
Compare the data sets using measures of center and variation.
Answer:
The measures of center of set B is greater and the variation of it greater than that of  the variation of set A.

Explanation:
Data of Set A = 10 10 20 20 20 30 30 30  30 40 40 50          Data of set B = 30 40 40 40 50 50 50 50 60 60 70 70
Order: 10 20 30 40 50                                                              Order: 30 40 50 60 70
Median of set A = 30                                                              Median of set B = 50
Mean of set A =  10+10 + 20 + 20 + 20 + 30 + 30 + 30 + 30 + 40 + 40 + 50  ÷ 12
=  330  ÷ 12
= 27.5
Variation of set A = (10 -27.5)^2 + (10-27.5)^2 + (20-27.5)^2 + (20-27.5)^2 + (20-27.5)^2 + (30-27.5)^2 + (30-27.5)^2 + (30-27.5)^2 + (30-27.5)^2 + (40-27.5)^2 + (40-27.5)^2 + (50-27.5)^2  ÷ 12
= (-17.5)^2 + (-17.5)^2 + (-7.5)^2 + (-7.5)^2 + (-7.5)^2 + (2.5)^2 + (2.5)^2 + (2.5)^2 + (2.5)^2 + (12.5)^2 + (12.5)^2 + (22.5)^2  ÷ 12
= 306. 25 + 306.25 + 56.25  + 56.25 + 56.25 + 6.25 + 6.25 +6.25 + 6.25 + 156.25 + 156.25 + 506.25 ÷ 12
=  1625  ÷ 12
= 135.42

Mean of set B= 30 + 40 + 40 + 40 + 50 + 50 + 50 + 50 + 60 + 60 + 70 + 70 ÷ 12
=  610  ÷ 12
= 50.83
Variation of set B = (30-50.83)^2 + (40-50.83)^2 + (40-50.83)^2 + (40-50.83)^2 + (50-50.83)^2 + (50-50.83)^2 + (50-50.83)^2 + (50-50.83)^2 + (60-50.83)^2 + (60-50.83)^2 + (70-50.83)^2 + (70-50.83)^2 ÷ 12
= (-20.83)^2 + (-10.83)^2 + (-10.83)^2 + (-10.83)^2 + (-0.83)^2 + (-0.83)^2 + (-0.83)^2 + (-0.83)^2 + (10.83)^2 + (10.83)^2 + (20.83)^2 + (20.83)^2  ÷ 12
= 433.89 +  117.29 + 117.29 + 117.29 + 0.6889  +  0.6889 + 0.6889 + 0.6889 + 117.29 + 117.29 +  433.89 + 433.89 ÷ 12
= 1890.8756 ÷ 12
= 157.57

Question 2.
Are the values of one data set significantly greater than the values of the other data set? Explain.
Answer:
Yes, the variation of center of set B is greater than the measure of set A and the variation of set B is greater than the variation of set A.

Explanation:
Median of set A = 30                                                              Median of set B = 50
Variation of set A = 135.42                                                    Variation of set B = 157.57

 

Solve the equation. Check your solution.
Question 3.
5b – 3 = 22
Answer:
b= 5.

Explanation:
5b – 3 = 22
5b =  22 + 3
5b = 25
b = 25/5
b= 5.

Check: substitute b =  5 in 5b – 3 = 22.
LHS => 5b – 3 = 22.
=> (5 x 5) -3 = 22
=> 25 – 3 = 22 = RHS

Question 4.
1.5d  + + 3 = – 4.5
Answer:
d = -5

Explanation:
1.5d  + 3 = – 4.5
=>1.5d = -4.5 -3
=> 1.5d = -7.5
=> d = -7.5 ÷ 1.5
=> d = – 5
Checking:
Substitute d = – 5 in Equation:
LHS:    1.5d  + 3
=> (1.5 × -5) + 3
=> -7.5 + 3
=> -4.5 = RHS

Question 5.
4 = 9z – 2
Answer:
z = 0.67

Explanation:
4 = 9z – 2
=> 4 + 2 = 9z
=> 6 = 9z
=>6 ÷ 9 = z
=> 0.67 = z
Checking:
Substitute z = 0.67 in Equation:
RHS:  9z – 2
=> (9 × 0.67) – 2
=> 6 – 2
= 4 = LHS

Concepts, Skills, & Problem Solving

USING RANDOM SAMPLES You want to compare the numbers of hours spent on recreation each week by teachers and non-teachers in your state. You take100 random samples of 15 teachers and 100 random samples of 15 non-teachers throughout the state and record the median value of each sample. The double box-and-whisker plot shows the distributions of sample medians. (See Exploration 1, p. 343.)
Big Ideas Math Solutions Grade 7 Chapter 8 Statistics 8.4 9

Question 6.
Are the samples likely to be representative of all teachers and non-teachers in your state?

Answer:
Yes, the samples are likely to be representative of all teachers and non-teachers in your state because it has been given in the question.

Question 7.
What can you conclude from the double box-and-whisker plot? Explain.

Answer:
The number of hours spent by Non-teachers is more than the teachers.

Explanation:
Median of Non-Teachers = 7
Median of Teachers = 5

Question 8.
COMPARING RANDOM SAMPLES
The double dot plot shows the weekly running habits of athletes at two colleges. Compare the samples using measures of center and variation. Can you determine which college’s athletes spend more time running? Explain.
Big Ideas Math Solutions Grade 7 Chapter 8 Statistics 8.4 10
Answer:
Data of College B has the  greater measure of center and in the measure of variation that of data of college B.
College B athletes spend more time running than college A.

Explanation:
Data of College A = 2  2  4  4  5  5  5  5  6  6  8  8
Order : 2 4 5 6 8
Median of Data A =  5
Variation of Data A = (2-5)^2 + (2-5)^2 + (4-5)^2 + (4-5)^2 + (5-5)^2 + (5-5)^2 + (5-5)^2 + (5-5)^2 + (6-5)^2 + (6-5)^2 + (8-5)^2 + (8-5)^2 ÷  12
= (-3)^2 + (-3)^2 + (-1)^2 + (-1)^2 + (0)^2 + (0)^2 + (0)^2 + (0)^2 + (1)^2 + (1)^2 + (3)^2 + (3)^2 ÷  12
= 9 + 9 + 1 + 1 + 0+ 0 + 0 + 0  + 1 + 1 + 9 + 9 ÷ 12
= 40 ÷ 12
= 3.33

Data of College B = 3  4  5  5  5  6  6  7  7  7  8  9
Order : 3 4 5 6  7 8
Median of Data A =  5 + 6 ÷ 2 = 11 ÷ 2 = 5.5
Variation of Data A = (3-5.5)^2 + (4-5.5)^2 + (5-5.5)^2 + (5-5.5)^2 + (5-5.5)^2 + (6-5.5)^2 + (6-5.5)^2 + (7-5.5)^2 + (7-5.5)^2 + (7-5.5)^2 + (8-5.5)^2 + (9-5.5)^2  ÷  12
= (-2.5)^2 + (-1.5)^2 + (0)^2 + (0)^2 + (0)^2 + (1.5)^2 + (1.5)^2 + (2.5)^2 + (2.5)^2 + (2.5)^2 + (3.5)^2 + (4.5)^2 ÷  12
= 6.25 + 2.25 + 0 + 0 + 0 + 2.25 + 2.25 + 6.25 + 6.25 + 6.25 + 12.25 + 20.25 ÷ 12
=  64.25  ÷ 12
= 5.35

Question 9.
USING MULTIPLE RANDOM SAMPLES
Two lakes each contain about 2000 fish. The double box-and-whisker plot shows the medians of 50 random samples of 14 fish lengths from each lake. Can you determine which lake contains longer fish? Explain.
Big Ideas Math Solutions Grade 7 Chapter 8 Statistics 8.4 11

Answer:
Lake A is the Lake contains longer fish because the measure in the variation is comparatively greater than the measure of variation of Lake B.

Explanation:
Data of Lake A = 6  7  8  9  10                    Data of Lake B =  3  4  5  6  7
Median of Lake A = 8                                 Median of Lake B = 5
IQR of Lake A = 9 – 7 = 2                          IQR of Lake B = 6 – 4 = 2

Question 10.
MODELING REAL LIFE
Two laboratories each produce 800 chemicals. A chemist takes 10 samples of 15 chemicals from each lab, and records the number that pass an inspection. Are the samples likely to be representative of all the chemicals for each lab? If so, which lab has more chemicals that will pass the inspection? Justify your answer.
Big Ideas Math Solutions Grade 7 Chapter 8 Statistics 8.4 12

Answer:
Yes, the samples are likely to be representative of all the chemicals for each lab.
Research Lab B is having the more chemicals that will pass the inspection.

Explanation:
Data of Research Lab A = 14 14 13 15 1515 15 13 14 12
Order : 12  13  14  15
Median of Research Lab A =13 +14 ÷ 2=27 ÷ 2 = 3.5
Mean of Research Lab A = 14 +14 +13+ 15 +15 +15 + 15 + 13 + 14 + 12 ÷ 10
=  140 ÷ 10
=    14

Data of Research Lab B = 12 9 10 14 12 11 14 11 11 15
Order : 9  10  11  12  14  15
Median of Research Lab B = 11+12 ÷ 2 = 23 ÷ 2 = 11.5
Mean of Research Lab B = 12  + 9 + 10 +14 + 12 +11 + 14 + 11 + 11 + 15 ÷ 10
= 119 ÷ 10
=  11.9
So, the variation of Research Lab A is having greater values and the measure of the center  Research Lab B is  greater than the center of the Research Lab A.

Question 11.
MODELING REAL LIFE
A farmer grows two types of corn seedlings. There are 1000 seedlings of each type. The double box-and-whisker plot represents the median growths of 50 random samples of 7 corn seedlings of each type. Compare the growths of each type of corn seedling. Justify your result.
Big Ideas Math Solutions Grade 7 Chapter 8 Statistics 8.4 13
Answer:
The growth of Type B type of corn seedling is more that of growth of Type B of corn seedling.

Explanation:
Data of Type A = 1.5  3  3.5  4  5.5                                                    Data of Type B = 2.5  5  6.5  7  9
Median of Type A = 3.5                                                                     Median of  Type B = 6.5
IQR of Type A = 4 – 3 = 1                                                                   IQR of Type B = 7 – 5 = 2
So, the Variation of Type B is having greater than the measure of Variation of Type A . The measure of the center  of Type B is  greater than the center of the Type A.

Question 12.
DIG DEEPER!
You want to compare the number of words per sentence in a sports magazine to the number of words per sentence in a political magazine.
a. The data represent random samples of the number of words in 10 sentences from each magazine. Compare the samples using measures of center and variation. Can you use the data to make a valid comparison about the magazines? Explain.
Sports magazine: 9, 21, 15, 14, 25, 26, 9, 19, 22, 30
Political magazine: 31, 22, 17, 5, 23, 15, 10, 20, 20, 17
b. The double box-and-whisker plot represents the means of 200 random samples of 20 sentences from each magazine. Compare the variability of the sample means to the variability of the sample numbers of words in part(a).
Big Ideas Math Solutions Grade 7 Chapter 8 Statistics 8.4 14
c. Make a conclusion about the numbers of words per sentence in each magazine.

Answer:
a) Variation of Sports magazine is having greater measure in variation and Median of  Sports magazine  is having greater value of center.

b) The variability of the sample is similar  to the variability of the sample numbers of words in part(a).

c) Number of words per sentence in political magazine are more than the number of words per sentence in  sports magazine.

Explanation:
a)
Data of Sports magazine =  9, 21, 15, 14, 25, 26, 9, 19, 22, 30
Order : 9 14 15 19  21 22 25  26 30
Median of  Sports magazine = 21
Mean of Sports magazine = 9 + 21 + 15 + 14 + 25 + 26 + 9 + 19 + 22 + 30 ÷ 10 = 190 ÷ 10 = 19
Variation of Sports magazine = (9-19)^2 + (21-19)^2 + (15-19)^2 + (14-19)^2 + (25-19)^2 + (26-19)^2 + (9-19)^2 + (19-19)^2 + (22-19)^2 + (30-19)^2  ÷ 10
= (-10)^2 + (2)^2 + (-4)^2 + (-5)^2 + (6)^2 + (7)^2 + (-10)^2 + (0)^2 + (3)^2 + (11)^2  ÷ 10
= 100 + 4 + 16 + 25 + 36 + 49 + 100 + 0 + 9 +121 ÷ 10
= 460 ÷ 10
= 46
Data of Political magazine = 31, 22, 17, 5, 23, 15, 10, 20, 20, 17
Order : 5  10 15  17  20  22  23  31
Median of Political magazine = 17 + 20 ÷ 2 = 37 ÷ 2 = 13.5
Mean of Political magazine = 31+ 22 +17 + 5 +23 + 15 +10 + 20 + 20 + 17 ÷ 10
= 180÷ 10
= 18
Variation of Sports magazine = (31-18)^2 + (22-18)^2 + (17-18)^2 + (5-18)^2 + (23-18)^2 + (15-18)^2 + (10-18)^2 + (20-18)^2 + (20-18)^2 + (17-18)^2  ÷ 10
= (13)^2 + (4)^2 + (-1)^2 + (-13)^2 + (5)^2 + (-3)^2 + (-8)^2 + (2)^2 + (2)^2 + (-1)^2  ÷ 10
= 169 + 16 + 1 + 169 + 25 + 9 + 64 + 4 + 4 + 1    ÷ 10
= 462 ÷ 10
= 46.2

b)
Data of Sports magazine = 16  19  20  21 23                  Data of Political magazine = 18  22  23  24  26
Median of Sports magazine = 20                                    Median of Political magazine = 23
IQR of Sports magazine = 21 – 9 = 2                               IQR of Political magazine = 24 -22 = 2
Mean of Sports magazine = 16 + 19 + 20 + 21 + 23  ÷ 5
= 99 ÷ 5  = 19.8
Mean of Political magazine = 18 + 22 + 23 + 24 + 26 ÷ 5
= 113 ÷ 5 = 22.6

c)
Data of Sports magazine = 16  19  20  21 23
Maximum used words in Political magazine = 16 to 23
=> 23 – 16 = 7
Data of Political magazine = 18  22  23  24  26
Maximum used words in Political magazine = 18 to 26.
=> 26 – 18 = 8

Question 13.
PROJECT
You want to compare the average amounts of time students in sixth, seventh, and eighth grade spend on homework each week.
a. Design an experiment involving random sampling that can help you make a comparison.
b. Perform the experiment. Can you make a conclusion about which grade spends the most time on homework? Explain your reasoning.
Answer:
a)Number of hours :
Total number of students of sixth spent = 26 + 33 +30 = 89
Total number of students of seventh spent = 32  + 46 + 40 = 118
Total number of students of Eighth spent = 42 +21 +25 = 88

b) Number of students who spent most time on homework  are the seventh grades students.

Explanation:
a) In a school, randomly chosen students are asked  about amounts of time students in sixth, seventh, and eighth grade spend on homework each week. The school has 1000 students.  Randomly 100 students are asked and three different sampling is taken.
Big-Ideas-Math-Book-7th-Grade-Answer-Key-Chapter-8-Statistics-PROJECT-13

b) Average of students spent on homework of sixth grades = 26 + 33 +30 ÷ 3 = 89 ÷ 3= 29.67
Average of students spent on homework of seventh grades = 32  + 46 + 40 ÷ 3 = 118 ÷ 3 = 39.33
Average of students spent on homework of eighth grades = 42 +21 +25 ÷ 3 = 88 ÷ 3 = 29.33

Statistics Connecting Concepts

Using the Problem-Solving Plan

Question 1.
In a city, 1500 randomly chosen residents are asked how many sporting events they attend each month. The city has 80,000 residents. Estimate the number of residents in the city who attend at least one sporting event each month.
Big Ideas Math Answer Key Grade 7 Chapter 8 Statistics cc 1
Understand the problem.
You are given the numbers of sporting events attended each month by a sample of 1500 residents. You are asked to make an estimate about the population, all residents of the city.
Make a plan.
The sample is representative of the population because it is selected at random and is large enough to provide accurate data. So, find the percent of people in the survey that attend at least one sporting event each month, and use the percent equation to make an estimate.
Solve and .check.
Use the plan to solve the problem. Then check your solution

Answer:
Number of residents attended one sports events = 30
Number of residents in the city who attend at least one sporting event each month = 30÷ 1500 × 80000
= 0.02 × 80000 = 1600

Explanation:
Number of residents attended zero sports events = 660
Number of residents attended zero sports events in the city = 660 ÷ 1500 = 0.44 × 80000 = 35200
Number of residents attended one or two sports events = 447
Number of residents attended one or two sports events in the city = 447 ÷ 1500 = 0.298 × 80000 =23840
Number of residents attended three or four sports events = 276
Number of residents attended three or four sports events in the city= 276 ÷ 1500 = 0.184 × 80000 = 14720
Number of residents attended five or more sports events = 87
Number  of residents attended five or more sports events in the city= 87 ÷ 1500 = 0.058 × 80000 =4640
Total number of residents sample taken = Number of residents attended zero sports events + Number of residents attended one or two sports events + Number of residents attended three or four sports events + Number of residents attended five or more sports events  + Number of residents attended one sports events
=>1500 = 660 + 447 + 276 + 87 + Number of residents attended one sports events
=> 1500 = 1470 + Number of residents attended one sports events
=> 1500 – 1470 = Number of residents attended one sports events
=> 30 = Number of residents attended one sports events
Check :
Total number of all events add = Total population in the city
LHS:
=> 35200 + 23840 + 14720 + 4640 + 1600
=> 80000
= RHS
Hence, LHS = RHS

Question 2.
The dot plots show the values in two data sets. Is the difference in the measures of center for the data sets significant?
Big Ideas Math Answer Key Grade 7 Chapter 8 Statistics cc 2
Answer:
Yes, the difference in the measures of center for the data set of A is not significant because the quotient is not less than 2.
No, the difference in the measures of center for the data set of B is significant because the quotient is less than 2.

Explanation:
Data of set A= – 6 – 5 -4 -4 – 4 -4 -3 -3 -3 -3
Order: -3 -4 -5 -6
Median of set A = -4 + – 5 ÷ 2 = -9 ÷ 2 = -4.5
Mean of set A = – 6 + -5 + -4 + -4 + -4 + -4 + -3 + -3 + -3 + -3 ÷10
= 35 ÷10 = 3.5
IQR of set A = -6 – (-4) = -6 + 4 = -2
Data of set B = -6 -6 -2 -2 -1 -1 -1 -1 0 0
Order : 0 -1 -2 -6
Median of set B = -1 + -2 ÷ 2 = -3 ÷ 2 = -1.5
Mean of set B = -6 + -6 + -2 + -2 + -1 + -1 + -1 + -1 + 0 + 0 ÷10
=  20  ÷10 = 2.0
IQR of set B = -6 – (-1) = -6 +1 = -5
Differences in medians ÷ IQR = -4.5 – (-1.5) ÷ -2 = -4.5 + 1.5 ÷ -2 = -3.0 ÷ -2 = 1.5
Differences in medians ÷ IQR = -4.5 – (-1.5) ÷ -5 = -4.5 + 1.5 ÷ -5 = -3.0 ÷ -5 = 0.6

Question 3.
You ask 60 randomly chosen students whether they support a later starting time for school. The table shows the results. Estimate the probability that at least two out of four randomly chosen students do not support a later starting time.
Big Ideas Math Answer Key Grade 7 Chapter 8 Statistics cc 3
Answer:

Explanation:

Performance Task

Estimating Animal Populations
At the beginning of the this chapter, you watched a STEAM Video called “Comparing Dogs.” You are now ready to complete the performance task related to this video, available at BigIdeasMath.com. Be sure to use the problem-solving plan as you work through the performance task.
Big Ideas Math Answer Key Grade 7 Chapter 8 Statistics cc 4

Statistics Chapter Review

Review Vocabulary

Write the definition and give an example of each vocabulary term.
Big Ideas Math Answer Key Grade 7 Chapter 8 Statistics cr 1

Answer:
Population: Population is an entire group of objects or people.
Sample: A Sample  is a part of the population.
Unbiased Sample: Unbiased Sample is a representative of a population. It is selected random and is a large enough to provide accurate data.
Biased Sample: Biased Sample is a not representative of a population. One or more parts of the population are favored over others.

Graphic Organizers
You can use a Definition and Example Chart to organize information about a concept. Here is an example of a Definition and Example Chart for the vocabulary term sample.
Big Ideas Math Answer Key Grade 7 Chapter 8 Statistics cr 2
Choose and complete a graphic organizer to help you study each topic.
Big Ideas Math Answer Key Grade 7 Chapter 8 Statistics cr 3
1. population
2. shape of a distribution
3. mean absolute deviation (MAD)
4. interquartile range
5. double box-and-whisker plot
6. double dot plot

Answer: Its a double box-and-whisker plot.
Big-Ideas-Math-Book-7th-Grade-Answer-Key-Chapter-8-Statistics-Graphic Organizers

Chapter Self-Assessment

As you complete the exercises, use the scale below to rate your understanding of the success criteria in your journal.
Big Ideas Math Answer Key Grade 7 Chapter 8 Statistics cr 4


8.1 Samples and Populations (pp. 325–330)

Learning Target: Understand how to use random samples to make conclusions about a population.

Question 1.
You want to estimate the number of students in your school whose favorite subject is biology. You survey the first 10 students who arrive at biology club. Determine whether the sample is biased or unbiased. Explain.
Answer:
The sample is a Biased Sample because the students in the biology club are not representative of a population. One or more parts of the students are favored over others.

Question 2.
You want to estimate the number of athletes who play soccer. Give an example of a biased sample. Give an example of an unbiased sample.
Big Ideas Math Answer Key Grade 7 Chapter 8 Statistics cr 5
Answer:
Example of Biased sample :
What is your favorite sport? Sample is chosen from people attending a soccer game.
Example of Unbiased sample :
What is your favorite sport? Sample is chosen by picking names out of a register of school.

Question 3.
You want to know how the residents of your town feel about building a new baseball stadium. You randomly survey 100 people who enter the current stadium. Eighty support building a new stadium, and twenty do not. So, you conclude that80% of the residents of your town support building a new baseball stadium. Is your conclusion valid? Explain.
Answer:
The conclusion is valid.

Explanation:
Number of randomly chosen = 100
Number of people support new stadium = 80
Number of people does not support new stadium = 20
Percentage of people support new stadium = 80 ÷100 × 100= 80%

Question 4.
Which sample is better for making an estimate? Explain.
Big Ideas Math Answer Key Grade 7 Chapter 8 Statistics cr 6

Answer:
Sample B is better for making an estimate about predicting the number of students in a school who like gym class because its a unbiased sample and we get accurate data out of it.

Question 5.
You ask 125 randomly chosen students to name their favorite beverage. There are 1500 students in the school. Predict the number of students in the school whose favorite beverage is a sports drink.
Big Ideas Math Answer Key Grade 7 Chapter 8 Statistics cr 7
Answer:
Number of students in the school whose favorite beverage is a sports drink = 696.

Explanation:
Number of students in the school whose favorite beverage is a sports drink = Favorite Beverage sport drink ÷ randomly chosen students × students in the school.
= 58 ÷ 125 ×1500
= 0.464 ×1500
= 696.

Question 6.
You want to know the number of students in your state who have summer jobs. Determine whether you should survey the population or a sample. Explain.
Answer:
You should survey the sample.

Explanation:
You should survey the sample because a sample is the representation of population and selected at random and is a large which gives you the accurate data.

8.2 Using Random Samples to Describe Populations (pp. 331–336)
Learning Target: Understand variability in samples of a population.

Question 7.
To pass a quality control inspection, the products at a factory must contain no critical defects, no more than 2.5% of products can contain major defects, and no more than 4% of products can contain minor defects. There are 40,000 products being shipped from a factory. Each inspector randomly samples 125 products. The table shows the results.
Big Ideas Math Answer Key Grade 7 Chapter 8 Statistics cr 8
a. Use each sample to make an estimate for the number of products with minor defects at the factory. Describe the center and the variation of the estimates.
b. Use the samples to make an estimate for the percent of products with minor defects, with major defects, and with critical defects at the factory. Does the factory pass inspection? Explain.

Answer:
a. The median of major defect is greater than the median of minor defect in the factory. The variation of major defects is greater than the measure of variation of minor defects in the factory.

b.  Yes, the factory pass inspection because not more than 2.5% of products can contain major defects, and not more than 4% of products can contain minor defects.

Explanation:
a. Number of products with minor defects at the factory = 5 6 3 6
Order: 3 5 6
Median of minor defects= 5
Mean of minor defects= 5+6+3+6 ÷ 4 = 20 ÷ 4 =5
Variation of minor defects= (5-5)^2 + (6-5)^2 + (3-5)^2 + (6-5)^2  ÷ 4
= (0)^2 + (1)^2 + (-2)^2 + (1)^2  ÷ 4
= 0 +  1 + 4 +1 ÷ 4
= 6÷ 4
= 1.5

Number of products with major defects at the factory = 2  1 3 5
Order: 1 2 3 5
Median of major defects= 2 + 3 ÷ 2 = 5 ÷ 2 = 2.5
Mean of major defects= 2 + 1+ 3 +5 ÷ 4 = 11 ÷ 4 = 2.75
Variation of major defects = (2-2.75)^2 + (1-2.75)^2 + (3-2.75)^2 + (5-2.75)^2  ÷ 4
= (-0.75)^2 + (-1.75)^2 + (0.25)^2 + (2.25)^2  ÷ 4
=  0.5625 + 3.0625 +0.625 + 5.0625 ÷ 4
=  9.3125 ÷ 4
= 2.33

Number of products with critical defects at the factory = 0 0 0 0
Median critical defects = 0
Mean critical defects = 0
Variation critical defects = 0

b. Percent of minor defects in the factory = 20 ÷ 125 = 0.16
Percent of major defects in the factory = 11 ÷125 = 0.088
Percent of critical defects in the factory = 0 ÷ 125 = 0

Question 8.
A scientist determines that 35% of packages of a food product contain a specific bacteria. Use technology to simulate choosing 100 random samples of 20 packages. How closely do the samples estimate the percent of all packages with the specific bacteria?

Answer:
Percent of bacteria in rice food product = 74 × 35 ÷ 100 = 2590 ÷ 100 = 25.9
Percent of bacteria in Wheat food product = 91 × 35 ÷ 100 = 3185 ÷ 100 = 31.85
Percent of bacteria in Milk food product = 78 × 35 ÷ 100 = 2730 ÷ 100 = 27.30
Percent of bacteria in Cereals food product = 57 × 35 ÷ 100 =  1995 ÷ 100 = 19.95

Explanation:
Big-Ideas-Math-Book-7th-Grade-Answer-Key-Chapter-8-Statistics-8.2-Using-Random-Samples-to-Describe-Populations-(pp. 331–336)8
Total number of Rice food product = 12 + 30 +32 = 74
Percent of bacteria in rice food product = 74 × 35 ÷ 100 = 2590 ÷ 100 = 25.9

Total number of Wheat food product = 34 + 29 + 28 = 91
Percent of bacteria in Wheat food product = 91 × 35 ÷ 100 = 3185 ÷ 100 = 31.85

Total number of Milk food product = 32 + 21 +25 = 78
Percent of bacteria in Milk food product = 78 × 35 ÷ 100 = 2730 ÷ 100 = 27.30

Total number of Cereals food product = 22 +20 +15 =
Percent of bacteria in Cereals food product = 57 × 35 ÷ 100 =  1995 ÷ 100 = 19.95

8.3 Comparing Populations (pp. 337–342)
Learning Target: Compare populations using measures of center and variation.

Question 9.
The double box-and-whisker plot represents the points scored per game by two football teams during the regular season.
Big Ideas Math Answer Key Grade 7 Chapter 8 Statistics cr 9
a. Compare the data sets using measures of center and variation.
b. Which team is more likely to score 28 points in a game?
Answer:
a. Team B has greater measures of center and both the team have same measures of variation.
b. Team A is most likely to score 28 points in a game.

Explanation:
a. Data of Team A =  0 6 14 22 28
Median of Team A =  14
IQR of Team A = 22 – 6 = 16
Data of Team B = 0  20  32  36  42
Median of Team B = 32
IQR of Team B = 36 – 20 = 16

b. In the data given, 28 points are given in team A not in Team B.

Question 10.
The dot plots show the ages of campers at two summer camps.
Big Ideas Math Answer Key Grade 7 Chapter 8 Statistics cr 10
a. Express the difference in the measures of center as a multiple of the measure of variation.
b. Are the ages of campers a tone camp significantly greater than at the other? Explain.

Answer:
a. Differences in medians ÷ IQR = 15 – 13 ÷ 2 = 2 ÷ 2 = 1
b. No, the ages of  campers are  significantly greater than at the other.

Explanation:
a. Data of Camp A =  13 13 13 14 14 14 14 15 15 15 15 15 15 16 16 16 16 17 17 17
Order: 13 14 15 16 17
Median of Camp A = 15
IQR of Camp A = 16 – 14 = 2
Data of Camp B = 11 11 11 12 12 12 12 13 13 13 13 13 13 14 14 14 14 15 15 15
Order: 11 12 13 14 15
Median of Camp B = 13
IQR of  Camp B = 14 – 12 = 2
Differences in medians ÷ IQR = 15 – 13 ÷ 2 = 2 ÷ 2 = 1

b. Because the quotient is less than 2, the difference in the medians is not significant.

8.4 Using Random Samples to Compare Populations (pp. 343–348)
Learning Target: Use random samples to compare populations.

Question 11.
The double dot plot shows the median gas mileages of 10 random samples of 50 vehicles for two car models. Compare the samples using measures of center and variation. Can you determine which car model has a better gas mileage? Explain.
Big Ideas Math Answer Key Grade 7 Chapter 8 Statistics cr 11
Answer:
The Variation of Model B measures are greater compared to the measures of  the Variation of Model A.
The measures of center of Model B is greater than the measures of Model A.
Model B cars has a better gas mileage than that of Model A cars.

Explanation:
Data of Model A =  18  19 19  20 20 20 21 21 22 27
Order: 18 19 20 21 22 27
Median of Model A =  20 + 21 ÷ 2 =41 ÷ 2 = 20.5
Mean of Model A = 18 + 19 + 19 + 20 + 20 + 20 + 21 + 21 +22 +27 ÷ 10
=  207 ÷ 10
= 20.7
Variation of Model A = (18-20.7)^2 + (19-20.7)^2 + (19-20.7)^2 + (20-20.7)^2 + (20-20.7)^2 + (20-20.7)^2 + (21-20.7)^2 + (21-20.7)^2 +(22-20.7)^2 + (27-20.7)^2  ÷ 10
= (-2.7)^2 + (-1.7)^2 + (-1.7)^2 + (-0.7)^2 + (-0.7)^2 + (-0.7)^2 + (0.3)^2 + (0.3)^2 +(1.3)^2 + (6.3)^2  ÷ 10
= 7.29 + 2.89 + 2.89 + 0.49 + 0.49 + 0.49 + 0.09 +0.09 + 1.69 + 39.69 ÷ 10
=  56.37  ÷ 10
= 5.637
Data of Model B = 20 21  27  29 29 29 30 30 31 32
Order: 20 21 27 29 30 31 32
Median of Model B = 29
Mean of Model B = 20 + 21 + 27 + 29 + 29 + 29 + 30 + 30 + 31 + 32 ÷ 10
=  278 ÷ 10
= 27.8
Variation of Model B = (20-27.8)^2 + (21-27.8)^2 + (27-27.8)^2 + (29-27.8)^2 + (29-27.8)^2 + (30-27.8)^2 + (30-27.8)^2 + (31-27.8)^2 +(31-27.8)^2 + (32-27.8)^2  ÷ 10
= (-7.8)^2 + (-6.8)^2 + (-0.8)^2 + (1.2)^2 + (1.2)^2 + (2.2)^2 + (2.2)^2 + (3.2)^2 +(3.2)^2 + (4.2)^2  ÷ 10
= 60.84 + 46.24 + 0.64  + 1.44 + 1.44 + 4.84 + 4.84 + 10.24 + 10.24 + 17.64 ÷ 10
= 158.4 ÷ 10
= 15.84

Question 12.
You compare the average amounts of time people in their twenties and thirties spend driving each week. The double box-and-whisker plot represents the medians of 100 random samples of 8 people from each age group. Can you determine whether one age group drives more than the other? Explain.
Big Ideas Math Answer Key Grade 7 Chapter 8 Statistics cr 12
Answer:
Because the variables are same, you can determine the one age group drives more than the other visual overlap by expressing the differences in the medians as the multiples of IQR.

Explanation:
Data of Twenties = 0 5 10 15 25
Median of Twenties = 10
IQR of Twenties = 15 – 5 = 10
Data of Thirties = 0 10 15 20  40
Median of Thirties = 15
IQR of Thirties = 20 – 10 =10
Differences in medians ÷ IQR = 15- 10 ÷10 = 5 ÷10 =0.5
Because the quotient is less than 2, the difference is not significant.

Statistics Practice Test

Question 1.
You want to estimate the number of students in your school who prefer to bring a lunch from home rather than buy one at school. You survey five students who are standing in the lunch line. Determine whether the sample biased or unbiased. Explain.

Answer:
The sample is biased.

Explanation:
The sample taken is biased because it does not gives you the accurate information of the number of students in your school prefer to bring a lunch from home rather than buy one at school. The sample size chosen is incorrect as taking the students in the lunch line..

Question 2.
You want to predict which candidate will likely be voted Seventh Grade Class President. There are 560 students in the seventh grade class. You randomly sample 3 different groups of 50 seventh-grade students. The results are shown.
Big Ideas Math Answers 7th Grade Chapter 8 Statistics pt 2
a. Use each sample to make an estimate for the number of students in seventh grade that vote for Candidate A.
b. Who do you expect to be voted Seventh Grade Class President? Explain.

Answer:
a. Estimation for the number of students who vote for Candidate A = 64.
b. Candidate B is to be expected to be voted Seventh Grade Class President.

Explanation:
a. Estimation for the number of students who vote for Candidate A = 27 + 22 +15 = 64.
Estimation for the number of students who vote for Candidate B = 23 + 28 + 35 = 86.
b. Candidate B is to be expected to be voted Seventh Grade Class President.

Question 3.
Of 60 randomly chosen students from a school surveyed, 16 chose the aquarium as their favorite field trip. There are 720 students in the school. Predict the number of students in the school who choose the aquarium as their favorite field trip.
Answer:
Number of students in the school who choose the aquarium as their favorite field trip = 192.

Explanation:
Number of students who chose the aquarium as their favorite field trip = 16
Number of students  chosen students from a school surveyed = 60
Total number of students in the school = 720
Number of students in the school who choose the aquarium as their favorite field trip = 720/60 × 16 = 192

Question 4.
The double box-and-whisker plot shows the ages of the viewers of two television shows in a small town.
Big Ideas Math Answers 7th Grade Chapter 8 Statistics pt 4
a. Compare the data sets using measures of center and variation.
b. Which show is more likely to have a 44-year-old viewer?

Answer:
a. The Show A  center is likely to be greater than the measures of the center of Show B. Variation of show B is greater than Variation of show A.
b.  Show A  is more likely to have a 44-year-old viewer.

Explanation:
Data of Show A = 35   40   45    50   55
Median of Show A = 45
Mean of Show A = 35  + 40 +  45  +  50 +  55 ÷ 5 = 225 ÷ 5 = 45
Variation of Show A = (35-45)^2 + (40-45)^2 + (45-45)^2 + (50-45)^2 +  (55-45)^2 ÷ 5
= (-10)^2 + (-5)^2 + (0)^2 + (5)^2 +  (10)^2 ÷ 5
= 100 + 25 + 0 + 25 +100 ÷ 5
= 250 ÷ 5
= 50.

Data of Show B = 20   25    35    40   55
Median of Show B = 35
Mean of Show B = 20 + 25  +  35  +  40 +  55 ÷ 5 = 175 ÷ 5 = 35.
Variation of show B = (20-35)^2 + (25-35)^2 + (35-35)^2 + (40-35)^2 +  (55-35)^2 ÷ 5
= (-15)^2 + (-10)^2 + (0)^2 + (5)^2 +  (5)^2 ÷ 5
= 225 + 100 + 0 + 25 + 25   ÷ 5
= 375 ÷ 5
= 75.

Question 5.
The double box-and-whisker plot shows the test scores for two French classes taught by the same teacher.
Big Ideas Math Answers 7th Grade Chapter 8 Statistics pt 5
a. Express the difference in the measures of center as a multiple of the measure of variation.
b. Are the scores for one class significantly greater than for the other? Explain.

Answer:
a. The difference of the Variation of Class A is similar to the Variation of Class A and the measures of the center of Class A  is comparably greater than the measures of the center of Class B.
b. Yes, the scores for one class significantly greater than for the other.

Explanation:
Data of Class A =  76  84  88  92  100                                          Data of Class B = 68  76  80   84   100
Median of Class A = 76 + 84 + 88 + 92 + 100 ÷ 5                     Median of Class B = 68 + 76 + 80 + 84  + 100 ÷ 5
=  440  ÷ 5 = 88                                                                                = 408  ÷ 5  = 81.6
IQR of Class A = 92 – 84 = 8                                                        IQR of Class B = 84 – 76 =  8

Question 6.
Two airplanes each hold about 400 pieces of luggage. The double dot plot shows a random sample of 8 pieces of luggage from each plane. Compare the samples using measures of center and variation. Can you determine which plane has heavier luggage? Explain.
Big Ideas Math Answers 7th Grade Chapter 8 Statistics pt 6
Answer:
The measures of center of Plane A  is greater than the measure of center of Plane B.
Plane A is likely to have heavier luggage than Plane B. The Variation of Plane A is greater than Variation of Plane B.

Explanation:
Data of Plane A =  28  36  36  38  38  40  40  40                     Data of plane B = 32  32  34  34  34  36  38 50
Order = 28 36 38 40                                                                Order = 32  34  36  38 50
Median of Plane A = 36 + 38 ÷ 2 = 74 ÷ 2 = 37                Median of Plane B  = 36
Variation of Plane A =  (28-37)^2 + (36-37)^2 + (36-37)^2 + (38-37)^2 +  (38-37)^2 + (40-37)^2 + (40-37)^2 +  (40-37)^2 ÷ 8
=(-9)^2 + (-1)^2 +  (-1)^2  + (1)^2 + (1)^2 +  (3)^2  + (3)^2 + (3)^2  ÷ 8
= 81 + 1 +1 +1 +1 +9 +9 +9  ÷ 8
= 112 ÷ 8
= 14.
Variation of Plane B =  (32-36)^2 + (32-36)^2 +  (34-36)^2  + (34-36)^2 + (34-36)^2 +  (36-36)^2  + (38-36)^2 + (50-36)^2 ÷ 8
= (-4)^2 + (-4)^2 +  (-2)^2 + (-2)^2 + (-2)^2 +  (0)^2  + (2)^2 + (4)^2  ÷ 8
= 16 + 16 + 4+ 4+ 4+ 0 +4 + 16  ÷ 8
=  64  ÷ 8
= 8.

Statistics Cumulative Practice

Big Ideas Math Answers 7th Grade Chapter 8 Statistics cp 1
Question 1.
Which of the ratios form a proportion?
A. 5 to 2 and 4 to 10
B. 2 : 3 and 7 : 8
C. 3 to 2 and 15 to 10
D. 12 : 8 and 8 : 42.
Answer:
C. 3 to 2 and 15 to 10 is the  ratios form a proportion

Explanation:
A) 5 to 2 and 4 to 10
B) 2 : 3 and 7 : 8
C) 3 to 2 and 15 to 10
=> 3 : 2 and 15 : 10
=> Divide by 5 on both sides
=> 3 : 2 and  3 : 2
D) 12 : 8 and 8 : 42.
=> Divide by 2 on both sides
=> 6 : 4 and 4 : 21
=> Divide by 2 on both sides
=> 3 : 2 and  2 :21

Question 2.
A student scored 600 the first time. she took the mathematics portion of a college entrance exam. The next time she took the exam, she scored 660. Her second score represents what percent increase over her first score?
F. 9.1%
G. 10%
H. 39.6%
I. 60%
Answer:
F = 9.1 %  is the Percentage increase in her second score over her first score.

Explanation:
Number of marks she scored in first attempt in the exam = 600
Number of marks she scored in second attempt in the exam = 660
Difference in the score = 660 – 600 = 60
Percentage increase in her second score over her first score = 60 ÷ 660 × 100
= 0.9090 × 100 = 9.1%

Question 3.
You ask 100 randomly chosen students to name their favorite food. There are 1250 students in the school. Based on this sample, what is the number of students in the school whose favorite food is chicken?
Big Ideas Math Answers 7th Grade Chapter 8 Statistics cp 3
A. 100
B. 225
C. 450
D. 475
Answer:
Number of students in the school whose favorite food is chicken = 100.

Explanation:
Total number of students in the school = 1250
Number of students randomly chosen = 100
Number of students in the school whose favorite food is chicken = 8 ÷ 100 × 1250
= 0.08 × 1250 = 100.

Question 4.
Which value of makes the equation p + 6 = 5 true?
F. – 1
G. 1
H. 11
I. 30
Answer:
F. – 1
p = -1 makes the equation p + 6 = 5 true.

Explanation:
F) Substitute p = -1 in Equation:
p + 6 = 5 = -1 + 6 = 5
G) Substitute p = 1 in Equation:
p + 6 = 5 = 1 + 6 = 7
H ) Substitute p = 11 in Equation:
p + 6 = 5 = 11 + 6 = 17
I) Substitute p = 30 in Equation:
p + 6 = 5 = 30 + 6 = 36

Question 5.
The table shows the costs for four cans of tomato soup. Which can has the lowest cost per ounce?
Big Ideas Math Answers 7th Grade Chapter 8 Statistics cp 5
A. Can A
B. Can B
C. Can C
D. Can D
Answer:
The lowest cost per ounce has D. Can D.

