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Engage NY Eureka Math 8th Grade Module 2 Lesson 6 Answer Key

Eureka Math Grade 8 Module 2 Lesson 6 Exercise Answer Key

Exercises 1–3

Exercise 1.
Show that (C) is implied by equation (5) of Lesson 4 when m>0, and explain why (C) continues to hold even when
m=0.
Answer:
Equation (5) says for any numbers x, y, (y≠0) and any positive integer n, the following holds: (\(\frac{x}{y}\))ⁿ= \(\frac{x^{n}}{y^{n}}\). So,
\(\left(\frac{1}{x}\right)^{m}\) = \(\frac{1^{m}}{x^{m}}\) By (\(\frac{x}{y}\))ⁿ=\(\frac{x^{n}}{y^{n}}\) for positive integer n and nonzero y (5)
= \(\frac{1}{x^{m}}\) Because 1^m=1
If m= 0, then the left side is
(\(\frac{x}{y}\))^m=(\(\frac{x}{y}\))⁰
=1 By definition of x⁰,
and the right side is
\(\frac{1}{x^{m}}\) =\(\frac{1}{x^{0}}\)
= \(\frac{1}{1}\) By definition of x⁰
=1.

Exercise 2.
Show that (B) is in fact a special case of (11) by rewriting it as (x^m)^-1=x^(-1)m for any whole number m, so that if b=m (where m is a whole number) and a=-1, (11) becomes (B).
Answer:
(B) says x^-m= \(\frac{1}{x^{m}}\).
The left side of (B), x^-m is equal to x^(-1)m).
The right side of (B), \(\frac{1}{x^{m}}\), is equal to (x^m)^-1 by the definition of (x^m)^-1 in Lesson 5.
Therefore, (B) says exactly that (x^m)^-1=x^(-1)m.

Exercise 3.
Show that (C) is a special case of (11) by rewriting (C) as (x^-1)^m=x^m(-1) for any whole number m. Thus, (C) is the special case of (11) when b=-1 and a=m, where m is a whole number.
Answer:
(C) says (\(\frac{1}{x}\))^m=\(\frac{1}{x^{m}}\) for any whole number m.
The left side of (C) is equal to
(\(\frac{1}{x}\))^m =(x^-1)^m By definition of x^-1,
and the right side of (C) is equal to
\(\frac{1}{x^{m}}\) = x^-m By definition of x^-m,
and the latter is equal to x^m(-1). Therefore, (C) says (x^-1)^m=x^m(-1) for any whole number m.

Exercise 4.
Proof of Case (iii): Show that when a<0 and b≥0, (x^b)^a=x^ab is still valid. Let a=-c for some positive integer c. Show that the left and right sides of (x^b)^a=x^ab are equal.
The left side is
(x^b)^a=(x^b)^-c
= \(\frac{1}{\left(x^{b}\right)^{c}}\) for any whole number m (B)
= \(\frac{1}{x^{c b}}\) . By (x^m)ⁿ=x^mn for all whole numbers m and n (A)
The right side is
x^ab=x^(-c)b
=x^-(cb)
= \(\frac{1}{x^{c b}}\). By x^-m= \(\frac{1}{x^{m}}\) for any whole number m (B)
So, the two sides are equal.

Eureka Math Grade 8 Module 2 Lesson 6 Problem Set Answer Key

Question 1.
You sent a photo of you and your family on vacation to seven Facebook friends. If each of them sends it to five of their friends, and each of those friends sends it to five of their friends, and those friends send it to five more, how many people (not counting yourself) will see your photo? No friend received the photo twice. Express your answer in exponential notation.

Answer:

The total number of people who viewed the photo is (5⁰+5¹+5²+5³)×7.