Explanation:
Cost of Can A cost per ounce= 1.95 ÷ 26 = 0.075
Cost of Can B  cost per ounce= 0.72 ÷ 8 = 0.09
Cost of Can C  cost per ounce=0.86 ÷ 10.75 = 0.08
Cost of Can D  cost per ounce= 2.32 ÷ 23.2 = 0.1

Question 6.
What value of y makes the equation – 3y = – 18 true?
Big Ideas Math Answers 7th Grade Chapter 8 Statistics cp 6
Answer:
y = 6 makes the equation – 3y = – 18 true.

Explanation:
Equation – 3y = – 18
=> y = -18 ÷ -3
Divide by 3 both sides.
=> y = 6

Question 7.
The double dot plot shows the values in two data sets. Which sentence best represents the difference in the measures of center as a multiple of the measure of variation?
Big Ideas Math Answers 7th Grade Chapter 8 Statistics cp 7
F. The difference of the means is about 3.3 times the MAD.
G. The difference of the means is about 3.8 times the MAD.
H. The difference of the means is 36 times the MAD.
I. The difference of the means is 48.7 times the MAD.

Answer:
G. The difference of the means is about 3.8 times the MAD.

Explanation:
Data of set A = 42  43  46  48   48  51  51  51  52  55
Mean of set A = 42 + 43 + 46 + 48 + 48 + 51 + 51 + 51 + 52 + 55 ÷ 10
= 487 ÷ 10
= 48.7
MAD of set A = |42−48.7∣ + |43−48.7∣ +|46−48.7∣ +|48−48.7∣ + |48-48.7∣ + |51-48.7∣ + |51-48.7∣ +|51-48.7∣ +|52-48.7∣ + |55-48.7∣ ÷ 10
=  |-6.7∣ + |-5.7∣ +|-2.7∣ +|-0.7∣ + |-0.7∣ + |2.3∣ + |2.3∣ +|2.3∣ +|3.3∣ + |6.3 ∣ ÷ 10
= 33 ÷ 10
= 3.3
Data of set B = 30  31  33  34  36  36  38  40  41  41
Mean of set B = 30 + 31 + 33 + 34 + 36 + 36 + 38 + 40 + 41 + 41 ÷ 10
= 360÷ 10
= 36.0
MAD of set B = |30-36∣ + |31-36∣ +|33-36∣ +|34-36∣ + |36-36∣ + |36-36∣ + |38-36∣ +|40-36∣ +|41-36∣ + |41-36∣ ÷ 10
= |-6∣ + |-5∣ +|-3∣ +|-2∣ + |0∣ + |0∣ + |2∣ +|4∣ +|5∣ + |5∣ ÷ 10
=  32  ÷ 10
= 3.2
Difference in Means ÷ MAD of set A= 48.7 – 36 ÷ 3.3 = 12.7 ÷ 3.3= 3.85
Difference in Means ÷ MAD of set B= 48.7 – 36 ÷ 3.2 = 12.7 ÷ 3.2 =3.97

Question 8.
What is the missing value in the ratio table?
Big Ideas Math Answers 7th Grade Chapter 8 Statistics cp 8
Answer:
The missing value in the ratio table is 20.

Explanation:
a) 2/3  = 5
=> 2 = 5 × 3
=> 2 = 15

b) 8/3 = X
8 = 3 × =  60
=> 60 ÷ 3
=> 20

Question 9.
You are selling tomatoes. What is the minimum number of 4pounds of tomatoes you need to sell to earn at least $44?
Big Ideas Math Answers 7th Grade Chapter 8 Statistics cp 9
A. 11
B. 12
C. 40
D. 176
Answer:
A. 11 is the minimum number of  tomatoes needed to sell to earn at least $44.

Explanation:
Cost of the tomatoes = $44
cost of the tomatoes per pound = $4
The minimum number of  tomatoes needed to sell to earn at least $44 = 44 ÷ 4 = 11.

Question 10.
You and a group of friends want to know how many students in your school prefer science. There are 900 students in your school. Each person randomly surveys 20 students. The table shows the results. Which subject do students at your school prefer?
Big Ideas Math Answers 7th Grade Chapter 8 Statistics cp 10
Part A Use each sample to make an estimate for the number of students in your school who prefer science.
Part B Describe the variation of the estimates.
Part C Use the samples to make one estimate for the number of students who prefer science in your school.

Answer:
Number of students who prefer science = 28
Mean = 5.6
Variation of the estimate  = 4.24

Explanation:
Number of students who prefer science = 6 + 7 + 8 +5 + 2 = 28
Mean of number of students who prefer science = 6 + 7 + 8 +5 + 2 ÷ 5 = 28 ÷ 5 = 5.6
Variation of the estimate  = (6-5.6)^2 + (7-5.6)^2 + (8-5.6)^2 + (5-5.6)^2 +  (2-5.6)^2  ÷ 5
= (-0.4)^2 + (1.4)^2 + (2.4)^2 + (0.6)^2 + (-3.6)^2  ÷ 5
=  0.16 +1.96 + 5.76 + 0.36 +  12.96  ÷ 5
=  21.2 ÷ 5
= 4.24

Conclusion:

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Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions

Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions

Here, on this page, we have shared the ultimate preparation guide for high school students on algebra 2 math concepts. The given material is Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions. You can access and download BIM Algebra 2 Ch 7 Answer key for free of cost. So, it helps you to practice efficiently and gain more subject knowledge. Get the exercise wise chapter 7 rational functions Big Ideas Math Algebra 2 Answers from the below pdf links and start your preparation by solving the questions from various given sources covered in the BIM Algebra 2 Chapter 7 Solution Key.

Big Ideas Math Book Algebra 2 Answer Key Chapter 7 Rational Functions

In the Big Ideas Math textbook solutions algebra 2 ch 7, you can find basic & strong fundamentals of Rational functons like inverse variation, graphing rational functions, adding, subtracting, mutliplying, and dividing rational expressions, and solving rational equations. By referring, practicing, and solving all the questions presented in the BigIdeas Math Book Algegra 2 Chapter 7 Answer Key, you can quickly learn the concepts and score highest rank in the exams. For students who are passionate about math skills should solve the problems in Topic-wise BIM Algebra 2 Rational functions solution key.

Rational Functions Maintaining Mathematical Proficiency

Evaluate.
Question 1.
\(\frac{3}{5}+\frac{2}{3}\)
Answer:

Question 2.
–\(\frac{4}{7}+\frac{1}{6}\)
Answer:

Question 3.
\(\frac{7}{9}-\frac{4}{9}\)
Answer:

Question 4.
\(\frac{5}{12}-\left(-\frac{1}{2}\right)\)
Answer:

Question 5.
\(\frac{2}{7}+\frac{1}{7}-\frac{6}{7}\)
Answer:

Question 6.
\(\frac{3}{10}-\frac{3}{4}+\frac{2}{5}\)
Answer:

Simplify.
Question 7.
\(\frac{\frac{3}{8}}{\frac{5}{6}}\)
Answer:

Question 8.
\(\frac{\frac{1}{4}}{-\frac{5}{7}}\)
Answer:

Question 9.
\(\frac{\frac{2}{3}}{\frac{2}{3}+\frac{1}{4}}\)
Answer:

Question 10.
ABSTRACT REASONING
For what value of x is the expression \(\frac{1}{x}\) undefined? Explain your reasoning.
Answer:

Rational Functions Mathematical Practices

Mathematically proficient students are careful about specifying units of measure and clarifying the relationship between quantities in a problem.

Monitoring Progress

Question 1.
You drive a car at a speed of 60 miles per hour. What is the speed in meters per second?
Answer:

Question 2.
A hose carries a pressure of 200 pounds per square inch. What is the pressure in kilograms per square centimeter?
Answer:

Question 3.
A concrete truck pours concrete at the rate of 1 cubic yard per minute. What is the rate in cubic feet per hour?
Answer:

Question 4.
Water in a pipe flows at a rate of 10 gallons per minute. What is the rate in liters per second?
Answer:

Lesson 7.1 Inverse Variation

Essential Question How can you recognize when two quantities vary directly or inversely?

EXPLORATION 1

Recognizing Direct VariationWork with a partner. You hang different weights from the same spring.
Big Ideas Math Algebra 2 Answer Key Chapter 7 Rational Functions 7.1 1
a. Describe the relationship between the weight x and the distance d the spring stretches from equilibrium. Explain why the distance is said to vary directly with the weight.
b. Estimate the values of d from the figure. Then draw a scatter plot of the data. What are the characteristics of the graph?
c. Write an equation that represents d as a function of x.
d. In physics, the relationship between d and x is described by Hooke’s Law. How would you describe Hooke’s Law?

EXPLORATION 2

Recognizing Inverse Variation
Work with a partner. The table shows the length x (in inches) and the width y (in inches) of a rectangle. The area of each rectangle is 64 square inches.
Big Ideas Math Algebra 2 Answer Key Chapter 7 Rational Functions 7.1 2
a. Copy and complete the table.
b. Describe the relationship between x and y. Explain why y is said to vary inversely with x.
c. Draw a scatter plot of the data. What are the characteristics of the graph?
d. Write an equation that represents y as a function of x.

Communicate Your Answer

Question 3.
How can you recognize when two quantities vary directly or inversely?
Answer:

Question 4.
Does the mapping rate of the wings of a bird vary directly or inversely with the length of its wings? Explain your reasoning.
Answer:

Monitoring Progress

Tell whether x and y show direct variation, inverse variation, or neither.
Question 1.
6x = y
Answer:

Question 2.
xy = −0.25
Answer:

Question 3.
y + x = 10
Answer:

Tell whether x and y show direct variation, inverse variation, or neither.
Question 4.
Big Ideas Math Algebra 2 Answer Key Chapter 7 Rational Functions 7.1 3
Answer:

Question 5.
Big Ideas Math Algebra 2 Answer Key Chapter 7 Rational Functions 7.1 4
Answer:

The variables x and y vary inversely. Use the given values to write an equation relating x and y. Then find y when x = 2.
Question 6.
x = 4, y = 5
Answer:

Question 7.
x = 6, y = −1
Answer:

Question 8.
x = \(\frac{1}{2}\), y = 16
Answer:

Question 9.
WHAT IF?
In Example 4, it takes a group of 10 volunteers 12 hours to build the playground. How long would it take a group of 15 volunteers?
Answer:

Inverse Variation 7.1 Exercises

Vocabulary and Core Concept Check
Question 1.
VOCABULARY
Explain how direct variation equations and inverse variation equations are different.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 7 Rational Functions 7.1 a 1

Question 2.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.
Big Ideas Math Algebra 2 Answer Key Chapter 7 Rational Functions 7.1 5
Answer:

(1): From  definition of inverse variation we have that y=a/x where in our case a=4.This implies that inverse variation equation is

Y=4/x.

(2):  Let now ration be constant a= 4. Then we get

y/x=4    => y =4x.

This implies that x and y show direct variation.

(3): From equation y=a/x and a=4 we have that

Y= 4/x

It implies that x and y show inverse variation.

(4): By equation x y =4 it implies that y=4/x

This equation means that x and y show inverse variation.

The different question is (2): where x and y show   direct variation

Monitoring Progress and Modeling with Mathematics

In Exercises 3–10, tell whether x and y show direct variation, inverse variation, or neither.
Question 3.
y = \(\frac{2}{x}\)
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 7 Rational Functions 7.1 a 3

Question 4.
xy = 12
Answer:

Given equation                    Solved for y                      Type of variation

xy=12                            Y=12/x                                 inverse variation

 

Question 5.
\(\frac{y}{x}\) = 8
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 7 Rational Functions 7.1 a 5

Question 6.
4x = y
Answer:

By definition ,x and y show direct variation

Question 7.
y = x + 4
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 7 Rational Functions 7.1 a 7

Question 8.
x + y = 6
Answer:

Given Equation                               Solved for  y                       Types of variation

x+y=6                                      Add -x to both sides          X and y shows no variation

Y= -x+6

Question 9.

8y = x
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 7 Rational Functions 7.1 a 9

Question 10.
xy = \(\frac{1}{5}\)
Answer:

Solve for y . Divide equation with x

Xy = 1/5
Y=1/5x=( 1/5)/x

inverse variation

 

In Exercises 11–14, tell whether x and y show direct variation, inverse variation, or neither.
Question 11.
Big Ideas Math Algebra 2 Answer Key Chapter 7 Rational Functions 7.1 6
Answer:

y/x    132/12=11    198/18 =11   235/23 =11   319/29  =11    374/34=11

The products are not constant , and the rations are constant .

X and y show direct variation .

 

Question 12.
Big Ideas Math Algebra 2 Answer Key Chapter 7 Rational Functions 7.1 7
Answer:

      xy      20.25      56.25     144           506.25         900
       y/ x       13.5/1.5=9 22.5/2.5=9 36/4=9 67.5/7.5=9 90/10=9

The products are not constant , and the rations are constant .

X and y show direct variation .

Question 13.
Big Ideas Math Algebra 2 Answer Key Chapter 7 Rational Functions 7.1 8
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 7 Rational Functions 7.1 a 13

Question 14.
Big Ideas Math Algebra 2 Answer Key Chapter 7 Rational Functions 7.1 9
Answer:

      xy       64       55          62         63          66
       y/ x       16/4=4   11/5=2.2 10/6.2=1.612 9/7=1.285 6/11=0.545

The products are not constant, and the ratios are not constant.

X and y show no variation .

 

In Exercises 15–22, the variables x and y vary inversely. Use the given values to write an equation relating x and y. Then find y when x = 3.
Question 15.
x = 5, y =−4
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 7 Rational Functions 7.1 a 15

Question 16.
x = 1, y = 9
Answer:

Y=a/x

Plug  x=1, y=9 into equation to determine a.

9=a/1

a =9.1=9

The inverse variation equation is y= 9/x

When x=3,

Y=9/3=3

Question 17.
x =−3, y = 8
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 7 Rational Functions 7.1 a 17

Question 18.
x = 7, y = 2
Answer:

Y=a/x

Plug  x=7, y=2 into equation to determine a.

7=a/2

a=7. 2=14

The inverse variation equation is y=14/x

When x=3

Y=14/3

Question 19.
x = \(\frac{3}{4}\), y = 28
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 7 Rational Functions 7.1 a 19

Question 20.
x = −4, y = –\(\frac{5}{4}\)
Answer:

Y=a/x

Plug  x= -4  ,  y=-5/4 into equation to determine a.

-5/4  = a/-4

a =-5/4  . (-4)=5

The inverse variation equation is y=5/x

When x=3

Y=5/3

Question 21.
x = −12, y = −\(\frac{6}{2}\)
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 7 Rational Functions 7.1 a 21

Question 22.
x = \(\frac{5}{3}\), y = −7
Answer:

Y=a/x

Plug  x= 5/3  ,  y=-7 into equation to determine a.

-7 =a/5/3

A=-7 . 5/3 = – 35/3

The inverse variation equation is y=-35/3/x

When x=3

Y=-35/3/3=-35/9

ERROR ANALYSIS In Exercises 23 and 24, the variables x and y vary inversely. Describe and correct the error in writing an equation relating x and y.
Question 23.
x = 8, y = 5
Big Ideas Math Algebra 2 Answer Key Chapter 7 Rational Functions 7.1 10
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 7 Rational Functions 7.1 a 23

Question 24.
x = 5, y = 2
Big Ideas Math Algebra 2 Answer Key Chapter 7 Rational Functions 7.1 11
Answer:

Every thing is correct except last row when they plug a = 10 into starting equation . it shoulds be like this

a=10

Xy=a

Xy=10, divide with x

Y=10/x

Question 25.
MODELING WITH MATHEMATICS
The number y of songs that can be stored on an MP3 player varies inversely with the average size x of a song. A certain MP3 player can store 2500 songs when the average size of a song is 4 megabytes (MB).
a. Make a table showing the numbers of songs that will fit on the MP3 player when the average size of a song is 2 MB, 2.5 MB, 3 MB, and 5 MB.
b. What happens to the number of songs as the average song size increases?
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 7 Rational Functions 7.1 a 25

Question 26.
MODELING WITH MATHEMATICS
When you stand on snow, the average pressure P (in pounds per square inch) that you exert on the snow varies inversely with the total area A (in square inches) of the soles of your footwear. Suppose the pressure is 0.43 pound per square inch when you wear the snowshoes shown. Write an equation that gives P as a function of A. Then find the pressure when you wear the boots shown.
Big Ideas Math Algebra 2 Answer Key Chapter 7 Rational Functions 7.1 12
Answer:

The general equation for inverse variation is

Y=a/x

Let y=P where P is average pressure   on snow and x =A is the total area .

If we wear the snowshoes then the   P is 0.43 and A=360. Using the equation about we have that

P=a/A

0.43=a/360

A=0.43.360=154.8

It implies that   P=154.8/A

Now  , if you boots then A=60. Using the equation about we have that

P=a/A

0.43=a/60

A=0.43.60=25.8

It implies that

P=25.8/A

 

Question 27.
PROBLEM SOLVING
Computer chips are etched onto silicon wafers. The table compares the area A (in square millimeters) of a computer chip with the number c of chips that can be obtained from a siliconwafer. Write a model that gives c as a function of A. Then predict the number of chips per wafer when the area of a chip is 81 square millimeters.
Big Ideas Math Algebra 2 Answer Key Chapter 7 Rational Functions 7.1 13
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 7 Rational Functions 7.1 a 27

Question 28.
HOW DO YOU SEE IT?
Does the graph of f represent inverse variation or direct variation? Explain your reasoning.
Big Ideas Math Algebra 2 Answer Key Chapter 7 Rational Functions 7.1 14
Answer:

As x increases, y increases as well, so product x. y cannot possibly be constant.

This means that x and y show direct variation .

Question 29.
MAKING AN ARGUMENT
You have enough money to buy 5 hats for $10 each or 10 hats for $5 each. Your friend says this situation represents inverse variation. Is your friend correct? Explain your reasoning.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 7 Rational Functions 7.1 a 29

Question 30.
THOUGHT PROVOKING
The weight w (in pounds) of an object varies inversely with the square of the distance d (in miles) of the object from the center of Earth. At sea level (3978 miles from the center of theEarth), an astronaut weighs 210 pounds. How much does the astronaut weigh 200 miles above sea level?
Answer:

Write the equation that represents  given variation :

W=a/d2 – there is d2 in denominator because w varies inversely   with square of the distance  .

Plug d=3978 and w=210 into equation to determine a.

210 = a/39782

A=210. 39782 = 3,323,141,640

Equation :

W = 3,323,141,640/d2

When d = 3978+200 = 4178:

W = 3,323,141,640/41782 = 190

Question 31.
OPEN-ENDED
Describe a real-life situation that can be modeled by an inverse variation equation.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 7 Rational Functions 7.1 a 31

Question 32.
CRITICAL THINKING
Suppose x varies inversely with y and y varies inversely with z. How does x vary with z? Justify your answer.
Answer:

Y=b/z

From first equation:

X=a/y, multiply with y

X y= a, divide  with x

Y=a/x

Substitute this y with y in equation with z:

Y=b/z

a/x = b/z, multiply with x z

a z=b z

x=a/b z

x varies directly with z.

Question 33.
USING STRUCTURE
To balance the board in the diagram, the distance (in feet) of each animal from the center of the board must vary inversely with its weight (in pounds). What is the distance of each animal from the fulcrum? Justify your answer.
Big Ideas Math Algebra 2 Answer Key Chapter 7 Rational Functions 7.1 15
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 7 Rational Functions 7.1 a 33

Maintaining Mathematical Proficiency Divide.
Question 34.
(x2 + 2x − 99) ÷ (x + 11)
Answer:

Question 35.
(3x4 − 13x2 − x3 + 6x − 30) ÷ (3x2 − x + 5)
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 7 Rational Functions 7.1 a 35

Graph the function. Then state the domain and range.
Question 36.
f(x) = 5x + 4
Answer:

First graph g(x)= 5x. Then translate it 4 units up  to obtain f(x).

On graph :

Red – g(x)

Blue – f(x)

Domain of f is all real number and range of f is (4, ∞)

Question 37.
g(x) = ex-1
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 7 Rational Functions 7.1 a 37

Question 38.
y = ln 3x – 6
Answer:

First graph g(x)=In 3x. Then translate it six units down  to obtain f(x).

On graph :

Red – g(x)

Blue – f(x)

Domain of f is all positive number real numbers,(0, ∞) and range of f is (-∞, ∞)- all real numbers.

Question 39.
ln(x) = 2 ln (x + 9)
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 7 Rational Functions 7.1 a 39

Lesson 7.2 Graphing Rational Functions

Essential Question What are some of the characteristics of the graph of a rational function?
The parent function for rational functions with a linear numerator and a linear denominator is
f(x) = \(\frac{1}{x}\). Parent function
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 1
The graph of this function, shown at the right, is a hyperbola.

EXPLORATION 1

Identifying Graphs of Rational Functions
Work with a partner. Each function is a transformation of the graph of the parent function f(x) = \(\frac{1}{x}\). Match the function with its graph. Explain your reasoning. Then describe the transformation.
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 2

Communicate Your Answer

Question 2.
What are some of the characteristics of the graph of a rational function?
Answer:

Question 3.
Determine the intercepts, asymptotes, domain, and range of the rational function g(x) = \(\frac{x-a}{x-b}\).
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 3
Answer:

Monitoring Progress

Question 1.
Graph g(x) = \(\frac{-6}{x}\). Compare the graph with the graph of f(x) = \(\frac{1}{x}\).
Answer:

Graph the function. State the domain and range.
Question 2.
y = \(\frac{3}{x}\) – 2
Answer:

Question 3.
y = \(\frac{-1}{x + 4}\)
Answer:

Question 4.
y = \(\frac{1}{x – 1}\) + 5
Answer:

Graph the function. State the domain and range.
Question 5.
f(x) = \(\frac{x-1}{x+3}\)
Answer:

Question 6.
f(x) = \(\frac{2x+1}{4x-2}\)
Answer:

Question 7.
f(x) = \(\frac{-3x+2}{-x-1}\)
Answer:

Question 8.
Rewrite g(x) = \(\frac{2x+3}{x+1}\) in the form g(x) = \(\frac{a}{x-h}\). Graph the function.
Describe the graph of g as a transformation of the graph of f(x) = \(\frac{a}{x}\).
Answer:

Question 9.
WHAT IF?
How do the answers in Example 5 change when the cost of the 3-D printer is $800?
Answer:

Graphing Rational Functions 7.2 Exercises

Vocabulary and Core Concept Check
Question 1.
COMPLETE THE SENTENCE
The function y = \(\frac{7}{x+2}\) has a(n) __________ of all real numbers except 3 and a(n) __________ of all real numbers except −4.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 a 1

Question 2.
WRITING
Is f(x) = \(\frac{-3 x+5}{2^{x}+1}\) a rational function? Explain your reasoning.
Answer:

Consider function f given by

f(x) =-3x+5/2x +1

We know that the rational function has form

Y=a x +b/ c x +d

Where a, b,c, d are constant such that c x+d ≠ 0

Since 2x is exponential function then the given function is not rational

Monitoring Progress and Modeling with Mathematics

In Exercises 3–10, graph the function. Compare the graph with the graph of f(x) = \(\frac{1}{x}\).
Question 3.
g(x) = \(\frac{3}{x}\)
Answer:

Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 a 3

Question 4.
g(x) = \(\frac{10}{x}\)
Answer:

Step:1

The function is of form g(x)= a/x, so the asymptotes are x=0 and y=0

Step:2

Make a table of value and plot the points.

Include both positive and negative values of x.

      x         -3        -2          -1         1        2           3
       y       -10/3      -5        -10       10       5         10/3

Step=3

On graph :

Red – f(x)= 1/x

Blue – g(x) = 10/x

Question 5.
g(x) = \(\frac{-5}{x}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 a 5

Question 6.
g(x) = \(\frac{-9}{x}\)
Answer:

Step:1

The function is of form g(x)= a/x, so the asymptotes are x=0 and y=0

Step:2

Make a table of value and plot the points.

Include both positive and negative values of x.

     x      -3       -2     -1      1       2                    3
     y        3      9/2       9      -9    -9/2       -3

Step=3

On graph :

Red – f(x)= 1/x

Blue – g(x) = -9/x

The graph of g lies further from axes than the graph of f. Both have the same asymptotes, domain and range . Graph of g lies in second and fourth quadrants  , and graph of f lies and third quadrants

Question 7.
g(x) = \(\frac{15}{x}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 a 7

Question 8.
g(x) = \(\frac{-12}{x}\)
Answer:

Step 1

The function is of form g(x)= a/x, so the asymptotes are x=0 and y=0

Step 2

Make a table of value and plot the points.

Include both positive and negative values of x.

     x      -3       -2     -1      1       2                    3
     y       4      6      12      -12    -6       -4

Step 3

On graph :

Red – f(x)= 1/x

Blue – g(x) = -12/x

The graph of g lies further from axes than the graph of f. Both have the same asymptotes, domain and range . Graph of g lies in second and fourth quadrants  , and graph of f lies and third quadrants.

Question 9.
g(x) = \(\frac{-0.5}{x}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 a 9

Question 10.
g(x) = \(\frac{0.1}{x}\)
Answer:

Step 1

The function is of form g(x)= a/x, so the asymptotes are x=0 and y=0

step 2

Make a table of value and plot the points.

Include both positive and negative values of x.

      x        -3        -2        -1       1          2         3
      y  -0.033      -0.05       -0.1     0.1       0.05       0.033

Step 3

On graph :

Red – f(x)= 1/x

Blue – g(x) = -0.1/x

The graph of g lies closer to axes than the graph of f . Both graphs lie in first and third quadrants and have the same asymptotes, domain and range

 

In Exercises 11–18, graph the function. State the domain and range.
Question 11.
g(x) = \(\frac{4}{x}\)+ 3
Answer:
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 a 11

Question 12.
y = \(\frac{2}{x}\) − 3
Answer:

Consider function f given by

F(x)=2/x   -3

We have that   x=0

Is vertical asymptote  .The horizontal asymptote is

Y=-3

Plot point to the left of the vertical   asymptote, such as (-1,-5),(-4,-7/2),

And (-2/5 , -8 ).Plot points to the right of the vertical asymptote, such as (2,-2),(1/2,1) and (4,-5/2).

The domain is all real number except x=0 and the range is all real numbers except y=-3

Question 13.
h(x) = \(\frac{6}{x-1}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 a 13

Question 14.
y = \(\frac{1}{x+2}\)
Answer:

Consider function f given by

f(x)=1/x+2

From x+2 =0 we get that  x =-2

is vertical asymptote  .The horizontal asymptote is

Y=0

Plot point to the left of the vertical   asymptote, such as (-3,-1),(-4,-1/2),

And (-7 , -1/5).Plot points to the right of the vertical asymptote, such as (-1/2,-2),(0,1/2) and (3,1/5).

The domain is all real number except x= -2 and the range is all real numbers except y=0

 

 

Question 15.
h(x) = \(\frac{-3}{x+2}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 a 15

Question 16.
f(x) = \(\frac{-2}{x-7}\)
Answer:

Consider function f given by

f(x)=-2/x-7

From x+2 =0 we get that  x = 7

is vertical asymptote  .The horizontal asymptote is

Y=0

Plot point to the left of the vertical   asymptote, such as (5,1),(3,1/2),

And (6 , 2).Plot points to the right of the vertical asymptote, such as (8,-2),(9,1) and (11,-1/2).

The domain is all real number except x=  7and the range is all real numbers except y=0

 

Question 17.
g(x) = \(\frac{-3}{x-4}\) − 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 a 17

Question 18.
y = \(\frac{10}{x+7}\) − 5
Answer:

Consider function f given by

f(x)=10/x+7   –  5

From x+2 =0 we get that  x = -7

is vertical asymptote  .The horizontal asymptote is

Y=-5

Plot point to the left of the vertical   asymptote, such as (-17,-6),(-12,-7),

And (-8 , -15).Plot points to the right of the vertical asymptote, such as (-5,0),(-2,-3) and (3,-4).

The domain is all real number except x=  -7and the range is all real numbers except y=-5

 

ERROR ANALYSIS In Exercises 19 and 20, describe and correct the error in graphing the rational function.
Question 19.
y = \(\frac{-8}{x}\)
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 4
Answer:
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 a 19

Question 20.
y = \(\frac{2}{x-1}\) – 2
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 5
Answer:

Vertical asymptote is incorrect.

It should be x-1=0

X=1

Correct   graph :

ANALYZING RELATIONSHIPS In Exercises 21–24, match the function with its graph. Explain your reasoning.
Question 21.
g(x) = \(\frac{2}{x-3}\) + 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 a 21

Question 22.
h(x) = \(\frac{2}{x+3}\) + 1
Answer:

The graph is A because the asymptotes are x=-3 and y=1

Question 23.
f(x) = \(\frac{2}{x-3}\) − 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 a 23

Question 24.
y = \(\frac{2}{x+3}\) − 1
Answer:

The graph is D  because the asymptotes are x=-3 and  y=1

Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 6.1

In Exercises 25–32, graph the function. State the domain and range.
Question 25.
f(x) = \(\frac{x+4}{x-3}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 a 25.1
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 a 25.2

Question 26.
y = \(\frac{x-1}{x+5}\)
Answer:

Consider function f given by

Y=x-1/x+5

We know that the rational function has form y= a x+ b/cx +d

Where a,b,c,d are constant such that cx+d 0 .

From x+5 =0 we get that  x = -5

is vertical asymptote   .The horizontal asymptote is

Y=a/c = 1/1 =1

Plot point to the left of the vertical   asymptote, such as (-11,2),(-8,3),

And(-7 , -4).Plot points to the right of the vertical asymptote, such as (-4,-5),(-2,-1) and (1,0).

The domain is all real number except x=  -5and the range is all real numbers except y=1

 

Question 27.
y = \(\frac{x+6}{4x-8}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 a 27.1
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 a 27.2

Question 28.
h(x) = \(\frac{8x+3}{2x-6}\)
Answer:

Question 29.
f(x) = \(\frac{-5x+2}{4x+5}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 a 29.1
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 a 29.2

Question 30.
g(x) = \(\frac{6x-1}{3x-1}\)
Answer:

Question 31.
h(x) = \(\frac{-5x}{-2x-3}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 a 31.1
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 a 31.2

Question 32.
y = \(\frac{-2x+1}{-x+10}\)
Answer:

In Exercises 33–40, rewrite the function in the form g(x) = \(\frac{a}{x-h}\) + k. Graph the function. Describe the graph of g as a transformation of the graph of f(x) = \(\frac{a}{x}\).
Question 33.
g(x) = \(\frac{5x+6}{x+1}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 a 33

Question 34.
g(x) = \(\frac{7x+4}{x-3}\)
Answer:

Question 35.
g(x) = \(\frac{2x-4}{x-5}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 a 35

Question 36.
g(x) = \(\frac{4x-11}{x-2}\)
Answer:

Question 37.
g(x) = \(\frac{x+18}{x-6}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 a 37

Question 38.
g(x) = \(\frac{x+2}{x-8}\)
Answer:

Question 39.
g(x) = \(\frac{7x-20}{x+13}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 a 39

Question 40.
g(x) = \(\frac{9x-3}{x+7}\)
Answer:

Question 41.
PROBLEM SOLVING
Your school purchases a math software program. The program has an initial cost of $500 plus $20 for each student that uses the program.
a. Estimate how many students must use the program for the average cost per student to fall to $30.
b. What happens to the average cost as more students use the program?
Answer:
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 a 41

Question 42.
PROBLEM SOLVING
To join a rock climbing gym, you must pay an initial fee of $100 and a monthly fee of $59.
a. Estimate how many months you must purchase a membership for the average cost per month to fall to $69.
b. What happens to the average cost as the number of months that you are a member increases?
Answer:

Question 43.
USING STRUCTURE
What is the vertical asymptote of the graph of the function y = \(\frac{2}{x+4}\) + 7?
A. x =−7
B. x = −4
C. x = 4
D. x = 7
Answer:
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 a 43

Question 44.
REASONING
What are the x-intercept(s) of the graph of the function y = \(\frac{x-5}{x^{2}-1}\)?
A. 1, −1
B. 5
C. 1
D. −5
Answer:

Question 45.
USING TOOLS
The time t (in seconds) it takes for sound to travel 1 kilometer can be modeled by
t = \(\frac{1000}{06T+331}\)
where T is the air temperature (in degrees Celsius).
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 6
a. You are 1 kilometer from a lightning strike. You hear the thunder 2.9 seconds later. Use a graph to fund the approximate air temperature.
b. Find the average rate of change in the time it takes sound to travel 1 kilometer as the air temperature increases from 0°C to 10°C.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 a 45

Question 46.
MODELING WITH MATHEMATICS
A business is studying the cost to remove a pollutant from the ground at its site. The function y = \(\frac{15x}{1.1-x}\) models the estimated cost y (in thousands of dollars) to remove x percent (expressed as a decimal) of the pollutant.
a. Graph the function. Describe a reasonable domain and range.
b. How much does it cost to remove 20% of the pollutant? 40% of the pollutant? 80% of the pollutant? Does doubling the percentage of the pollutant removed double the cost? Explain.
Answer:

USING TOOLS In Exercises 47–50, use a graphing calculator to graph the function. Then determine whether the function is even, odd, or neither.
Question 47.
h(x) = \(\frac{6}{x^{2}+1}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 a 47

Question 48.
f(x) = \(\frac{2 x^{2}}{x^{2}-9}\)
Answer:

Question 49.
y = \(\frac{x^{3}}{3 x^{2}+x^{4}}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 a 49

Question 50.
f(x) = \(\frac{4 x^{2}}{2 x^{3}-x}\)
Answer:

Question 51.
MAKING AN ARGUMENT
Your friend claims it is possible for a rational function to have two vertical asymptotes. Is your friend correct? Justify your answer.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 a 51

Question 52.
HOW DO YOU SEE IT?
Use the graph of f to determine the equations of the asymptotes. Explain.
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 7
Answer:

Question 53.
DRAWING CONCLUSIONS
In what line(s) is the graph of y = \(\frac{1}{x}\) symmetric? What does this symmetry tell you about the inverse of the function f(x) = \(\frac{1}{x}\)?
Answer:
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 a 53

Question 54.
THOUGHT PROVOKING
There are four basic types of conic sections: parabola, circle, ellipse, and hyperbola. Each of these can be represented by the intersection of a double-napped cone and a plane. The intersections for a parabola, circle, and ellipse are shown below.Sketch the intersection for a hyperbola.
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 8
Answer:

Question 55.
REASONING
The graph of the rational function f is a hyperbola. The asymptotes of the graph of f intersect at (3, 2). The point (2, 1) is on the graph. Find another point on the graph. Explain your reasoning.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 a 55

Question 56.
ABSTRACT REASONING
Describe the intervals where the graph of y = \(\frac{a}{x}\) is increasing or decreasing when (a) a > 0 and (b) a < 0. Explain your reasoning.
Answer:

Question 57.
PROBLEM SOLVING
An Internet service provider charges a $50 installation fee and a monthly fee of $43. The table shows the average monthly costs y of a competing provider for x months of service. Under what conditions would a person choose one provider over the other? Explain your reasoning.
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 9
Answer:
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 a 57

Question 58.
MODELING WITH MATHEMATICS
The Doppler effect occurs when the source of a sound is moving relative to a listener, so that the frequency f(in hertz) heard by the listener is different from the frequency fs(in hertz) at the source. In both equations below, r is the speed (in miles per hour) of the sound source.
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 10
a. An ambulance siren has a frequency of 2000 hertz. Write two equations modeling the frequencies heard when the ambulance is approaching and when the ambulance is moving away.
b. Graph the equations in part (a) using the domain 0 ≤ r ≤ 60.
c. For any speed r, how does the frequency heard for an approaching sound source compare with the frequency heard when the source moves away?
Answer:

Maintaining Mathematical Proficiency

Factor the polynomial.
Question 59.
4x2 − 4x− 80
Answer:
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 a 59

Question 60.
3x2 − 3x − 6
Answer:

Question 61.
2x2 − 2x − 12
Answer:
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 a 61

Question 62.
10x2 + 31x − 14
Answer:

Simplify the expression.
Question 63.
32 • 34
Answer:
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 a 63

Question 64.
21/2 • 23/5
Answer:

Question 65.
\(\frac{6^{5 / 6}}{6^{1 / 6}}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 a 65

Question 66.
\(\frac{6^{8}}{6^{10}}\)
Answer:

Rational Functions Study Skills: Analyzing Your Errors

7.1–7.2 What Did You Learn?