Question 2.
Show directly, without using (11), that (1.27^-36)^-85=1.27^-36.85.
Answer:
(1.27^-36)^-85= (\(\frac{1}{1.27^{36}}\))^-85 By definition
= \(\frac{1}{\left(1.27^{36}\right)^{85}}\) By \(\left(\frac{1}{x}\right)\)^m = \(\frac{1}{x^{m}}\) for any whole number m(c)
= \(\frac{1}{1.27^{36 \cdot 85}}\) By (x^m)ⁿ = m^mn for whole numbers m and (7)
= 1.27^-36.85 By x^-m=\(\frac{1}{x^{m}}\) for any whole number m(B)

Question 3.
Show directly that (\(\frac{2}{13}\))^-127∙(\(\frac{2}{13}\))^-56=(\(\frac{2}{13}\))^-183.
Answer:

Question 4.
Prove for any nonzero number x, x^-127∙x^-56=x^-183.
Answer:
x^-127∙x^-56 =\(\frac{1}{x^{127}}\)∙\(\frac{1}{x^{56}}\) By definition
= \(\frac{1}{x^{127} \cdot x^{56}}\) By the product formula for complex fractions
=\(\frac{1}{x^{127+56}}\) By x^m∙xⁿ=x^m+n for whole numbers m and n (6)
=\(\frac{1}{x^{183}}\)
= x^-183 By x^-m= \(\frac{1}{x^{m}}\)^m for any whole number m (B)

Question 5.
Prove for any nonzero number x, x^-m∙x^-n=x^-m-n for positive integers m and n.
Answer:
x^-m∙x^-n= \(\frac{1}{x^{m}}\)∙\(\frac{1}{x^{n}}\) By definition
= \(\frac{1}{x^{m \cdot x^{n}}}\) By the product formula for complex fractions
= \(\frac{1}{x^{m+n}}\) By x^m∙xⁿ =x^m+n for whole numbers m and n(6)
= x^-(m+n) By x^-m = \(\frac{1}{x^{m}}\) for any whole number m(B)
= x^-m-n

Question 6.
Which of the preceding four problems did you find easiest to do? Explain.
Answer:
Students will likely say that x^-m∙x^-n=x^-m-n (Problem 5) was the easiest problem to do. It requires the least amount of writing because the symbols are easier to write than decimal or fraction numbers.

Question 7.
Use the properties of exponents to write an equivalent expression that is a product of distinct primes, each raised to an integer power.
Answer:

Eureka Math Grade 8 Module 2 Lesson 6 Exit Ticket Answer Key

Question 1.
Show directly that for any nonzero integer x, x^-5∙x^-7=x^-12.
Answer:
x^-5∙x^-7 =\(\frac{1}{x^{5}}\)\(\frac{1}{x^{7}}\) By x^-m= \(\frac{1}{x^{m}}\) for any whole number m (B)
= \(\frac{1}{x^{5} \cdot x^{7}}\) By the product formula for complex fractions
=\(\frac{1}{x^{5+7}}\) By x^m∙xⁿ=x^m+n for whole numbers m and n (6)
=\(\frac{1}{x^{12}}\)
= x^-12 By x^-m= \(\frac{1}{x^{m}}\) for any whole number m (B)

Question 2.
Show directly that for any nonzero integer x, (x^-2)^-3=x⁶.
Answer:
(x^-2)^-3 =\(\frac{1}{\left(x^{-2}\right)^{3}}\) By x^-m= \(\frac{1}{x^{m}}\) for any whole number m (B)
=\(\frac{1}{x^{-(2 \cdot 3)}}\) By case (ii) of (11)
=\(\frac{1}{x^{-6}}\)
= x⁶ By x^-m= \(\frac{1}{x^{m}}\) for any whole number m (B)

Engage NY Eureka Math Grade 6 Module 5 Lesson 8 Answer Key

Eureka Math Grade 6 Module 5 Lesson 8 Example Answer Key

Question 1.
Plot and connect the points A(3, 2), B(3, 7), and C(8, 2). Name the shape, and determine the area of the polygon.

Answer:

Right Triangle

A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) (5 units)(5 units)
A = \(\frac{1}{2}\) (25 units²)
A = 12.5 units²

Question 2.
Plot and connect the points E(-8, 8), F(-2, 5), and G(-7, 2). Then give the best name for the polygon, and determine the area.
Answer:

The shape is a triangle.