Core Vocabulary
inverse variation, p. 360
constant of variation, p. 360
rational function, p. 366

Core Concepts
Section 7.1
Inverse Variation, p. 360
Writing Inverse Variation Equations, p. 361

Section 7.2
Parent Function for Simple Rational Functions, p. 366
Graphing Translations of Simple Rational Functions, p. 367

Mathematical Practices
Question 1.
Explain the meaning of the given information in Exercise 25 on page 364.
Answer:

Question 2.
How are you able to recognize whether the logic used in Exercise 29 on page 364 is correct or flawed?
Answer:

Question 3.
How can you evaluate the reasonableness of your answer in part (b) of Exercise 41 on page 371?
Answer:

Question 4.
How did the context allow you to determine a reasonable domain and range for the function in Exercise 46 on page 371?
Answer:

Study Skills: AnalyzingYour Errors

Study Errors
What Happens: You do not study the right material or you do not learn it well enough to remember it on a test without resources such as notes.
How to Avoid This Error: Take a practice test. Work with a study group. Discuss the topics on the test with your teacher. Do not try to learn a whole chapter’s worth of material in one night.
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions 7.2 11

Rational Functions 7.1–7.2 Quiz

Tell whether x and y show direct variation, inverse variation, or neither. Explain your reasoning.
Question 1.
x + y = 7
Answer:

Question 2.
\(\frac{2}{5}\)x = y
Answer:

Question 3.
xy = 0.45′
Answer:

Question 4.
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions q 1
Answer:

Question 5.
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions q 2
Answer:

Question 6.
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions q 3
Answer:

Question 7.
The variables x and y vary inversely, and y= 10 when x= 5. Write an equation that relates x and y. Then find y when x = −2.
Answer:

Match the equation with the correct graph. Explain your reasoning.
Question 8.
f(x) = \(\frac{3}{x}\) + 2
Answer:

Question 9.
y = \(\frac{-2}{x+3}\) − 2
Answer:

Question 10.
h(x) = \(\frac{2x+2}{3x+1}\)
Answer:

Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions q 4

Question 11.
Rewrite g(x) = \(\frac{2x+9}{x+8}\) in the form g(x) = \(\frac{a}{x-h}\). Graph the function. Describe the graph of g as a transformation of the graph of f(x) = \(\frac{a}{x}\).
Answer:

Question 12.
The time t (in minutes) required to empty a tank varies inversely with the pumping rate r (in gallons per minute). The rate of a certain pump is 70 gallons per minute. It takes the pump 20 minutes to empty the tank. Complete the table for the times it takes the pump to empty a tank for the given pumping rates.
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions q 5
Answer:

Question 13.
A pitcher throws 16 strikes in the first 38 pitches. The table shows how a pitcher’s strike percentage changes when the pitcher throws x consecutive strikes after the first 38 pitches. Write a rational function for the strike percentage in terms of x. Graph the function. How many consecutive strikes must the pitcher throw to reach a strike percentage of 0.60?
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions q 6
Answer:

Lesson 7.3 Multiplying and Dividing Rational Expressions

Essential Question How can you determine the excluded values in a product or quotient of two rational expressions?
You can multiply and divide rational expressions in much the same way that you multiply and divide fractions. Values that make the denominator of an expression zero are excluded values
Big Ideas Math Algebra 2 Solutions Chapter 7 Rational Functions 7.3 1

EXPLORATION 1

Multiplying and Dividing Rational Expressions
Work with a partner. Find the product or quotient of the two rational expressions. Then match the product or quotient with its excluded values. Explain your reasoning.
Big Ideas Math Algebra 2 Solutions Chapter 7 Rational Functions 7.3 2

EXPLORATION 2

Writing a Product or Quotient
Big Ideas Math Algebra 2 Solutions Chapter 7 Rational Functions 7.3 3
Work with a partner. Write a product or quotient of rational expressions that has the given excluded values. Justify your answer.
a. −1
b. −1 and 3
c. −1, 0, and 3

Communicate Your Answer

Question 3.
How can you determine the excluded values in a product or quotient of two rational expressions?
Answer:

Question 4.
Is it possible for the product or quotient of two rational expressions to have no excluded values? Explain your reasoning. If it is possible, give an example.
Answer:

Monitoring Progress

Simplify the rational expression, if possible.
Question 1.
\(\frac{2(x+1)}{(x+1)(x+3)}\)
Answer:

Question 2.
\(\frac{x+4}{x^{2}-16}\)
Answer:

Question 3.
\(\frac{4}{x(x+2)}\)
Answer:

Question 4.
\(\frac{x^{2}-2 x-3}{x^{2}-x-6}\)
Answer:

Find the product.
Question 5.
\(\frac{3 x^{5} y^{2}}{8 x y} \cdot \frac{6 x y^{2}}{9 x^{3} y}\)
Answer:

Question 6.
\(\frac{2 x^{2}-10 x}{x^{2}-25} \cdot \frac{x+3}{2 x^{2}}\)
Answer:

Question 7.
\(\frac{x+5}{x^{3}-1}\)
Answer:

Question 8.
\(\) • (x2 + x + 1)
Answer:

Find the quotient.
Question 9.
\(\frac{4 x}{5 x-20} \div \frac{x^{2}-2 x}{x^{2}-6 x+8}\)
Answer:

Question 10.
\(\frac{2 x^{2}+3 x-5}{6 x}\) ÷ (2x2 + 5x)
Answer:

Multiplying and Dividing Rational Expressions 7.3 Exercises

Vocabulary and Core Concept Check
Question 1.
WRITING
Describe how to multiply and divide two rational expressions.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 7 Rational Functions 7.3 a 1

Question 2.
WHICH ONE DOESN’T BELONG?
Which rational expression does not belong with the other three? Explain your reasoning.
Big Ideas Math Algebra 2 Solutions Chapter 7 Rational Functions 7.3 4
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–10, simplify the expression, if possible.
Question 3.
\(\frac{2 x^{2}}{3 x^{2}-4 x}\)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 7 Rational Functions 7.3 a 3

Question 4.
\(\frac{7 x^{3}-x^{2}}{2 x^{3}}\)
Answer:

Question 5.
\(\frac{x^{2}-3 x-18}{x^{2}-7 x+6}\)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 7 Rational Functions 7.3 a 5

Question 6.
\(\frac{x^{2}+13 x+36}{x^{2}-7 x+10}\)
Answer:

Question 7.
\(\frac{x^{2}+11 x+18}{x^{3}+8}\)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 7 Rational Functions 7.3 a 7

Question 8.
\(\frac{x^{2}-7 x+12}{x^{3}-27}\)
Answer:

Question 9.
\(\frac{32 x^{4}-50}{4 x^{3}-12 x^{2}-5 x+15}\)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 7 Rational Functions 7.3 a 9

Question 10.
\(\frac{3 x^{3}-3 x^{2}+7 x-7}{27 x^{4}-147}\)
Answer:

In Exercises 11–20, find the product.
Question 11.
\(\frac{4 x y^{3}}{x^{2} y} \cdot \frac{y}{8 x}\)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 7 Rational Functions 7.3 a 11

Question 12.
\(\frac{48 x^{5} y^{3}}{y^{4}} \cdot \frac{x^{2} y}{6 x^{3} y^{2}}\)
Answer:

Question 13.
\(\frac{x^{2}(x-4)}{x-3} \cdot \frac{(x-3)(x+6)}{x^{3}}\)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 7 Rational Functions 7.3 a 13

Question 14.
\(\frac{x^{3}(x+5)}{x-9} \cdot \frac{(x-9)(x+8)}{3 x^{3}}\)
Answer:

Question 15.
\(\frac{x^{2}-3 x}{x-2} \cdot \frac{x^{2}+x-6}{x}\)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 7 Rational Functions 7.3 a 15

Question 16.
\(\frac{x^{2}-4 x}{x-1} \cdot \frac{x^{2}+3 x-4}{2 x}\)
Answer:

Question 17.
\(\frac{x^{2}+3 x-4}{x^{2}+4 x+4} \cdot \frac{2 x^{2}+4 x}{x^{2}-4 x+3}\)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 7 Rational Functions 7.3 a 17

Question 18.
\(\frac{x^{2}-x-6}{4 x^{3}} \cdot \frac{2 x^{2}+2 x}{x^{2}+5 x+6}\)
Answer:

Question 19.
\(\frac{x^{2}+5 x-36}{x^{2}-49}\) • (x2 – 11x + 28)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 7 Rational Functions 7.3 a 19

Question 20.
\(\frac{x^{2}-x-12}{x^{2}-16}\) • (x2 – 2x + 8)
Answer:

Question 21.
ERROR ANALYSIS
Describe and correct the error in simplifying the rational expression.
Big Ideas Math Algebra 2 Solutions Chapter 7 Rational Functions 7.3 5
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 7 Rational Functions 7.3 a 21

Question 22.
ERROR ANALYSIS
Describe and correct the error in finding the product.
Big Ideas Math Algebra 2 Solutions Chapter 7 Rational Functions 7.3 6
Answer:

Question 23.
USING STRUCTURE
Which rational expression is in simplified form?
Big Ideas Math Algebra 2 Solutions Chapter 7 Rational Functions 7.3 7
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 7 Rational Functions 7.3 a 23

Question 24.
COMPARING METHODS
Find the product below by multiplying the numerators and denominators, then simplifying. Then find the product by simplifying each expression, then multiplying. Which method do you prefer? Explain.
\(\frac{4 x^{2} y}{2 x^{3}} \cdot \frac{12 y^{4}}{24 x^{2}}\)
Answer:

Question 25.
WRITING
Compare the function f(x) = \(\frac{(3 x-7)(x+6)}{(3 x-7)}\) to the function g(x) = x + 6.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 7 Rational Functions 7.3 a 25

Question 26.
MODELING WITH MATHEMATICS
Write a model in terms of x for the total area of the base of the building.
Big Ideas Math Algebra 2 Solutions Chapter 7 Rational Functions 7.3 8
Answer:

In Exercises 27–34, find the quotient.
Question 27.
\(\frac{32 x^{3} y}{y^{8}} \div \frac{y^{7}}{8 x^{4}}\)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 7 Rational Functions 7.3 a 27

Question 28.
\(\frac{2 x y z}{x^{3} z^{3}} \div \frac{6 y^{4}}{2 x^{2} z^{2}}\)
Answer:

Question 29.
\(\frac{x^{2}-x-6}{2 x^{4}-6 x^{3}} \div \frac{x+2}{4 x^{3}}\)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 7 Rational Functions 7.3 a 29

Question 30.
\(\frac{2 x^{2}-12 x}{x^{2}-7 x+6} \div \frac{2 x}{3 x-3}\)
Answer:

Question 31.
\(\frac{x^{2}-x-6}{x+4}\) ÷ (x2 – 6x + 9)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 7 Rational Functions 7.3 a 31

Question 32.
\(\frac{x^{2}-5 x-36}{x+2}\) ÷ (x2 – 18x + 81)
Answer:

Question 33.
\(\frac{x^{2}+9 x+18}{x^{2}+6 x+8} \div \frac{x^{2}-3 x-18}{x^{2}+2 x-8}\)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 7 Rational Functions 7.3 a 33

Question 34.
\(\frac{x^{2}-3 x-40}{x^{2}+8 x-20} \div \frac{x^{2}+13 x+40}{x^{2}+12 x+20}\)
Answer:

In Exercises 35 and 36, use the following information. Manufacturers often package products in a way that uses the least amount of material. One measure of the efficiency of a package is the ratio of its surface area S to its volume V. The smaller the ratio, the more efficient the packaging.
Question 35.
You are examining three cylindrical containers.
a. Write an expression for the efficiency ratio \(\frac{S}{V}\) of a cylinder.
b. Find the efficiency ratio for each cylindrical can listed in the table. Rank the three cans according to efficiency.
Big Ideas Math Algebra 2 Solutions Chapter 7 Rational Functions 7.3 9
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 7 Rational Functions 7.3 a 35

Question 36.
PROBLEM SOLVING
A popcorn company is designing a new tin with the same square base and twice the height of the old tin.
Big Ideas Math Algebra 2 Solutions Chapter 7 Rational Functions 7.3 10
a. Write an expression for the efficiency ratio \(\frac{S}{V}\) of each tin.
b. Did the company make a good decision by creating the new tin? Explain.
Answer:

Question 37.
MODELING WITH MATHEMATICS
The total amount I (in millions of dollars) of healthcare expenditures and the residential population P (in millions) in the United States can be modeled by
I = \(\frac{171,000 t+1,361,000}{1+0.018 t}\) and P = 2.96t + 278.649
where t is the number of years since 2000. Find a model M for the annual healthcare expenditures per resident. Estimate the annual healthcare expenditures per resident in 2010.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 7 Rational Functions 7.3 a 37

Question 38.
MODELING WITH MATHEMATICS
The total amount I (in millions of dollars) of school expenditures from prekindergarten to a college level and the enrollment P(in millions) in prekindergarten through college in the United States can be modeled by
I = \(\frac{17.913 t+709,569}{1-0.028 t}\) and P = 0.5906t + 70.219
where t is the number of years since 2001. Find a model M for the annual education expenditures per student. Estimate the annual education expenditures per student in 2009.
Big Ideas Math Algebra 2 Solutions Chapter 7 Rational Functions 7.3 11
Answer:

Question 39.
USING EQUATIONS
Refer to the population model P in Exercise 37.
a. Interpret the meaning of the coefficient of t.
b. Interpret the meaning of the constant term.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 7 Rational Functions 7.3 a 39

Question 40.
HOW DO YOU SEE IT?
Use the graphs of f and g to determine the excluded values of the functions h(x) = (fg)(x) and k(x) = (\(\frac{f}{g}\)) (x). Explain your reasoning.
Big Ideas Math Algebra 2 Solutions Chapter 7 Rational Functions 7.3 12
Answer:

Question 41.
DRAWING CONCLUSIONS
Complete the table for the function y = \(\frac{x+4}{x^{2}-16}\) . Then use the trace feature of a graphing calculator to explain the behavior of the function at x = −4.
Big Ideas Math Algebra 2 Solutions Chapter 7 Rational Functions 7.3 13
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 7 Rational Functions 7.3 a 41

Question 42.
MAKING AN ARGUMENT
You and your friend are asked to state the domain of the expression below.
\(\frac{x^{2}+6 x-27}{x^{2}+4 x-45}\)
Your friend claims the domain is all real numbers except 5. You claim the domain is all real numbers except −9 and 5. Who is correct? Explain.
Answer:

Question 43.
MATHEMATICAL CONNECTIONS
Find the ratio of the perimeter to the area of the triangle shown.
Big Ideas Math Algebra 2 Solutions Chapter 7 Rational Functions 7.3 14
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 7 Rational Functions 7.3 a 43

Question 44.
CRITICAL THINKING
Find the expression that makes the following statement true. Assume x ≠ −2 and x ≠ 5.
Big Ideas Math Algebra 2 Solutions Chapter 7 Rational Functions 7.3 15
ans;

USING STRUCTURE In Exercises 45 and 46, perform the indicated operations.
Question 45.
\(\frac{2 x^{2}+x-15}{2 x^{2}-11 x-21}\) • (6x + 9) ÷ \(\frac{2 x-5}{3 x-21}\)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 7 Rational Functions 7.3 a 45

Question 46.
(x3 + 8) • \(\frac{x-2}{x^{2}-2 x+4} \div \frac{x^{2}-4}{x-6}\)
Answer:

Question 47.
REASONING
Animals that live in temperatures several degrees colder than their bodies must avoid losing heat to survive. Animals can better conserve body heat as their surface area to volume ratios decrease. Find the surface area to volume ratio of each penguin shown by using cylinders to approximate their shapes. Which penguin is better equipped to live in a colder environment? Explain your reasoning.
Big Ideas Math Algebra 2 Solutions Chapter 7 Rational Functions 7.3 16
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 7 Rational Functions 7.3 a 47

Question 48.
THOUGHT PROVOKING
Is it possible to write two radical functions whose product when graphed is a parabola and whose quotient when graphed is a hyperbola? Justify your answer.
Answer:

Question 49.
REASONING
Find two rational functions f and g that have the stated product and quotient.
(fg)(x) = x2, (\(\left(\frac{f}{g}\right)\)) (x) = \(\frac{(x-1)^{2}}{(x+2)^{2}}\)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 7 Rational Functions 7.3 a 49

Maintaining Mathematical Proficiency

Solve the equation. Check your solution.
Question 50.
\(\frac{1}{2}\)x + 4 = \(\frac{3}{2}\)x + 5
Answer:

Question 51.
\(\frac{1}{3}\)x − 2 = \(\frac{3}{4}\)x
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 7 Rational Functions 7.3 a 51

Question 52.
\(\frac{1}{4}\)x − \(\frac{3}{5}\) = \(\frac{9}{2}\)x − \(\frac{4}{5}\)
Answer:

Question 53.
\(\frac{1}{2}\)x + \(\frac{1}{3}\) = \(\frac{3}{4}\)x − \(\frac{1}{5}\)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 7 Rational Functions 7.3 a 53

Write the prime factorization of the number. If the number is prime, then write prime.
Question 54.
42
Answer:

Question 55.
91
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 7 Rational Functions 7.3 a 54

Question 56.
72
Answer:

Question 57.
79
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 7 Rational Functions 7.3 a 57

Lesson 7.4 Adding and Subtracting Rational Expressions

Essential Question
How can you determine the domain of the sum or difference of two rational expressions?
You can add and subtract rational expressions in much the same way that you add and subtract fractions.
\(\frac{x}{x+1}+\frac{2}{x+1}=\frac{x+2}{x+1}\) Sum of rational expressions
\(\frac{1}{x}-\frac{1}{2 x}=\frac{2}{2 x}-\frac{1}{2 x}=\frac{1}{2 x}\) Difference of rational expressions

EXPLORATION 1

Adding and Subtracting Rational Expressions
Work with a partner. Find the sum or difference of the two rational expressions. Then match the sum or difference with its domain. Explain your reasoning.
Big Ideas Math Answer Key Algebra 2 Chapter 7 Rational Functions 7.4 1

EXPLORATION 2

Writing a Sum or Difference
Big Ideas Math Answer Key Algebra 2 Chapter 7 Rational Functions 7.4 2
Work with a partner. Write a sum or difference of rational expressions that has the given domain. Justify your answer.
a. all real numbers except −1
b. all real numbers except −1 and 3
c. all real numbers except −1, 0, and 3

Communicate Your Answer

Question 3.
How can you determine the domain of the sum or difference of two rational expressions?
Answer:

Question 4.
Your friend found a sum as follows. Describe and correct the error(s).
\(\frac{x}{x+4}+\frac{3}{x-4}=\frac{x+3}{2 x}\)
Answer:

Monitoring Progress

Find the sum or difference.
Question 1.
\(\frac{8}{12 x}-\frac{5}{12 x}\)
Answer:

Question 2.
\(\frac{2}{3 x^{2}}+\frac{1}{3 x^{2}}\)
Answer:

Question 3.
\(\frac{4 x}{x-2}-\frac{x}{x-2}\)
Answer:

Question 4.
\(\frac{2 x^{2}}{x^{2}+1}+\frac{2}{x^{2}+1}\)
Answer:

Question 5.
Find the least common multiple of 5x3 and 10x2 − 15x.
Answer:

Find the sum or difference.
Question 6.
\(\frac{3}{4 x}-\frac{1}{7}\)
Answer:

Question 7.
\(\frac{1}{3 x^{2}}+\frac{x}{9 x^{2}-12}\)
Answer:

Question 8.
\(\frac{x}{x^{2}-x-12}+\frac{5}{12 x-48}\)
Answer:

Question 9.
Rewrite g(x) = \(\frac{2 x-4}{x-3}\) in the form g(x) = \(\frac{a}{x-h}\) + k. Graph the function. Describe the graph of g as a transformation of the graph of f(x) = \(\frac{a}{x}\).
Answer:

Simplify the complex fraction.
Question 10.
\(\frac{\frac{x}{6}-\frac{x}{3}}{\frac{x}{5}-\frac{7}{10}}\)
Answer:

Question 11.
\(\frac{\frac{2}{x}-4}{\frac{2}{x}+3}\)
Answer:

Question 12.
\(\frac{\frac{3}{x+5}}{\frac{2}{x-3}+\frac{1}{x+5}}\)
Answer:

Adding and Subtracting Rational Expressions 7.4 Exercises

Vocabulary and Core Concept Check
Question 1.
COMPLETE THE SENTENCE
A fraction that contains a fraction in its numerator or denominator is called a(n) __________.
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 7 Rational Functions 7.4 a 1

Question 2.
WRITING
Explain how adding and subtracting rational expressions is similar to adding and subtracting numerical fractions.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–8, find the sum or difference.
Question 3.
\(\frac{15}{4 x}+\frac{5}{4 x}\)
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 7 Rational Functions 7.4 a 3

Question 4.
\(\frac{x}{16 x^{2}}-\frac{4}{16 x^{2}}\)
Answer:

Question 5.
\(\frac{9}{x+1}-\frac{2 x}{x+1}\)
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 7 Rational Functions 7.4 a 5

Question 6.
\(\frac{3 x^{2}}{x-8}+\frac{6 x}{x-8}\)
Answer:

Question 7.
\(\frac{5 x}{x+3}+\frac{15}{x+3}\)
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 7 Rational Functions 7.4 a 7

Question 8.
\(\frac{4 x^{2}}{2 x-1}-\frac{1}{2 x-1}\)

In Exercises 9–16, find the least common multiple of the expressions.
Question 9.
3x, 3(x − 2)
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 7 Rational Functions 7.4 a 9

Question 10.
2x2, 4x+ 12
Answer:

Question 11.
2x, 2x(x − 5)
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 7 Rational Functions 7.4 a 11

Question 12.
24x2, 8x2 − 16x
Answer:

Question 13.
x2 − 25, x − 5
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 7 Rational Functions 7.4 a 13

Question 14.
9x2 − 16, 3x2 + x − 4
Answer:

Question 15.
x2 + 3x − 40, x − 8
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 7 Rational Functions 7.4 a 15

Question 16.
x2 − 2x − 63, x + 7
Answer:

ERROR ANALYSIS In Exercises 17 and 18, describe and correct the error in finding the sum.
Question 17.
Big Ideas Math Answer Key Algebra 2 Chapter 7 Rational Functions 7.4 3
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 7 Rational Functions 7.4 a 17

Question 18.
Big Ideas Math Answer Key Algebra 2 Chapter 7 Rational Functions 7.4 4
Answer:

In Exercises 19–26, find the sum or difference.
Question 19.
\(\frac{12}{5 x}-\frac{7}{6 x}\)
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 7 Rational Functions 7.4 a 19

Question 20.
\(\frac{8}{3 x^{2}}+\frac{5}{4 x}\)
Answer:

Question 21.
\(\frac{3}{x+4}-\frac{1}{x+6}\)
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 7 Rational Functions 7.4 a 21

Question 22.
\(\frac{9}{x-3}+\frac{2 x}{x+1}\)
Answer:

Question 23.
\(\frac{12}{x^{2}+5 x-24}+\frac{3}{x-3}\)
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 7 Rational Functions 7.4 a 23

Question 24.
\(\frac{x^{2}-5}{x^{2}+5 x-14}-\frac{x+3}{x+7}\)
Answer:

Question 25.
\(\frac{x+2}{x-4}+\frac{2}{x}+\frac{5 x}{3 x-1}\)
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 7 Rational Functions 7.4 a 25

Question 26.
\(\frac{x+3}{x^{2}-25}-\frac{x-1}{x-5}+\frac{3}{x+3}\)
Answer:

REASONING In Exercises 27 and 28, tell whether the statement is always, sometimes, or never true. Explain.
Question 27.
The LCD of two rational expressions is the product of the denominators.
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 7 Rational Functions 7.4 a 27

Question 28.
The LCD of two rational expressions will have a degree greater than or equal to that of the denominator with the higher degree.
Answer:

Question 29.
ANALYZING EQUATIONS
How would you begin to rewrite the function g(x) = \(\frac{4 x+1}{x+2}\) to obtain the form g(x) = a \(\frac{a}{x-h}\) + k?
Big Ideas Math Answer Key Algebra 2 Chapter 7 Rational Functions 7.4 5
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 7 Rational Functions 7.4 a 29

Question 30.
ANALYZING EQUATIONS
How would you begin to rewrite the function g(x) = \(\frac{x}{x-5}\) to obtain the form g(x) = \(\frac{a}{x-h}\) + k?
Big Ideas Math Answer Key Algebra 2 Chapter 7 Rational Functions 7.4 6
Answer:

In Exercises 31–38, rewrite the function g in the form g(x) = \(\frac{a}{x-h}\) + k. Graph the function. Describe the graph of g as a transformation of the graph of f(x) = \(\frac{a}{x}\).
Question 31.
g(x) = \(\frac{5 x-7}{x-1}\)
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 7 Rational Functions 7.4 a 31

Question 32.
g(x) = \(\frac{6 x+4}{x+5}\)
Answer:

Question 33.
g(x) = \(\frac{12 x}{x-5}\)
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 7 Rational Functions 7.4 a 33

Question 34.
g(x) = \(\frac{8 x}{x+13}\)
Answer:

Question 35.
g(x) = \(\frac{2 x+3}{x}\)
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 7 Rational Functions 7.4 a 35

Question 36.
g(x) = \(\frac{4 x-6}{x}\)
Answer:

Question 37.
g(x) = \(\frac{3 x+11}{x-3}\)
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 7 Rational Functions 7.4 a 37

Question 38.
g(x) = \(\frac{7 x-9}{x+10}\)
Answer:

In Exercises 39–44, simplify the complex fraction.
Question 39.
\(\frac{\frac{x}{3}-6}{10+\frac{4}{x}}\)
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 7 Rational Functions 7.4 a 39

Question 40.
\(\frac{15-\frac{2}{x}}{\frac{x}{5}+4}\)
Answer:

Question 41.
\(\frac{\frac{1}{2 x-5}-\frac{7}{8 x-20}}{\frac{x}{2 x-5}}\)
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 7 Rational Functions 7.4 a 41

Question 42.
\(\frac{\frac{16}{x-2}}{\frac{4}{x+1}+\frac{6}{x}}\)
Answer:

Question 43.
\(\frac{\frac{1}{3 x^{2}-3}}{\frac{5}{x+1}-\frac{x+4}{x^{2}-3 x-4}}\)
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 7 Rational Functions 7.4 a 43

Question 44.
\(\frac{\frac{3}{x-2}-\frac{6}{x^{2}-4}}{\frac{3}{x+2}+\frac{1}{x-2}}\)
Answer:

Question 45.
PROBLEM SOLVING
The total time T (in hours) needed to fly from New York to Los Angeles and back can be modeled by the equation below, where dis the distance (in miles) each way, a is the average airplane speed (in miles per hour), and j is the average speed (in miles per hour) of the jet stream. Simplify the equation. Then find the total time it takes to fly 2468 miles when a= 510 miles per hour and j = 115 miles per hour.
Big Ideas Math Answer Key Algebra 2 Chapter 7 Rational Functions 7.4 7
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 7 Rational Functions 7.4 a 45

Question 46.
REWRITING A FORMULA
The total resistance Rt of two resistors in a parallel circuit with resistances R1 and R2 (in ohms) is given by the equation shown. Simplify the complex fraction. Then find the total resistance when R1 = 2000 ohms and R2 = 5600 ohms.
Big Ideas Math Answer Key Algebra 2 Chapter 7 Rational Functions 7.4 8
Answer:

Question 47.
PROBLEM SOLVING
You plan a trip that involves a 40-mile bus ride and a train ride. The entire trip is 140 miles. The time (in hours) the bus travels is y1 = \(\frac{40}{x}\), where x is the average speed (in miles per hour) of the bus. The time (in hours) the train travels is y2= \(\frac{100}{x+30}\).Write and simplify a model that shows the total time y of the trip.
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 7 Rational Functions 7.4 a 47

Question 48.
PROBLEM SOLVING
You participate in a sprint triathlon that involves swimming, bicycling, and running. The table shows the distances (in miles) and your average speed for each portion of the race.
Big Ideas Math Answer Key Algebra 2 Chapter 7 Rational Functions 7.4 9
a. Write a model in simplified form for the total time (in hours) it takes to complete the race.
b. How long does it take to complete the race if you can swim at an average speed of 2 miles per hour? Justify your answer.
Answer:

Question 49.
MAKING AN ARGUMENT
Your friend claims that the least common multiple of two numbers is always greater than each of the numbers. Is your friend correct? Justify your answer.
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 7 Rational Functions 7.4 a 49

Question 50.
HOW DO YOU SEE IT?
Use the graph of the function f(x) = \(\frac{a}{x-h}\) + k to determine the values of h and k.
Big Ideas Math Answer Key Algebra 2 Chapter 7 Rational Functions 7.4 10
Answer:

Question 51.
REWRITING A FORMULA
You borrow P dollars to buy a car and agree to repay the loan over t years at a monthly interest rate of i (expressed as a decimal). Your monthly payment M is given by either formula below.
M = \(\frac{P i}{1-\left(\frac{1}{1+i}\right)^{12 t}}\) or M = \(\frac{P i(1+i)^{12 t}}{(1+i)^{12 t}-1}\)
a. Show that the formulas are equivalent by simplifying the first formula.
b. Find your monthly payment when you borrow$15,500 at a monthly interest rate of 0.5% and repay the loan over 4 years.
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 7 Rational Functions 7.4 a 51

Question 52.
THOUGHT PROVOKING
Is it possible to write two rational functions whose sum is a quadratic function? Justify your answer.
Answer:

Question 53.
USING TOOLS
Use technology to rewrite the function g(x) = \(\frac{(97.6)(0.024)+x(0.003)}{12.2+x}\) in the form g(x) = \(\frac{a}{x-h}\) + k. Describe the graph of g as a transformation of the graph of f(x) = \(\frac{a}{x}\).
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 7 Rational Functions 7.4 a 53

Question 54.
MATHEMATICAL CONNECTIONS
Find an expression for the surface area of the box.
Big Ideas Math Answer Key Algebra 2 Chapter 7 Rational Functions 7.4 11
Answer:

Question 55.
PROBLEM SOLVING
You are hired to wash the new cars at a car dealership with two other employees. You take an average of 40 minutes to wash a car (R1 = 1/40 car per minute). The second employee washes a car in x minutes. The third employee washes a car in x + 10 minutes.
a. Write expressions for the rates that each employee can wash a car.
b. Write a single expression R for the combined rate of cars washed per minute by the group.
c. Evaluate your expression in part (b) when the second employee washes a car in 35 minutes. How many cars per hour does this represent? Explain your reasoning.
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 7 Rational Functions 7.4 a 55

Question 56.
MODELING WITH MATHEMATICS
The amount A(in milligrams) of aspirin in a person’s bloodstream can be modeled by
A = \(\frac{391 t^{2}+0.112}{0.218 t^{4}+0.991 t^{2}+1}\)
where t is the time (in hours) after one dose is taken.
Big Ideas Math Answer Key Algebra 2 Chapter 7 Rational Functions 7.4 12
a. A second dose is taken 1 hour after the first dose. Write an equation to model the amount of the second dose in the bloodstream.
b. Write a model for the total amount of aspirin in the bloodstream after the second dose is taken.
Answer:

Question 57.
FINDING A PATTERN
Find the next two expressions in the pattern shown. Then simplify all five expressions. What value do the expressions approach?
Big Ideas Math Answer Key Algebra 2 Chapter 7 Rational Functions 7.4 13
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 7 Rational Functions 7.4 a 57

Maintaining Mathematical Proficiency

Solve the system by graphing.
Question 58.
y = x2 + 6
y = 3x + 4
Answer:

Question 59.
2x2 − 3x − y = 0
\(\frac{5}{2}\)x − y = \(\frac{9}{4}\)
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 7 Rational Functions 7.4 a 59

Question 60.
3 = y − x2 − x
y = −x2 − 3x − 5
Answer:

Question 61.
y= (x + 2)2 − 3
y = x2 + 4x + 5
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 7 Rational Functions 7.4 a 61

Lesson 7.5 Solving Rational Equations

Essential Question How can you solve a rational equation?

EXPLORATION 1

Solving Rational Equations
Work with a partner. Match each equation with the graph of its related system of equations. Explain your reasoning. Then use the graph to solve the equation.
Big Ideas Math Answers Algebra 2 Chapter 7 Rational Functions 7.5 1

EXPLORATION 2

Solving Rational Equations
Work with a partner. Look back at the equations in Explorations 1(d) and 1(e). Suppose you want a more accurate way to solve the equations than using a graphical approach.
a. Show how you could use a numerical approach by creating a table. For instance, you might use a spreadsheet to solve the equations.
Big Ideas Math Answers Algebra 2 Chapter 7 Rational Functions 7.5 2
b. Show how you could use an analytical approach. For instance, you might use the method you used to solve proportions.

Communicate Your Answer

Question 3.
How can you solve a rational equation?
Answer:

Question 4.
Use the method in either Exploration 1 or 2 to solve each equation.
a. \(\frac{x+1}{x-1}=\frac{x-1}{x+1}\)
b. \(\frac{1}{x+1}=\frac{1}{x^{2}+1}\)
c. \(\frac{1}{x^{2}-1}=\frac{1}{x-1}\)
Answer:

Monitoring Progress

Solve the equation by cross multiplying. Check your solution(s).
Question 1.
\(\frac{3}{5 x}=\frac{2}{x-7}\)
Answer:

Question 2.
\(\frac{-4}{x+3}=\frac{5}{x-3}\)
Answer:

Question 3.
\(\frac{1}{2 x+5}=\frac{x}{11 x+8}\)
Answer:

Solve the equation by using the LCD. Check your solution(s).
Question 4.
\(\frac{15}{x}+\frac{4}{5}=\frac{7}{x}\)
Answer:

Question 5.
\(\frac{3 x}{x+1}-\frac{5}{2 x}=\frac{3}{2 x}\)
Answer:

Question 6.
\(\frac{4 x+1}{x+1}=\frac{12}{x^{2}-1}+3\)
Answer:

Solve the equation. Check your solution(s).
Question 7.
\(\frac{9}{x-2}+\frac{6 x}{x+2}=\frac{9 x^{2}}{x^{2}-4}\)
Answer:

Question 8.
\(\frac{7}{x-1}-5=\frac{6}{x^{2}-1}\)
Answer:

Question 9.
Consider the function f(x) = \(\frac{1}{x}\) − 2. Determine whether the inverse of f is a function. Then find the inverse.
Answer:

Question 10.
WHAT IF?
How do the answers in Example 6 change when c = \(\frac{50 m+800}{m}\)?
Answer:

Solving Rational Equations 7.5 Exercises

Vocabulary and Core Concept Check
Question 1.
WRITING
When can you solve a rational equation by cross multiplying? Explain.
Answer:
Big Ideas Math Answers Algebra 2 Chapter 7 Rational Functions 7.5 a 1

Question 2.
WRITING
A student solves the equation \(\frac{4}{x-3}=\frac{x}{x-3}\) and obtains the solutions 3 and 4. Are either of these extraneous solutions? Explain.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–10, solve the equation by cross multiplying. Check your solution(s).
Question 3.
\(\frac{4}{2 x}=\frac{5}{x+6}\)
Answer:
Big Ideas Math Answers Algebra 2 Chapter 7 Rational Functions 7.5 a 3

Question 4.
\(\frac{9}{3 x}=\frac{4}{x+2}\)
Answer:

Question 5.
\(\frac{6}{x-1}=\frac{9}{x+1}\)
Answer:
Big Ideas Math Answers Algebra 2 Chapter 7 Rational Functions 7.5 a 5

Question 6.
\(\frac{8}{3 x-2}=\frac{2}{x-1}\)
Answer:

Question 7.
\(\frac{x}{2 x+7}=\frac{x-5}{x-1}\)
Answer:
Big Ideas Math Answers Algebra 2 Chapter 7 Rational Functions 7.5 a 7

Question 8.
\(\frac{-2}{x-1}=\frac{x-8}{x+1}\)
Answer:

Question 9.
\(\frac{x^{2}-3}{x+2}=\frac{x-3}{2}\)
Answer:
Big Ideas Math Answers Algebra 2 Chapter 7 Rational Functions 7.5 a 9

Question 10.
\(\frac{-1}{x-3}=\frac{x-4}{x^{2}-27}\)
Answer:

Question 11.
USING EQUATIONS
So far in your volleyball practice, you have put into play 37 of the 44 serves you have attempted. Solve the equation \(\frac{90}{100}=\frac{37+x}{44+x}\) to find the number of consecutive serves you need to put into play in order to raise your serve percentage to 90%.
Big Ideas Math Answers Algebra 2 Chapter 7 Rational Functions 7.5 3
Answer:
Big Ideas Math Answers Algebra 2 Chapter 7 Rational Functions 7.5 a 11

Question 12.
USING EQUATIONS
So far this baseball season, you have 12 hits out of 60 times at-bat. Solve the equation 0.360 = \(\frac{12+x}{60+x}\) to find the number of consecutive hits you need to raise your batting average to 0.360.
Answer:

Question 13.
MODELING WITH MATHEMATICS
Brass is an alloy composed of 55% copper and 45% zinc by weight. You have 25 ounces of copper. How many ounces of zinc do you need to make brass?
Answer:
Big Ideas Math Answers Algebra 2 Chapter 7 Rational Functions 7.5 a 13

Question 14.
MODELING WITH MATHEMATICS
You have 0.2 liter of an acid solution whose acid concentration is 16 moles per liter. You want to dilute the solution with water so that its acid concentration is only 12 moles per liter. Use the given model to determine how many liters of water you should add to the solution.
Big Ideas Math Answers Algebra 2 Chapter 7 Rational Functions 7.5 4
Answer:

USING STRUCTURE In Exercises 15–18, identify the LCD of the rational expressions in the equation.
Question 15.
\(\frac{x}{x+3}+\frac{1}{x}=\frac{3}{x}\)
Answer:
Big Ideas Math Answers Algebra 2 Chapter 7 Rational Functions 7.5 a 15

Question 16.
\(\frac{5 x}{x-1}-\frac{7}{x}=\frac{9}{x}\)
Answer:

Question 17.
\(\frac{2}{x+1}+\frac{x}{x+4}=\frac{1}{2}\)
Answer:
Big Ideas Math Answers Algebra 2 Chapter 7 Rational Functions 7.5 a 17

Question 18.
\(\frac{4}{x+9}+\frac{3 x}{2 x-1}=\frac{10}{3}\)
Answer:

In Exercises 19–30, solvethe equation by using the LCD. Check your solution(s).
Question 19.
\(\frac{3}{2}+\frac{1}{x}\) = 2
Answer:
Big Ideas Math Answers Algebra 2 Chapter 7 Rational Functions 7.5 a 19

Question 20.
\(\frac{2}{3 x}+\frac{1}{6}=\frac{4}{3 x}\)
Answer:

Question 21.
\(\frac{x-3}{x-4}\) + 4 = \(\frac{3 x}{x}\)
Answer:
Big Ideas Math Answers Algebra 2 Chapter 7 Rational Functions 7.5 a 21

Question 22.
\(\frac{2}{x-3}+\frac{1}{x}=\frac{x-1}{x-3}\)
Answer:

Question 23.
\(\frac{6 x}{x+4}\) + 4 = \(\frac{2 x+2}{x-1}\)
Answer:
Big Ideas Math Answers Algebra 2 Chapter 7 Rational Functions 7.5 a 23

Question 24.
\(\frac{10}{x}\) + 3 = \(\frac{x+9}{x-4}\)
Answer:

Question 25.
\(\frac{18}{x^{2}-3 x}-\frac{6}{x-3}=\frac{5}{x}\)
Answer:
Big Ideas Math Answers Algebra 2 Chapter 7 Rational Functions 7.5 a 25

Question 26.
\(\frac{10}{x^{2}-2 x}+\frac{4}{x}=\frac{5}{x-2}\)
Answer:

Question 27.
\(\frac{x+1}{x+6}+\frac{1}{x}=\frac{2 x+1}{x+6}\)
Answer:
Big Ideas Math Answers Algebra 2 Chapter 7 Rational Functions 7.5 a 27

Question 28.
\(\frac{x+3}{x-3}+\frac{x}{x-5}=\frac{x+5}{x-5}\)
Answer:

Question 29.
\(\frac{5}{x}\) – 2 = \(\frac{2}{x+3}\)
Answer:
Big Ideas Math Answers Algebra 2 Chapter 7 Rational Functions 7.5 a 29

Question 30.
\(\frac{5}{x^{2}+x-6}\) = 2 + \(\frac{x-3}{x-2}\)
Answer:

ERROR ANALYSIS In Exercises 31 and 32, describe and correct the error in the first step of solving the equation.
Question 31.
Big Ideas Math Answers Algebra 2 Chapter 7 Rational Functions 7.5 5
Answer:
Big Ideas Math Answers Algebra 2 Chapter 7 Rational Functions 7.5 a 31

Question 32.
Big Ideas Math Answers Algebra 2 Chapter 7 Rational Functions 7.5 6
Answer:

Question 33.
PROBLEM SOLVING
You can paint a room in 8 hours. Working together, you and your friend can paint the room in just 5 hours.
a. Let t be the time (in hours) your friend would take to paint the room when working alone. Copy and complete the table.
(Hint: (Work done) = (Work rate) × (Time))
Big Ideas Math Answers Algebra 2 Chapter 7 Rational Functions 7.5 7
b. Explain what the sum of the expressions represents in the last column. Write and solve an equation to find how long your friend would take to paint the room when working alone.
Answer:
Big Ideas Math Answers Algebra 2 Chapter 7 Rational Functions 7.5 a 33

Question 34.
PROBLEM SOLVING
You can clean a park in 2 hours. Working together, you and your friend can clean the park in just 1.2 hours.a. Let t be the time (in hours) your friend would take to clean the park when working alone. Copy and complete the table.
(Hint: (Work done) = (Work rate) × (Time))
Big Ideas Math Answers Algebra 2 Chapter 7 Rational Functions 7.5 8
b. Explain what the sum of the expressions represents in the last column. Write and solve an equation to find how long your friend would take to clean the park when working alone.
Answer:

Question 35.
OPEN-ENDED
Give an example of a rational equation that you would solve using cross multiplication and one that you would solve using the LCD. Explain your reasoning.
Answer:
Big Ideas Math Answers Algebra 2 Chapter 7 Rational Functions 7.5 a 35

Question 36.
OPEN-ENDED
Describe a real-life situation that can be modeled by a rational equation. Justify your answer.
Answer:

In Exercises 37–44, determine whether the inverse of f is a function. Then find the inverse.
Question 37.
f(x) = \(\frac{2}{x-4}\)
Answer:
Big Ideas Math Answers Algebra 2 Chapter 7 Rational Functions 7.5 a 37

Question 38.
f(x) = \(\frac{7}{x+6}\)
Answer:

Question 39.
f(x) = \(\frac{3}{x}\) – 2
Answer:
Big Ideas Math Answers Algebra 2 Chapter 7 Rational Functions 7.5 a 39

Question 40.
f(x) = \(\frac{5}{x}\) – 6
Answer:

Question 41.
f(x) = \(\frac{4}{11-2 x}\)
Answer:
Big Ideas Math Answers Algebra 2 Chapter 7 Rational Functions 7.5 a 41

Question 42.
f(x) = \(\frac{8}{9+5 x}\)
Answer:

Question 43.
f(x) = \(\frac{1}{x^{2}}\) + 4
Answer:
Big Ideas Math Answers Algebra 2 Chapter 7 Rational Functions 7.5 a 43

Question 44.
f(x) = \(\frac{1}{x^{4}}\) – 7
Answer:

Question 45.
PROBLEM SOLVING
The cost of fueling your car for 1 year can be calculated using this equation:
Big Ideas Math Answers Algebra 2 Chapter 7 Rational Functions 7.5 9
Last year you drove9000 miles, paid $3.24 per gallon of gasoline, and spent a total of $1389 on gasoline. Find the fuel-efficiency rate of your car by (a) solving an equation, and (b) using the inverse of the function.
Big Ideas Math Answers Algebra 2 Chapter 7 Rational Functions 7.5 10
Answer:
Big Ideas Math Answers Algebra 2 Chapter 7 Rational Functions 7.5 a 45

Question 46.
PROBLEM SOLVING
The recommended percent p (in decimal form) of nitrogen (by volume) in the air that a diver breathes is given by p = \(\frac{105.07}{d+33}\), where d is the depth (in feet) of the diver. Find the depth when the air contains 47% recommended nitrogen by (a) solving an equation, and (b) using the inverse of the function.
Answer:

USING TOOLS In Exercises 47–50, use a graphing calculator to solve the equation f(x) = g(x).
Question 47.
f(x) = \(\frac{2}{3}\)x, g(x) = x
Answer:
Big Ideas Math Answers Algebra 2 Chapter 7 Rational Functions 7.5 a 47

Question 48.
f(x) = −\(\frac{3}{5x}\), g(x) = −x
Answer:

Question 49.
f(x) = \(\frac{1}{x}\) + 1, g(x) = x2
Answer:
Big Ideas Math Answers Algebra 2 Chapter 7 Rational Functions 7.5 a 49

Question 50.
f(x) = \(\frac{2}{x}\) + 1, g(x) = x2 + 1
Answer:

Question 51.
MATHEMATICAL CONNECTIONS
Golden rectangles are rectangles for which the ratio of the width w to the length ℓ is equal to the ratio of ℓ to ℓ+w. The ratio of the length to the width for these rectangles is called the golden ratio. Find the value of the golden ratio using a rectangle with a width of 1 unit.
Big Ideas Math Answers Algebra 2 Chapter 7 Rational Functions 7.5 11
Answer:
Big Ideas Math Answers Algebra 2 Chapter 7 Rational Functions 7.5 a 51

Question 52.
HOW DO YOU SEE IT?
Use the graph to identify the solution(s) of the rational equation \(\frac{4(x-1)}{x-1}=\frac{2 x-2}{x+1}\). Explain your reasoning.
Big Ideas Math Answers Algebra 2 Chapter 7 Rational Functions 7.5 12
Answer:

USING STRUCTURE In Exercises 53 and 54, find the inverse of the function. (Hint: Try rewriting the function by using either inspection or long division.)
Question 53.
f(x) = \(\frac{3 x+1}{x-4}\)
Answer:
Big Ideas Math Answers Algebra 2 Chapter 7 Rational Functions 7.5 a 53

Question 54.
f(x) = \(\frac{4 x-7}{2 x+3}\)
Answer:

Question 55.
ABSTRACT REASONING
Find the inverse of rational functions of the form y = \(\frac{a x+b}{c x+d}\). Verify your answer is correct by using it to find the inverses in Exercises 53 and 54.
Answer:
Big Ideas Math Answers Algebra 2 Chapter 7 Rational Functions 7.5 a 55

Question 56.
THOUGHT PROVOKING
Is it possible to write a rational equation that has the following number of solutions? Justify your answers.
a. no solution
b. exactly one solution
c. exactly two solutions
d. infinitely many solutions
Answer:

Question 57.
CRITICAL THINKING
Let a be a nonzero real number. Tell whether each statement is always true, sometimes true, or never true. Explain your reasoning.
a. For the equation \(\frac{1}{x-a}=\frac{x}{x-a}\), x=a is an extraneous solution.
b. The equation \(\frac{3}{x-a}=\frac{x}{x-a}\) has exactly one solution.
c. The equation \(\frac{1}{x-a}=\frac{2}{x+a}+\frac{2 a}{x^{2}-a^{2}}\) has no solution.
Answer:
Big Ideas Math Answers Algebra 2 Chapter 7 Rational Functions 7.5 a 57

Question 58.
MAKING AN ARGUMENT
Your friend claims that it is not possible for a rational equation of the form \(\frac{x-a}{b}=\frac{x-c}{d}\), where b≠ 0 and d≠ 0, to have extraneous solutions. Is your friend correct? Explain your reasoning.
Answer:

Maintaining Mathematical Proficiency

Is the domain discrete or continuous? Explain. Graph the function using its domain.
Question 59.
The linear function y = 0.25x represents the amount of money y (in dollars) of x quarters in your pocket. You have a maximum of eight quarters in your pocket.
Answer:
Big Ideas Math Answers Algebra 2 Chapter 7 Rational Functions 7.5 a 59

Question 60.
A store sells broccoli for $2 per pound. The total cost t of the broccoli is a function of the number of pounds p you buy.
Answer:

Evaluate the function for the given value of x.
Question 61.
f(x) = x3 − 2x + 7; x = −2
Answer:
Big Ideas Math Answers Algebra 2 Chapter 7 Rational Functions 7.5 a 61

Question 62.
g(x) = −2x4 + 7x3 + x − 2; x = 3
Answer:

Question 63.
h(x) = −x3 + 3x2 + 5x; x = 3
Answer:
Big Ideas Math Answers Algebra 2 Chapter 7 Rational Functions 7.5 a 63

Question 64.
k(x) = −2x3 − 4x2 + 12x − 5; x = −5
Answer:

Rational Functions Performance Task: Circuit Design

7.3–7.5 What Did You Learn?