Area of Square
A = s²
A = (6 units)²
A = 36 units²

Area of Triangle 1

A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) (1 unit) (6 units)
A = \(\frac{1}{2}\) (6 units²)
A = 3 units²

Area of Triangle 2

A = \(\frac{1}{2}\)bh
A = \(\frac{1}{2}\) (6 units)(3 units)
A = \(\frac{1}{2}\) (18 units²)
A = 9 units²

Area of Triangle 3

A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) (5 units) (3 units)
A = \(\frac{1}{2}\) (15 units²)
A = 7.5 units²

Total Area of Triangle
A = 36 units² – 3 units² – 9 units² – 7.5 units²
A = 16.5 units²

Question 3.
Plot and connect the following points: K(-9, -7), L(-4, -2), M(-1, -5), and N(-5, -5). Give the best name for the polygon, and determine the area.
Answer:

This polygon has 4 sides and has no pairs of parallel sides.
Therefore, the best name for this shape is a quadrilateral.
To determine the area, I will separate the shape into two
triangles.
Area of Triangle 1
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) (6 units) (3 units)
A = \(\frac{1}{2}\) (18 units²)
A = 9 units²

Area of Triangle 2
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) (2 units) (2 units)
A = \(\frac{1}{2}\) (4 units²)
A = 2 units²

Total Area = 9 units² + 2 units²
Total Area = 11 units²

Question 4.
Plot and connect the following points: P(1, -4), Q(5, -2), R(9, -4), S(7, -8), and T(3, -8). Give the best name for the polygon, and determine the area.
Answer:

To determine the area, I will decompose this pentagon into four smaller shapes.

Area of Shape 1
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) (8 units)(2 units)
A = \(\frac{1}{2}\) (16 units²)
A = 8 units²

Area of Shape 2 and Shape 4
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) (4 units)(2 units)
A = \(\frac{1}{2}\) (8 units²)
A = 4 units²

Because there are two of the same triangle, that makes a total of 8 units².

Area of Shape 3
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) (4 units) (4 units)
A = 16 units²
Total Area = 8 units² + 8 units² + 16 units²
Total Area = 32 units²

Question 5.
Two of the coordinates of a rectangle are A (3, 7) and B (3, 2). The rectangle has an area of 30 square units. Give the possible locations of the other two vertices by identifying their coordinates. (Use the coordinate plane to draw and check your answer.)

Answer:

One possible location of the other two vertices is (9, 2) and (9, 7). Using these coordinates will result in a distance, or side length, of 6 units. Since the height is 5 units, 5 units × 6 units = 30 units².

Another possible location of the other two vertices is (-3, 2) and (-3, 7). Using these coordinates will result in a distance, or side length, of 6 units. Since the height is 5 units, 5 units × 6 units = 30 units².

Eureka Math Grade 6 Module 5 Lesson 8 Exercise Answer Key

Exercise:

For Exercises 1 and 2, plot the points, name the shape, and determine the area of the shape. Then write an expression that could be used to determine the area of the figure. Explain how each part of the expression corresponds to the situation.

Question 1.
A(4, 6), B(8, 6), C(10, 2), D(8, -3), E(5, -3), and F(2, 2)
Answer:

This shape is a hexagon.

Area of Shape 1
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) (2 units)(4 units)
A = \(\frac{1}{2}\) (8 units2)
A = 4 units²

Area of Shape 2
A = \(\frac{1}{2}\) bh
A = (1 unit) (4 units)
A = 4 units²

Area of Shape 3
A = \(\frac{1}{2}\) bh
A = (3 units) (9 units)
A = 27 units²

Area of Shape 4
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) (3 units)(5 units)
A = \(\frac{1}{2}\) (15 units2)
A = 7.5 units²

Area of Shape 5
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) (2 units)(9 units)
A = \(\frac{1}{2}\) (18 units2)
A = 9 units²

Total Area = 4 units² + 4 units² + 27 units² + 7.5 units² + 9 units²

Expression:
\(\frac{1}{2}\) (2) (4) + (1) (4) + (3) (9) + \(\frac{1}{2}\) (3)(5) +\(\frac{1}{2}\) (2)(9).
Each term represents the area of a section of the hexagon. They must be added together to get the total.