Core Vocabulary
Big Ideas Math Answers Algebra 2 Chapter 7 Rational Functions 7.5 13

Core Concepts
Section 7.3
Simplifying Rational Expressions, p. 376
Multiplying Rational Expressions, p. 377
Dividing Rational Expressions, p. 378

Section 7.4
Adding or Subtracting with Like Denominators, p. 384
Adding or Subtracting with Unlike Denominators, p. 384
Simplifying Complex Fractions, p. 387

Section 7.5
Solving Rational Equations by Cross Multiplying, p. 392
Solving Rational Equations by Using the Least Common Denominator, p. 393
Using Inverses of Functions, p. 395

Mathematical Practices
Question 1.
In Exercise 37 on page 381, what type of equation did you expect to get as your solution? Explain why this type of equation is appropriate in the context of this situation.
Answer:

Question 2.
Write a simpler problem that is similar to Exercise 44 on page 382. Describe how to use the simpler problem to gain insight into the solution of the more complicated problem in Exercise 44.
Answer:

Question 3.
In Exercise 57 on page 390, what conjecture did you make about the value the given expressions were approaching? What logical progression led you to determine whether your conjecture was correct?
Answer:

Question 4.
Compare the methods for solving Exercise 45 on page 397. Be sure to discuss the similarities and differences between the methods as precisely as possible.
Answer:

Performance Task: Circuit Design

A thermistor is a resistor whose resistance varies with temperature. Thermistors are an engineer’s dream because they are inexpensive, small, rugged, and accurate. The one problem with thermistors is their responses to temperature are not linear. How would you design a circuit that corrects this problem?
To explore the answers to these questions and more, go to BigIdeasMath.com.
Big Ideas Math Answers Algebra 2 Chapter 7 Rational Functions 7.5 14

Rational Functions Chapter Review

7.1 Inverse Variation (pp. 359–364)

Tell whether x and y show direct variation, inverse variation, or neither.
Question 1.
xy = 5
Answer:

Question 2.
5y = 6x
Answer:

Question 3.
15 = \(\frac{x}{y}\)
Answer:

Question 4.
y − 3 = 2x
Answer:

Question 5.
Big Ideas Math Algebra 2 Answer Key Chapter 7 Rational Functions cr 5
Answer:

Question 6.
Big Ideas Math Algebra 2 Answer Key Chapter 7 Rational Functions cr 6
Answer:

The variables x and y vary inversely. Use the given values to write an equation relating x and y. Then find y when x = -3.
Question 7.
x = 1, y = 5
Answer:

Question 8.
x = −4, y =−6
Answer:

Question 9.
x = \(\frac{5}{2}\),y = 18
Answer:

Question 10.
x = −12, y = \(\frac{2}{3}\)
Answer:

7.2 Graphing Rational Functions (pp. 365–372)

Graph the function. State the domain and range.
Question 11.
y = \(\frac{4}{x-3}\)
Answer:

Question 12.
y = \(\frac{1}{x+5}\) + 2
Answer:

Question 13.
f(x) = \(\frac{3 x-2}{x-4}\)
Answer:

7.3 Multiplying and Dividing Rational Expressions (pp. 375–382)

Find the product or quotient.
Question 14.
\(\frac{80 x^{4}}{y^{3}} \cdot \frac{x y}{5 x^{2}}\)
Answer:

Question 15.
\(\frac{x-3}{2 x-8} \cdot \frac{6 x^{2}-96}{x^{2}-9}\)
Answer:

Question 16.
\(\frac{16 x^{2}-8 x+1}{x^{3}-7 x^{2}+12 x} \div \frac{20 x^{2}-5 x}{15 x^{3}}\)
Answer:

Question 17.
\(\frac{x^{2}-13 x+40}{x^{2}-2 x-15}\) ÷ (x2 – 5x – 24)
Answer:

7.4 Adding and Subtracting Rational Expressions (pp. 383–390)

Find the sum or difference.
Question 18.
\(\frac{5}{6(x+3)}+\frac{x+4}{2 x}\)
Answer:

Question 19.
\(\frac{5 x}{x+8}+\frac{4 x-9}{x^{2}+5 x-24}\)
Answer:

Question 20.
\(\frac{x+2}{x^{2}+4 x+3}-\frac{5 x}{x^{2}-9}\)
Answer:

Rewrite the function in the form g(x) = \(\frac{a}{x-h}\) h + k. Graph the function. Describe the graph of gas a transformation of the graph of f(x) = \(\frac{a}{x}\).
Question 21.
g(x) = \(\frac{5 x+1}{x-3}\)
Answer:

Question 22.
g(x) = \(\frac{4 x+2}{x+7}\)
Answer:

Question 23.
g(x) = \(\frac{9 x-10}{x-1}\)
Answer:

Question 24.
Let f be the focal length of a thin camera lens, p be the distance between the lens and an object being photographed, and q be the distance between the lens and the film. For the photograph to be in focus, the variables should satisfy the lens equation to the right. Simplify the complex fraction.
Big Ideas Math Algebra 2 Answer Key Chapter 7 Rational Functions cr 24
Answer:

7.5 Solving Rational Equations (pp. 391–398)

Solve the equation. Check your solution(s).
Question 25.
\(\frac{5}{x}=\frac{7}{x+2}\)
Answer:

Question 26.
\(\frac{8(x-1)}{x^{2}-4}=\frac{4}{x+2}\)
Answer:

Question 27.
\(\frac{2(x+7)}{x+4}\) – 2 = \(\frac{2 x+20}{2 x+8}\)
Answer:

Determine whether the inverse of f is a function. Then find the inverse.
Question 28.
f(x) = \(\frac{3}{x+6}\)
Answer:

Question 29.
f(x) = \(\frac{10}{x-7}\)
Answer:

Question 30.
f(x) = \(\frac{1}{x}\) + 8
Answer:

Question 31.
At a bowling alley, shoe rentals cost $3 and each game costs $4. The average cost c (in dollars) of bowlingn games is given by c = \(\frac{4 n+3}{n}\). Find how many games you must bowl for the average cost to fall to $4.75 by (a) solving an equation, and (b) using the inverse of a function.
Answer:

Rational Functions Chapter Test

The variables x and y vary inversely. Use the given values to write an equation relating x and y. Then find y when x = 4.
Question 1.
x = 5, y = 2
Answer:

Question 2.
x = −4, y = \(\frac{7}{2}\)
Answer:

Question 3.
x = \(\frac{3}{4}\), y = \(\frac{5}{8}\)
Answer:

The graph shows the function y = \(\frac{1}{x-h}\) + k. Determine whether the value of each constant h and k is positive, negative, or zero. Explain your reasoning.
Question 4.
Big Ideas Math Algebra 2 Answer Key Chapter 7 Rational Functions ct 4
Answer:

Question 5.
Big Ideas Math Algebra 2 Answer Key Chapter 7 Rational Functions ct 5
Answer:

Question 6.
Big Ideas Math Algebra 2 Answer Key Chapter 7 Rational Functions ct 6
Answer:

Perform the indicated operation.
Question 7.
\(\frac{3 x^{2} y}{4 x^{3} y^{5}} \div \frac{6 y^{2}}{2 x y^{3}}\)
Answer:

Question 8.
\(\frac{3 x}{x^{2}+x-12}-\frac{6}{x+4}\)
Answer:

Question 9.
\(\frac{x^{2}-3 x-4}{x^{2}-3 x-18} \cdot \frac{x-6}{x+1}\)
Answer:

Question 10.
\(\frac{4}{x+5}+\frac{2 x}{x^{2}-25}\)
Answer:

Question 11.
Let g(x) = \(\frac{(x+3)(x-2)}{x+3}\). Simplify g(x). Determine whether the graph of f(x) =x− 2 and the graph of g are different. Explain your reasoning.
Answer:

Question 12.
You start a small beekeeping business. Your initial costs are $500 for equipment and bees. You estimate it will cost $1.25 per pound to collect, clean, bottle, and label the honey. How many pounds of honey must you produce before your average cost per pound is $1.79? Justify your answer.
Answer:

Question 13.
You can use a simple lever to lift a 300-pound rock. The force F (in foot-pounds) needed to lift the rock is inversely related to the distance d (in feet) from the pivot point of the lever. To lift the rock, you need 60 pounds of force applied to a lever with a distance of 10 feet from the pivot point. What force is needed when you increase the distance to 15feet from the pivot point? Justify your answer.
Big Ideas Math Algebra 2 Answer Key Chapter 7 Rational Functions ct 13
Answer:

Question 14.
Three tennis balls fit tightly in a can as shown.
a. Write an expression for the height h of the can in terms of its radius r. Then rewrite the formula for the volume of a cylinder in terms of r only.
b. Find the percent of the can’s volume that is not occupied by tennis balls.
Big Ideas Math Algebra 2 Answer Key Chapter 7 Rational Functions ct 14
Answer:

Rational Functions Cumulative Assessment

Question 1.
Which of the following functions are shown in the graph? Select all that apply. Justify your answers.
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions ca 1
Answer:

Question 2.
You step onto an escalator and begin descending. After riding for 12 feet, you realize that you dropped your keys on the upper floor and walk back up the escalator to retrieve them. The total time T of your trip down and up the escalator is given by
T = \(\frac{12}{s}+\frac{12}{w-s}\)
where s is the speed of the escalator and w is your walking speed. The trip took 9 seconds, and you walk at a speed of 6 feet per second. Find two possible speeds of the escalator.
Answer:

Question 3.
The graph of a rational function has asymptotes that intersect at the point (4, 3). Choose the correct values to complete the equation of the function.
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions ca 3
Answer:

Question 4.
The tables below give the amounts A (in dollars) of money in two different bank accounts over time t (in years).
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions ca 4
a. Determine the type of function represented by the data in each table.
b. Provide an explanation for the type of growth of each function.
c. Which account has a greater value after 10 years? after 15 years? Justify your answers.
Answer:

Question 5.
Order the expressions from least to greatest. Justify your answer.(HSN-RN.A.1)
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions ca 5
Answer:

Question 6.
A movie grosses $37 million after the first week of release. The weekly gross sales y decreases by 30% each week. Write an exponential decay function that represents the weekly gross sales in week x. What is a reasonable domain and range in this situation? Explain your reasoning.(HSF-LE.A.2)
Answer:

Question 7.
Choose the correct relationship among the variables in the table. Justify your answer by writing an equation that relates p, q, and r. (HSA-CED.A.2)
Big Ideas Math Algebra 2 Answers Chapter 7 Rational Functions ca 6
A. The variable p varies directly with the difference of q and r.
B. The variable r varies inversely with the difference of p and q.
C. The variable q varies inversely with the sum of p and r.
D. The variable p varies directly with the sum of q and r.
Answer:

Question 8.
You have taken five quizzes in your history class, and your average score is 83 points. You think you can score 95 points on each remaining quiz. How many quizzes do you need to take to raise your average quiz score to 90 points? Justify your answer. (HSA-REI.A.2)
Answer:

Big Ideas Math Answers Grade 3 Chapter 2 Multiplication Facts and Strategies

Big Ideas Math Answers Grade 3 Chapter 2

Hi guys!!! Are you in the search of various opportunities to increase your problem-solving skills? We are giving the best solution to it, just practice questions from Big Ideas Math Grade 3 Chapter 2 Multiplication Facts and Strategies Textbook. This BIM Book 3rd Grade 2nd Chapter Multiplication Facts and Strategies Answer Key is designed in a fun learning way like mathematical practices through engaging activities like vocabulary exercises, games, and practice tests. Hence, download Big Ideas Math Answers Grade 3 Chapter 2 Multiplication Facts and Strategies PDF to prepare effectively by clearing all your doubts while practicing.

Big Ideas Math Grade 3 Answer Key Chapter 2 Multiplication Facts and Strategies

Access the lessonwise BIM Book Grade 3 Answer Key available via quick links and use them as a reference whenever you have queries and unable to solve a question. Elementary School Big Ideas Math 3rd Grade 2nd Chapter Multiplication Facts and Strategies Solutions makes it easy to learn the concert of multiplication and their strategies. Feel free to prepare any topic from 2nd chapter for your assessments or homework.

The lessons in Big Ideas Math Book 3rd Grade Answers Chapter 2 Multiplication Facts and Strategies are Multiply by 2, Multiply by 5, Multiply by 10, Multiply by 0 or 1, and Use the Distributive Property. Just tap on the link to navigate to the place where you can see the solutions of each topic.

Lesson 1 – Multiply by 2

Lesson 2 – Multiply by 5

Lesson 3 – Multiply by 10

Lesson 4 – Multiply by 0 or 1

Lesson 5 – Use the Distributive Property

Lesson 6 – Problem Solving: Multiplication

Performance Task

Lesson 2.1 Multiply by 2

Explore and Grow

Model 3 × 2 using equal groups.
Big Ideas Math Answer Key Grade 3 Chapter 2 Multiplication Facts and Strategies 2.1 1
3 × 2 = ____

Answer: 6
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-2-Multiplication-Facts-and-Strategies-2.1-1
2 + 2 + 2 = 6
Multiplication is 3 × 2 = 6

Structure
How can you use the model to find 4 × 2?

Think and Grow Multiply by 2

Answer: 4 × 2 = 8
Big Ideas Math Grade 3 Answer Key Chapter 2 Multiplication Facts and Strategies
2 + 2 + 2 + 2 =8
4 × 2 = 8

Example Find 6 × 2.
Model six groups of two.
Big Ideas Math Answer Key Grade 3 Chapter 2 Multiplication Facts and Strategies 2.1 2

A multiple of a number is the product of that number and any other counting number.

Answer:
There are 6 groups and 2 counters in each group.
So add all the counters in each group.
2 + 2 + 2 + 2 + 2 + 2 = 12
Multiply the number of groups and the number of counters in each group.
6 × 2 = 12

Example
Complete the table and the statements.
Big Ideas Math Answer Key Grade 3 Chapter 2 Multiplication Facts and Strategies 2.1 3

Answer:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-2-Multiplication-Facts-and-Strategies-2.1-3

Show and Grow

Question 1.
Complete the equations for the model.
Big Ideas Math Answer Key Grade 3 Chapter 2 Multiplication Facts and Strategies 2.1 4

Answer:
2 + 2 + 2 + 2 = 8
4 × 2 = 8

Find the product
Question 2.
7 × 2 = ____
Answer: 14

Explanation:
2 + 2 + 2 + 2 + 2 + 2 + 2 = 14
7 × 2 = 14

Question 3.
2 × 3 = _____
Answer: 6

Explanation:
2 + 2 + 2 = 6
2 × 3 = 6

Apply and Grow: Practice

Complete the equations for the model
Question 4.
Big Ideas Math Answer Key Grade 3 Chapter 2 Multiplication Facts and Strategies 2.1 5
Answer: 2 + 2 + 2 + 2 + 2 = 10
2 × 5 = 10

Explanation:
By seeing the above figure we can say that there are 2 counters in 5 groups.
Add the number of counters in each group.
That means 2 + 2 + 2 + 2 + 2 = 10
Multiply the number of groups and number of counters in each group.
2 × 5 = 10

Question 5.
Big Ideas Math Answer Key Grade 3 Chapter 2 Multiplication Facts and Strategies 2.1 6
Answer: 9 + 9 = 18
9 × 2 = 18

Explanation:
There are 9 counters in each group
Total number of groups = 2
Add the number of groups
9 + 9 = 18
Multiply the number of groups and number of counters in each group
9 × 2 = 18

Find the product
Question 6.
Big Ideas Math Answer Key Grade 3 Chapter 2 Multiplication Facts and Strategies 2.1 7
Answer: 4
Multiply the two numbers 2 and 2.
2 × 2 = 4

Question 7.
Big Ideas Math Answer Key Grade 3 Chapter 2 Multiplication Facts and Strategies 2.1 8
Answer: 6
Multiply the two numbers 3 and 2.
3 × 2 = 6

Question 8.
Big Ideas Math Answer Key Grade 3 Chapter 2 Multiplication Facts and Strategies 2.1 9
Answer: 20

Explanation:
Multiply the two numbers 2 and 10.
2 × 10 = 20

Question 9.
Big Ideas Math Answer Key Grade 3 Chapter 2 Multiplication Facts and Strategies 2.1 10
Answer: 2
Any number multiplied with 1 will be the same number.
1 × 2 = 2

Question 10.
Big Ideas Math Answer Key Grade 3 Chapter 2 Multiplication Facts and Strategies 2.1 11
Answer: 12

Explanation:
Multiply the two numbers 6 and 2.
6 × 2 = 12

Question 11.
Big Ideas Math Answer Key Grade 3 Chapter 2 Multiplication Facts and Strategies 2.1 12
Answer: 16

Explanation:
Multiply the two numbers 2 and 8.
2 × 8 = 16

Question 12.
Big Ideas Math Answer Key Grade 3 Chapter 2 Multiplication Facts and Strategies 2.1 13
Answer: 8

Explanation:
Multiply the two numbers with 4 and 2.
4 × 2 = 8

Question 13.
Big Ideas Math Answer Key Grade 3 Chapter 2 Multiplication Facts and Strategies 2.1 14
Answer: 14

Explanation:
Multiply the two numbers 7 and 2.
7 × 2 = 14

Find the missing factor.
Question 14.
2 × ___ = 14
Answer: 7

Explanation:
Let the missing factor be x
2 × x = 14
x = 14/2 = 7
X = 7
Thus the missing factor is 7.

Question 15.
2 × ____ = 20
Answer: 10

Explanation:
Let the missing factor be y
2 × y = 20
y = 20/2
y = 10
Therefore the missing factor y is 10.

Question 16.
5 × ____ = 10
Answer: 2

Explanation:
Let a be the missing factor
5 × a = 10
a = 10/5
a = 2
Thus the missing factor is 2.

Question 17.
How many gloves are in 9 pairs of gloves?
Answer: 18

Explanation:
Given that there are 9 pairs of gloves
Each pair contains 2 gloves
9 × 2 = 18
There are 18 gloves in 9 pairs.

Question 18.
Structure
How are the models similar? How are they different?
Big Ideas Math Answer Key Grade 3 Chapter 2 Multiplication Facts and Strategies 2.1 15
Answer: There are 8 counters in 2 groups. Counters are similar but the groups are different.

Think and Grow: Modeling Real Life

A sled dog team must have at least 6 rows of 2 dogs. A musher has 15 dogs. Does he have enough dogs for a team?
Multiplication equation:
Big Ideas Math Answer Key Grade 3 Chapter 2 Multiplication Facts and Strategies 2.1 16
The musher ______ have enough dogs for a team.

Answer:
Given,
A sled dog team must have at least 6 rows of 2 dogs.
6 × 2 = 12
A musher has 15 dogs.
15 – 12 = 3 dogs
Therefore the musher has enough dogs for a team.

Show and Grow

Question 19.
Your teacher asks you to make 2 rows of 8 chairs. There are18 chairs. Do you have enough chairs?
Answer: Yes

Explanation:
Given,
Your teacher asks you to make 2 rows of 8 chairs. There are18 chairs.
2 × 8 = 16
Thus 16 chairs are enough to arrange 8 chairs in 2 rows.

Question 20.
You have 5 pairs of socks. Do you have enough socks to make 12 sock puppets?
Big Ideas Math Answer Key Grade 3 Chapter 2 Multiplication Facts and Strategies 2.1 17
DIG DEEPER!
You want to make as many sock puppets as you can with the socks that you have. You need 2 googly eyes for each puppet. How many googly eyes do you need?
Answer: 10

Explanation:
Given that, there are 5 pairs of socks.
You need 2 googly eyes for each puppet.
5 × 2 = 10
Thus you need 10 googly eyes.

Multiply by 2 Homework & Practice 2.1

Complete the equations for the model.
Question 1.
Big Ideas Math Answer Key Grade 3 Chapter 2 Multiplication Facts and Strategies 2.1 18
Answer:
Observe the figure given above,
There are 3 groups and 2 counters in each group.
Add the number of counters in each group.
2 + 2 + 2 = 6
Multiply the number of groups and number of counters in each group.
3 × 2 = 6

Question 2.
Big Ideas Math Answer Key Grade 3 Chapter 2 Multiplication Facts and Strategies 2.1 19
Answer:
There are 2 groups and 6 counters in each group.
Add the number of counters in each group.
6 + 6 = 12
Multiply the number of groups and number of counters in each group
2 × 6 = 12

Find the product.
Question 3.
Big Ideas Math Answer Key Grade 3 Chapter 2 Multiplication Facts and Strategies 2.1 20
Answer: 8

Explanation:
Multiply the two numbers 4 and 2.
4 × 2 = 8

Question 4.
Big Ideas Math Answer Key Grade 3 Chapter 2 Multiplication Facts and Strategies 2.1 21
Answer: 20

Explanation:
Multiply the two numbers 10 and 2.
10 × 2 = 20

Question 5.
Big Ideas Math Answer Key Grade 3 Chapter 2 Multiplication Facts and Strategies 2.1 22
Answer: 16

Explanation:
Multiply the two numbers 2 and 8.
2 × 8 = 16

Question 6.
Big Ideas Math Answer Key Grade 3 Chapter 2 Multiplication Facts and Strategies 2.1 23
Answer: 18

Explanation:
Multiply the two numbers 2 and 9.
2 × 9 = 18

Question 7.
Big Ideas Math Answer Key Grade 3 Chapter 2 Multiplication Facts and Strategies 2.1 24
Answer: 14

Explanation:
Multiply the two numbers 7 and 2.
7 × 2 = 14

Question 8.
Big Ideas Math Answer Key Grade 3 Chapter 2 Multiplication Facts and Strategies 2.1 25
Answer: 4
Multiply the two numbers 2 and 2.
2 × 2 = 4

Question 9.
Big Ideas Math Answer Key Grade 3 Chapter 2 Multiplication Facts and Strategies 2.1 26
Answer: 10

Explanation:
Multiply the two numbers 5 and 2.
5 × 2 = 10

Question 10.
Big Ideas Math Answer Key Grade 3 Chapter 2 Multiplication Facts and Strategies 2.1 27
Answer: 2

Explanation:
Multiply the two numbers 2 and 1.
2 × 1 = 2

Find the missing factor.
Question 11.
2 × ___ = 16
Answer: 8

Explanation:
Let the missing factor be x.
2 × x = 16
x = 16/2
x = 8
Thus the missing factors is 8.

Question 12.
2 × ___ = 18
Answer: 9

Explanation:
Let the missing factor be a.
2 × a = 18
a = 18/2
a = 9
Thus the missing factor is 9.

Question 13.
____ × 3 = 6
Answer: 2

Explanation:
Let the missing factor be y.
y × 3 = 6
y = 6/3 = 2
y = 2
Therefore the missing factor is 2.

Question 14.
A pendulum swings once every 2 seconds. How long will it take for the pendulum to swing 4 times?
Answer: 8

Explanation:
Given,
A pendulum swings once every 2 seconds.
2 × 4 = 8
x = 4/2 = 2

Question 15.
Repeated Reasoning
Complete the multiplication table.
Big Ideas Math Answer Key Grade 3 Chapter 2 Multiplication Facts and Strategies 2.1 28
Answer:

Big-Ideas-Math-Answer-Key-Grade-3-Chapter-2-Multiplication-Facts-and-Strategies-2.1-28

Question 16.
You buy a shirt that costs $7 and a pair of pants that costs 2 times as much. Write a multiplication equation to show the cost of the pants.
_____ × _____ = ____
How much money do you spend in all? _____
Answer: $7 × 2 = $14

Explanation:
Given,
You buy a shirt that costs $7 and a pair of pants that costs 2 times as much.
Multiply the cost of the shirt and the cost of the pants.
7 × 2 = 14

Question 17.
Modeling Real Life
A plumber needs to install handles in 9 sinks. Each sink has a hot water handle and a cold water handle. He has 19 handles. Does he have enough handles?
Answer: Yes

Explanation:
Given that,
A plumber needs to install handles in 9 sinks.
Each sink has a hot water handle and a cold water handle.
The total number of handles = 19 handles.
9 × 2 = 18
Thus he has enough handles.

Question 18.
Modeling Real Life
You have 6 pairs of chopsticks. Do you have enough chopsticks for 4 people if they each get a pair?
Big Ideas Math Answer Key Grade 3 Chapter 2 Multiplication Facts and Strategies 2.1 28

DIG DEEPER!
You want to make wrappers for each pair of chop sticks that you have. You need 2 pieces of paper for each wrapper.How many pieces of paper do you need?
Answer: 24

Explanation:
Given that there are 6 pairs of chopsticks.
Each pair has 2 chopsticks.
6 × 2 = 12
You need 2 pieces of paper for each wrapper.
12 × 2 = 24
Thus I need 24 pieces of paper for 12 chopsticks.

Review & Refresh

Compare
Question 19.
Big Ideas Math Answer Key Grade 3 Chapter 2 Multiplication Facts and Strategies 2.1 30
Answer: >
923 is greater than 854.

Question 20.
Big Ideas Math Answer Key Grade 3 Chapter 2 Multiplication Facts and Strategies 2.1 31
Answer: >
621 is greater than 63.

Question 21.
Big Ideas Math Answer Key Grade 3 Chapter 2 Multiplication Facts and Strategies 2.1 32
Answer: <
746 is less than 752.

Lesson 2.2 Multiply by 5

Explore and Grow

Question 1.
Model 3 × 5 on the number line
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.2 1
3 × 5 = ____
Answer: 15

Question 2.
Model 4 × 5 on the number line
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.2 2
4 × 5 = ____
Answer: 20

Structure
Compare your models. How are the models the same? How are they different?
Answer:

Think and Grow: Multiply by 5

Example
Find 8 × 5.
Skip count by 5s eight times.
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.2 3

Answer: 40

Explanation:
The count starts from 0.
The counters jump for every 5s.
There are 8 jumps of 5
8 × 5 = 40

Example
Complete the table and the statements.
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.2 4
• Multiples of 5 end in 5s Facts ____ or _____ .
• An odd number times 5 is always an ______ number.
• An even number times 5 is always an ______ number.

Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-2-Multiplication-Facts-and-Strategies-2.2-4
Multiples of 5 end in 5s Facts 0 or 5.
• An odd number times 5 is always an odd number.
• An even number times 5 is always an even number.

Show and Grow

Question 1.
Complete the model and the equation.
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.2 5
Answer: 30

Big-Ideas-Math-Answers-3rd-Grade-Chapter-2-Multiplication-Facts-and-Strategies-2.2-5

Explanation:
The count starts from 0.
The counters jump for every 5s.
There are 6 jumps of 5.
6 × 5 = 30

Find the product.
Question 2.
9 × 5 = ____
Answer: 45

Explanation:
Multiply the two numbers 9 and 5.
9 × 5 = 45

Question 3.
5 × 4 = _____
Answer: 20

Explanation:
Multiply the two numbers 5 and 4.
5 × 4 = 20

Apply and Grow: Practice

Complete the model and the equation
Question 4.
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.2 6
Answer: 15

Big-Ideas-Math-Answers-3rd-Grade-Chapter-2-Multiplication-Facts-and-Strategies-2.2-6

Explanation:
The count starts from 0.
The counters jump for every 5s.
There are 3 jumps of 5.
3 × 5 = 15

Question 5.
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.2 7
Answer: 5 × 2 = 10

Big-Ideas-Math-Answers-3rd-Grade-Chapter-2-Multiplication-Facts-and-Strategies-2.2-7

Find the product
Question 6.
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.2 8
Answer: 20

Explanation:
Multiply the two numbers 4 and 5.
4 × 5 = 20

Question 7.
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.2 9
Answer: 50

Explanation:
Multiply the two numbers 5 and 10.
5 × 10 = 20

Question 8.
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.2 10
Answer: 40

Explanation:
Multiply the two numbers 8 and 5.
8 × 5 = 40

Question 9.
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.2 11
Answer: 45

Explanation:
Multiply the two numbers 9 and 5.
9 × 5 = 45

Question 10.
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.2 12
Answer: 30

Explanation:
Multiply the two numbers 6 and 5.
6 × 5 = 30

Question 11.
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.2 13
Answer: 5

Explanation:
Multiply the two numbers 5 and 1.
5 × 1 = 5

Question 12.
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.2 14
Answer: 25

Explanation:
Multiply the two numbers 5 and 5.
5 × 5 = 25

Question 13.
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.2 15
Answer: 35

Explanation:
Multiply the two numbers 5 and 7.
5 × 7 = 35

Find the missing factor.
Question 14.
5 × ____ = 35
Answer: 7

Explanation:
Let the missing factor be a.
5 × a = 35
a = 35/5 = 7
a = 7
Thus the missing factor is 7.

Question 15.
____ × 5 = 50
Answer: 10

Explanation:
Let the missing factor be x.
x × 5 = 50
x = 50/5
x = 10
Thus the missing factor is 10.

Question 16.
10 = 2 × ____
Answer: 5

Explanation:
Let the missing factor be p.
10 = 2 × p
p = 10/2 = 5
p = 5
Thus the missing factor is 5.

Question 17.
Number Sense
Which numbers are multiples of 5? Think: How do you know?
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.2 16

Answer: The multiples of 5 are 50, 20, 5, 85.

Question 18.
DIG DEEPER!
Use each number to complete the Venn diagram
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.2 17
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.2 18
Answer:
Big-Ideas-Math-Answers-3rd-Grade-Chapter-2-Multiplication-Facts-and-Strategies-2.2-18

Think and Grow: Modeling Real Life

A summer camp is in session 5 days each week for 8 weeks. Each day,1 camper is chosen to lead the camp song. There are 35 campers. Can each camper lead the camp song?
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.2 19
Multiplication equation:
Each camper ____ lead the camp song.

Answer: Each camper can lead the camp song.

Explanation:
Given,
A summer camp is in session 5 days each week for 8 weeks. Each day,1 camper is chosen to lead the camp song. There are 35 campers.
5 days – 1 week
? – 8 weeks
8 × 5 = 40 days
There are 35 campers.
Thus Each camper can lead the camp song.

Show and Grow

Question 19.
Your teacher has 5 packs of spinning toys. There are 5 toys in each pack. There are 27 students in your class. Does every student get a spinning toy?
Answer: No

Explanation:
Given,
Your teacher has 5 packs of spinning toys. There are 5 toys in each pack. There are 27 students in your class.
5 × 5 = 25
27 – 25 = 2
Thus 2 students will not get a spinning toy.

Question 20.
You recycle 9 bottles and receive 5¢ for each bottle. You spend 25¢ on a pack of gum. How many cents do you have left?
DIG DEEPER!
How many more bottles do you need to recycle to buy another pack of gum?
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.2 20
Answer: 20 cents

Explanation:
You recycle 9 bottles and receive 5¢ for each bottle. You spend 25¢ on a pack of gum.
9 × 5 = ¢45
¢45 – ¢25 = ¢20

Multiply by 5 Homework & Practice 2.2

Complete the model and the equation.
Question 1.
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.2 21
Answer: 5 × 4 = 20

Explanation:
The count starts from 0.
The count jumps for every 5.
From the figure, we can see there are 4 jumps of 5.
5 × 4 = 20

Question 2.
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.2 22
Answer: 5 × 1 = 5

Big-Ideas-Math-Answers-3rd-Grade-Chapter-2-Multiplication-Facts-and-Strategies-2.2-22

Find the product
Question 3.
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.2 23
Answer: 15

Explanation:
Multiply the two numbers 5 and 3.
5 × 3 = 15

Question 4.
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.2 24
Answer: 50

Explanation:
Multiply the two numbers 5 and 10.
5 × 10 = 50

Question 5.
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.2 25
Answer: 40

Explanation:
Multiply the two numbers 8 and 5.
8 × 5 = 40

Question 6.
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.2 26
Answer: 10

Explanation:
Multiply the two numbers 5 and 2
5 × 2 = 10

Question 7.
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.2 27
Answer: 45

Explanation:
Multiply the two numbers 5 and 9.
5 × 9 = 45

Question 8.

Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.2 28
Answer: 35

Explanation:
Multiply the two numbers 5 and 7.
5 × 7 = 35

Question 9.
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.2 29
Answer: 25

Explanation:
Multiply the two numbers 5 and 5.
5 × 5 = 25

Question 10.
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.2 30
Answer: 30

Explanation:
Multiply the two numbers 5 and 6.
5 × 6 = 30

Find the missing factor.
Question 11.
5 × ___ = 20
Answer: 4

Explanation:
Let the missing factor be t.
5 × t = 20
t = 20/5
t = 4
Thus the missing factor is 4.

Question 12.
45 = 5 × ____
Answer: 9

Explanation:
Let the missing factor be a.
45 = 5 × a
a = 45/5
a = 9

Question 13.
____ × 5 = 15
Answer: 3

Explanation:
Let the missing factor be x.
x × 5 = 15
x = 15/5
x = 3

Question 14.
There are 8 teams in a basketball tournament. Each team has 5 players. How many players are in the tournament?
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.2 31
Answer: 40

Explanation:
Given,
There are 8 teams in a basketball tournament. Each team has 5 players.
8 × 5 = 40
Therefore there are 40 players in the tournament.

Question 15.
Reasoning
Newton has some nickels. He says they have a total value of 14 cents. Explain how you know that Newton is incorrect.
Answer:
Given,
Newton has some nickels. He says they have a total value of 14 cents.
Convert from cents to nickels.
1 nickel = 5 cents
14 cents = 2.8 nickels

Question 16.
Number Sense
Tell whether the number is a multiple of 2, 5, or both. Explain.
25 & 16
Answer:
25 is a multiple of 5.
16 is a multiple of 2.

Question 17.
Modeling Real Life
A firefighter visits 5 houses. Each house has 3 smoke detectors that need a new battery.She has 18 batteries. Does every smoke detector get a new battery?
Answer:
Given,
A firefighter visits 5 houses.
Each house has 3 smoke detectors that need a new battery.
She has 18 batteries.
5 × 3 = 15
18 – 15 = 3
So, every smoke detector gets a new battery.

Question 18.
Modeling Real Life
Descartes earns $5 for each lawn that he mows. He mows 7 lawns. He spends $20 on a video game. How much money does he have left?

DIG DEEPER!
How many more lawns does Descartes have to mow to buy another video game?
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.2 32
Answer:
Given,
Descartes earns $5 for each lawn that he mows. He mows 7 lawns.
5 × 7 = $35
He spends $20 on a video game.
$35 – $20 = $15

Review & Refresh

Question 19.
Count by tens.
____, 20, ____, 40, ____, ____, 70, ____, ____, _____

Answer: 10, 20, 30, 40, 50, 60, 70, 80, 90, 100.

Lesson 2.3 Multiply by 10

Question 1.
Explore 6 × 10 on the tape diagram.
Big Ideas Math Answers Grade 3 Chapter 2 Multiplication Facts and Strategies 2.3 1
6 × 10 = ___
Answer: 60
Big-Ideas-Math-Answers-Grade-3-Chapter-2-Multiplication-Facts-and-Strategies-2.3-1

Structure
How can you see the model to find 7 × 10?
Answer: 70
Big-Ideas-Math-Answers-Grade-3-Chapter-2-Multiplication-Facts-and-Strategies-2.3-1
10 + 10 + 10 + 10 + 10 + 10 + 10 = 70
7 × 10 = 70

Think and Grow: Multiply by 10

Example
Find 4 × 10.
Model four groups of ten.
Big Ideas Math Answers Grade 3 Chapter 2 Multiplication Facts and Strategies 2.3 2

Example
Complete the table and the statements
Big Ideas Math Answers Grade 3 Chapter 2 Multiplication Facts and Strategies 2.3 3

Show and Grow

Question 1.
Complete the model and the equations.
Big Ideas Math Answers Grade 3 Chapter 2 Multiplication Facts and Strategies 2.3 4
Answer:
10 + 10 + 10 = 30
3 × 10 = 30

Find the product
Question 2.
8 × 10 = ____
Answer: 80
Explanation:
10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 = 80
Thus 8 × 10 = 80

Question 3.
10 × 5 = ____
Answer: 50

Explanation:
5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 = 50
Thus 10 × 5 = 50

Apply and Grow: Practice

Complete the model and the equations.
Question 4.
Big Ideas Math Answers Grade 3 Chapter 2 Multiplication Facts and Strategies 2.3 5
Answer:
10 + 10 + 10 + 10 + 10 = 50
5 × 10 = 50

Question 5.
Big Ideas Math Answers Grade 3 Chapter 2 Multiplication Facts and Strategies 2.3 6
Answer: 20

Explanation:
10 + 10 = 20
2 jumps of 10.
2 × 10 = 20

Find the product
Question 6.
Big Ideas Math Answers Grade 3 Chapter 2 Multiplication Facts and Strategies 2.3 7
Answer: 40

Explanation:
Multiply the two numbers 4 and 10.
4 × 10 = 40

Question 7.
Big Ideas Math Answers Grade 3 Chapter 2 Multiplication Facts and Strategies 2.3 8
Answer: 100

Explanation:
Multiply the two numbers 10 and 10.
10 × 10 = 100

Question 8.
Big Ideas Math Answers Grade 3 Chapter 2 Multiplication Facts and Strategies 2.3 9
Answer: 70

Explanation:
Multiply the two numbers 10 and 7
10 × 7 = 70

Question 9.
Big Ideas Math Answers Grade 3 Chapter 2 Multiplication Facts and Strategies 2.3 10
Answer: 90

Explanation:
Multiply the two numbers 9 and 10.
9 × 10 = 90

Question 10.
Big Ideas Math Answers Grade 3 Chapter 2 Multiplication Facts and Strategies 2.3 11
Answer: 60

Explanation:
Multiply the two numbers 6 and 10.
6 × 10 = 60

Question 11.
Big Ideas Math Answers Grade 3 Chapter 2 Multiplication Facts and Strategies 2.3 12
Answer: 20

Explanation:
Multiply the two numbers 2 and 10.
2 × 10 = 20

Question 12.
Big Ideas Math Answers Grade 3 Chapter 2 Multiplication Facts and Strategies 2.3 13
Answer: 30

Explanation:
Multiply the two numbers 10 and 3.
10 × 3 = 30

Question 13.
Big Ideas Math Answers Grade 3 Chapter 2 Multiplication Facts and Strategies 2.3 14
Answer: 10

Explanation:
Multiply the two numbers 10 and 1.
10 × 1 = 10

Tell Whether the product is even or odd.
Question 14.
6 × 2 = ____
Answer: Even number

Explanation:
6 × 2 = 12
An even number is an integer that can be divided by two and remain an integer or has no remainder.
12 is an even number.

Question 15.
5 × 3 = ___
Answer: Odd number

Explanation:
5 × 3 = 15
An odd number is an integer when divided by two, either leaves a remainder or the result is a fraction.
15 is an odd number.

Question 16.
8 × 10 = _____
Answer: Even number

Explanation:
8 × 10 = 80
An even number is an integer that can be divided by two and remain an integer or has no remainder.
80 is an even number.

Question 17.
5 × 10 = ____
Answer: Even number

Explanation:
5 × 10 = 50
An even number is an integer that can be divided by two and remain an integer or has no remainder.
50 is an even number.

Question 18.
10 × 7 = _____
Answer: Even number

Explanation:
7 × 10 = 70
An even number is an integer that can be divided by two and remain an integer or has no remainder.
70 is an even number.

Question 19.
5 × 7 = ____
Answer: Odd number

Explanation:
5 × 7 = 35
An odd number is an integer when divided by two, either leaves a remainder or the result is a fraction.
35 is an odd number.