The first term is the area of triangle 1 on the left.
The second term is the area of rectangle 2.
The third term is the area of the large rectangle 3.
The fourth term is the area of triangle 4 on the left.
The fifth term is the area of triangle 5 on the right.

Question 2.
X (-9, 6), Y (-2, -1), and Z (-8, -7)
Answer:

This shape is a triangle.

Area of Outside Rectangle
A = lw
A = (7 units) (13 units)
A = 91 units²

Area of Triangle 1

A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) (7 units)(7 units)
A = \(\frac{1}{2}\) (49 units²)
A = 24.5 units²

Area of Triangle 2

A = \(\frac{1}{2}\)bh
A = \(\frac{1}{2}\) (6 units)(6 units)
A = \(\frac{1}{2}\) (36 units²)
A = 18 units²

Area of Triangle 3
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) (13 units)(1 unit)
A = \(\frac{1}{2}\) (13 units²)
A = 6.5 units²

Total Area = 91 units² – 24.5 units² – 18 units² – 6.5 units²
Total Area = 42 units²

Expression:

(7)(13) – \(\frac{1}{2}\) (7)(7) – \(\frac{1}{2}\) (6)(6) – \(\frac{1}{2}\) (13)(1)

The first term in the expression represents the area of the rectangle that goes around the outside of the triangle.
The next three terms represent the areas that need to be subtracted from the rectangle so that we are only left with the given triangle.
The second term is the area of the top right triangle.
The third term is the area of the bottom right triangle.
The fourth term is the area of the triangle on the left.

Question 3.
A rectangle with vertices located at (-3, 4) and (5, 4) has an area of 32 square units. Determine the location of the other two vertices.
Answer:
The other two points could be located at (-3, 8) and (5, 8) or (-3, 0) and (5, 0).

Question 4.
Challenge: A triangle with vertices located at (-2, -3) and (3, -3) has an area of 20 square units. Determine one possible location of the other vertex.
Answer:
Answers will vary. Possible solutions include points that are 8 units from the base. (-2, 5) or (3, -11).

Eureka Math Grade 6 Module 5 Lesson 8 Problem Set Answer Key

Plot the points for each shape, determine the area of the polygon, and then write an expression that could be used to determine the area of the figure. Explain how each part of the expression corresponds to the situation.

Question 1.
A(1, 3), B(2, 8), C(8, 8), D(10, 3), and E(5, -2)

Answer:

Area of Triangle 1
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) (6 units) (5 units)
A = \(\frac{1}{2}\) (30 units²)
A = 15 units²

Area of Triangle 2
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) (9 units)(5 units)
A = \(\frac{1}{2}\) (45 units²)
A = 22.5 units²

Area of Triangle 3
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) (9 units)(5 units)
A = \(\frac{1}{2}\) (45 units²)
A = 22.5 units²

Pentagon total area = 15 units² + 22.5 units² + 22.5 units²
Total Area = 60 units²

Expression:
\(\frac{1}{2}\) (6)(5) + \(\frac{1}{2}\) (9)(5) + \(\frac{1}{2}\) (9)(5)

Each term in the expression represents the area of one of the triangular pieces that fits inside the pentagon. They are all added together to form the complete figure.

Question 2.
X(-10, 2), Y(-3, 6), and Z(-6, -5)

Answer:

Area of Outside Rectangle
A = lw
A = (11 units) (7 units)
A = 77 units²

Area of Top Triangle
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) (7 units)(4 units)
A = \(\frac{1}{2}\) (28 units²)
A = 14 units²

Area of Bottom Left Triangle
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) (4 units)(7 units)
A = \(\frac{1}{2}\) (28 units²)
A = 14 units²

Area of Bottom Right Triangle
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) (3 units)(11 units)
A = \(\frac{1}{2}\) (33 units²)
A = 16.5 units²

Area of center triangle = 77 units² – 14 units² – 14 units² – 16.5 units²
Area of center triangle = 32.5 units²
Expression:
(11)(7) – \(\frac{1}{2}\) (7)(4) – \(\frac{1}{2}\) (4)(7) – \(\frac{1}{2}\) (3)(11)
The first term in the expression represents the area of the rectangle that would enclose the triangle. Then the three terms after represent the areas of the triangles that need to be removed from the area of the rectangle to find the area of the given triangle.

Question 3.
E(5, 7), F(9, -5), and G(1, -3)

Answer:

Area of Triangle on the Left
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) (11 units)(4 units)
A = 22 units

Area of Triangle on the Right
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) (11 units)(4 units)
A = 22 units²
Total Area = 22 units² + 22 units² = 44 units²

Expression:
\(\frac{1}{2}\) (11)(4) + \(\frac{1}{2}\) (11)(4)
Each term in the expression represents the area of a triangle that makes up the total area. The first term is the area of the triangle on the left, and the second term is the area of the triangle on the right.

Question 4.
Find the area of the triangle In Problem 3 using a different method. Then, compare the expressions that can be used for both solutions in Problems 3 and 4.
Answer:

Area of Rectangle
A = lw
A = (12 units) (8 units)
A = 96 units²

Area of Triangle on Top Left
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) (4 units)(10 units)
A = 20 units²

Area of Triangle on Bottom Left
A =\(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) (8 units)(2 units)
A = 8 units²

Area of Triangle on Right
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) (4 units)(12 units)
A = 24 units²

Total Area = 96 units² – 20 units² – 8 units² – 24 units²
Total Area = 44 units²

Expression: (12)(8) – \(\frac{1}{2}\) (4)(10) – \(\frac{1}{2}\) (8)(2) – \(\frac{1}{2}\) (4)(12)
The first term in the expression is the area of a rectangle around the outside of the figure. Then we subtracted all of the extra areas with the next three terms.

The two expressions are different because of the way we divided up the figure. In the first expression, we split the shape into two triangles that had to be added together to get the whole. In the second expression, we enclosed the triangle inside a new figure and then had to subtract the extra area.

Question 5.
Two vertices of a rectangle are (8, -5) and (8, 7). If the area of the rectangle is 72 square units, name the possible location of the other two vertices.
Answer:
(2, -5) and (2, 7) or (14, -5) and (14, 7)

Question 6.
A triangle with two vertices located at (5, -8) and (5, 4) has an area of 48 square units. Determine one possible location of the other vertex.
Answer:
Answers will vary. Possible solutions include points that are 8 units from the base such as (13, -2) or (-3, -2).

Eureka Math Grade 6 Module 5 Lesson 8 Exit Ticket Answer Key

Question 1.
Determine the area of both polygons on the coordinate plane, and explain why you chose the methods you used. Then write an expression that could be used to determine the area of the figure. Explain how each part of the expression corresponds to the situation.

Answer:

Methods to calculate the answer will vary.

#1 Area of shape a
A = lw
A = (2 units) (6 units)
A = 12 units²

Area of shape b
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) (4 units)(2 units)
A = \(\frac{1}{2}\) (8 units²)
A = 4 units²

Total Area = 12 units² + 4 units² = 16 units²
Explanations will vary depending on the method chosen.

Expression: (2)(6) + latex]\frac{1}{2}[/latex] (4)(2)

The first term represents the area of the rectangle on the left, which makes up part of the figure.
The second term represents the area of the triangle on the right that completes the figure.

#2 Area of outside rectangle
A = lw
A = (7 units) (6 units)
A = 42 units²

Area of shape c
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) (6 units)(3 units)
A = \(\frac{1}{2}\) (18 units²)
A = 9 units²

Area of shape d
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) (7 units)(3 units)
A = \(\frac{1}{2}\) (21 units²)
A = 10.5 units²

Area of shape e
A = \(\frac{1}{2}\) bh
A = \(\frac{1}{2}\) (1 units)(6 units)
A = \(\frac{1}{2}\) (6 units²)
A = 3 units²

Total Area = 42 units² – 9 units² – 10.5 units² – 3 units²
Total Area = 19.5 units²
Explanations will vary depending on method chosen.