Question 20.
There are 10 millimeters in 1 centimeter. The width of a cell phone is 6 centimeters. What is the width of the phone in millimeters?
Answer: 60 millimeters

Explanation:
Given,
There are 10 millimeters in 1 centimeter. The width of a cell phone is 6 centimeters
Convert from centimeters to millimeters.
1 cm = 10 mm
6 cm = 6 × 10mm = 60mm
Thus the width of the phone in millimeters is 60mm.

Question 21.
Patterns
Complete the table using the words even or odd
Big Ideas Math Answers Grade 3 Chapter 2 Multiplication Facts and Strategies 2.3 32
Answer:
Big-Ideas-Math-Answers-Grade-3-Chapter-2-Multiplication-Facts-and-Strategies-2.3-32

Think and Grow: Modeling Real Life

Newton has 8 dimes. Descartes has 75¢. Who has more money?
Multiplication equation:
Big Ideas Math Answers Grade 3 Chapter 2 Multiplication Facts and Strategies 2.3 16
_____ has more money.

Explanation:
Given,
Newton has 8 dimes. Descartes has 75¢.
1 dime =  ¢10
8 dimes = 8 × ¢10 = ¢80
¢80 – ¢75 = ¢5
Thus Newton has more money than Descartes.

Show and Grow

Question 22.
You have 10 nickels. Your friend has 46¢. Who has more money?
Answer: I have more money

Explanation:
Convert from nickels to cents.
1 Nickel = ¢5
10 Nickels = 10 × ¢5 = ¢50
¢50 – ¢46 = ¢4
Thus I have more money.

Question 23.
You have 34 trading cards. You buy 3 packs of cards. Each pack has 10 cards. How many trading cards do you have now?
Answer: 74 trading cards

Explanation:
You have 34 trading cards. You buy 3 packs of cards. Each pack has 10 cards.
1 pack = 10 cards
3 packs = 3 × 10 cards = 30 cards
34 cards + 30 cards = 74 cards

Question 24.
Your teacher has 10 calculators. Each calculator uses 4 batteries. The batteries come in packs of 10. How many packs of batteries does your teacher buy? Explain.
Big Ideas Math Answers Grade 3 Chapter 2 Multiplication Facts and Strategies 2.3 17
Answer: 4 packs

Explanation:
Given,
Your teacher has 10 calculators. Each calculator uses 4 batteries. The batteries come in packs of 10
1 calculator = 4 batteries
10 calculators = 10 × 4 batteries = 40 batteries
Each pack contains 10 batteries
Thus the teacher needs to buy 4 packs of batteries.

Multiply by 10 Homework & Practice 2.3

Complete the model and the equation
Question 1.
Big Ideas Math Answers Grade 3 Chapter 2 Multiplication Facts and Strategies 2.3 18
Answer:
10 + 10 + 10 + 10 + 10 = 50
5 × 10 = 50

Question 2.
Big Ideas Math Answers Grade 3 Chapter 2 Multiplication Facts and Strategies 2.3 19
Answer:
10 + 10 + 10 = 30
3 × 10 = 30

Find the product
Question 3.
Big Ideas Math Answers Grade 3 Chapter 2 Multiplication Facts and Strategies 2.3 20
Answer: 40

Explanation:
Multiply the two numbers 10 and 4.
First, multiply 4 with 0
Next multiply 4 with 1
4 × 0 = 0
4 × 1 = 4
10 × 4 = 40

Question 4.
Big Ideas Math Answers Grade 3 Chapter 2 Multiplication Facts and Strategies 2.3 21
Answer: 80

Explanation:
Multiply the two numbers 8 and 10.
8 × 10 = 80

Question 5.
Big Ideas Math Answers Grade 3 Chapter 2 Multiplication Facts and Strategies 2.3 22
Answer: 100

Explanation:
Multiply the two numbers 10 and 10.
10 × 10 = 100

Question 6.
Big Ideas Math Answers Grade 3 Chapter 2 Multiplication Facts and Strategies 2.3 23
Answer: 20

Explanation:
Multiply the two numbers 10 and 2.
10 × 2 = 20

Question 7.
Big Ideas Math Answers Grade 3 Chapter 2 Multiplication Facts and Strategies 2.3 24
Answer: 90

Explanation:
Multiply the two numbers 10 and 9.
10 × 9 = 90

Question 8.
Big Ideas Math Answers Grade 3 Chapter 2 Multiplication Facts and Strategies 2.3 25
Answer: 70

Explanation:
Multiply the two numbers 10 and 7.
10 × 7 = 70

Question 9.
Big Ideas Math Answers Grade 3 Chapter 2 Multiplication Facts and Strategies 2.3 26
Answer: 10

Explanation:
Multiply the two numbers 10 and 1.
10 × 1 = 10

Question 10.
Big Ideas Math Answers Grade 3 Chapter 2 Multiplication Facts and Strategies 2.3 27
Answer: 50

Explanation:
Multiply the two numbers 10 and 5.
10 × 5 = 50

Find the missing factor
Question 11.
10 × ____ = 90
Answer: 9

Explanation:
Let the missing factor be r.
10 × r = 90
r = 90/10
r = 9
Thus the missing factor is 9.

Question 12.
60 = 10 × ____
Answer: 6

Explanation:
Let the missing factor be s.
60 = 10 × s
s = 60/10
s = 6
Thus the missing factor is 6.

Question 13.
____ × 10 = 70
Answer: 7

Explanation:
Let the missing factor be t.
t × 10 = 70
t = 70/10
t = 7
Thus the missing factor is 7.

Question 14.
Structure
You buy a thank-you card for $1. You pay using all dimes. How many dimes do you use?
Answer: 10

Explanation:
Given,
You buy a thank-you card for $1. You pay using all dimes.
Convert from the dollar to dime.
1 dollar = 10 dime
Thus you use 10 dimes.

Question 15.
Number Sense
In bowling, knocking down all 10 pins at once is called a strike. You roll strikes in your first two turns and knock down 4 pins in your third turn. How many pins do you knock down in all?
Big Ideas Math Answers Grade 3 Chapter 2 Multiplication Facts and Strategies 2.3 28
Answer: 34 pins

Explanation:
Given,
In bowling, knocking down all 10 pins at once is called a strike.
You roll strikes in your first two turns and knock down 4 pins in your third turn.
1 turn = 10 pins
2 turns = 20 pins
20 + 4 = 24 pins
10 + 24 = 34 pins

Question 16.
Reasoning
Ten years is called a decade. Your neighbor is 70 years old. For how many decades has he been alive? Explain.
Answer: 7 decades

Explanation:
Given,
Ten years is called a decade. Your neighbor is 70 years old.
1 decade = 10 years
? = 70 years
70 × 1/10 = 7
Thus he has been alive for 7 decades.

Question 17.
Modeling Real Life
You have 2 dimes. Your friend has a 24¢. Who has more money?
Answer: My friend has more money

Explanation:
Given,
You have 2 dimes. Your friend has 24¢
1 dime = ¢10
2 dimes = 2 × ¢10 = ¢20
¢24 – ¢20 = ¢4

Question 18.
Modeling Real Life
Your teacher has 35 balloons. She buys 4 packs of balloons. Each pack has 10 balloons. How many balloons does she have now?
Answer: 75 balloons

Explanation:
Given,
Your teacher has 35 balloons.
She buys 4 packs of balloons.
Each pack has 10 balloons
1 pack = 10 balloons
4 packs = 4 × 10 balloons = 40 balloons
35 + 40 = 75 balloons

Review & Refresh

Find the difference.
Question 19.
Big Ideas Math Answers Grade 3 Chapter 2 Multiplication Facts and Strategies 2.3 29
Answer: 302

4 8 9
-1 8 7
3 0 2

Question 20.
Big Ideas Math Answers Grade 3 Chapter 2 Multiplication Facts and Strategies 2.3 30
Answer: 328

Big Ideas Math Grade 3 Chapter 2 img-2

Question 21.
Big Ideas Math Answers Grade 3 Chapter 2 Multiplication Facts and Strategies 2.3 31
Answer: 269

Big Ideas Math Grade 3 Chapter 2 img-1

Lesson 2.4 Multiply by 0 or 1

Explore and Grow

Question 1.
Draw an array to find the products
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.4 1
Answer:

Big-Ideas-Math-Solutions-Grade-3-Chapter-2-Multiplication-Facts-and-Strategies-2.4-1

Question 2.
What pattern do you see?
Answer: We can see the multiplication pattern in the above question.

Repeated Reasoning
How can you see the pattern to find 5 × 0?
5 × 0 = _____
Answer: 0
Any number multiplied by 0 is always 0.

Think and Grow: The Multiplication Property of 0 and 1

Multiplication Property of Zero:The product of any number and 0 is 0.
Multiplication Property of One:The product of any number and 1 is that number.
Example
There are 4 groups with 1 counter in each group. How many counters are there in all?
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.4 2

Example
There are 4 groups with 0 counters in each group. How many counters are there in all?
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.4 3
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.4 4

Show and Grow

Write an equation for the model.

Question 1.
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.4 5
Answer: 7 × 0 = 0
There are no counters in the groups. So, the equation for the above figure is 7 × 0 = 0.

Question 2.
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.4 6
Answer: 3 × 1 = 3
There is 1 counter in each group. Thus the equation for the above figure is 3 × 1 = 3

Find the product.
Question 3.
8 × 1 = _____
Answer: 8

Explanation:
Multiply 8 by 1.
Any number multiplied by 1 is always itself.
8 × 1 = 8

Question 4.
6 × 0 = _____
Answer: 0

Explanation:
Multiply 6 by 0.
Any number multiplied by 0 is always 0.
6 × 0 = 0

Apply and Grow: Practice

Complete the equation for the model.

Question 5
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.4 7
Answer: 0

Explanation:
There are 0 counters in the group.
Number of groups = 5
Number of counters = 0
5 × 0 = 0

Question 6.
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.4 8
Answer: 2

Explanation:
There are 2 counters.
Number of counters in each group = 1
Number of groups = 2
2 × 1 = 2

Find the product.
Question 7
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.4 9
Answer: 1

Explanation:
Multiply 1 by 1
Any number multiplied by 1 is always itself.
1 × 1 = 1

Question 8.
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.4 10
Answer: 0

Explanation:
Any number multiplied by 0 is always 0.
2 × 0 =0

Question 9.
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.4 11
Answer: 0

Explanation:
Any number multiplied by 0 is always 0.
0 × 0 =0

Question 10.
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.4 12
Answer: 9

Explanation:
Any number multiplied by 1 is always itself.
1 × 9 = 9

Find the missing factor.
Question 11.
10 × ____ = 0
Answer: 0

Explanation:
Let the missing factor be x
10 × x = 0
x = 0/10
x = 0
Thus the missing factor is 0

Question 12.
____ × 1 = 5
Answer: 5

Explanation:
Let the missing factor be y.
y × 1 = 5
y = 5/1
y = 5
Thus the missing factor is 5.

Question 13.
1 × _____ = 8
Answer: 8

Explanation:
Let the missing factor be z.
1 × z = 8
z = 8/1
z = 8

Compare.
Question 14.
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.4 13
Answer: >

Explanation:
1 × 4 = 4
6 × 0 = 0
4 is greater than 0.
4 > 0
Thus 1 × 4 > 6 × 0

Question 15.
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.4 14
Answer: =

Explanation:
9 × 0 = 0
7 × 0 =0
0 is equal to 0.
Thus 9 × 0 = 7 × 0

Question 16.
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.4 15
Answer: <

Explanation:
3 × 1 = 3
6 × 1 = 6
3 is less than 6.
3 < 6
Thus 3 × 1 < 6 × 1

Question 17.
Logic
Complete each statement with always, sometimes never.
The product of any number and 0 is ______ 0.
The product of any number and 1 is ________ that number.
Answer:
The product of any number and 0 is always 0.
The product of any number and 1 is the same that number.

Question 18.
Structure
Use the model to complete the statements. Which two properties are shown?
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.4 16
Answer:
There are 5 groups of 1 = 1 group of 5
5 × 1 = 1 × 5
5 = 5

Think and Grow: Modeling Real Life

There are 4 performers riding unicycles and 3 performers using balance boards in a talent show.How many wheels are used in the talent show?
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.4 17
Multiplication equations:
Addition equation:
There are ______ wheels used in the talent show.

Answer: 4 wheels are used in the talent show.

Show and Grow

Question 19.
A is a performance by 1 person. A is a performance by 2 people. There are 5 students performing solos and 0 students performing duets at a school concert. How many students perform a solo or a duet?
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.4 18
Answer: 5

Explanation:
Given,
A is a performance by 1 person. A is a performance by 2 people.
There are 5 students performing solos and 0 students performing duets at a school concert.
5 + 0 = 5
Therefore 5 students performing a solo or duet.

Question 20.
A group of 6 adults and 9 students attend a school concert. What is the total cost for the group?
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.4 19
DIG DEEPER!
How much money would it cost the group to attend the concert if the tickets cost $1 for all ages? Explain.
Answer:
Given that,
A group of 6 adults and 9 students attend a school concert.
1 adult = $1
6 adults = 6 × $1 = $6
There is no ticket price for students.
$6 + 0 = $6
If the ticket price is $1 for all the students.
9 + 6 = 15
15 × $1 = $15
Thus the ticket price for all the students is $15.

Multiply by 0 or 1 Homework & Practice 2.4

Complete the equation for the model.
Question 1.
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.4 20
Answer: 0

Explanation:
Number of groups = 3
Number of counters in each group = 0
The equation for the model is 3 × 0 = 0

Question 2.
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.4 21
Answer: 8

Explanation:
Number of groups = 8
Number of counters in each group = 1
The equation for the model is 8 × 1 = 8

Find the product
Question 3.
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.4 22
Answer: 10

Explanation:
Multiply the two numbers 10 and 1.
Any number multiplied by 1 is the same number.
10 × 1 = 10

Question 4.
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.4 23
Answer: 0

Explanation:
Multiply the two numbers 6 and 0.
Any number multiplied by 0 is always 0.
6 × 0 = 0

Question 5.
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.4 24
Answer: 7

Explanation:
Multiply the two numbers 1 and 7.
Any number multiplied by 1 is the same number.
1 × 7 = 7

Question 6.
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.4 25
Answer: 0

Explanation:
Multiply the two numbers 1 and 0.
Any number multiplied by 0 is always 0.
0 × 1 = 0

Find the missing factor
Question 7.
_____ × 4 = 0
Answer: 0

Explanation:
Let the missing factor be x
x × 4 = 0
x = 0/4
x = 0
Thus the missing factor is 0

Question 8.
1 × ____ = 2
Answer: 2

Explanation:
Let the missing factor be x
1 × x = 2
x = 2/1
x = 2
Thus the missing factor is 2.

Question 9.
9 × ____ = 9
Answer: 1

Explanation:
Let the missing factor be x
9 × x = 9
x = 9/9
x = 1
Thus the missing factor is 1.

Compare
Question 10.
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.4 26
Answer: =

Explanation:
Any number multiplied by 0 is always 0.
0 × 9 = 10 × 0

Question 11.
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.4 27
Answer: >

Explanation:
Any number multiplied by 0 is always 0.
Any number multiplied by 1 is the same number.
4 × 1 = 4
8 × 0 = 0
4 > 0
Thus 4 × 1 > 8 × 0

Question 12.
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.4 28
Answer: =

Explanation:
Any number multiplied by 1 is the same number.
6 × 1 = 1 × 6

Question 13.
Logic
Which equation is true?
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.4 29
Answer: All the equations are false.
i. 8 × 0 = 8 + 0
0 = 8 (false)
ii. 3 × 1 = 3 + 1
3 = 4 (false)
iii. 6 + 1 = 6 × 1
7 = 6 (false)
iv. 4 × 1 = 4 × 0
4 = 0 (false)

Question 14.
Writing
How are the problems the same? How are they different?
5 × 1 = 5  5 + 0 = 5
Answer:
The answers for the problems are same. But the difference is the first equation is multiplication and the second equation is addition.

Question 15.
Modeling Real Life
U.S. presidents can serve two 4-year terms. A president serves the first term, but is not reelected. How many years does the president serve?
Answer: 8 years.

Explanation:
Given,
U.S. presidents can serve two 4-year terms. A president serves the first term, but is not reelected.
2 × 4 = 8
Thus the president serves 8 years.

Question 16.
Modeling Real Life
There are 5 adults and 8 children in a group. What is the total cost for the group to ride the Ferris wheel?
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.4 30
DIG DEEPER!
How much money would it cost the group toride the Ferris wheel if the tickets cost $1 for all ages? Explain.
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.4 31
Answer: $13

Explanation:
The ticket price for adult is $1
The ticket price for student is free
The total price is 5 × $1 = $5
The ticket price for an adult is $1
The ticket price for the student is free
Total = 5 + 8 = 13
The total price is 13 × $1 = $13

Review & Refresh

Question 17.
Use the array to fill in the blanks.
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.4 32
Answer:
3 rows 6 columns
3 × 6 = 18

Lesson 2.5 Use the Distributive Property

Explore and Grow

Break apart the 4 × 7 array into two smaller arrays. Write an equation for each new array.
Big Ideas Math Answer Key Grade 3 Chapter 2 Multiplication Facts and Strategies 2.5 1
Equation: _____
Equation: ____

Structure
Find the sum of your two products. Compare the sum to the total number of objects in the original array. What do you notice?
Answer:
Equation: 2 × 7
Equation: 2 × 7

Explanation:
By using the distributive property we can find the sum of the two products of the above array.
Distribute 4 to 2 and 2.
4 × 7 = (2 + 2) × 7
= (2 × 7) + (2 × 7)

Think and Grow: Use the Distributive Property with Addition

Distributive Property (with addition)
Big Ideas Math Answer Key Grade 3 Chapter 2 Multiplication Facts and Strategies 2.5 2

Big Ideas Math Answer Key Grade 3 Chapter 2 Multiplication Facts and Strategies 2.5 3

Show and Grow
Question 1.
Use the Distributive Property to show two different ways to find 2 × 4
Big Ideas Math Answer Key Grade 3 Chapter 2 Multiplication Facts and Strategies 2.5 4
Answer:
Big-Ideas-Math-Answer-Key-Grade-3-Chapter-2-Multiplication-Facts-and-Strategies-2.5-4
In the first figure distribute 4 to 2 and 2.
In the second figure distribute 2 to 1 and 1.

Apply and Grow: Practice

Use the Distributive Property to find the product
Question 2.
Big Ideas Math Answer Key Grade 3 Chapter 2 Multiplication Facts and Strategies 2.5 5
Answer:
Distribute 4 to 2 and 2.
4 × 7 = (2 + 2) × 7
4 × 7 = (2 × 7) + (2 × 7)
4 × 7 = 14 + 14
4 × 7 = 28

Question 3.
Big Ideas Math Answer Key Grade 3 Chapter 2 Multiplication Facts and Strategies 2.5 6
Answer:
Distribute 6 to 3 and 3.
6 × 6 = 6 × (3 + 3)
6 × 6 = (6 × 3) + (6 × 3)
6 × 6 =  18 + 18
6 × 6 = 36

Use the Distributive Property to fill in the blanks.

Question 4.
7 × 4 = 7 × (___ + 2)
= (7 × ___) + (____ × 2)
= ___ + ____
= ____
Answer: 2

Explanation:
Distribute 4 to 2 and 2.
7 × 4 = 7 × (2 + 2)
= (7 × 2) + (7 × 2)
= 14 + 14
= 28

Question 5.
3 × 6 = (___ + ___) × 6
= (___ × 6) + (____ × 6)
= ___ + ____
= ____
Answer:
3 × 6 = (1 + 2) × 6
= (1 × 6) + (2 × 6)
= 6 + 12
= 18

Question 6.
Which one Doesn’t Belong? Which expression cannot be used to find 3 × 7?
Big Ideas Math Answer Key Grade 3 Chapter 2 Multiplication Facts and Strategies 2.5 7
Answer: 3 × (7 + 1)

Explanation:
i. 3 × (5 + 2) = 3 × 7
ii. 3 × (7 + 1) = 3 × 8
iii. (2 + 1) × 7 = 3 × 7
iv. (1 + 2) × 7 = 3 × 7
The second expression does not belong to 3 × 7.

Question 7.
Writing
Explain how you can use the Distributive Property to find 5 × 12.
Answer:
Given expression 5 × 12
First, distribute 12 to 6 and 6.
5 × 12 = 5 × (6 + 6)
5 × 12 = (5 × 6) + (5 × 6)
5 × 12 = 30 + 30
5 × 12 = 60

Think and Grow: Modeling Real Life

A scooter rental center has 6 rows of scooters. There are 9 scooters in each row. Can 50 people each rent a scooter at the same time?
Multiplication expression:
Distributive Property:
50 people _______ each rent a scooter at the same time.

Answer: 50 people can each rent a scooter at the same time.

Explanation:
Given,
A scooter rental center has 6 rows of scooters. There are 9 scooters in each row.
6 × 9 = (3 + 3) × 9
6 × 9 = (3 × 9) + (3 × 9)
6 × 9 = 27 + 27
6 × 9 = 54
54 – 50 = 4
Thus 50 people can each rent a scooter at the same time.

Show and Grow

Question 8.
A roller-coaster train has 9 rows of seats. There are 4 seats in each row. Can a group of 38 students ride the roller coaster at the same time?
Big Ideas Math Answer Key Grade 3 Chapter 2 Multiplication Facts and Strategies 2.5 8
Answer:
Given,
A roller-coaster train has 9 rows of seats. There are 4 seats in each row.
We can find the product by using the distributive property.
9 × 4 = 9 × (2 + 2)
9 × 4 = (9 × 2) + (9 × 2)
9 × 4 = 18 + 18
9 × 4 = 36
We have to find whether a group of 38 students ride the roller coaster at the same time.
36 – 38 = -2
Thus a group of 38 students cannot ride the roller coaster at the same time.

Question 9.
DIG DEEPER!
A joke book has 20 pages. There are 5 jokes on each page. You read 16 jokes. How many jokes do you have left to read?
Answer:
Given that,
A joke book has 20 pages.
There are 5 jokes on each page.
Total jokes = 20 × 5 = 100 jokes
You read 16 jokes.
100 – 16 = 84 jokes
∴ 84 jokes are left to read.

Use the Distributive Property Homework & Practice 2.5

Use the Distributive Property to find the product
Question 1.
Big Ideas Math Answer Key Grade 3 Chapter 2 Multiplication Facts and Strategies 2.5 9
Answer:
Distribute 4 to 2 and 2.
4 × 4 = 4 × (2 + 2)
4 × 4 = (4 × 2) + (4 × 2)
4 × 4 = 8 + 8
4 × 4 = 16

Question 2.
Big Ideas Math Answer Key Grade 3 Chapter 2 Multiplication Facts and Strategies 2.5 10
Answer:
Distribute 3 to 1 and 2.
3 × 9 = (1 + 2) × 9
3 × 9 = (1 × 9) + (2 × 9)
3 × 9 = 9 + 18
3 × 9 = 27

Use the Distributive Property to fill in the blanks.
Question 3.
8 × 6 = ___ × (5 + ____)
= (8 × 5) + (____ × ____)
= ___ + ___
= ____
Answer:
8 × 6 = 8 × (5 + 1)
= (8 × 5) + (8 × 1)
= 40 + 8
= 48

Question 4.
7 × 7 = 7 × (5 + ____)
= (____ × 5) + (7 × ____)
= ____ + ____
= ____
Answer:
7 × 7 = 7 × (5 + 2)
= (7 × 5) + (7 × 2)
= 35 + 14
= 49

Question 5.
Structure
Which ways can you break apart the array?
Big Ideas Math Answer Key Grade 3 Chapter 2 Multiplication Facts and Strategies 2.5 11
Big Ideas Math Answer Key Grade 3 Chapter 2 Multiplication Facts and Strategies 2.5 12
Answer:
(7 × 5) + (7 × 1)
(5 × 6) + (2 + 6)

Explanation:
Distribute 6 to 5 and 1.
7 × 6 = 7 × (5 + 1)
7 × 6 = (7 × 5) + (7 × 1)
7 × 6 = 35 + 7
7 × 6 = 42
Distribute 7 to 5 and 2.
7 × 6 = (5 + 2) × 6
7 × 6 = (5 × 6) + (2 + 6)
7 × 6 = 30 + 12
7 × 6 = 42

Question 6.
Logic
Your friend breaks apart an array into a 4 × 5 array and 4 × 2 array.Draw a picture to show the original array.
Answer:
Bigideas Math Grade 3 Chapter 2 img-3

Question 7.
DIG DEEPER!
Is the equation true? Does it demonstrate the Distributive Property? Explain.?
Big Ideas Math Answer Key Grade 3 Chapter 2 Multiplication Facts and Strategies 2.5 13
Answer:
The above equation is true.
4 × 8 = (4 × 5) + (4 × 2) + (4 × 1)
4 × 8 = 20 + 8 + 4
4 × 8 = 32

Question 8.
Modeling Real Life
Newton sleeps 9 hours each night. How many hours does he sleep in 1 week?
Big Ideas Math Answer Key Grade 3 Chapter 2 Multiplication Facts and Strategies 2.5 14
Answer:
Given,
Newton sleeps 9 hours each night.
1 week = 7 days
9 × 7 = 63 hours
Thus Newton sleeps 63 hours in 1 week.

Question 9.
Modeling Real Life
There are 3 lanes of cars sitting at a red light. Each lane has 6 cars in it. How many cars are sitting at the red light?
Answer:
Given that,
There are 3 lanes of cars sitting at a red light. Each lane has 6 cars in it.
3 × 6 = 3 × (3 + 3)
3 × 6 = 3 × 3 + 3 × 3
3 × 6 = 9 + 9
3 × 6 = 18
Thus 18 cars are sitting at the red light.

Review & Refresh

Question 10.
Some kids are at an arcade. Seventeen of them leave. There are 39 left. How many kids were at the arcade to start?
Answer:
Given,
Some kids are at an arcade. Seventeen of them leave. There are 39 left.
39 + 17 = 56

Lesson 2.6 Problem Solving: Multiplication

Explore and Grow

Use any strategy to solve
You, Newton, and Descartes each have 4 marbles. How many marbles are there in all?
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.6 1
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.6 2
Answer:
Given that,
You, Newton, and Descartes each have 4 marbles
3 × 4 = 12 marbles
Thus there are 12 marbles in all.

Repeated Reasoning
Explain to your partner how you solved the problem.
Answer:
Given that,
Each person has 4 marbles
1 = 4 marbles
3 = 4 × 3 = 12 marbles
By this, we can say that there are 12 marbles in all.

Think and Grow: Using the Problem-Solving PlanHow

Example
A bookshelf has 5 shelves. There are 8 comic books on each shelf. You take 3 of them to read. How many comic books are left on the shelves?

Understand the Problem

What do you know?
• There are ______ shelves.
• There are _____ comic books on each shelf.
• You take _____ comic books to read.

Answer:
• There are 5 shelves.
• There are 8 comic books on each shelf.
• You take 3 comic books to read.

What do you need to find?
• You need to find how many _____ are left on the shelves after you take _____ of them.

Answer: • You need to find how many comic books are left on the shelves after you take 3 of them.

Make a Plan

How will you solve?
• Multiply _____ by ______ to find how many ______ are on the shelves.
• Then subtract ______ from the product.

Answer:
• Multiply the number of shelves by the numbers of books on each shelf to find how many ______ are on the shelves.
• Then subtract the number of books read from the product.

Solve
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.6 3
There are ______ comic books left on the shelves.

Answer:
5 × 8 = 40
40 – 3 = 37
There are 37 comic books left on the shelves

Show and Grow

Question 1.
A photo album has 10 pages. Each page holds 6 photos. You put 52 photos in the album. How many more photos can you put in the album?
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.6 4
Answer:
Given,
A photo album has 10 pages. Each page holds 6 photos.
You put 52 photos in the album.
10 × 6 = 60 photos
60 – 52 = 8 photos
Thus you can put 8 more photos in the album.

Apply and Grow: Practice

Question 2.
Your classroom has 5 groups of desks with 5 desks in each group. There are 22 students in your class. What information do you know that would help you find how many empty desks there are?
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.6 5
Answer:
Given,
Your classroom has 5 groups of desks with 5 desks in each group.
5 × 5 = 25 desks
There are 22 students in your class.
25 – 22 = 3
Thus there are 3 empty desks in the classroom.

Question 3.
You buy 2 boxes of cherry gelatin, 4 boxes of strawberry gelatin, and 3 boxes of orange gelatin. Each box contains 2 packets of gelatin. How many packets of gelatin do you buy?
Answer:
Given,
You buy 2 boxes of cherry gelatin, 4 boxes of strawberry gelatin, and 3 boxes of orange gelatin. Each box contains 2 packets of gelatin.
1 box – 2 packets of gelatin
2 box – 4 packets of gelatin
4 boxes – 4 × 2 = 8 packets of gelatin
3 boxes – 3 × 2 = 6 packets of gelatin
4 + 8 + 6 = 18 packets of gelatin.
Therefore you can buy 18 packets of gelatin.

Question 4.
A pack of gum has 5 pieces. You have 3 packs of gum and give4 friends each 1 piece. How many pieces of gum do you have left?
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.6 6
Answer:
Given,
A pack of gum has 5 pieces. You have 3 packs of gum.
Total number of packs = 5 × 3 = 15 pieces
and give4 friends each 1 piece.
15 – 4 = 11 pieces
Thus 11 pieces of gum have left.

Think and Grow: Modeling Real Life

Descartes has four $10 bills, seven $5 bills, and one $2 bill. How much money does he have in all?
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.6 7
Understand the problem:
Make a plan:
Solve:
Descartes has $ _____.

Answer:
Given,
Descartes has four $10 bills, seven $5 bills, and one $2 bill.
4 × $10 = $40
7 × $5 = $35
1 × $2 = $2
$40 + $35 + $2 = $72

Show and Grow

Question 5.
A teacher has 10 boxes of dry-erase markers. Five boxes have4 markers each, 2 boxes have 8 markers each, and 3 boxes have10 markers each. How many dry-erase markers does the teacher have in all?
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.6 8
Answer:
Given,
A teacher has 10 boxes of dry-erase markers. Five boxes have4 markers each, 2 boxes have8 markers each, and 3 boxes have10 markers each.
5 boxes – 4 markers
2 boxes – 8 markers
3 boxes – 10 markers
Add all the markers in all 10 boxes.
4 + 8 + 10 = 22 markers
Therefore the teacher has 22 markers in all.

Question 6.
DIG DEEPER!
You sell candles as a fundraiser. You earn $2 for each small candle you sell, $5 for each medium candle, and $10 for each large candle. You pay $8 to have the candles shipped. How much money do you raise?
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.6 9
Answer:
Given,
You sell candles as a fundraiser. You earn $2 for each small candle you sell, $5 for each medium candle, and $10 for each large candle. You pay $8 to have the candles shipped.
7 × $2 = $14
9 × $5 = $45
6 × $10 = 60
$119 – $8 = $111

Problem Solving: Multiplication Homework & Practice 2.6

Question 1.
You buy 2 books and 3 magazines. Each book costs $8 and each magazine costs $2. How much money do you spend in all?
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.6 10
Answer:
Given,
You buy 2 books and 3 magazines. Each book costs $8 and each magazine costs $2.
1 book – $8
2 books – 2 × $8 = $16
1 magazine – $2
3 magazines – 3 × $2 = $6
$16 + $6 = $22
Thus you spend $22 in all.

Question 2.
In a game, teams earn 10 points for each correct answer and lose 5 points for each incorrect answer. Your team answers 7 questions correctly and 3 questions incorrectly. How many points does your team have?
Answer:
Given,
In a game, teams earn 10 points for each correct answer and lose 5 points for each incorrect answer.
Your team answers 7 questions correctly and 3 questions incorrectly.
1 – 10 points
7 – 10 × 7 = 70 points
1 wrong answer – 5 points
3 – 5 × 3 = 15 points
70 – 15 = 55 points

Question 3.
An origami cube requires 3 pieces of orange paper, 2 pieces of yellow paper, and 1 piece of blue paper. You make 5 cubes. How many pieces of paper do you need?
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.6 11
Answer:
Given,
An origami cube requires 3 pieces of orange paper, 2 pieces of yellow paper, and 1 piece of blue paper.
You make 5 cubes.
Each cube needs 3 + 2 + 1 = 6 pieces of paper.
5 cubes – 6 × 5 = 30 pieces of papers
Thus you need 30 pieces of paper to make 5 cubes.

Question 4.
Writing
Write and solve your own word problem that involves multiplication.
Answer:
To make a cuboid you need 3 pieces of yellow paper, 3 pieces of pink paper. You need to make 7 cuboids. How many pieces of paper are required to make 7 cuboids?
Each cuboid needs 3 + 3 = 6 pieces of paper.
7 × 6 = 42 pieces of paper
Thus you need 42 pieces of paper to make 7 cuboids.

Question 5.
Modeling Real Life
A group of students orders 6 small, 5 medium, and 3 large smoothies. The students pay with five $10 bills. How much change do they receive?
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.6 12
Answer:
Given,
A group of students orders 6 small, 5 medium, and 3 large smoothies.
The students pay with five $10 bills.
1 small – $2
6 small – 6 × $2 = $12
1 Medium – $4
5 medium – 5 × $4 = $20
1 large – $5
3 large – 3 × $5 = $15
12 + 20 + 15 = $47
$50 – $47 = $3
Thus the students receive $3 change.

Review & Refresh

Question 6.
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.6 13
Answer:
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.6 13
6      2      7

Question 7.
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.6 14
Answer:
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.6 14
9     2       2

Question 8.
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.6 15
Answer:
Big Ideas Math Answers 3rd Grade Chapter 2 Multiplication Facts and Strategies 2.6 15
7       5       0

Multiplication Facts and Strategies Performance Task

Question 1.
Three students make a model of our solar system.
a. Student A buys foam spheres to make the Sun and the planets. He pays with a $10 bill. What is his change?
Big Ideas Math Answers Grade 3 Chapter 2 Multiplication Facts and Strategies 2.7 1

Answer:
Given,
Student A buys foam spheres to make the Sun and the planets.
$10 – $1 = $9
Thus the change is $9
b. Student B buys wooden rods to attach each planet to the sun. She pays with two $5 bills. What is her change?
Big Ideas Math Answers Grade 3 Chapter 2 Multiplication Facts and Strategies 2.7 2

Answer:
Given,
Student B buys a wooden rod for each planet and sun.
Cost of the wooden rod is $1
9 planets and 1 sun
9 × 1 + 1 × 1 = 9 + 1 = $10
Thus the change for student B is 0.

c.Student C spends $13 on 3 paintbrushes and some bottles of paint. How many bottles of paint does Student C buy?
Big Ideas Math Answers Grade 3 Chapter 2 Multiplication Facts and Strategies 2.7 3

Answer:
Cost of 1 paintbrush = $1
3 paintbrushes cost = 3 × $1 = $3
Cost of bottle of paint = $2
For $10 you get 5 bottles of paint.
Thus Student C buys 5 bottles of paint.
d. What is the total cost of the project? If the 3 students divide the cost equally, how much would they each spend?
Answer:
The total cost of the project = $1 + $ 13 + $10 = $24
$24/3 = $8

Multiplication Facts and Strategies Activity

Three in a Row: Multiplication
Directions:
1. Players take turns.
2. On your turn, spin both spinners. Multiply the two numbers and cover the product.
3.The first player to get three counters in a row, horizontally, vertically, or diagonally, wins!
Big Ideas Math Answers Grade 3 Chapter 2 Multiplication Facts and Strategies 2.7 4
Big Ideas Math Answers Grade 3 Chapter 2 Multiplication Facts and Strategies 2.7 5

Answer:
Game A:
10 × 1, 2 × 2, 1 × 2
2 × 3, 5 × 1, 10 × 2
2 × 4, 10 × 5, 0 × 3
Game B:
1 × 3, 10 × 4, 10 × 5
1 × 1, 0 × 4, 10 × 3
5 × 3, 5 × 5, 10 × 1

Multiplication Facts and Strategies Chapter Practice

2.1 Multiply by 2

Find the Product
Question 1.
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.9 1
Answer: 14

Explanation:
Multiply the number by 2.
7 × 2 = 14

Question 2.
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.9 2
Answer: 6

Explanation:
Multiply the number by 2.
Multiply the two numbers 3 and 2.
3 × 2 = 6

Question 3.
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.9 3
Answer: 4

Explanation:
Multiply the number by 2.
Multiply the two numbers 2 and 2.
2 × 2 = 4

Question 4.
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.9 4
Answer: 18

Explanation:
Multiply the number by 2.
Multiply the two numbers 2 and 9.
2 × 9 = 18

Find the missing factor.
Question 5.
2 × ___ = 2
Answer: 1

Explanation:
Let the missing factor be x.
2 × x = 2
x = 2/2 = 1
x = 1
Thus the missing is 1.

Question 6.
2 × ___ = 20
Answer: 10

Explanation:
Let the missing factor be x.
2 × x = 20
x = 20/2 = 10
x = 10
Thus the missing factor is 10.

Question 7.
____ × 4 = 8
Answer: 2

Explanation:
Let the missing factor be b.
b × 4 = 8
b = 8/4
b = 2
Thus the missing factor is 2.

2.2 Multiply by 5

Question 8.
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.9 5
Answer: 15

Explanation:
Multiply the number by 5.
5 × 3 = 15

Question 9.
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.9 6
Answer: 50

Explanation:
Multiply the number by 5.
5 × 10 = 50

Question 10.
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.9 7
Answer: 40

Explanation:
Multiply the number by 5.
8 × 5 = 40

Question 11.
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.9 8
Answer: 10

Explanation:
Multiply the number by 5.
5 × 2 = 10

Question 12.
Number Sense
Lightning strikes our planet 6 times every second. How many lightning strikes occur in 5 seconds?
Answer: 30 seconds

Explanation:
Given,
Lightning strikes our planet 6 times every second.
Lightning strikes for 5 seconds is
6 × 5 = 30 seconds

2.3 Multiply by 10

Find the product
Question 13.
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.9 9
Answer: 80

Explanation:
Multiply the product by 10.
10 × 8 = 80

Question 14.
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.9 10
Answer: 100

Explanation:
Multiply the product by 10.
10 × 10 = 100

Question 15.
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.9 11
Answer: 40

Explanation:
Multiply the product by 10.
10 × 4 = 40

Question 16.
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.9 12
Answer: 70

Explanation:
Multiply the product by 10.
10 × 7 = 70

2.4 Multiply by 0 and 1

Find the product
Question 17.
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.9 13
Answer: 6

Explanation:
Any number multiplied by 1 will be the same number.
1 × 6 = 6

Question 18.
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.9 14
Answer: 0

Explanation:
Any number multiplied by 0 will be always 0.
0 × 3 = 0

Question 19.
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.9 15
Answer: 0

Explanation:
Any number multiplied by 0 will be always 0.
1 × 0 = 0

Question 20.
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.9 16
Answer: 4

Explanation:
Any number multiplied by 1 will be the same number.
4 × 1 = 4

Compare
Question 21.
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.9 17
Answer: =

Explanation:
Any number multiplied by 0 will be always 0.
0 × 10 = 0
9 × 0 = 0
0 = 0
Thus 0 × 10 = 9 × 0

Question 22.
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.9 18
Answer: >

Explanation:
Any number multiplied by 0 will be always 0.
Any number multiplied by 1 will be the same number.
2 × 1 = 2
5 × 0 = 0
2 > 0
Thus 2 × 1 > 5 × 0

Question 23.
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.9 19
Answer: <

Explanation:
Any number multiplied by 1 will be the same number.
7 × 1 = 7
1 × 8 = 8
7 < 8
Thus 7 × 1 < 1 × 8

2.5 Use the Distributive Property

Use the Distributive Property to fill in the blanks.

Question 24.
7 × 3 = (____ + 2) × 3
= (____ × 3) + (2 × ____)
= ____ + ____
= ____

Answer:

7 × 3 = (5 + 2) × 3
= (5 × 3) + (2 × 3)
= 15 + 6
= 21

Question 25.
6 × 6 = 6 × (___ + ____)
= (6 × ____) + (6 × ____)
= ____ + ____
= ____

Answer:
6 × 6 = 6 × (3 + 3)
= (6 × 3) + (6 × 3)
= 18 + 18
= 36

2.5 Problem Solving: Multiplication

Question 26.
You place your trading card collection into an album that has 5 pages. You put 9 cards on each page. Some of the cards are the same, so you give 3 cards to a friend. How many trading cards do you have left?
Answer: 30

Explanation:
Given,
You place your trading card collection into an album that has 5 pages. You put 9 cards on each page. Some of the cards are the same, so you give 3 cards to a friend.
9 × 5 = 45 cards
If 3 cards are the same, then 5 × 3 = 15
45 – 15 = 30
Thus 30 trading cards are left.

Question 27.
Modeling Real Life
In a game, you start with ten $1 bills, five $5 bills, and two $10 bills. How much money do you start with in all?
Big Ideas Math Solutions Grade 3 Chapter 2 Multiplication Facts and Strategies 2.9 20
Answer: $55

Explanation:
In a game, you start with ten $1 bills, five $5 bills, and two $10 bills.
10 $1 bills = 10 × $1 = $10
5 $5 bills = 5 × $ 5 = $25
2 $10 bills = 2 × $10 = $20
10 + 25 + 20 = $55

Final Words:

We wish that we have given useful information regarding BIM Book Grade 3 Chapter 2 Multiplication Facts and Strategies. So, students have to download Big Ideas Math Grade 3 Chapter 2 Multiplication Facts and Strategies Answer Key pdf for free of cost. If you have any queries regarding the solutions, you can write your question in the below provided comment box. Bookmark our site to get solutions for the remaining chapters of Big Ideas Math Grade 3 Book.

Big Ideas Math Geometry Answers Chapter 10 Circles

Big Ideas Math Geometry Answers Chapter 10 Circles

Gain more subject knowledge on Circles concepts with the help of Big Ideas Math Geometry Answers Chapter 10 Circles guide. Shared guide of BIM Textbook Solutions Geometry Ch 10 Circles helps students understand the concepts quickly and concentrate on other math concepts too. Want to get a good grip on the high school-related circle topics? Then, make use of this BigIdeas Math Geometry Answers Ch 10 Circles Pdf for free. Subject experts provided this BIM geometry 10th chapter solution key based on the latest syllabus and common core standards curriculum guidelines.

Big Ideas Math Book Geometry Answer Key Chapter 10 Circles

Presented Geometry Chapter 10 Circles Big Ideas Math Answers are easy to understand and free to access at any time and anywhere. So, students are advised to access online or download the lesson-wise BIM Textbook Answers of Geometry Ch 10 Circles in Pdf format. Check out each and every lesson of circles as a part of your preparation and clear your queries within no time.

Common Core Curriculum High School Big Ideas Math Book Geometry Ch 10 Solution Key are prepared as per the latest textbooks. Questions included in the BIM Modeling Real Life Geometry Answers are from Exercises, Chapter Tests, Review Tests, Cumulative Practice, Assessment Tests, etc.

Circles Maintaining Mathematical Proficiency

Find the Product.

Question 1.
(x + 7) (x + 4)

Answer:
(x + 7) (x + 4) = x² + 14x + 28

Explanation:
(x + 7) (x + 4) = x(x + 7) + 7(x + 4)
= x² + 7x + 7x + 28
= x² + 14x + 28

Question 2.
(a + 1) (a – 5)

Answer:
(a + 1) (a – 5) = a² – 4a – 5

Explanation:
(a + 1) (a – 5) = a(a – 5) + 1(a – 5)
= a² – 5a + a – 5
= a² – 4a – 5

Question 3.
(q – 9) (3q – 4)

Answer:
(q – 9) (3q – 4) = 3q² – 31q + 32

Explanation:
(q – 9) (3q – 4) = q(3q – 4) – 9(3q – 4)Exercise 10.3 Using Chords

= 3q² – 4q – 27q + 32
= 3q² – 31q + 32

Question 4.
(2v – 7) (5v + 1)

Answer:
(2v – 7) (5v + 1) = 10v² – 33v – 7

Explanation:
(2v – 7) (5v + 1) = 2v(5v + 1)- 7(5v + 1)
= 10v² + 2v – 35v – 7
= 10v² – 33v – 7

Question 5.
(4h + 3) (2 + h)

Answer:
(4h + 3) (2 + h) = 4h² + 11h + 6

Explanation:
(4h + 3) (2 + h) = 4h(2 + h) + 3(2 + h)
= 8h + 4h² + 6 + 3h
= 4h² + 11h + 6

Question 6.
(8 – 6b) (5 – 3b)

Answer:
(8 – 6b) (5 – 3b) = 18b² – 54b + 40

Explanation:
(8 – 6b) (5 – 3b) = 8(5 – 3b) – 6b(5 – 3b)
= 40 – 24b – 30b + 18b²
= 18b² – 54b + 40

Solve the equation by completing the square. Round your answer to the nearest hundredth, if necessary.

Question 7.
x2 – 2x = 5

Answer:
The solutions are x = √6 + 1, x = 1 – √6

Explanation:
x² – 2x = 5
x² – 2x + 1² = 5 + 1²
(x – 1)² = 6
x – 1 = ±√6
x = ±√6 + 1
The solutions are x = √6 + 1, x = -√6 + 1

Question 8.
r2 + 10r = -7

Answer:
The solutions are r = √18 – 5, r = 5 – √18

Explanation:
r2 + 10r = -7
r² + 10r + 5² = -7 + 5²
(r + 5)² = -7 + 25 = 18
r + 5 = ±√18
r = ±√18 – 5
The solutions are r = √18 – 5, r = 5 – √18

Question 9.
w2 – 8w = 9

Answer:
The solutions are w = 9, w = -1

Explanation:
w2 – 8w = 9
w2 – 8w + 4² = 9 + 4²
(w – 4)² = 9 + 16 = 25
w – 4 = ±5
w = 5 + 4, w = -5 + 4
w = 9, w = -1
The solutions are w = 9, w = -1

Question 10.
p2 + 10p – 4 = 0

Answer:
The solutions are p = √29 – 5, p = 5 – √29

Explanation:
p2 + 10p = 4
p² + 10p + 5² = 4 + 5²
(p + 5)² = 4 + 25
(p + 5)² = 29
p + 5 = ±√29
p = ±√29 – 5
The solutions are p = √29 – 5, p = 5 – √29

Question 11.
k2 – 4k – 7 = 0

Answer:
The solutions are k = √11 + 2, k = 2 – √11

Explanation:
k² – 4k= 7
k² – 4k + 2² = 7 + 4
(k – 2)² = 11
k – 2 = ±√11
k = √11 + 2, k = 2 – √11
The solutions are k = √11 + 2, k = 2 – √11

Question 12.
– z2 + 2z = 1

Answer:
The solutions are z = 1

Explanation:
-z² + 2z = 1
z² – 2z = -1
z² – 2z + 1 = -1 + 1
(z – 1)² = 0
z = 1
The solutions are z = 1

Question 13.
ABSTRACT REASONING
write an expression that represents the product of two consecutive positive odd integers. Explain your reasoning.

Answer:
Let us take two consecutive odd integers are x and (x + 2)
The product of two consecutive odd integers is x • (x + 2)
x(x + 2) = x² + 2x

Circles Mathematical Practices

Monitoring progress

Let ⊙A, ⊙B, and ⊙C consist of points that are 3 units from the centers.

Circles Maintaining Mathematical Proficiency Find the Product. Question 1. (x + 7) (x + 4) Answer: Question 2. (a + 1) (a - 5) Answer: Question 3. (q - 9) (3q - 4) Answer: Question 4. (2v - 7) (5v + 1) Answer: Question 5. (4h + 3) (2 + h) Answer: Question 6. (8 - 6b) (5 - 3b) Answer: Solve the equation by completing the square. Round your answer to the nearest hundredth, if necessary. Question 7. x<sup>2</sup> - 2x = 5 Answer: Question 8. r<sup>2</sup> + 10r = -7 Answer: Question 9. w<sup>2</sup> - 8w = 9 Answer: Question 10. p<sup>2</sup> + 10p - 4 = 0 Answer: Question 11. k<sup>2</sup> - 4k - 7 = 0 Answer: Question 12. - z<sup>2</sup> + 2z = 1 Answer: Question 13. ABSTRACT REASONING write an expression that represents the product of two consecutive positive odd integers. Explain your reasoning. Answer: Circles Mathematical Practices Monitoring progress Let ⊙A, ⊙B, and ⊙C consist of points that are 3 units from the centers. im - 1 Question 1. Draw ⊙C so that it passes through points A and B in the figure at the right. Explain your reasoning. Answer: Question 2. Draw ⊙A, ⊙B, and OC so that each is tangent to the other two. Draw a larger circle, ⊙D, that is tangent to each of the other three circles. Is the distance from point D to a point on ⊙D levss than, greater than, or equal to 6? Explain. Answer:

Question 1.
Draw ⊙C so that it passes through points A and B in the figure at the right. Explain your reasoning.

Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 1

Question 2.
Draw ⊙A, ⊙B, and OC so that each is tangent to the other two. Draw a larger circle, ⊙D, that is tangent to each of the other three circles. Is the distance from point D to a point on ⊙D less than, greater than, or equal to 6? Explain.

Answer:

10.1 Lines and Segments that Intersect Circles

Exploration 1

Lines and Line Segments That Intersect Circles

Big Ideas Math Geometry Answers Chapter 10 Circles 2

Work with a partner: The drawing at the right shows five lines or segments that intersect a circle. Use the relationships shown to write a definition for each type of line or segment. Then use the Internet or some other resource to verify your definitions.
Chord: _________________
Secant: _________________
Tangent: _________________
Radius: _________________
Diameter: _________________

Answer:
Chord: A chord of a circle is a straight line segment whose endpoints both lie on a circular arc.
Secant: A straight line that intersects a circle in two points is called a secant line.
Tangent: Tangent line is a line that intersects a curved line at exactly one point.
Radius: It is the distance from the centre of the circle to any point on the circle.
Diameter: It the straight that joins two points on the circle and passes through the centre of the circle.

Exploration 2

Using String to Draw a Circle

Work with a partner: Use two pencils, a piece of string, and a piece of paper.

a. Tie the two ends of the piece of string loosely around the two pencils.
Answer:

b. Anchor one pencil of the paper at the center of the circle. Use the other pencil to draw a circle around the anchor point while using slight pressure to keep the string taut. Do not let the string wind around either pencil.
Big Ideas Math Geometry Answers Chapter 10 Circles 3
Answer:

c. Explain how the distance between the two pencil points as you draw the circle is related to two of the lines or line segments you defined in Exploration 1.
REASONING ABSTRACTLY
To be proficient in math, you need to know and flexibly use different properties of operations and objects.
Answer:

Communicate Your Answer

Question 3.
What are the definitions of the lines and segments that intersect a circle?
Answer:

Question 4.
Of the five types of lines and segments in Exploration 1, which one is a subset of another? Explain.
Answer:

Question 5.
Explain how to draw a circle with a diameter of 8 inches.
Answer:

Lesson 10.1 Lines and Segments that Intersect Circles

Monitoring progress

Question 1.
In Example 1, What word best describes \(\overline{A G}\)? \(\overline{C B}\)?

Answer:
\(\overline{A G}\) is secant because it is a line that intersects the circle at two points.
\(\overline{C B}\) is the radius as it is the distance from the centre to the point of a circle.

Question 2.
In Example 1, name a tangent and a tangent segment.

Answer:
\(\overline{D E}\) is the tangent of the circle
\(\overline{D E}\) is the tangent segment of the circle.

Tell how many common tangents the circles have and draw them. State whether the tangents are external tangents or internal tangents.

Question 3.
Big Ideas Math Geometry Answers Chapter 10 Circles 4

Answer:
4 tangents.
Big Ideas Math Geometry Answers Chapter 10 Circles 2
A tangent is a line segment that intersects the circle at exactly one point. Internal tangents are the lines that intersect the segments joining the centres of two circles. External tangents are the lines that do not cross the segment joining the centres of the circles.
Blue lines represent the external tangents and red lines represent the internal tangents.

Question 4.
Big Ideas Math Geometry Answers Chapter 10 Circles 5

Answer:
One tangent.
Big Ideas Math Geometry Answers Chapter 10 Circles 3
One external tangent.

Question 5.
Big Ideas Math Geometry Answers Chapter 10 Circles 6

Answer:
No tangent.
It is not possible to draw a common tangent for this type of circle.

Question 6.
Is \(\overline{D E}\) tangent to ⊙C?
Big Ideas Math Geometry Answers Chapter 10 Circles 7

Answer:
Use the converse of Pythagorean theorem i.e 2² = 3² + 4²
4 = 9 + 16
By the tangent line to the circle theorem, \(\overline{D E}\) is not a tangent to ⊙C

Question 7.
\(\overline{S T}\) is tangent to ⊙Q.
Find the radius of ⊙Q.
Big Ideas Math Geometry Answers Chapter 10 Circles 8

Answer:
The radius of ⊙Q is 7 units.

Explanation:
By using the Pythagorean theorem
(18 + r)² = r² + 24²
324 + 36r + r² = r² + 576
36r = 576 – 324
36r = 252
r = 7

Question 8.
Points M and N are points of tangency. Find the value(s) of x.
Big Ideas Math Geometry Answers Chapter 10 Circles 9

Answer:
The values of x are 3 or -3.

Explanation:
x² = 9
x = ±3

Exercise 10.1 Lines and Segments that Intersect Circles

Vocabulary and Core Concept Check

Question 1.
WRITING
How are chords and secants alike? How are they different?
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 1

Question 2.
WRITING
Explain how you can determine from the context whether the words radius and diameter are referring to segments or lengths.

Answer:
Radius and diameter are the lengths of the line segments that pass through the centre of a circle. Radius is half of the diameter.

Question 3.
COMPLETE THE SENTENCE
Coplanar circles that have a common center are called ____________ .
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 3

Question 4.
WHICH ONE DOESNT BELONG?
Which segment does not belong with the other three? Explain your reasoning.
chord radius tangent diameter

Answer:
A chord, a radius and a diamter are segments and they intersect a circle in two points. A tangent is a line that intersects a circle at one point.

Monitoring Progress and Modeling with Mathematics

In Exercises 5 – 10, use the diagram.

Big Ideas Math Geometry Answers Chapter 10 Circles 10

Question 5.
Name the circle.
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 5

Question 6.
Name two radii.

Answer:
The name of the two radii is CD and AC.

Question 7.
Name two chords.
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 7

Question 8.
Name a diameter.

Answer:
The name of diameter is AD

Question 9.
Name a secant.
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 9

Question 10.
Name a tangent and a point of tangency

Answer:
GE is the tangent and F is the point of tangency.

In Exercises 11 – 14, copy the diagram. Tell how many common tangents the circles have and draw them.

Question 11.
Big Ideas Math Geometry Answers Chapter 10 Circles 11
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 11

Question 12.
Big Ideas Math Geometry Answers Chapter 10 Circles 12

Answer:
No common tangent because two circles do not intersect at one point.

Question 13.
Big Ideas Math Geometry Answers Chapter 10 Circles 13
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 13

Question 14.
Big Ideas Math Geometry Answers Chapter 10 Circles 14

Answer:
One common tangent.
Big Ideas Math Geometry Answers Chapter 10 Circles 3

In Exercises 15 – 18, tell whether the common tangent is internal or external.

Question 15.
Big Ideas Math Geometry Answers Chapter 10 Circles 15
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 15

Question 16.
Big Ideas Math Geometry Answers Chapter 10 Circles 16

Answer:
The common tangent is the internal tangent because it intersects the segment that joins the centres of two circles.

Question 17.
Big Ideas Math Geometry Answers Chapter 10 Circles 17
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 17

Question 18.
Big Ideas Math Geometry Answers Chapter 10 Circles 18
Answer:
The common tangent is the internal tangent because it intersects the segment that joins the centres of two circles.

In Exercises 19 – 22, tell whether \(\overline{A B}\) is tangent to ⊙C. Explain your reasoning.

Question 19.
Big Ideas Math Geometry Answers Chapter 10 Circles 19
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 19

Question 20.
Big Ideas Math Geometry Answers Chapter 10 Circles 20

Answer:
Use the converse of the Pythagorean theorem
18² _____________ 15² + 9²
324 _____________ 225 + 81
324 ≠ 304
△ ACB is not a right angled triangle.
So, \(\overline{A B}\) is not tangent to ⊙C at B.

Question 21.
Big Ideas Math Geometry Answers Chapter 10 Circles 21
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 21

Question 22.
Big Ideas Math Geometry Answers Chapter 10 Circles 22

Answer:
Use the converse of the Pythagorean theorem
8² _____________ 12² + 16²
64 _____________ 144 + 256
64 ≠ 400
△ ACB is not a right angled triangle.
So, \(\overline{A B}\) is not tangent to ⊙C at B.

In Exercises 23 – 26, point B is a point of tangency. Find the radius r of ⊙C.

Question 23.
Big Ideas Math Geometry Answers Chapter 10 Circles 23
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 23

Question 24.
Big Ideas Math Geometry Answers Chapter 10 Circles 24

Answer:
(r + 6)² = r² + 9²
r² + 12r + 36 = r² + 81
12r = 81 – 36
12r = 45
r = \(\frac { 15 }{ 4 } \)
Therefore, the radius of ⊙C is \(\frac { 15 }{ 4 } \)

Question 25.
Big Ideas Math Geometry Answers Chapter 10 Circles 25
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 25

Question 26.
Big Ideas Math Geometry Answers Chapter 10 Circles 26

Answer:
(r + 18)² = r² + 30²
r² + 36r + 324 = r² + 900
36r = 900 – 324
36r = 576
r = 16
Therefore, the radius of ⊙C is 16

CONSTRUCTION
In Exercises 27 and 28. construct ⊙C with the given radius and point A outside of ⊙C. Then construct a line tangent to ⊙C that passes through A.

Question 27.
r = 2 in.
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 27

Question 28.
r = 4.5 cm

Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 5

In Exercises 29 – 32, points B and D are points of tangency. Find the value(s) of x.

Question 29.
Big Ideas Math Geometry Answers Chapter 10 Circles 27
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 29

Question 30.
Big Ideas Math Geometry Answers Chapter 10 Circles 28

Answer:
3x + 10 = 7x – 6
7x – 3x = 10 + 6
4x = 16
x = 4

Question 31.
Big Ideas Math Geometry Answers Chapter 10 Circles 29
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 31

Question 32.
Big Ideas Math Geometry Answers Chapter 10 Circles 30

Answer:
2x + 5 = 3x² + 2x – 7
3x² = 5 + 7
3x² = 12
x² = 4
x = ±2

Question 33.
ERROR ANALYSIS
Describe and correct the error in determining whether \(\overline{X Y}\) is tangent to ⊙Z.
Big Ideas Math Geometry Answers Chapter 10 Circles 31
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 33

Question 34.
ERROR ANALYSIS
Describe and correct the error in finding the radius of ⊙T.
Big Ideas Math Geometry Answers Chapter 10 Circles 32

Answer:
39² = 36² + 15²
So, 15 is the diameter.
The radius is \(\frac { 15 }{ 2 } \).

Question 35.
ABSTRACT REASONING
For a point outside of a circle, how many lines exist tangent to the circle that pass through the point? How many such lines exist for a point on the circle? inside the circle? Explain your reasoning.
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 35

Question 36.
CRITICAL THINKING
When will two lines tangent to the same circle not intersect? Justify your answer.

Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 6
Using tangent line to circle theorem, it follow that the angle between tangent and radius is a right angle. Let’s draw these tangents at the two ends of the same diameter. We can observe a diameter AD like a transverzal of these tangents.

Question 37.
USING STRUCTURE
Each side of quadrilateral TVWX is tangent to ⊙Y. Find the perimeter of the quadrilateral.
Big Ideas Math Geometry Answers Chapter 10 Circles 33
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 37

Question 38.
LOGIC
In ⊙C, radii \(\overline{C A}\) and \(\overline{C B}\) are perpendicular. Big Ideas Math Geometry Answers Chapter 10 Circles 34 are tangent to ⊙C.

a. Sketch ⊙C, \(\overline{C A}\), \(\overline{C B}\), Big Ideas Math Geometry Answers Chapter 10 Circles 34.
Answer:

b. What type of quadrilateral is CADB? Explain your reasoning.
Answer:

Question 39.
MAKING AN ARGUMENT
Two hike paths are tangent to an approximately circular pond. Your class is building a nature trail that begins at the intersection B of the bike paths and runs between the bike paths and over a bridge through the center P of the pond. Your classmate uses the Converse of the Angle Bisector Theorem (Theorem 6.4) to conclude that the trail must bisect the angle formed by the bike paths. Is your classmate correct? Explain your reasoning.
Big Ideas Math Geometry Answers Chapter 10 Circles 35
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 39

Question 40.
MODELING WITH MATHEMATICS
A bicycle chain is pulled tightly so that \(\overline{M N}\) is a common tangent of the gears. Find the distance between the centers of the gears.
Big Ideas Math Geometry Answers Chapter 10 Circles 36

Answer:
height h = 4.3 – 1.8
h = 2.5
x² = MN² + h²
x² = 17.6² + 2.5²
x² = 316.01
x = 17.8
Therefore, the distance between the centre of the gear is 17.8 in.

Question 41.
WRITING
Explain why the diameter of a circle is the longest chord of the circle.
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 41

Question 42.
HOW DO YOU SEE IT?
In the figure, \(\vec{P}\)A is tangent to the dime. \(\vec{P}\)C is tangent to the quarter, and \(\vec{P}\)B is a common internal tangent. How do you know that \(\overline{P A} \cong \overline{P B} \cong \overline{P C}\)
Big Ideas Math Geometry Answers Chapter 10 Circles 37
Answer:

Question 43.
PROOF
In the diagram, \(\overline{R S}\) is a common internal tangent to ⊙A and ⊙B. Prove that \(\frac{\Lambda C}{B C}=\frac{R C}{S C}\)
Big Ideas Math Geometry Answers Chapter 10 Circles 38
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 43.1
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 43.2

Question 44.
THOUGHT PROVOKING
A polygon is circumscribed about a circle when every side of the polygon is tangent to the circle. In the diagram. quadrilateral ABCD is circumscribed about ⊙Q. Is it always true that AB + CD = AD + BC? Justify your answer.
Big Ideas Math Geometry Answers Chapter 10 Circles 39
Answer:

Question 45.
MATHEMATICAL CONNECTIONS
Find the values of x and y. Justify your answer.
Big Ideas Math Geometry Answers Chapter 10 Circles 40
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 45

Question 46.
PROVING A THEOREM
Prove the External Tangent Congruence Theorem (Theorem 10.2).
Big Ideas Math Geometry Answers Chapter 10 Circles 41
Given \(\overline{S R}\) and \(\overline{S T}\) are tangent to ⊙P.
Prove \(\overline{S R} \cong \overline{S T}\)

Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 7
∠PRS and ∠PTS are the right angles. So the legs of circles are congruent.
Therefore, \(\overline{S R} \cong \overline{S T}\)

Question 47.
PROVING A THEOREM
Use the diagram to prove each part of the biconditional in the Tangent Line to Circle Theorem (Theorem 10.1 ).
Big Ideas Math Geometry Answers Chapter 10 Circles 42
a. Prove indirectly that if a line is tangent to a circle, then it is perpendicular to a radius. (Hint: If you assume line m is not perpendicular to \(\overline{Q P}\), then the perpendicular segment from point Q to line m must intersect line m at some other point R.)
Ghen Line m is tangent to ⊙Q at point P.
Prove m ⊥ \(\overline{Q P}\)
b. Prove indirectly that if a line is perpendicular to a radius at its endpoint, then the line is tangent to the circle.
Gien m ⊥ \(\overline{Q P}\)
Prove Line m is tangent to ⊙Q.
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 47.1
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 47.2

Question 48.
REASONING
In the diagram, AB = AC = 12, BC = 8, and all three segments are Langent to ⊙P. What is the radius of ⊙P? Justify your answer.
Big Ideas Math Geometry Answers Chapter 10 Circles 43

Answer:

Maintaining Mathematical Proficiency

Find the indicated measure.

Question 49.
m∠JKM
Big Ideas Math Geometry Answers Chapter 10 Circles 44
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 49

Question 50.
AB
Big Ideas Math Geometry Answers Chapter 10 Circles 45

Answer:
AC = AB + BC
10 = AB + 7
AB = 10 – 7
AB = 3

10.2 Finding Arc Measures

Exploration 1

Measuring Circular Arcs

Work with a partner: Use dynamic geometry software to find the measure of \(\widehat{B C}\). Verify your answers using trigonometry.

a.
Big Ideas Math Answers Geometry Chapter 10 Circles 46
Points
A(0, 0)
B(5, 0)
C(4, 3)

Answer:
30 degrees

b.
Big Ideas Math Answers Geometry Chapter 10 Circles 47
Points
A(0, 0)
B(5, 0)
C(3, 4)

Answer:
60 degrees

c.
Big Ideas Math Answers Geometry Chapter 10 Circles 48
Points
A(0, 0)
B(4, 3)
C(3, 4)

Answer:
15 degrees

d.
Big Ideas Math Answers Geometry Chapter 10 Circles 49
Points
A(0, 0)
B(4, 3)
C(- 4, 3)

Answer:
90 degrees

Communicate Your Answer

Question 2.
How are circular arcs measured?
Answer:

Question 3.
Use dynamic geometry software to draw a circular arc with the given measure.
USING TOOLS STRATEGICALLY
To be proficient in math, you need to use technological tools to explore and deepen your understanding of concepts.
a. 30°

Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 8

b. 45°

Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 9

c. 60°
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10

d. 90°

Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 11

Lesson 10.2 Finding Arc Measures

Monitoring Progress

Identify the given arc as a major arc, minor arc, or semicircle. Then find the measure of the arc.

Big Ideas Math Answers Geometry Chapter 10 Circles 50

Question 1.
\(\widehat{T Q}\)

Answer:
\(\widehat{T Q}\) is a minor arc.
\(\widehat{T Q}\) = 120°

Question 2.
\(\widehat{Q R T}\)

Answer:
\(\widehat{Q R T}\)

Question 3.
\(\widehat{T Q R}\) is a major arc.
\(\widehat{Q R T}\) = QR + RS + ST
RS = 360° – (60 + 120 + 80)
= 360 – 260 = 100°
So, \(\widehat{Q R T}\) = 60° + 100° + 80°
\(\widehat{Q R T}\) = 240°

Answer:

Question 4.
\(\widehat{Q S}\)

Answer:
\(\widehat{Q S}\) = QR + RS
= 60 + 100 = 160°
Therefore, \(\widehat{Q S}\) = 160° and it is a minor arc.

Question 5.
\(\widehat{T S}\)

Answer:
\(\widehat{T S}\) = 80° and it is a minor arc.

Question 6.
\(\widehat{R S T}\)

Answer:
\(\widehat{R S T}\) = 100 + 80 = 180
Therefore, \(\widehat{R S T}\) = 180° and it is a minor arc.

Tell whether the red arcs are congruent. Explain why or why not.

Question 7.
Big Ideas Math Answers Geometry Chapter 10 Circles 51

Answer:
\(\widehat{A B}\), \(\widehat{C D}\) are congruent as they measure same radius and same arc length.

Question 8.
Big Ideas Math Answers Geometry Chapter 10 Circles 52

Answer:
\(\widehat{M N}\), \(\widehat{P Q}\) are not congruent as they measure different radius.

Exercise 10.2 Finding Arc Measures

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
Copy and complele:
If ∠ACB and ∠DCE are congruent central angles of ⊙C, then \(\widehat{A B}\) and \(\widehat{D E}\) arc.
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.2 Ans 1

Question 2.
WHICH ONE DOESNT BELONG?
Which circle does not belong with the other three? Explain your reasoning.
Big Ideas Math Answers Geometry Chapter 10 Circles 53

Answer:
We know that 1 ft = 12 in
So, the fourth circle does not belong to the other three as its diameter is different.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6, name the red minor arc and find its measure. Then name the blue major arc and find its measure.

Question 3.
Big Ideas Math Answers Geometry Chapter 10 Circles 54
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.2 Ans 3

Question 4.
Big Ideas Math Answers Geometry Chapter 10 Circles 55

Answer:
The minor arc \(\widehat{E F}\) = 68° and major arc \(\widehat{F G E}\) = 360° – 68° = 292°.

Question 5.
Big Ideas Math Answers Geometry Chapter 10 Circles 56
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.2 Ans 5

Question 6.
Big Ideas Math Answers Geometry Chapter 10 Circles 57

Answer:
The minor arc is \(\widehat{M N}\) = 170°, major arc \(\widehat{N P M}\) = 360° – 170° = 190°.

In Exercises 7 – 14. identify the given arc as a major arc, minor arc, or semicircle. Then find the measure of the arc.

Big Ideas Math Answers Geometry Chapter 10 Circles 58

Question 7.
\(\widehat{B C}\)
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.2 Ans 7

Question 8.
\(\widehat{D C}\)

Answer:
\(\widehat{D C}\) is a minor arc and it has a measure of 65°.

Question 9.
\(\widehat{E D}\)
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.2 Ans 9

Question 10.
\(\widehat{A E}\)

Answer:
\(\widehat{A E}\) is a minor arc and it has a measure of 70°.

Question 11.
\(\widehat{E A B}\)
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.2 Ans 11

Question 12.
\(\widehat{A B C}\)

Answer:
\(\widehat{A B C}\) is a semicircle and it has a measure of 180°.

Question 13.
\(\widehat{B A C}\)
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.2 Ans 13

Question 14.
\(\widehat{E B D}\)

Answer:
\(\widehat{E B D}\) is a major arc and it has a measure of 315°.

In Exercises 15 and 16, find the measure of each arc.

Question 15.
Big Ideas Math Answers Geometry Chapter 10 Circles 59

a. \(\widehat{J L}\)

b. \(\widehat{K M}\)

c. \(\widehat{J L M}\)

d. \(\widehat{J M}\)
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.2 Ans 15

Question 16.
Big Ideas Math Answers Geometry Chapter 10 Circles 60

a. \(\widehat{R S}\)

Answer:
\(\widehat{R S}\) = \(\widehat{Q R S}\) – \(\widehat{Q R}\)
= 180 – 42
= 138°
So, \(\widehat{R S}\) = 138°

b. \(\widehat{Q R S}\)

Answer:
\(\widehat{Q R S}\) = 180°

c. \(\widehat{Q S T}\)

Answer:
\(\widehat{Q S T}\) = \(\widehat{Q R S}\) + \(\widehat{S T}\)
= 180 + 42 = 222
So, \(\widehat{Q S T}\) = 222°

d. \(\widehat{Q T}\)

Answer:
\(\widehat{Q T}\) = 360 – (42 + 138 + 42)
= 360 – (222)
= 138°
\(\widehat{Q T}\) = 138°

Question 17.
MODELING WITH MATHEMATICS
A recent survey asked high school students their favorite type of music. The results are shown in the circle graph. Find each indicated arc measure.
Big Ideas Math Answers Geometry Chapter 10 Circles 61
a. m\(\widehat{A E}\)
b. m\(\widehat{A C E}\)
c. m\(\widehat{G D C}\)
d. m\(\widehat{B H C}\)
e. m\(\widehat{F D}\)
f. m\(\widehat{F B D}\)
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.2 Ans 17

Question 18.
ABSTRACT REASONING
The circle graph shows the percentages of students enrolled in fall Sports at a high school. Is it possible to find the measure of each minor arc? If so, find the measure 0f the arc for each category shown. If not, explain why it is not possible.
Big Ideas Math Answers Geometry Chapter 10 Circles 62

Answer:
Soccer angle = 30% of 360 = 108°
Volleyball angle = 15% of 360 = 54°
Cross-country angle = 20% of 360 = 72°
None angle = 15% of 360 = 54°
Football angle = 20% of 360 = 72°

In Exercises 19 – 22, tell whether the red arcs are congruent. Explain why or why not.

Question 19.
Big Ideas Math Answers Geometry Chapter 10 Circles 63
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.2 Ans 19

Question 20.
Big Ideas Math Answers Geometry Chapter 10 Circles 64

Answer:
\(\widehat{L P}\) and \(\widehat{M N}\) are not congruet because they are not in the same circle.

Question 21.
Big Ideas Math Answers Geometry Chapter 10 Circles 65
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.2 Ans 21

Question 22.
Big Ideas Math Answers Geometry Chapter 10 Circles 66

Answer:
\(\widehat{R S Q}\), \(\widehat{F G H}\) are not congruent because those two circles have different radii.

MATHEMATICAL CONNECTIONS
In Exercises 23 and 24. find the value of x. Then find the measure of the red arc.

Question 23.
Big Ideas Math Answers Geometry Chapter 10 Circles 67
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.2 Ans 23

Question 24.
Big Ideas Math Answers Geometry Chapter 10 Circles 68

Answer:
4x + 6x + 7x + 7x = 360
24x = 360
x = 15
m\(\widehat{R S T}\) = 6(15) + 7(15)
= 90 + 105 = 195°
So, m\(\widehat{R S T}\) = 195°

Question 25.
MAKING AN ARGUMENT
Your friend claims that any two arcs with the same measure are similar. Your cousin claims that an two arcs with the same measure are congruent. Who is correct? Explain.
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.2 Ans 25

Question 26.
MAKING AN ARGUMENT
Your friend claims that there is not enough information given to find the value of x. Is your friend correct? Explain your reasoning.
Big Ideas Math Answers Geometry Chapter 10 Circles 69

Answer:
My friend is wrong.
4x + x + x + 4x = 360
10x = 360
x = 36°

Question 27.
ERROR ANALYSIS
Describe and correct the error in naming the red arc.
Big Ideas Math Answers Geometry Chapter 10 Circles 70
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.2 Ans 27

Question 28.
ERROR ANALYSIS
Describe and correct the error in naming congruent arc.
Big Ideas Math Answers Geometry Chapter 10 Circles 71

Answer:
\(\widehat{J K}\), \(\widehat{N P}\) are not congruent because those two arcs are from different circles.

Question 29.
ATTENDING TO PRECISION
Two diameters of ⊙P are \(\widehat{A B}\) and \(\widehat{C D}\). Find m\(\widehat{A C D}\) and m\(\widehat{A C}\) when m\(\widehat{A D}\) = 20°.
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.2 Ans 29

Question 30.
REASONING
In ⊙R, m\(\widehat{A B}\) = 60°, m\(\widehat{B C}\) = 25°. m\(\widehat{C D}\) = 70°, and m\(\widehat{D E}\) = 20°. Find two possible measures of \(\widehat{A E}\).

Answer:
\(\widehat{A E}\) = 360 – (\(\widehat{A B}\) + \(\widehat{B C}\) + \(\widehat{C D}\) + \(\widehat{D E}\))
= 360 – (60 + 25 + 70 + 20)
= 360 – (175)
= 185
\(\widehat{A E}\) = \(\widehat{A B}\) + \(\widehat{B C}\) + \(\widehat{C D}\) + \(\widehat{D E}\)
= 60 + 25 + 70 + 20 = 175
So, the two possibilities of \(\widehat{A E}\) are 185°, 175°

Question 31.
MODELING WITH MATHEMATICS
On a regulation dartboard, the outermost circle is divided into twenty congruent sections. What is the measure of each arc in this circle?
Big Ideas Math Answers Geometry Chapter 10 Circles 72
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.2 Ans 31

Question 32.
MODELING WITH MATHEMATICS
You can use the time zone wheel to find the time in different locations across the world. For example, to find the time in Tokyo when it is 4 P.M. in San Francisco, rotate the small wheel until 4 P.M. and San Francisco line up, as shown. Then look at Tokyo to see that it is 9 A.M. there.
Big Ideas Math Answers Geometry Chapter 10 Circles 73
a. What is the arc measure between each time zone 0n the wheel?

Answer:
As the circle is divided into 24 sectors, each time zone angle = \(\frac { 360 }{ 24 } \) = 15°

b. What is the measure of the minor arc from the Tokyo zone to the Anchorage zone?

Answer:
The measure of the minor arc from the Tokyo zone to the Anchorage zone = 15 + 15 + 15 + 15 + 15 + 15
= 90°

c. If two locations differ by 180° on the wheel, then it is 3 P.M. at one location when it is _________ at the other location.

Answer:
Kuwaiti city.

Question 33.
PROVING A THEOREM
Write a coordinate proof of the Similar Circles Theorem (Theorem 10.5).
Given ⊙O with center O(0, 0) and radius r.
⊙A with center A(a, 0) and radius s
Prove ⊙O ~ ⊙A
Big Ideas Math Answers Geometry Chapter 10 Circles 74
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.2 Ans 33

Question 34.
ABSTRACT REASONING
Is there enough information to tell whether ⊙C ≅ ⊙D? Explain your reasoning.
Big Ideas Math Answers Geometry Chapter 10 Circles 75

Answer:
Both circles ⊙C and ⊙D have the same radius so those circles are congruent.

Question 35.
PROVING A THEOREM
Use the diagram to prove each part of the biconditional in the Congruent Circles Theorem (Theorem 10.3).
Big Ideas Math Geometry Solutions Chapter 10 Circles 297
a. Given \(\overline{A C} \cong \overline{B D}\)
Prove ⊙A ≅ ⊙B
b. Given ⊙A ≅ ⊙B
prove \(\overline{A C} \cong \overline{B D}\)
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.2 Ans 35

Question 36.
HOW DO YOU SEE IT?
Are the circles on the target similar or congruent? Explain your reasoning.
Big Ideas Math Answers Geometry Chapter 10 Circles 76
Answer:

Question 37.
PROVING A THEOREM
Use the diagram to prove each part of the biconditional in the Congruent Central Angles Theorem (Theorem 10.4).
Big Ideas Math Answers Geometry Chapter 10 Circles 77
a. Given ∠ABC ≅ ∠DAE
Prove \(\widehat{B C}\) ≅ \(\widehat{D E}\)
b. Given \(\widehat{B C}\) ≅ \(\widehat{D E}\)
Prove ∠ABC ≅ ∠DAE
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.2 Ans 37

Question 38.
THOUGHT PROVOKING
Write a formula for the length of a circular arc. Justify your answer.

Answer:
The formula to find the length of a circular arc is radius x angle.

Maintaining Mathematical Proficiency

Find the value of x. Tell whether the side lengths form a Pythagorean triple.

Question 39.
Big Ideas Math Answers Geometry Chapter 10 Circles 78
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.2 Ans 39

Question 40.
Big Ideas Math Answers Geometry Chapter 10 Circles 81

Answer:
x² = 13² + 13²
= 169 + 169
= 338
x = 13√2

Question 41
Big Ideas Math Answers Geometry Chapter 10 Circles 79
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.2 Ans 41

Question 42.
Big Ideas Math Answers Geometry Chapter 10 Circles 80

Answer:
14² = x² + 10²
196 = x² + 100
x² = 196 – 100
x² = 96
x = 4√6

10.3 Using Chords

Exploration 1

Drawing Diameters

Work with a partner: Use dynamic geometry software to construct a circle of radius 5 with center at the origin. Draw a diameter that has the given point as an endpoint. Explain how you know that the chord you drew is a diameter.
a. (4, 3)
b. (0, 5)
c. (-3, 4)
d. (-5, 0)

Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 12

Exploration 2

Writing a Conjecture about Chords

Work with a partner. Use dynamic geometry software to construct a chord \(\overline{B C}\) of a circle A. Construct a chord on the perpendicular bisector of \(\overline{B C}\). What do you notice? Change the original chord and the circle several times. Are your results always the same? Use your results to write a conjecture.
LOOKING FOR STRUCTURE
To be proficient in math, you need to look closely to discern a pattern or structure.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 82
Answer:

Exploration 3

A Chord Perpendicular to a Diameter

Work with a partner. Use dynamic geometry software to construct a diameter \(\overline{B C}\) of a circle A. Then construct a chord \(\overline{D E}\) perpendicular to \(\overline{B C}\) at point F. Find the lengths DF and EF. What do you notice? Change the chord perpendicular to \(\overline{B C}\) and the circle several times. Do you always get the same results? Write a conjecture about a chord that is perpendicular to a diameter of a circle.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 83
Answer:

Communicate Your Answer

Question 4.
What are two ways to determine when a chord is a diameter of a circle?

Answer:
If a chord passes through the centre of the circle, then it is the diameter of a circle.
The longest chord of the circle is the diameter of a circle.

Lesson 10.3 Using Chords

Monitoring Progress

In Exercises 1 and 2, use the diagram of ⊙D.

Big Ideas Math Geometry Answer Key Chapter 10 Circles 84

Question 1.
If m\(\widehat{A B}\) = 110°. find m\(\widehat{B C}\).

Answer:
Because AB and BC are congruent chords in congruent circles, the corresponding minor arcs \(\widehat{A B}\), \(\widehat{B C}\) are congruent by the congruent corresponding chords theorem.
So, \(\widehat{A B}\) = \(\widehat{B C}\)
\(\widehat{B C}\) = 110°

Question 2.
If m\(\widehat{A C}\) = 150° find m\(\widehat{A B}\).

Answer:
\(\widehat{A C}\) = 360 – (\(\widehat{A B}\) + \(\widehat{B C}\))
150 = 360 – 2(\(\widehat{A B}\))
2(\(\widehat{A B}\)) = 360 – 150 = 210
\(\widehat{A B}\) = 105°

In Exercises 3 and 4. find the indicated length or arc measure.

Big Ideas Math Geometry Answer Key Chapter 10 Circles 85

Question 3.
CE

Answer:
CE = 5 + 5
= 10 units

Question 4.
m\(\widehat{C E}\)

Answer:
m\(\widehat{C E}\) = 9x + 180 – x = 180 – 8x
m\(\widehat{C E}\) = 180 – 8x

Question 5.
In the diagram, JK = LM = 24, NP = 3x, and NQ = 7x – 12. Find the radius of ⊙N
Big Ideas Math Geometry Answer Key Chapter 10 Circles 86

Answer:

Exercise 10.3 Using Chords

Vocabulary and Core Concept Check

Question 1.
WRITING
Describe what it means to bisect a chord.
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.3 Ans 1

Question 2.
WRITING
Two chords of a circle are perpendicular and congruent. Does one of them have to be a diameter? Explain your reasoning.

Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 14
Imagine a line segment of length 3 units, AB.
Big Ideas Math Geometry Answers Chapter 10 Circles 15
A second congruent segment of length 3 that is perpendicular to AB called CD.
Big Ideas Math Geometry Answers Chapter 10 Circles 16
Circumscribe both these line segments and note that AB and CD are now chords.
While both chords are perpendicular and congruent, neither chord is a diameter. Thus, it is possible to have two chords of this type with neither one diameter of the circle.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6, find the measure of the red arc or chord in ⊙C.

Question 3.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 87
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.3 Ans 3

Question 4.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 88

Answer:
Arc length = radius x angle
= 5 x 34 = 170

Question 5.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 89
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.3 Ans 5

Question 6.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 90

Answer:
As the two circles radius is the same and the angle is the same so the chord length is 11 units.

In Exercise 7-10, find the value of x.

Question 7.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 91
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.3 Ans 7

Question 8.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 92

Answer:
By the perpendicular bisector theorem RS = ST
x = 40°

Question 9.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 93
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.3 Ans 9

Question 10.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 94

Answer:
5x + 2 = 7x – 12
7x – 5x = 2 + 12
2x = 14
x = 7

Question 11.
ERROR ANALYSIS
Describe and correct the error in reasoning.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 95
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.3 Ans 11

Question 12.
PROBLEM SOLVING
In the cross section of the submarine shown, the control panels are parallel and the same length. Describe a method you can use to find the center of the cross section. Justify your method.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 96

Answer:

In Exercises 13 and 14, determine whether \(\overline{A B}\) is a diameter of the circle. Explain your reasoning.

Question 13.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 97
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.3 Ans 13

Question 14.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 98

Answer:
5² = 3² + x²
25 = 9 + x²
x² = 25 – 9
x = 4
So, AB is not diameter of the circle.

In Exercises 15 and 16, find the radius of ⊙Q.

Question 15.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 99
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.3 Ans 15

Question 16.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 100

Answer:
AD = BC
4x + 4 = 6x – 6
6x – 4x = 4 + 6
2x = 10
x = 5
BC = 6(5) – 6 = 30 – 6 = 24
QC² = 5² + 12²
= 25 + 144 = 169
QC = 13
Therefore, the radius is 13.

Question 17.
PROBLEM SOLVING
An archaeologist finds part of a circular plate. What was the diameter of the plate to the nearest tenth of an inch? Justify your answer.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 101
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.3 Ans 17

Question 18.
HOW DO YOU SEE IT?
What can you conclude from each diagram? Name a theorem that justifies your answer.
a.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 102

Answer:
Perpendicular chord bisector converse theorem.

b.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 103

Answer:
Congruent Corresponding Chords theorem

c.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 104

Answer:
Perpendicular chord bisector theorem

d.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 105

Answer:
Equidistant chords theorem

Question 19.
PROVING A THEOREM
Use the diagram to prove each part of the biconditional in the Congruent Corresponding Chords Theorem (Theorem 10.6).
Big Ideas Math Geometry Answer Key Chapter 10 Circles 106
a. Given \(\overline{A B}\) and \(\overline{C D}\) are congruent chords.
Prove \(\widehat{A B} \cong \widehat{C D}\)
b. Given \(\widehat{A B} \cong \widehat{C D}\)
Prove \(\overline{A B}\) ≅ \(\overline{C D}\)
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.3 Ans 19

Question 20.
MATHEMATICAL CONNECTIONS
In ⊙P, all the arcs shown have integer measures. Show that x must be even.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 109

Answer:

Question 21.
REASONING
In ⊙P. the lengths of the parallel chords are 20, 16, and 12. Find m\(\widehat{A B}\). Explain your reasoning.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 109
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.3 Ans 21

Question 22.
PROVING A THEOREM
Use congruent triangles to prove the Perpendicular Chord Bisector Theorem (Theorem 10.7).
Big Ideas Math Geometry Answer Key Chapter 10 Circles 109
Given \(\overline{E G}\) is a diameter of ⊙L.
\(\overline{E G}\) ⊥ \(\overline{D F}\)
Prove \(\overline{D C}\) ≅ \(\overline{F C}\), \(\widehat{D G} \cong \widehat{F G}\)

Answer:
Let L be the centre of the circle
draw any chord DF on the circle
As DF passes through LG.
The length of DC is the same as FC.

Question 23.
PROVING A THEOREM
Write a proof of the Perpendicular Chord Bisector Converse (Theorem 10.8).
Big Ideas Math Geometry Answer Key Chapter 10 Circles 110
Given \(\overline{Q S}\) is a perpendicular bisector of \(\overline{R T}\).
Prove \(\overline{Q S}\) is a diameter of the circle L.
(Hint: Plot the center L and draw △LPT and △LPR.)
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.3 Ans 23

Question 24.
THOUGHT PROVOKING
Consider two chords that intersect at point P. Do you think that \(\frac{A P}{B P}=\frac{C P}{D P}\)? Justify your answer.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 111

Answer:

Question 25.
PROVING A THEOREM
Use the diagram with the Equidistant Chords Theorem (Theorem 10.9) to prove both parts of the biconditional of this theorem.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 112
\(\overline{A B}\) ≅ \(\overline{C D}\) if and only if EF = EG
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.3 Ans 25

Question 26.
MAKING AN ARGUMENT
A car is designed so that the rear wheel is only partially visible below the body of the car. The bottom edge of the panel is parallel [o the ground. Your friend claims that the point where the tire touches the ground bisects \(\widehat{A B}\). Is your friend correct? Explain your reasoning.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 113

Answer:

Maintaining Mathematical Proficiency

Find the missing interior angle measure.

Question 27.
Quadrilateral JKLW has angle measures m∠J = 32°, m∠K = 25°, and m∠L = 44°. Find m∠M.
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.3 Ans 27

Question 28.
Pentagon PQRST has angle measures m∠P = 85°, m∠Q = 134°, m∠R = 97°, and m∠S =102°.
Find m∠T.

Answer:
The sum of interior angles of a pentagon = 540°
m∠T = 540 – (85 + 134 + 97 + 102)
= 540 – 418 = 122
m∠T = 122°.

10.1 – 10.3 Quiz

In Exercises 1 – 6, use the diagram. (Section 10.1)

Big Ideas Math Geometry Answer Key Chapter 10 Circles 114

Question 1.
Name the circle.

Answer:
The circle has a chord, diameter and tangent.

Question 2.
Name a radius.

Answer:
NP is the radius of the circle.

Question 3.
Name a diameter.

Answer:
KN is the diameter of the circle.

Question 4.
Name a chord.

Answer:
JL is the chord

Question 5.
Name a secant.

Answer:
SN is the secant

Question 6.
Name a tangent.

Answer:
QR is the tangent.

Find the value of x.

Question 7.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 115

Answer:
(9 + x)² = x² + 15²
81 + 18x + x² = x² + 225
18x = 225 – 81
18x = 144
x = 8

Question 8.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 116
Answer:
6x – 3 = 3x + 18
6x – 3x = 18 + 3
3x = 21
x = 7

Identify the given arc as a major arc, minor arc, or semicircle. Then find the measure of the arc.

Big Ideas Math Geometry Answer Key Chapter 10 Circles 117

Question 9.
\(\widehat{A E}\)

Answer:
\(\widehat{A E}\) = 180 – 36
= 144
So, \(\widehat{A E}\) = 144°

Question 10.
\(\widehat{B C}\)

Answer:
\(\widehat{B C}\) = 180 – (67 + 70)
= 180 – 137 = 43
So, \(\widehat{B C}\) = 43°

Question 11.
\(\widehat{A C}\)

Answer:
\(\widehat{A C}\) = 43 + 67 = 110°

Question 12.
\(\widehat{A C D}\)

Answer:
\(\widehat{A C D}\) = 180°

Question 13.
\(\widehat{A C E}\)

Answer:
\(\widehat{A C E}\) = 180 + 36 = 216°

Question 14.
\(\widehat{B E C}\)

Answer:
\(\widehat{B E C}\) = 70 + 36 + 43 = 149°

Tell whether the red arcs are congruent. Explain why or why not.

Question 15.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 118

Answer:
As two chords pass through the centre of the circle. Those two red arcs are congruent.

Question 16.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 119
Answer:
Red arcs are not congruent because the radius of the two circles is different.

Question 17.
Find the measure of the red arc in ⊙Q.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 120

Answer:

Question 18.
In the diagram. AC = FD = 30, PG = x + 5, and PJ = 3x – 1. Find the radius of ⊙P.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 121

Answer:

Question 19.
A circular clock can be divided into 12 congruent sections.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 122

a. Find the measure of each arc in this circle.
Answer:
The measure of each arc = \(\frac { 360 }{ 12 } \) = 30°

b. Find the measure of the minor arc formed by the hour and minute hands when the times is 7:00.
Answer:
When the time is 7:00 the minute hand is at 12 and hour hand is at 7 and so the minor arc is subtended by 12 – 7 = 5 of these sections and so the angle subtended is 30 x 5 = 150°

c. Find a time at which the hour and minute hands form an arc that is congruent to the arc in part (b).
Answer:
A minor arc is equal to 150° can be formed by multiple placements of the hour and the minute hand. One of them can be the time 5:00 when the minute hand is at 12 and the hour hand is at 5.

10.4 Inscribed Angles and Polygons

Exploration 1

Inscribed Angles and Central Angles

work with a partner: Use dynamic geometry software.

Sample
Big Ideas Math Geometry Solutions Chapter 10 Circles 123

a. Construct an inscribed angle in a circle. Then construct the corresponding central angle.
Answer:

b. Measure both angles. How is the inscribed angle related to its intercepted arc?
Answer:

c. Repeat parts (a) and (b) several times. Record your results in a table. Write a conjecture about how an inscribed angle is related to its intercepted arc.
ATTENDING TO PRECISION
To be proficient in math, you need to communicate precisely with others.
Answer:

Exploration 2

A Quadrilateral with Inscribed Angles

work with a partner: Use dynamic geometry software.

Sample
Big Ideas Math Geometry Solutions Chapter 10 Circles 124

a. Construct a quadrilateral with each vertex on a circle.
Answer:

b. Measure all four angles. What relationships do you notice?
Answer:

c. Repeat parts (a) and (b) several times. Record your results in a table. Then write a conjecture that summarizes the data.
Answer:

Communicate Your Answer

Question 3.
How are inscribed angles related to their intercepted arcs? How are the angles of an inscribed quadrilateral related to each other?
Answer:

Question 4.
Quadrilateral EFGH is inscribed in ⊙C. and m ∠ E = 80°. What is m ∠ G? Explain.

Answer:
m ∠ E + m ∠ H = 80 + 80 = 160°
m ∠ E + m ∠ H + m ∠ G + m ∠ F = 360
160° + m ∠ G + m ∠ F = 360
m ∠ G + m ∠ F = 360 – 160 = 200
m ∠ G = 100°

Lesson 10.4 Inscribed Angles and Polygons

Monitoring Progress

Find the measure of the red arc or angle.

Question 1.
Big Ideas Math Geometry Solutions Chapter 10 Circles 125

Answer:
m∠G = \(\frac { 90 }{ 2 } \) = 45°

Question 2.
Big Ideas Math Geometry Solutions Chapter 10 Circles 126

Answer:
\(\widehat{T V}\) = 2 • 38 = 76°

Question 3.
Big Ideas Math Geometry Solutions Chapter 10 Circles 127

Answer:
m∠W = 72°

Find the value of each variable.

Question 4.
Big Ideas Math Geometry Solutions Chapter 10 Circles 128

Answer:
x° = 90°
y° = 180 – (40 + 90) = 180 – 130
y° = 50°

Question 5.
Big Ideas Math Geometry Solutions Chapter 10 Circles 129

Answer:
∠B + ∠D = 180
∠B + 82 = 180
x° = 98°
∠C + ∠A = 180
68 + y° = 180
y° = 112°

Question 6.
Big Ideas Math Geometry Solutions Chapter 10 Circles 130

Answer:
∠S + ∠U = 180°
c + 2c – 6 = 180
3c = 186
c = 62°
∠T + ∠V = 180°
10x + 8x = 180
18x = 180
x = 10°

Question 7.
In Example 5, explain how to find locations where the left side of the statue is all that appears in your camera’s field of vision.
Answer:

Exercise 10.4 Inscribed Angles and Polygons

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
If a circle is circumscribed about a polygon, then the polygon is an ___________ .
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 1

Question 2.
DIFFERENT WORDS, SAME QUESTION
Which is different?
Find “both” answers.
Big Ideas Math Geometry Solutions Chapter 10 Circles 131

Find m∠ABC.
Answer:
m∠ABC = 60°

Find m∠AGC.
Answer:
m∠AGC = 180 – (25 + 25)
= 180 – 50 = 130°

Find m∠AEC.
Answer:
m∠AEC = 180 – (50 + 50)
= 180 – 100 = 80°

Find m∠ADC.
Answer:
m∠ADC = 180 – (25 + 50)
= 180 – 75 = 105°

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 8, find the indicated measure.

Question 3.
m∠A
Big Ideas Math Geometry Solutions Chapter 10 Circles 132
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 3

Question 4.
m∠G
Big Ideas Math Geometry Solutions Chapter 10 Circles 133
Answer:
m∠G = 360 – (70 + 120)
= 360 – 190 = 170°

Question 5.
m ∠ N
Big Ideas Math Geometry Solutions Chapter 10 Circles 134
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 5

Question 6.
m\(\widehat{R S}\)
Big Ideas Math Geometry Solutions Chapter 10 Circles 135
Answer:
m\(\widehat{R S}\) = 2 • 67 = 134°

Question 7.
m\(\widehat{V U}\)
Big Ideas Math Geometry Solutions Chapter 10 Circles 136
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 7

Question 8.
m\(\widehat{W X}\)
Big Ideas Math Geometry Solutions Chapter 10 Circles 137
Answer:
m\(\widehat{W X}\) = \(\frac { 75 }{ 2 } \) = 37.5

In Exercises 9 and 10, name two pairs of congruent angles.

Question 9.
Big Ideas Math Geometry Solutions Chapter 10 Circles 138
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 9

Question 10.
Big Ideas Math Geometry Solutions Chapter 10 Circles 139
Answer:
m∠W = m∠Z, m∠X = m∠Y

In Exercises 11 and 12, find the measure of the red arc or angle.

Question 11.
Big Ideas Math Geometry Solutions Chapter 10 Circles 140
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 11

Question 12.
Big Ideas Math Geometry Solutions Chapter 10 Circles 141
Answer:
\(\widehat{P S}\) = 2 • 40 = 80

In Exercises 13 – 16, find the value of each variable.

Question 13.
Big Ideas Math Geometry Solutions Chapter 10 Circles 142
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 13

Question 14.
Big Ideas Math Geometry Solutions Chapter 10 Circles 143
Answer:
m∠E + m∠G = 180
m + 60 = 180
m = 120°
m∠D + m∠F = 180
60 + 2k = 180
k = 60°

Question 15.
Big Ideas Math Geometry Solutions Chapter 10 Circles 144
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 15

Question 16.
Big Ideas Math Geometry Solutions Chapter 10 Circles 145
Answer:
3x° = 90°
x° = 30°
2y° + 90° + 34° = 180°
2y° + 124° = 180°
2y° = 56°
y° = 28°

Question 17.
ERROR ANALYSIS
Describe and correct the error in finding m\(\widehat{B C}\).
Big Ideas Math Geometry Solutions Chapter 10 Circles 146
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 17

Question 18.
MODELING WITH MATHEMATICS
A carpenter’s square is an L-shaped tool used to draw right angles. You need to cut a circular piece of wood into two semicircles. How can you use the carpenter’s square to draw a diameter on the circular piece of wood?
Big Ideas Math Geometry Solutions Chapter 10 Circles 147
Answer:
Recall that when a right triangle is inscribed in a circle, then the hypotenuse is the diameter of the circle. Simply use the carpenter’s square to inscribe it into the circle. The hypotenuse formed by both legs of the square should provide a diameter for the circle.

MATHEMATICAL CONNECTIONS
In Exercises 19 – 21, find the values of x and y. Then find the measures of the interior angles of the polygon.

Question 19.
Big Ideas Math Geometry Solutions Chapter 10 Circles 148
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 19

Question 20.
Big Ideas Math Geometry Solutions Chapter 10 Circles 149
Answer:
∠B + ∠C = 180
14x + 4x = 180
18x = 180°
x = 10°
∠A + ∠D = 180
9y + 24y = 180
33y = 180°
y = 5.45°
∠A = 130.9°, ∠B = 40°, ∠C = 140°, ∠D = 49°

Question 21.
Big Ideas Math Geometry Solutions Chapter 10 Circles 150
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 21

Question 22.
MAKING AN ARGUMENT
Your friend claims that ∠PTQ ≅ ∠PSQ ≅ ∠PRQ. Is our friend correct? Explain your reasoning.
Big Ideas Math Geometry Solutions Chapter 10 Circles 151
Answer:
Yes, my friend is correct.
∠PTQ ≅ ∠PSQ ≅ ∠PRQ is correct according to the inscribed angles of a circle theorem.

Question 23.
CONSTRUCTION
Construct an equilateral triangle inscribed in a circle.
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 23

Question 24.
CONSTRUTION
The side length of an inscribed regular hexagon is equal to the radius of the circumscribed circle. Use this fact to construct a regular hexagon inscribed in a circle.

Answer:
As the side length is equal to the radius. Draw a line representing the radius and draw a chord different chords in the form of hexagons of the radius of the circle.

REASONING
In Exercises 25 – 30, determine whether a quadrilateral of the given type can always be inscribed inside a circle. Explain your reasoning.

Question 25.
Square
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 25

Question 26.
rectangle
Answer:
yes, angles are right angles.

Question 27.
parallelogram
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 27

Question 28.
kite
Answer:
No.

Question 29.
rhombus
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 29

Question 30.
isosceles trapezoid
Answer:
Yes, the opposite angles are always supplementary.

Question 31.
MODELING WITH MATHEMATICS
Three moons, A, B, and C, are in the same circular orbit 1,00,000 kilometers above the surface of a planet. The planet is 20,000 kilometers in diameter and m∠ABC = 90°. Draw a diagram of the situation. How far is moon A from moon C?
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 31

Question 32.
MODELING WITH MATHEMATICS
At the movie theater. you want to choose a seat that has the best viewing angle, so that you can be close to the screen and still see the whole screen without moving your eyes. You previously decided that seat F7 has the best viewing angle, but this time someone else is already sitting there. Where else can you sit so that your seat has the same viewing angle as seat F7? Explain.
Big Ideas Math Geometry Solutions Chapter 10 Circles 152
Answer:

Question 33.
WRITING
A right triangle is inscribed in a circle, and the radius of the circle is given. Explain how to find the length of the hypotenuse.
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 33

Question 34.
HOW DO YOU SEE IT?
Let point Y represent your location on the soccer field below. What type of angle is ∠AYB if you stand anywhere on the circle except at point A or point B?
Big Ideas Math Geometry Solutions Chapter 10 Circles 153
Answer:

Question 35.
WRITING
Explain why the diagonals of a rectangle inscribed in a circle are diameters of the circle.
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 35

Question 36.
THOUGHT PROVOKING
The figure shows a circle that is circumscribed about ∆ABC. Is it possible to circumscribe a circle about any triangle? Justify your answer.
Big Ideas Math Geometry Solutions Chapter 10 Circles 154
Answer:
Yes.

Question 37.
PROVING A THEOREM
If an angle is inscribed in ⊙Q. the center Q can be on a side of the inscribed angle, inside the inscribed angle, or outside the inscribed angle. Prove each case of the Measure of an Inscribed Angle Theorem (Theorem 10. 10).

a. Case 1
Given ∠ABC is inscribed in ⊙Q
Let m∠B = x°
Center Q lies on \(\overline{B C}\).
Prove m∠ABC = \(\frac{1}{2}\)m\(\widehat{A C}\)
(Hint: Show that ∆AQB is isosceles. Then write m\(\widehat{A C}\) in terms of x.)
Big Ideas Math Geometry Solutions Chapter 10 Circles 155
b. Case 2
Use the diagram and auxiliary line to write Given and Prove statements for Case 2. Then write a proof
Big Ideas Math Geometry Solutions Chapter 10 Circles 156
c. Case 3
Use the diagram and auxiliary line to write Given and Prove statements for Case 3. Then write a proof.
Big Ideas Math Geometry Solutions Chapter 10 Circles 157
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 37.1
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 37.2
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 37.3

Question 38.
PROVING A THEOREM
Write a paragraph proof of the Inscribed Angles of a Circle Theorem (Theorem 10.11). First, draw a diagram and write Given and Prove statements.
Answer:
If two inscribed angles of a circle intercept the same arc, then the angles are congruent.

Question 39.
PROVING A THEOREM
The Inscribed Right Triangle Theorem (Theorem 10.12) is written as a conditional statement and its converse. Write a plan for proof for each statement.
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 39

Question 40.
PROVING A THEOREM
Copy and complete the paragraph proof for one part of the Inscribed Quadrilateral Theorem (Theorem 10. 13).
Given ⊙C with inscribed quadrilateral DEFG
Prove m ∠ D + m ∠ F = 180°,
m ∠ E + m ∠ G = 180°
Big Ideas Math Geometry Solutions Chapter 10 Circles 158
By the Arc Addition Postulate (Postulate 10. 1),
m\(\widehat{E F G}\) + ________ = 360° and m\(\widehat{F G D}\) + m\(\widehat{D E F}\) = 360°.
Using the ___________ Theorem. m\(\widehat{E D G}\) = 2m ∠ F, m\(\widehat{E F G}\) = 2m∠D, m\(\widehat{D E F}\) = 2m∠G, and m\(\widehat{F G D}\) = 2m ∠ E. By the Substitution Property of Equality, 2m∠D + ________ = 360°, So _________ . Similarly, __________ .
Answer:
m\(\widehat{E F G}\) + m\(\widehat{E D F}\) = 360° and m\(\widehat{F G D}\) + m\(\widehat{D E F}\) = 360°.
Using the the measure of an inscribed angle Theorem. m\(\widehat{E D G}\) = 2m ∠ F, m\(\widehat{E F G}\) = 2m∠D, m\(\widehat{D E F}\) = 2m∠G, and m\(\widehat{F G D}\) = 2m ∠ E. By the Substitution Property of Equality, 2m∠D + 2m∠G = 360°.

Question 41.
CRITICAL THINKING
In the diagram, ∠C is a right angle. If you draw the smallest possible circle through C tangent to \(\overline{A B}\), the circle will intersect \(\overline{A C}\) at J and \(\overline{B C}\) at K. Find the exact length of \(\overline{J K}\).
Big Ideas Math Geometry Solutions Chapter 10 Circles 159
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 41

Question 42.
CRITICAL THINKING
You are making a circular cutting board. To begin, you glue eight 1-inch boards together, as shown. Then you draw and cut a circle with an 8-inch diameter from the boards.
Big Ideas Math Geometry Solutions Chapter 10 Circles 160
a. \(\overline{F H}\) is a diameter of the circular cutting board. Write a proportion relating GJ and JH. State a theorem in to justify your answer.
Answer:
Each board is 1 inch and FJ spans 6 boards.
\(\overline{F H}\) = 6 inches

b. Find FJ, JH, and GJ. What is the length of the cutting board seam labeled \(\overline{G K}\)?
Answer:
Each board is 1 inch and JH spans 2 boards.
JH = 2 inches
Equation is \(\frac { JH }{ GJ } \) = \(\frac { GJ }{ FJ } \)
\(\frac { 2 }{ GJ } \) = \(\frac { GJ }{ 6 } \)
12 = GJ²
GJ = 2√3
GK = 2(GJ)
GK = 4√3
So, FJ = 6, JH = 2, JG = 2√3, GK = 4√3

Maintaining Mathematical Proficiency

Solve the equation. Check your solution.

Question 43.
3x = 145
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 43

Question 44.
\(\frac{1}{2}\)x = 63
Answer:
x = 63 • 2
x = 126

Question 45.
240 = 2x
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 45

Question 46.
75 = \(\frac{1}{2}\)(x – 30)
Answer:
75 • 2 = x – 30
150 + 30 = x
x = 180

10.5 Angle Relationships in Circles

Exploration 1

Angles Formed by a Chord and Tangent Line

Work with a partner: Use dynamic geometry software.

Sample
Big Ideas Math Answer Key Geometry Chapter 10 Circles 161

a. Construct a chord in a circle. At one of the endpoints of the chord. construct a tangent line to the circle.
Answer:

b. Find the measures of the two angles formed by the chord and the tangent line.
Answer:

c. Find the measures of the two circular arcs determined by the chord.
Answer:

d. Repeat parts (a) – (c) several times. Record your results in a table. Then write a conjecture that summarizes the data.
Answer:

Exploration 2

Angles Formed by Intersecting Chords

Work with a partner: Use dynamic geometry software.

sample
Big Ideas Math Answer Key Geometry Chapter 10 Circles 162

a. Construct two chords that intersect inside a circle.
Answer:

b. Find the measure of one of the angles formed by the intersecting chords.
Answer:

c. Find the measures of the arcs intercepted h the angle in part (b) and its vertical angle. What do you observe?
Answer:

d. Repeat parts (a) – (c) several times. Record your results in a table. Then write a conjecture that summarizes the data.
CONSTRUCTING VIABLE ARGUMENTS
To be proficient in math, you need to understand and use stated assumptions, definitions, and previously established results.
Answer:

Communicate Your Answer

Question 3.
When a chord intersects a tangent line or another chord, what relationships exist among the angles and arcs formed?
Answer:

Question 4.
Line m is tangent to the circle in the figure at the left. Find the measure of ∠1.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 163
Answer:
m∠1 = \(\frac { 1 }{ 2 } \) • 148
m∠1 = 74°

Question 5.
Two chords intersect inside a circle to form a pair of vertical angles with measures of 55°. Find the sum of the measures of the arcs intercepted by the two angles.
Answer:
The sum of the measures of the arcs intercepted by the two angles = \(\frac { 1 }{ 2 } \) • 55
= 27.5

Lesson 10.5 Angle Relationships in Circles

Monitoring Progress

Line m is tangent to the circle. Find the indicated measure.

Question 1.
m ∠ 1
Big Ideas Math Answer Key Geometry Chapter 10 Circles 164
Answer:
m ∠ 1 = \(\frac { 1 }{ 2 } \) • 210
m ∠ 1 = 105°

Question 2.
m\(\widehat{R S T}\)
Big Ideas Math Answer Key Geometry Chapter 10 Circles 165
Answer:
m\(\widehat{R S T}\) = 2 • 98 = 196°
m\(\widehat{R S T}\) = 196°

Question 3.
m\(\widehat{X Y}\)
Big Ideas Math Answer Key Geometry Chapter 10 Circles 166
Answer:
m\(\widehat{X Y}\) = \(\frac { 1 }{ 2 } \) • 80
m\(\widehat{X Y}\) = 40°

Find the value of the variable.

Question 4.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 167
Answer:
y° = \(\frac { 1 }{ 2 } \) • (102 + 95)
= 98.5°

Question 5.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 168
Answer:
a° = 2 • 30° + 44°
= 60° + 44° = 104°
So, a° = 104°.

Find the value of x.

Question 6.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 169
Answer:
x° = 180° – 120°
x° = 60°

Question 7.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 170
Answer:
50° = 180° – x°
x° = 180° – 50°
x° = 130°

Question 8.
You are on top of Mount Rainier on a clear day. You are about 2.73 miles above sea level at point B. Find m\(\widehat{C D}\), which represents the part of Earth that you can see.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 171
Answer:
CB and BD are tangents, CB is perpendicular to AB and CD is perpendicular to AD by the tangent line to circle theorem.
△ABC is similar to △ABD by the hypotenuese leg congruence theorem.
∠CBA is similar to ∠ABD. So, m∠CBA = 74.5°, m∠CBD = 2 • 74.5° = 149°
m∠CBD = 180° – m∠CAD
m∠CBD = 180° – CD
149° = 180° – CD
CD = 31°
The part of earth you can see

Exercise 10.5 Angle Relationships in Circles

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
Points A, B, C, and D are on a circle, and Big Ideas Math Answer Key Geometry Chapter 10 Circles 172 intersects Big Ideas Math Answer Key Geometry Chapter 10 Circles 173 at point P.
If m∠APC = \(\frac{1}{2}\)(m\(\widehat{B D}\) – m\(\widehat{A C}\)). then point P is _________ the circle.
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 1

Question 2.
WRITING
Explain how to find the measure of a circumscribed angle.
Answer:
A circumscribed angle is the angle made by two intersecting tangent lines to a circle. Draw lines from the circle centre to the point of tangency. The angle between the radius and tangent line is 90°. The sum of angles of a quadrilateral is 360°. Angles between radii and tangent lines is 180°. The angle at two tangent lines meet is circumscribed angle.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6, line t is tangent to the circle. Find the indicated measure.

Question 3.
m\(\widehat{A B}\)
Big Ideas Math Answer Key Geometry Chapter 10 Circles 174
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 3

Question 4.
m\(\widehat{D E F}\)
Big Ideas Math Answer Key Geometry Chapter 10 Circles 175
Answer:
m\(\widehat{D E F}\) = 2(117°) = 234°

Question 5.
m < 1
Big Ideas Math Answer Key Geometry Chapter 10 Circles 176
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 5

Question 6.
m ∠ 3
Big Ideas Math Answer Key Geometry Chapter 10 Circles 177
Answer:
m ∠ 3 = ½ • 140 = 70°

In Exercises 7 – 14, find the value of x.

Question 7.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 178
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 7

Question 8.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 179
Answer:
x° = ½ • (30 + 2x – 30)

Question 9.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 180
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 9

Question 10.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 181
Answer:
34° = ½ (3x – 2 – (x + 6))
34° = ½ (3x – 2 – x – 6)
34° = ½ (2x – 8)
34° = x – 4
x° = 34 + 4
x° = 38°

Question 11.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 182
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 11

Question 12.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 183
Answer:
6x – 11 = 2  • 125
6x = 250 + 11
6x = 261
x° = 43.5°

Question 13.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 184
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 13

Question 14.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 185

Answer:
17x° = 75°
x° = 4.41°

ERROR ANALYSIS
In Exercises 15 and 16, describe and correct the error in finding the angle measure.

Question 15.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 186
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 15

Question 16.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 187
Answer:
m∠1 = ½ (122 – 70)
= ½ (52) = 26
So, m∠1 = 26°

In Exercises 17 – 22, find the indicated angle measure. justify your answer.

Big Ideas Math Answer Key Geometry Chapter 10 Circles 188

Question 17.
m ∠ 1
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 17

Question 18.
m ∠ 2
Answer:
m ∠ 2 = 60°

Explanation:
m ∠ 3 =30°, So, m ∠ 2 = 180° – (90° + 30°)
= 180° – 120° = 60°
Therefore, m ∠ 2 = 60°

Question 19.
m ∠ 3
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 19

Question 20.
m ∠ 4
Answer:
m ∠ 4 = 90°

Question 21.
m ∠ 5
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 21

Question 22.
m ∠ 6
Answer:
m ∠ 6 = 180° – (60° + 30° + 30°) = 180° – 120°
m ∠ 6 = 60°

Question 23.
PROBLEM SOLVING
You are flying in a hot air balloon about 1.2 miles above the ground. Find the measure of the arc that represents the part of Earth you can see. The radius of Earth is about 4000 miles.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 189
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 23

Question 24.
PROBLEM SOLVING
You are watching fireworks over San Diego Bay S as you sail away in a boat. The highest point the fireworks reach F is about 0.2 mile above the bay. Your eyes E are about 0.01 mile above the water. At point B you can no longer see the fireworks because of the curvature of Earth. The radius of Earth is about 4000 miles, and \(\overline{F E}\) is tangent to Earth at point T. Find m\(\widehat{s B}\). Round your answer to the nearest tenth.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 190
Answer:

Question 25.
MATHEMATICAL CONNECTIONS
In the diagram, \(\vec{B}\)A is tangent to ⊙E. Write an algebraic expression for m\(\widehat{C D}\) in terms of x. Then find m\(\widehat{C D}\).
Big Ideas Math Answer Key Geometry Chapter 10 Circles 191
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 25

Question 26.
MATHEMATICAL CONNECTIONS
The circles in the diagram are concentric. Write an algebraic expression for c in terms of a and b.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 192
Answer:
a° = ½(c° – b°)

Question 27.
ABSTRACT REASONING
In the diagram. \(\vec{P}\)L is tangent to the circle, and \(\overline{K J}\) is a diameter. What is the range of possible angle measures of ∠LPJ? Explain your reasoning.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 193
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 27

Question 28.
ABSTRACT REASONING
In the diagram, \(\overline{A B}\) is an chord that is not a diameter of the circle. Line in is tangent to the circle at point A. What is the range of possible values of x? Explain your reasoning. (The diagram is not drawn to scale.)
Big Ideas Math Answer Key Geometry Chapter 10 Circles 194
Answer:
The possible values of x are less than 180°.

Question 29.
PROOF
In the diagram Big Ideas Math Answer Key Geometry Chapter 10 Circles 195 and Big Ideas Math Answer Key Geometry Chapter 10 Circles 196 are secant lines that intersect at point L. Prove that m∠JPN > m∠JLN.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 197
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 29

Question 30.
MAKING AN ARGUMENT
Your friend claims that it is possible for a circumscribed angle to have the same measure as its intercepted arc. Is your friend correct? Explain your reasoning.

Answer:
Yes, when the circumscribed angle measures 90°, the central angle measures 90°, so the intercepted arc also measures 90°.

Question 31.
REASONING
Points A and B are on a circle, and t is a tangent line containing A and another point C.
a. Draw two diagrams that illustrate this situation.
b. Write an equation for m\(\widehat{A B}\) in terms of m∠BAC for each diagram.
c. For what measure of ∠BAC can you use either equation to find m\(\widehat{A B}\)? Explain.
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 31

Question 32.
REASONING
∆XYZ is an equilateral triangle inscribed in ⊙P. AB is tangent to ⊙P at point X, \(\overline{B C}\) is tangent to ⊙P at point Y. and \(\overline{A C}\) is tangent to ⊙P at point Z. Draw a diagram that illustrates this situation. Then classify ∆ABC by its angles and sides. Justify your answer.
Answer:

Question 33.
PROVING A THEOREM
To prove the Tangent and Intersected Chord Theorem (Theorem 10. 14), you must prove three cases.
a. The diagram shows the case where \(\overline{A B}\) contains the center of the circle. Use the Tangent Line to Circle Theorem (Theorem 10.1) to write a paragraph proof for this case.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 198
b. Draw a diagram and write a proof for the case where the center of the circle is in the interior of ∠CAB.
c. Draw a diagram and write a proof for the case where the center of the circle is in the exterior of ∠CAB.
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 33.1
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 33.2

Question 34.
HOW DO YOU SEE IT?
In the diagram, television cameras are Positioned at A and B to record what happens on stage. The stage is an arc of ⊙A. You would like the camera at B to have a 30° view of the stage. Should you move the camera closer or farther away? Explain your reasoning.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 199

Answer:
25° = ½(80° – 30°) = ½(50°)
So, you should move the camera closer.

Question 35.
PROVING A THEOREM
Write a proof of the Angles Inside the Circle Theorem (Theorem 10.15).
Big Ideas Math Answer Key Geometry Chapter 10 Circles 200
Given Chords \(\overline{A C}\) and \(\overline{B D}\) intersect inside a circle.
Prove m ∠ 1 = \(\frac{1}{2}\)(m\(\widehat{D C}\) + m\(\widehat{A B}\))
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 35

Question 36.
THOUGHT PROVOKING
In the figure, Big Ideas Math Answer Key Geometry Chapter 10 Circles 201 and Big Ideas Math Answer Key Geometry Chapter 10 Circles 202 are tangent to the circle. Point A is any point on the major are formed by the endpoints of the chord \(\overline{B C}\). Label all congruent angles in the figure. Justify your reasoning.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 203

Answer:
m∠CPB = ½(CAB – CB)

Question 37.
PROVING A THEOREM
Use the diagram below to prove the Angles Outside the Circle Theorem (Theorem 10.16) for the case of a tangent and a secant. Then copy the diagrams for the other two cases on page 563 and draw appropriate auxiliary segments. Use your diagrams to prove each case.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 204
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 37.1
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 37.2
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 37.3

Question 38.
PROVING A THEOREM
Prove that the Circumscribed Angle Theorem (Theorem 10.17) follows from the Angles Outside the Circle Theorem (Theorem 10.16).
Answer:

In Exercises 39 and 40, find the indicated measure(s). justify your answer

Question 39.
Find m ∠ P when m\(\widehat{W Z Y}\) = 200°
Big Ideas Math Answer Key Geometry Chapter 10 Circles 205
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 39

Question 40.
Find m\(\widehat{A B}\) and m\(\widehat{E D}\)
Big Ideas Math Answer Key Geometry Chapter 10 Circles 206

Answer:
m\(\widehat{E D}\) = ½ (115°) = 57.5°
∠GJA = 30°

Maintaining Mathematical Proficiency

Solve the equation.

Question 41.
x2 + x = 12
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 41

Question 42.
x2 = 12x + 35

Answer:

Explanation:
x² = 12x + 35
x = \(\frac { 12 ± √(144 + 140)  }{ 2 } \)
x = \(\frac { 12 ± √284  }{ 2 } \)
x = \(\frac { 12 + √284  }{ 2 } \), \(\frac { 12 – √284  }{ 2 } \)

Question 43.
– 3 = x2 + 4x
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 43

10.6 Segment Relationships in Circles

Exploration 1

Segments Formed by Two Intersecting Chords

Work with a partner: Use dynamic geometry software.

Sample
Big Ideas Math Geometry Answers Chapter 10 Circles 207

a.
Construct two chords \(\overline{B C}\) and \(\overline{D E}\) that intersect in the interior of a circle at a point F.
Answer:

b.
Find the segment lengths BE, CF, DF, and EF and complete the table. What do you observe?
Big Ideas Math Geometry Solutions Chapter 10 Circles 298
Answer:

c. Repeat parts (a) and (b) several times. Write a conjecture about your results.
REASONING ABSTRACTLY
To be proficient in math, you need to make sense of quantities and their relationships in problem situations.
Answer:

Exploration 2

Secants Intersecting Outside a Circle

Work with a partner: Use dynamic geometry software.

Sample
Big Ideas Math Geometry Answers Chapter 10 Circles 208

a. Construct two secant Big Ideas Math Geometry Answers Chapter 10 Circles 209 and Big Ideas Math Geometry Answers Chapter 10 Circles 210 that intersect at a point B outside a circle, as shown.
Answer:

b. Find the segment lengths BE, BC, BF, and BD. and complete the table. What do you observe?
Big Ideas Math Geometry Solutions Chapter 10 Circles 299
Answer:

c. Repeat parts (a) and (b) several times. Write a conjecture about your results.
Answer:

Communicate Your Answer

Question 3.
What relationships exist among the segments formed by two intersecting chords or among segments of two secants that intersect outside a circle?

Question 4.
Find the segment length AF in the figure at the left.
Big Ideas Math Geometry Answers Chapter 10 Circles 211

Answer:

Explanation:
EA • AF = AD • AC
18 • AF = 9 • 8
AF = 4

Lesson 10.6 Segment Relationships in Circles

Monitoring Progress

Find the value of x.

Question 1.
Big Ideas Math Geometry Answers Chapter 10 Circles 212

Answer:
x = 8

Explanation:
4 • 6 = 3 • x
3x = 24
x = 8

Question 2.
Big Ideas Math Geometry Answers Chapter 10 Circles 213

Answer:
x = 5

Explanation:
2 • x + 1 = 4 • 3
x + 1 = 6
x = 5

Question 3.
Big Ideas Math Geometry Answers Chapter 10 Circles 214

Answer:
x = \(\frac { 54 }{ 5 } \)

Explanation:
6 • 9 = 5 • x
54 = 5x
x = \(\frac { 54 }{ 5 } \)

Question 4.
Big Ideas Math Geometry Answers Chapter 10 Circles 215

Answer:
x = \(\frac { 3 ± √37 }{ 2 } \)

Explanation:
3 • x + 2 = x + 1 • x – 1
3x + 6 = x² – 1
x² – 3x – 7 = 0
x = \(\frac { 3 ± √(9 + 28) }{ 2 } \)
x = \(\frac { 3 ± √37 }{ 2 } \)

Question 5.
Big Ideas Math Geometry Answers Chapter 10 Circles 216

Answer:
x = ±√3

Explanation:
x² = 3 • 1
x² = 3
x = ±√3

Question 6.
Big Ideas Math Geometry Answers Chapter 10 Circles 217

Answer:
x = \(\frac { 49 }{ 5 } \)

Explanation:
7² = 5 • x
49 = 5x
x = \(\frac { 49 }{ 5 } \)

Question 7.
Big Ideas Math Geometry Answers Chapter 10 Circles 218

Answer:
x = 14.4

Explanation:
12² = 10x
144 = 10x
x = 14.4

Question 8.
WHAT IF?
In Example 4, CB = 35 feet and CE = 14 feet. Find the radius of the tank.

Answer:
The radius of the tank is 36.75

Explanation:
CB² = CE ⋅ CD
35² = 14 ⋅ (2r + 14)
1225 = 28r + 196
28r = 1029
r = 36.75

Exercise 10.6 Segment Relationships in Circles

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
The part of the secant segment that is outside the circle is called a(n) ______________ .
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.6 Ans 1

Question 2.
WRITING
Explain the difference between a tangent segment and a secant segment.

Answer:
A tangent segment intersects the circle at only one point. It actually doesn’t go through the circle. If a ball is rolling on a table top, then it would be the tangent. A secant segment intersects the circle in two points. It goes through the circle. In a tangent, no part is in the interior of the circle. In a secant, there is a part in the interior called a chord.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6, find the value of x.

Question 3.
Big Ideas Math Geometry Answers Chapter 10 Circles 219
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.6 Ans 3

Question 4.
Big Ideas Math Geometry Answers Chapter 10 Circles 220

Answer:
x = 23

Explanation:
10 • 18 = 9 • (x – 3)
20 = x – 3
x = 23

Question 5.
Big Ideas Math Geometry Answers Chapter 10 Circles 221
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.6 Ans 5

Question 6.
Big Ideas Math Geometry Answers Chapter 10 Circles 222

Answer:
x = 5

Explanation:
2x • 12 = 15 • (x + 3)
24x = 15x + 45
9x = 45
x = 5

In Exercises 7 – 10, find the value of x.

Question 7.
Big Ideas Math Geometry Answers Chapter 10 Circles 223
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.6 Ans 7

Question 8.
Big Ideas Math Geometry Answers Chapter 10 Circles 224

Answer:
x = \(\frac { 35 }{ 4 } \)

Explanation:
5 • 7 = 4 • x
4x = 35
x = \(\frac { 35 }{ 4 } \)

Question 9.
Big Ideas Math Geometry Answers Chapter 10 Circles 225
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.6 Ans 9

Question 10.
Big Ideas Math Geometry Answers Chapter 10 Circles 226

Answer:
x = 30

Explanation:
45 • x = 50 • 27
45x = 1350
x = 30

In Exercises 11 – 14. find the value of x.

Question 11.
Big Ideas Math Geometry Answers Chapter 10 Circles 227
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.6 Ans 11

Question 12.
Big Ideas Math Geometry Answers Chapter 10 Circles 228

Answer:
x = 48

Explanation:
24² = 12x
576 = 12x
x = 48

Question 13.
Big Ideas Math Geometry Answers Chapter 10 Circles 229
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.6 Ans 13

Question 14.
Big Ideas Math Geometry Answers Chapter 10 Circles 230

Answer:
x = 1.5

Explanation:
3 = 2x
x = 1.5

Question 15.
ERROR ANALYSIS
Describe and correct the error in finding CD.
Big Ideas Math Geometry Answers Chapter 10 Circles 231
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.6 Ans 15

Question 16.
MODELING WITH MATHEMATICS
The Cassini spacecraft is on a mission in orbit around Saturn until September 2017. Three of Saturn’s moons. Tethys. Calypso, and Teleslo. have nearly circular orbits of radius 2,95,000 kilometers. The diagram shows the positions of the moons and the spacecraft on one of Cassini’s missions. Find the distance DB from Cassini to Tethys when \(\overline{A D}\) is tangent to the circular orbit.
Big Ideas Math Geometry Answers Chapter 10 Circles 232

Answer:
BD = 579493 km

Explanation:
(203,000)² = 83000x
x = 496493
BC = 496493
BD = 496493 + 83000 = 579493

Question 17.
MODELING WITH MATHEMATICS
The circular stone mound in Ireland called Newgrange has a diameter of 250 feet. A passage 62 feet long leads toward the center of the mound. Find the perpendicular distance x from the end of the passage to either side of the mound.
Big Ideas Math Geometry Answers Chapter 10 Circles 233
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.6 Ans 17

Question 18.
MODELING WITH MATHEMATICS
You are designing an animated logo for our website. Sparkles leave point C and move to the Outer circle along the segments shown so that all of the sparkles reach the outer circle at the same time. Sparkles travel from point C to point D at 2 centimeters per second. How fast should sparkles move from point C to point N? Explain.
Big Ideas Math Geometry Answers Chapter 10 Circles 234

Answer:
5.33 should sparkles move from point C to point N.

Explanation:
4 • 8 = 6 • x
x = 5.33

Question 19.
PROVING A THEOREM
Write a two-column proof of the Segments of Chords Theorem (Theorem 10.18).

Plan for Proof:
Use the diagram to draw \(\overline{A C}\) and \(\overline{D B}\). Show that ∆EAC and ∆EDB are similar. Use the fact that corresponding side lengths in similar triangles are proportional.
Big Ideas Math Geometry Answers Ch
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.6 Ans 19.1
Big Ideas Math Geometry Answers Chapter 10 Circles 10.6 Ans 19.2

Question 20.
PROVING A THEOREM
Prove the Segments of Secants Theorem (Theorem 10.19). (Hint: Draw a diagram and add auxiliary line segments to form similar triangles.)
Answer:

Question 21.
PROVING A THEOREM
Use the Tangent Line to Circle Theorem (Theorem 10. 1) to prove the Segments of Secants and Tangents Theorem (Theorem 10.20) for the special case when the secant segment Contains the center of the circle.
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.6 Ans 21

Question 22.
PROVING A THEOREM
Prove the Segments of Secants and Tangents Theorem (Theorem 10.20). (Hint: Draw a diagram and add auxiliary line segments to form similar triangles.)
Answer:

Question 23.
WRITING EQUATIONS
In the diagram of the water well, AB, AD, and DE are known. Write an equation for BC using these three measurements.
Big Ideas Math Geometry Answers Chapter 10 Circles 236
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.6 Ans 23

Question 24.
HOW DO YOU SEE IT?
Which two theorems would you need to use to tind PQ? Explain your reasoning.
Big Ideas Math Geometry Answers Chapter 10 Circles 237

Answer:

Question 25.
CRITICAL THINKING
In the figure, AB = 12, BC = 8, DE = 6, PD = 4, and a is a point of tangency. Find the radius of ⊙P.
Big Ideas Math Geometry Answers Chapter 10 Circles 238
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.6 Ans 25

Question 26.
THOUGHT PROVOKING
Circumscribe a triangle about a circle. Then, using the points of tangency, inscribe a triangle in the circle. Must it be true that the two triangles are similar? Explain your reasoning.

Answer:

Maintaining Mathematical Proficiency

Solve the equation by completing the square.

Question 27.
x2 + 4x = 45
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.6 Ans 27

Question 28.
x2 – 2x – 1 = 8

Answer:
x = 1 + √10, x = 1 – √10

Explanation:
x² – 2x – 1 = 8
x² – 2x – 9 = 0
x = \(\frac { 2 ± √(4 + 36) }{ 2 } \)
x = \(\frac { 2 ± √40 }{ 2 } \)
x = 1 + √10, x = 1 – √10

Question 29.
2x2 + 12x + 20 = 34
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.6 Ans 29

Question 30.
– 4x2 + 8x + 44 = 16

Answer:
x = 1 + √8, x = 1 – √8

Explanation:
– 4x² + 8x + 44 = 16
4x² – 8x – 28 =0
x² – 2x – 7 = 0
x = \(\frac { 2 ± √(4 + 28) }{ 2 } \)
x = \(\frac { 2 ± √32 }{ 2 } \)
x = 1 ± √8

10.7 Circles in the Coordinate Plane

Exploration 1

The Equation of a Circle with Center at the Origin

Work with a partner: Use dynamic geometry software to Construct and determine the equations of circles centered at (0, 0) in the coordinate plane, as described below.
Big Ideas Math Geometry Solutions Chapter 10 Circles 300
a. Complete the first two rows of the table for circles with the given radii. Complete the other rows for circles with radii of your choice.

Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 18

b. Write an equation of a circle with center (0, 0) and radius r.

Answer:
x²  + y²  = r²

Explanation:
(x – 0)² + (y – 0)²  = r²
x²  + y²  = r²

Exploration 2

The Equation of a Circle with Center (h, k)

Work with a partner: Use dynamic geometry software to construct and determine the equations of circles of radius 2 in the coordinate plane, as described below.
Big Ideas Math Geometry Solutions Chapter 10 Circles 301
a. Complete the first two rows of the table for circles with the given centers. Complete the other rows for circles with centers of your choice.

Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 17

b. Write an equation of a circle with center (h, k) and radius 2.

Answer:
(x – h)² + (y – k)² = 4

c. Write an equation of a circle with center (h, k) and radius r.

Answer:
(x – h)² + (y – k)² = r²

Exploration 3

Deriving the Standard Equation of a Circle

Work with a partner. Consider a circle with radius r and center (h, k).

Write the Distance Formula to represent the distance d between a point (x, y) on the circle and the center (h, k) of the circle. Then square each side of the Distance Formula equation.

How does your result compare with the equation you wrote in part (c) of Exploration 2?

MAKING SENSE OF PROBLEMS
To be proficient in math, you need to explain correspondences between equations and graphs.
Big Ideas Math Answers Geometry Chapter 10 Circles 239

Answer:
(x – h)² + (y – k)² = r²

Communicate Your Answer

Question 4.
What is the equation of a circle with center (h, k) and radius r in the coordinate plane?

Answer:
(x – h)² + (y – k)² = r²

Question 5.
Write an equation of the circle with center (4, – 1) and radius 3.

Answer:
x² + y² – 8x + 2y + 8 = 0

Explanation:
(x – 4)² + (y + 1)² = 9
x² – 8x + 16 + y² + 2y + 1 = 9
x² + y² – 8x + 2y = 9 – 17
x² + y² – 8x + 2y + 8 = 0

Lesson 10.7 Circles in the Coordinate Plane

Monitoring Progress

Write the standard equation of the circle with the given center and radius.

Question 1.
center: (0, 0), radius: 2.5

Answer:
x² + y² = 6.25

Explanation:
(x – 0)² + (y – 0)² = 2.5²
x² + y² = 6.25

Question 2.
center: (- 2, 5), radius: 7

Answer:
(x + 2)² + (y – 5)² = 49

Explanation:
(x + 2)² + (y – 5)² = 7²
(x + 2)² + (y – 5)² = 49

Question 3.
The point (3, 4) is on a circle with center (1, 4). Write the standard equation of the circle.

Answer:
(x – 1)² + (y – 4)² = 4

Explanation:
r = √(3 – 1)² + (4 – 4)²
= √(2)²
r = 2
(x – 1)² + (y – 4)² = 2²
(x – 1)² + (y – 4)² = 4

Question 4.
The equation of a circle is x2 + y2 – 8x + 6y + 9 = 0. Find the center and the radius of the circle. Then graph the circle.

Answer:
The center of the circle (4, -3) and radius is 4.

Explanation:
x² + y² – 8x + 6y + 9 = 0
x² – 8x + 16 + y² + 6y + 9 = 16
(x – 4)² + (y + 3)² = 4²
The center of the circle (4, -3) and radius is 4.

Question 5.
Prove or disprove that the point (1, √5 ) lies on the circle centered at the origin and containing the point (0, 1).

Answer:
Disproved.

Explanation:
We consider the circle centred at the origin and containing the point (0, 1).
Therefore, we can conclude that the radius of the circle r = 1, let O (0, 0) and B (1, √5). So the distance between two points is
OB = √(1 – 0) + (√5 – 0)² = √(1 + 5) = √6
As the radius of the given circle is 1 and distance of the point B from its centre is √6. So we can conclude that point does lie on the given circle.

Question 6.
why are three seismographs needed to locate an earthquake’s epicentre?

Answer:

Exercise 10.7 Circles in the Coordinate Plane

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
What is the standard equation of a circle?
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 1

Question 2.
WRITING
Explain why knowing the location of the center and one point on a circle is enough to graph the circle.

Answer:
If we know the location of the center and one point on the circle, we can graph a circle because the distance from the center to the point is called the radius.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 8, write the standard equation of the circle.

Question 3.
Big Ideas Math Answers Geometry Chapter 10 Circles 240
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 3

Question 4.
Big Ideas Math Answers Geometry Chapter 10 Circles 241

Answer:
x² + y² = 36

Explanation:
The center is (0, 0) and the radius is 6
(x – h)² + (y – k)² = r²
(x – 0)² + (y – 0)² = 6²
x² + y² = 36

Question 5.
a circle with center (0, 0) and radius 7
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 5

Question 6.
a circle with center (4, 1) and radius 5

Answer:
(x – 4)² + (y – 1)² = 25

Explanation:
(x – h)² + (y – k)² = r²
(x – 4)² + (y – 1)² = 5²
(x – 4)² + (y – 1)² = 25

Question 7.
a circle with center (- 3, 4) and radius 1
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 7

Question 8.
a circle with center (3, – 5) and radius 7

Answer:
(x – 3)² + (y + 5)² = 49

Explanation:
(x – h)² + (y – k)² = r²
(x – 3)² + (y + 5)² = 7²
(x – 3)² + (y + 5)² = 49

In Exercises 9 – 11, use the given information to write the standard equation of the circle.

Question 9.
The center is (0, 0), and a point on the circle is (0, 6).
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 9

Question 10.
The center is (1, 2), and a point on the circle is (4, 2).

Answer:
x² + y² = 9

Explanation:
r = √(x – h)² + (y – k)²
= √(4 – 1)² + (2 – 2)²
= √3²
r = 3
(x – h)² + (y – k)² = r²
(x – 0)² + (y – 0)² = 3²
x² + y² = 9

Question 11.
The center is (0, 0). and a point on the circle is (3, – 7).
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 11

Question 12.
ERROR ANALYSIS
Describe and correct the error in writing the standard equation of a circle.
Big Ideas Math Answers Geometry Chapter 10 Circles 242

Answer:
(x – h)² + (y – k)² = r²
(x + 3)² + (y + 5)² = 3²
(x + 3)² + (y + 5)² = 9

In Exercises 13 – 18, find the center and radius of the circle. Then graph the circle.

Question 13.
x2 + y2 = 49
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 13

Question 14.
(x + 5)2 + (y – 3)2 = 9

Answer:
Center is (-5, 3) and rdaius is 3.

Explanation:
For the equation (x + 5)2 + (y – 3)2 = 9, center is (-5, 3) and rdaius is 3.

Question 15.
x2 + y2 – 6x = 7
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 15

Question 16.
x2 + y2 + 4y = 32

Answer:
The center is (0, -2) and radius is 6

Explanation:
x2 + y2 + 4y = 32
x² + y² + 4y + 4 = 32 + 4
x² + (y + 2)² = 36
(x – 0)² + (y – (-2))² = 6²
The center is (0, -2) and radius is 6
Big Ideas Math Geometry Answers Chapter 10 Circles 21

Question 17.
x2 + y2 – 8x – 2y = – 16
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 17

Question 18.
x2 + y2 + 4x + 12y = – 15

Answer:
The center is (-2, -6) and radius is 5

Explanation:
x2 + y2 + 4x + 12y = – 15
x² + 4x + 4 + y² + 12y + 36 = -15 + 36 + 4
(x + 2)² + (y + 6)² = 5²
The center is (-2, -6) and radius is 5

In Exercises 19 – 22, prove or disprove the statement.

Question 19.
The point (2, 3) lies on the circle centered at the origin with radius 8.
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 19

Question 20.
The point (4, √5) lies on the circle centered at the origin with radius 3.

Answer:
The point (4, √5) does not lie on the circle.

Explanation:
r² = (x – h)² + (y – k)²
3² ______________ (4 – 0)² + (√5 – 0)²
9 ______________ 16 + 5
9 ≠ 21
Because the radius is 3 and the distance between center and the point is more than the radius. So the point does not lie on the circle.

Question 21.
The point (√6, 2) lies on the circle centered at the origin and containing the point (3, – 1).
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 21

Question 22.
The point (√7, 5) lies on the circle centered at the origin and containing the point (5, 2).

Answer:
The point (√7, 5) does not lie on the circle.

Explanation:
r² = (x – h)² + (y – k)²
= (√7 – 0)² + (5 – 0)² = 7 + 25 = 32
r = 5.65
(5.65)² ______________ (5 – 0)² + (2 – 0)²
32 ______________ 25 + 4
32 ≠ 29
Because the radius is 3 and the distance between center and the point is more than the radius. So the point does not lie on the circle.

Question 23.
MODELING WITH MATHEMATICS
A City’s commuter system has three zones. Zone I serves people living within 3 miles of the city’s center. Zone 2 serves those between 3 and 7 miles from the center. Zone 3 serves those over 7 miles from the center.
Big Ideas Math Answers Geometry Chapter 10 Circles 243
a. Graph this Situation on a coordinate plane where each unit corresponds to 1 mile. Locate the city’s center at the origin.
b. Determine which zone serves people whose homes are represented by the points (3, 4), (6, 5), (1, 2), (0.3). and (1, 6).
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 23.1
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 23.2

Question 24.
MODELING WITH MATHEMATICS
Telecommunication towers can be used to transmit cellular phone calls. A graph with units measured in kilometers shows towers at points (0, 0), (0, 5), and (6, 3). These towers have a range of about 3 kilometers.
a. Sketch a graph and locate the towers. Are there any locations that may receive calls from more than one tower? Explain your reasoning.
Answer:

b. The center of City A is located at (- 2, 2.5), and the center of City B is located at (5, 4). Each city has a radius of 1.5 kilometers. Which city seems to have better cell phone coverage? Explain your reasoning.

Answer:
There are three towers at points (0, 0), (0, 5) and (6, 3) with range of about 3 km
a. Let’s sketch the graph to locate towers. Draw the points (0, 0), (0, 5) and (6, 3). Then draw three circles with centers at these points and radii 3.
There are locations that can receive calls from more that one tower because circles with centers (0, 0) and (0, 5) overlap. Locations in their intersection can receive calls from two towers.
The city A has a center located at (-2, 2.5) and city B has a center located at (5, 4). Both cities have radius 1.5 km
Let’s draw the city A as a circle with center (-2, 2.5) and radius 1.5 and city B with center (5, 4) and radius 1.5.
From the graph we can conclude that the city B has better cell phone coverage because parts of city A do not have coverage.

Question 25.
REASONING
Sketch the graph of the circle whose equation is x2 + y2 = 16. Then sketch the graph of the circle after the translation (x, y) → (x – 2, y – 4). What is the equation of the image? Make a conjecture about the equation of the image of a circle centered at the origin after a translation m units to the left and n units down.
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 25

Question 26.
HOW DO YOU SEE IT?
Match each graph with its equation.

a. Big Ideas Math Answers Geometry Chapter 10 Circles 244 A. x2 + (y + 3) 2 = 4
b. Big Ideas Math Answers Geometry Chapter 10 Circles 245 B. (x – 3) 2 + y2 = 4
c. Big Ideas Math Answers Geometry Chapter 10 Circles 246 C. (x + 3) 2 + y2 = 4
d. Big Ideas Math Answers Geometry Chapter 10 Circles 247 D. x2 + (y – 3) 2 = 4

Answer:
a ➝ C, b ➝ A, c ➝ D, d ➝ B

Question 27.
USING STRUCTURE
The vertices of ∆XYZ are X(4, 5), Y(4, 13), and Z(8, 9). Find the equation of the circle circumscribed about ∆XYZ. Justify your answer.
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 27.1
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 27.2

Question 28.
THOUGHT PROVOKING
A circle has center (h, k) and contains point (a, b). Write the equation of the line tangent to the circle at point (a, b).

Answer:
y – b = \(\frac { h – a }{ b – k } \)(x – a)

Explanation:
It is given a circle with center C(h, k. A circle with point A(a, b). We have to write the equation of a tangent that intersects the circle at point A
By the tangent line to circle theorem, a tangent is perpendicular to the radius. Two lines are perpendicular if and only if their slopes are negative reciprocals. So, find the equation of the line AC to know its slope.
The equation of the line which has two points (a, b), (c, d) is y – b = \(\frac { b – c }{ a – d } \)(x – a)
The equation of the line which has two points A(a, b) and C(h, k) is
y – b = \(\frac { b – k }{ a – h } \)(x – a)
Therefore, the slope of the line through A nad C is \(\frac { b – k }{ a – h } \)
Hence the slope of the tangent is –\(\frac { a – h }{ b – k } \) = \(\frac { h – a }{ b – k } \)
Use the equation of the line y = kx + n through the point (a, b)
y – b = k(x – a) to find the equation of the tangent
The equation of the tangent with slope \(\frac { h – a }{ b – k } \) and through the point A(a, b) is y – b = \(\frac { h – a }{ b – k } \)(x – a)

MATHEMATICAL CONNECTIONS
In Exercises 29 – 32, use the equations to determine whether the line is a tangent, a secant a secant that contains the diameter, or name of these. Explain your reasoning.

Question 29.
Circle: (x – 4)2 + (y – 3)2 = 9
Line: y = 6
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 29.1

Question 30.
Circle: (x + 2)2 + (y – 2)2 = 16
Line: y = 2x – 4

Answer:
The line is a secant line.

Explanation:
(x + 2)2 + (y – 2)2 = 16, y = 2x – 4
(x + 2)2 + (2x – 4 – 2)2 = 16
x² + 4x + 4 + (2x – 6)² = 16
x² + 4x + 4 + 4x² – 24x + 36 = 16
5x² – 20x + 40 – 16 = 0
5x² – 20x + 24 = 0
x = \(\frac { 20 ±√-80 }{ 10 } \)
x = 2, y = 2 • 2 – 4 = 0, (2, 0)
The system has two solutions and point does not lie on the line. So the line is a secant line.

Question 31.
Circle: (x – 5)2 + (y + 1)2 = 4
Line: y = \(\frac{1}{5}\)x – 3\
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 31.1
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 31.2
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 31.3

Question 32.
Circle: (x + 3)2 + (y – 6)2 = 25
Line: y = –\(\frac{4}{3}\)x + 2

Answer:
The line is a secant line.

Explanation:
(x + 3)2 + (y – 6)2 = 25, y = –\(\frac{4}{3}\)x + 2
(x + 3)2 + (-\(\frac{4}{3}\)x + 2 – 6)2 = 25
x² + 6x + 9 + \(\frac { 16x² }{ 9 } \) + \(\frac { 32x }{ 3 } \) + 16 = 25
\(\frac { 25x² }{ 9 } \) + \(\frac { 50x }{ 3 } \) = 0
x(25x + 150) = 0
x = 0 or x = -6
y = 2, y = 10
(0, 2) and (-6, 10)
d = √(0 + 6)² + (2 – 10)²
= √(36 + 64)
= 10 ≠ 5
The system has two solutions and point does not lie on the line. So the line is a secant line.

MAKING AN ARGUMENT
Question 33.
Your friend claims that the equation of a circle passing through the points (- 1, 0) and (1, 0) is x2 – 2yk + y2 = 1 with center (0, k). Is your friend correct? Explain your reasoning.
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 33

Question 34.
REASONING
Four tangent circles are centered on the x-axis. The radius of ⊙A is twice the radius of ⊙O, The radius of ⊙B is three times the radius of ⊙O, The radius of ⊙C is four times the radius of ⊙O, All circles have integer radii, and the point (63, 16) is On ⊙C. What is the equation of ⊙A? Explain your reasoning.
Big Ideas Math Answers Geometry Chapter 10 Circles 248
Answer:

Maintaining Mathematical Proficiency

Identify the arc as a major arc, minor arc, or semicircle. Then find the measure of the arc.

Big Ideas Math Answers Geometry Chapter 10 Circles 249

Question 35.
\(\widehat{R S}\)
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 35

Question 36.
\(\widehat{P R}\)

Answer:
\(\widehat{P R}\) is a right angle
\(\widehat{P R}\) = 25 + 65 = 90°

Question 37.
\(\widehat{P R T}\)
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 37

Question 38.
\(\widehat{S T}\)

Answer:
\(\widehat{S T}\) is a major arc
\(\widehat{S T}\) = 360 – (90 + 65 +25 + 53) = 127°

Question 39.
\(\widehat{R S T}\)
Answer:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 39

Question 40.
\(\widehat{Q S}\)

Answer:
\(\widehat{Q S}\) is a minor arc
\(\widehat{Q S}\) = 25 + 53 = 78°

Circles Review

10.1 Lines and Segments That Intersect Circles

Tell whether the line, ray, or segment is best described as a radius, chord, diameter, secant, or tangent of ⊙P.

Big Ideas Math Geometry Answer Key Chapter 10 Circles 250

Question 1.
\(\overline{P K}\)

Answer:
\(\overline{P K}\) is radius

Question 2.
\(\overline{N M}\)

Answer:
\(\overline{N M}\) is chord

Question 3.
\(\vec{J}\)L

Answer:
\(\vec{J}\)L is tangent

Question 4.
\(\overline{K N}\)

Answer:
\(\overline{K N}\) is diameter

Question 5.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 251

Answer:
NL is secant

Question 6.
\(\overline{P N}\)

Answer:
\(\overline{P N}\) is radius

Tell whether the common tangent is internal or external.

Question 7.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 252

Answer:
Internal common tangent

Question 8.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 253

Answer:
External common tangent

Points Y and Z are points of tangency. Find the value of the variable.

Question 9.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 254

Answer:
a = \(\frac { 3 ± 33 }{ 18 } \)

Explanation:
3a = 9a² – 30
9a² – 3a – 30 = 0
a = \(\frac { 3 ± 33 }{ 18 } \)

Question 10.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 255

Answer:
c = 2

Explanation:
2c² + 9c + 6 = 9c + 14
2c² – 8 = 0
c² – 4 = 0
c = 2

Question 11.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 256

Answer:
r = 12

Explanation:
(3 + r)² = r² + 9²
9 + 6r + r² = r² + 81
6r = 72
r = 12

Question 12.
Tell whether \(\overline{B D}\) is tangent to ⊙C. Explain.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 257

Answer:
\(\overline{B D}\) is not tangent to ⊙C

Explanation:
52² = 10² + 48²
2704 = 100 + 2304
So, \(\overline{B D}\) is not tangent to ⊙C

10.2 Finding Arc Measures

Use the diagram above to find the measure of the indicated arc.

Question 13.
\(\widehat{K L}\)

Answer:
\(\widehat{K L}\) = 100°

Explanation:
\(\widehat{K L}\) = ∠KPL = 100°

Question 14.
\(\widehat{L M}\)

Answer:
\(\widehat{L M}\) = 60°

Explanation:
\(\widehat{L M}\) = 180° – 120°
= 60°

Question 15.
\(\widehat{K M}\)

Answer:
\(\widehat{K M}\) = 160°

Explanation:
\(\widehat{K M}\) = 100° + 60°
= 160°

Question 16.
\(\widehat{K N}\)

Answer:
\(\widehat{K N}\) = 80°

Explanation:
\(\widehat{K N}\) = 360 – (120 + 100 + 60)
= 360 – 280 = 80°

Tell whether the red arcs are congruent. Explain why or why not.

Question 17.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 258

Answer:
\(\widehat{S T}\), \(\widehat{Y Z}\) are not congruent.

Explanation:
\(\widehat{S T}\), \(\widehat{Y Z}\) are not congruent. Because both arcs are from different circles and having different radii.

Question 18.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 259

Answer:
\(\widehat{A B}\), \(\widehat{E F}\) are congruent.

Explanation:
\(\widehat{A B}\), \(\widehat{E F}\) are congruent. Because those circles have same radii.

10.3 Using Chords

Find the measure of \(\widehat{A B}\).

Question 19.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 260

Answer:
\(\widehat{A B}\) = 61°

Explanation:
\(\widehat{A B}\) = 61°
If ED = AB, then \(\widehat{A B}\) = \(\widehat{E D}\)

Question 20.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 261

Answer:
\(\widehat{A B}\) = 65°

Explanation:
\(\widehat{A B}\) = \(\widehat{A D}\)
So, \(\widehat{A B}\) = 65°

Question 21.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 262

Answer:
\(\widehat{A B}\) = 91°

Explanation:
\(\widehat{A B}\) = \(\widehat{E D}\)
So, \(\widehat{A B}\) = 91°

Question 22.
In the diagram. QN = QP = 10, JK = 4x, and LM = 6x – 24. Find the radius of ⊙Q.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 263

Answer:
The radius of ⊙Q is 26.

Explanation:
6x – 24 = 4x
6x – 4x = 24
2x = 24
x = 12
ML = 6(12) – 24 = 48
JN = \(\frac { 48 }{ 2 } \) = 24
JQ² = JN² + NQ²
= 24² + 10² = 576 + 100
JQ = 26
Therefore, the radius of ⊙Q is 26

10.4 Inscribed Angles and Polygons

Find the value(s) of the variable(s).

Question 23.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 264

Answer:
x° = 80°

Explanation:
x° = 2 • 40° = 80°

Question 24.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 265

Answer:
q° = 100°, r° = 20°

Explanation:
q° + 80° = 180°
q° = 100°
4r° + 100 = 180°
4r° = 80°
r° = 20°

Question 25.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 266

Answer:
d° = 5°

Explanation:
14d° = 70°
d° = 5°

Question 26.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 267

Answer:
y° = 30°, z° = 10°

Explanation:
3y° = 90°
y° = 30°
50° + 90° + 4z° = 180°
4z° = 40°
z° = 10°

Question 27.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 268

Answer:
m° = 44°
n° = 39°

Explanation:
m° = 44°
n° = 39°

Question 28.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 269

Answer:
c° = 28°

Explanation:
c° = ½ • 56 = 28

10.5 Angle Relationships in Circles

Find the value of x.

Question 29.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 270

Answer:
x° = 250°

Explanation:
x° = 250°

Question 30.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 271

Answer:
x° = 106°

Explanation:
x° = ½(152 + 60)
= ½(212) = 106°

Question 31.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 272

Answer:
x° = 28°

Explanation:
x° = ½(96 – 40)
= ½(56) = 28°

Question 32.
Line l is tangent to the circle. Find m\(\widehat{X Y Z}\).
Big Ideas Math Geometry Answer Key Chapter 10 Circles 273

Answer:
m\(\widehat{X Y Z}\) = 240°

Explanation:
m\(\widehat{X Y Z}\) = 2(120)
= 240°

10.6 Segment Relationships in Circles

Find the value of x.

Question 33.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 274

Answer:
x = 8

Explanation:
3 • x = 4 • 6
x = 8

Question 34.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 275

Answer:
x = 3

Explanation:
(x + 3) • x = (6 – x) • 2x
x + 3 = 12 – 2x
3x = 9
x = 3

Question 35.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 276

Answer:
x = 18

Explanation:
12² = 8 • x
144 = 8x
x = 18

Question 36.
A local park has a circular ice skating rink. You are standing at point A, about 12 feet from the edge of the rink. The distance from you to a point of tangency on the rink is about 20 feet. Estimate the radius of the rink.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 277

Answer:
Estimated radius of the rink is 10 ft

Explanation:
20² = 12 • (2r + 12)
400 = 24r + 144
256 = 24r
r = 10.66
Therefore, estimated radius of the rink is 10 ft

10.7 Circles in the Coordinate Plane

Write the standard equation of the circle shown.

Question 37.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 278

Answer:
(x – 4)² + (y + 1)² = 12.25

Explanation:
(x – 4)² + (y + 1)² = 3.5²
(x – 4)² + (y + 1)² = 12.25

Question 38.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 279

Answer:
(x – 8)² + (y – 5)² = 36

Explanation:
(x – 8)² + (y – 5)² = 6²
(x – 8)² + (y – 5)² = 36

Question 39.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 280

Answer:
x² + y² = 4

Explanation:
(x – 0)² + (y – 0)² = 2²
x² + y² = 4

Write the standard equation of the circle with the given center and radius.

Question 40.
center: (0,0), radius: 9

Answer:
x² + y² = 81

Explanation:
(x – 0)² + (y – 0)² = 9²
x² + y² = 81

Question 41.
center: (- 5, 2), radius: 1.3

Answer:
(x + 5)² + (y – 2)² = 1.69

Explanation:
(x + 5)² + (y – 2)² = 1.3²
(x + 5)² + (y – 2)² = 1.69

Question 42.
center: (6, 21), radius: 4

Answer:
(x – 6)² + (y – 21)² = 16

Explanation:
(x – 6)² + (y – 21)² = 4²
(x – 6)² + (y – 21)² = 16

Question 43.
center: (- 3, 2), radius: 16

Answer:
(x + 3)² + (y – 2)² = 256

Explanation:
(x + 3)² + (y – 2)² = 16²
(x + 3)² + (y – 2)² = 256

Question 44.
center: (10, 7), radius: 3.5

Answer:
(x – 10)² + (y – 7)² = 12.25

Explanation:
(x – 10)² + (y – 7)² = 3.5²
(x – 10)² + (y – 7)² = 12.25

Question 45.
center: (0, 0), radius: 5.2

Answer:
x² + y² = 27.04

Explanation:
(x – 0)² + (y – 0)² = 5.2²
x² + y² = 27.04

Question 46.
The point (- 7, 1) is on a circle with center (- 7, 6). Write the standard equation of the circle.

Answer:
(x + 7)² + (y – 6)² = 25

Explanation:
r² = (-7 + 7)² + (6 – 1)²
= 5²
r = 5
And, centre is (-7, 6)
The standard equation of a circle is (x – (-7))² + (y – 6)² = 5²
(x + 7)² + (y – 6)² = 25

Question 47.
The equation of a circle is x2 + y2 – 12x + 8y + 48 = 0. Find the center and the radius of the circle. Then graph the circle.

Answer:
The radius of the circle is 2, the centre is (6, -4)

Explanation:
x2 + y2 – 12x + 8y + 48 = 0
x² – 12x + 36 + y² + 8y + 16 = 4
(x – 6)² + (y + 4)² = 2²
So, the radius of the circle is 2, the centre is (6, -4)
Big Ideas Math Geometry Answers Chapter 10 Circles 23

Question 48.
Prove or disprove that the point (4, – 3) lies on the circle centred at the origin and containing
the point (- 5, 0).

Answer:
The point (4, – 3) lies on the circle.

Explanation:
Use the distance formula to find the radius of the circle with cente (0, 0) and a point (-5, 0)
r = √(-5 – 0)² + (0 – 0)² = 5
The distance from the point (4, -3) to the center (0, 0)
d = √(4 – 0)² + (-3 – 0)² = √(16 +9) = 5
Since the radius of the circle is 5, the point lies on the circle.

Circles Chapter Test

Find the measure of each numbered angle in ⊙P. Justify your answer.

Question 1.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 281

Answer:
m∠1 = 72.5°
m∠2 = 145°

Explanation:
m∠1 = \(\frac { 145 }{ 2 } \)
= 72.5°
m∠2 = 145°

Question 2.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 282

Answer:
m∠1 = 60°, m∠2 = 90°

Explanation:
A tangent is perpendicualr to diameter. So, m∠2 = 90°
m∠1 = 60°

Question 3.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 283

Answer:
m∠1 = 48°

Explanation:
m∠1 = \(\frac { 96 }{ 2 } \) = 48°

Question 4.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 284

Answer:

Use the diagram.

Big Ideas Math Geometry Answer Key Chapter 10 Circles 285

Question 5.
AG = 2, GD = 9, and BG = 3. Find GF.

Answer:

Question 6.
CF = 12, CB = 3, and CD = 9. Find CE.

Answer:

Question 7.
BF = 9 and CB = 3. Find CA

Answer:

Question 8.
Sketch a pentagon inscribed in a circle. Label the pentagon ABCDE. Describe the relationship between each pair of angles. Explain your reasoning.

a. ∠CDE and ∠CAE

Answer:

b. ∠CBE and ∠CAE

Answer:

Find the value of the variable. Justify your answer.

Question 9.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 286

Answer:
x = 5

Explanation:
5x – 4 = 3x + 6
5x – 3x = 6 + 4
2x = 10
x = 5

Question 10.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 287

Answer:
r = 9

Explanation:
(6 + r)² = 12² + r²
36 + 12r + r² = 144 + r²
12r = 108
r = 9

Question 11.
Prove or disprove that the point (2√2, – 1) lies on the circle centered at (0, 2) and containing the point (- 1, 4).

Answer:
Disproved

Explanation:
We consider the circle centred at the A(0, 2) and containing the point B(-1, 4).
AB = √(-1 – 0)² + (4 – 2)² = √1 + 4 = √5
The distance between centre A(0, 2) and P(2√2, – 1) is
AP = √(2√2 – 0)² + (-1 – 2)² = √8 + 9 = √17
AB ≠ AP
So, the point (2√2, – 1) dies not lie on the circle.

Prove the given statement.

Question 12.
\(\widehat{S T} \cong \widehat{R Q}\)
Big Ideas Math Geometry Answer Key Chapter 10 Circles 288

Answer:

Question 13.
\(\widehat{J M} \cong \widehat{L M}\)
Big Ideas Math Geometry Answer Key Chapter 10 Circles 289

Answer:

Question 14.
\(\widehat{D G} \cong \widehat{F G}\)
Big Ideas Math Geometry Answer Key Chapter 10 Circles 290

Answer:

Question 15.
A bank of lighting hangs over a stage. Each light illuminates a circular region on the stage. A coordinate plane is used to arrange the lights, using a corner of the stage as the origin. The equation (x – 13)2 + (y – 4)2 = 16 represents the boundary of the region illuminated by one of the lights. Three actors stand at the points A(11, 4), B(8, 5), and C(15, 5). Graph the given equation. Then determine which actors are illuminated by the light.

Answer:
The equation (x – 13)² + (y – 4)²= 16 represents the standard equation of the circle with center (13, 4) and radius 4
Graph the circle with center S(13, 4), radius 4. Then graph the points A(11,4), B (8, 5), C(15,5) which represents the places where the actors stand.
Big Ideas Math Geometry Answers Chapter 10 Circles 24
From the graph, we can see that points A and C inside the circle and point B is outside the circle. Therefore, actors who stand at points A and C are illuminated by the light

Question 16.
If a car goes around a turn too quickly, it can leave tracks that form an arc of a circle. By finding the radius of the circle, accident investigators can estimate the speed of the car.
Answer:

Big Ideas Math Geometry Answer Key Chapter 10 Circles 291

a. To find the radius, accident investigators choose points A and B on the tire marks. Then the investigators find the midpoint C of \(\overline{A B}\). Use the diagram to find the radius r of the circle. Explain why this method works.

Answer:
The radius r of the circle = 155.71 ft

Explanation:
Given that,
AC = 130 ft, CD = 70 ft
CE = (r – 70) ft
r² = a² + b²
r²= 130²+ (r – 70)²
r² = 16900 + r² – 140r + 4900
140r = 21,800
r = 155.71 ft

b. The formula S = 3.87√fr can be used to estimate a car’s speed in miles per hour, where f is the coefficient of friction and r is the radius of the circle in feet. If f = 0.7, estimate the car’s speed in part (a).

Answer:
The estimated car’s speed is 39.67 miles per hour

Explanation:
S = 3.87√fr
S = 3.8 √(0.7 x 155.71)
S = 3.8 √108.997
S = 3.8 x 10.44
S = 39.67

Circles Cumulative Assessment

Question 1.
Classify each segment as specifically as possible.
Big Ideas Math Geometry Solutions Chapter 10 Circles 292
a. \(\overline{B G}\)

Answer:
\(\overline{B G}\) is a chord

b. \(\overline{C D}\)

Answer:
\(\overline{C D}\) is radius.

c. \(\overline{A D}\)

Answer:
\(\overline{A D}\) is diameter.

d. \(\overline{F E}\)

Answer:
\(\overline{F E}\) is a chord

Question 2.
Copy and complete the paragraph proof.

Big Ideas Math Geometry Solutions Chapter 10 Circles 293

Given Circle C with center (2, 1) and radius 1,
Circle D with center (0, 3) and radius 4
Prove Circle C is similar to Circle D.

Map Circle C to Circle C’ by using the _________ (x, y) → _________ so that Circle C’ and Circle D have the same center at (____, _____). Dilate Circle C’ using a cellIer of dilation (_____, _____) and a scale factor of _____ . Because there is a _________ transformation that maps Circle C to Circle D, Circle C is __________ Circle D.

Answer:
Map Circle C to Circle C’ by using the scale factor (x, y) → (0, 3) so that Circle C’ and Circle D have the same center at (2, 1). Dilate Circle C’ using a cellIer of dilation (2, 1) and a scale factor of circles. Because there is a transformation that maps Circle C to Circle D, Circle C is similar to Circle D.

Question 3.
Use the diagram to write a proof.
Big Ideas Math Geometry Solutions Chapter 10 Circles 294
Given ∆JPL ≅ ∆NPL
\(\overline{P K}\) is an altitude of ∆JPL
\(\overline{P M}\) is an altitude ∆NPL
Prove ∆PKL ~ ∆NMP

Answer:
∆JPL is similar to ∆NPL and PK is the altitude of ∆JPL and PM is an altitude of ∆NPL
Altitude is a line drawn from one vertex to the opposite site. It is perpendicular to the side.
So, ∆PKL is similar to ∆NMP

Question 4.
The equation of a circle is x² + y² + 14x – 16y + 77 = 0. What are the center and radius of the circle?
(A) center: (14, – 16). radius: 8.8
(B) center: (- 7, 8), radius: 6
(C) center (- 14, 16), radius: 8.8
(D) center: (7, – 8), radius: 5.2

Answer:
(B) center: (- 7, 8), radius: 6

Explanation:
x² + y² + 14x – 16y + 77 = 0
x² + 14x + 49 + y² – 16y + 64 = 36
(x + 7)² + (y – 8)² = 6²
So, the centre is (-7, 8) and radius is 6.

Question 5.
The coordinates of the vertices of a quadrilateral are W(- 7, – 6), X(1, – 2), Y(3, – 6) and Z(- 5, – 10). Prove that quadrilateral WXYZ is a rectangle.

Answer:
Proved

Explanation:
Find the distance of WY and ZX
WY = √(-7 – 3)² + (-6 + 6)² = √(-10)² = 10
ZX = √(1 + 5)² + (-2 + 10)² = √6² + 8² = 10
WY = ZX, the diagonals are congruent
Use the slope formula to find the slopes of diagonals
Slope of WY = \(\frac { -6 + 6 }{ -7 – 3 } \) = 0
Slope of ZX = \(\frac { -2 + 10 }{ 1 + 5 } \) = \(\frac { 4 }{ 3 } \)
Because the product of slopes of diagonals is 0, the diagonals are not perpendicular
Therefore, the quadrilateral WXYZ is a rectangle.

Question 6.
Which angles have the same measure as ∠ACB? Select all that apply.
im – 295
∠DEF ∠JGK ∠KGL ∠LGM ∠MGJ
∠QNR ∠STV ∠SWV ∠VWU ∠XYZ
Answer:
∠VWU

Question 7.
Classify each related conditional statement based on the conditional statement
“If you are a soccer player. then you are an athlete.”
a. If you are not a soccer player, then you are not an athlete.

Answer:
False

b. If you are an athlete, then you are a soccer player.

Answer:
False

c. You are a soccer player if and only if you are an athlete.

Answer:
True

d. If you are not an athlete, then you are not a soccer player.

Answer:
False

Question 8.
Your friend claims that the quadrilateral shown can be inscribed in a circle. Is your friend correct? Explain our reasoning.
Big Ideas Math Geometry Solutions Chapter 10 Circles 296

Answer:
If the sum of any two angles is 180°, then the quadrilateral is inscribed in a circle.
So, 70° + 110° = 180°, 110° + 70°= 180°
So, my friend is correct.