Expression:
(7)(6) – \(\frac{1}{2}\) (6)(3) – \(\frac{1}{2}\) (7)(3) – \(\frac{1}{2}\) (1)(6)

The first term in the expression is the area of a rectangle that goes around the triangle.
Each of the other terms represents the triangles that need to be subtracted from the rectangle so that we are left with just the figure in the center.

Engage NY Eureka Math Grade 6 Module 5 Lesson 7 Answer Key

Eureka Math Grade 6 Module 5 Lesson 7 Example Answer Key

Question 1.
Determine the lengths of the given line segments by determining the distance between the two endpoints.

Answer:

Eureka Math Grade 6 Module 5 Lesson 7 Exercise Answer Key

Question 1.
complete the table using the diagram on the coordinate plane.

Answer:

Extension:

For each problem below, write the coordinates of two points that are S units apart with the segment connecting these points having the following characteristics.

Question a.
The segment is vertical.
Answer:
Answers may vary. One possible solution is (2, 1) and (2,6).

Question b.
The segment intersects the x-axis.
Answer:
Answers may vary. One possible solution is (3,-4) and (3, 1).

Question c.
The segment intersects the y-axis.
Answer:
Answers may wary. One possible solution is (-4, 3) and (1, 3).

Question d.
The segment is vertical and lies above the x-axis.
Answer:
Answers may wary. One possible solution is (-3, 5) and (-3, 10).

Eureka Math Grade 6 Module 5 Lesson 7 Problem Set Answer Key

Question 1.
Given the pairs of points, determine whether the segment that joins them is horizontal, vertical, or neither.
a. X(3, 5) and Y(-2, 5) _____
Answer:
Horizontal

b. M(-4, 9) and N(4, -9) ____
Answer:
Neither

c. E(-7, 1) and F(-7, 4) _____
Answer:
Vertical

Question 2.
Complete the table using absolute value to determine the lengths of the line segments.

Answer:

Question 3.
Complete the table using the diagram and absolute value to determine the lengths of the line segments.

Answer:

Question 4.
Complete the table using the diagram and absolute value to determine the lengths of the line segments.

Answer:

Question 5.
Name two points in different quadrants that form a vertical line segment that is 8 units in length.
Answer:
Answers will vary. One possible solution is (2, 5) and (2, -3).

Question 6.
Name two points in the same quadrant that form a horizontal line segment that is 5 units in length.
Answer:
Answers will vary. One possible solution is (-4, -11) and (-9, -11).

Eureka Math Grade 6 Module 5 Lesson 7 Exit Ticket Answer Key

Question 1.
Use absolute value to show the lengths of \(\overline{A B}\), \(\overline{B C}\), \(\overline{C D}\), \(\overline{D E}\),and \(\overline{E F}\).

Answer:

Eureka Math Grade 6 Module 5 Lesson 7 Addition of Decimals Answer Key

Addition of Decimals II – Round 1

Directions: Evaluate each expression.

Question 1.
2.5 + 4 =
Answer:
6.5

Question 2.
2.5 + 0.4 =
Answer:
2.9

Question 3.
2.5 + 0.04 =
Answer:
2.54

Question 4.
2.5 + 0.004 =
Answer:
2.504

Question 5.
2.5 + 0.0004 =
Answer:
2.5004

Question 6.
6 + 1.3 =
Answer:
7.3

Question 7.
0.6 + 1.3 =
Answer:
1.9

Question 8.
0.06 + 1.3 =
Answer:
1.36

Question 9.
0.006 + 1.3 =
Answer:
1.306

Question 10.
0.0006 + 1.3 =
Answer:
1.3006

Question 11.
0.6 + 13 =
Answer:
13.6

Question 12.
7 + 0.2 =
Answer:
7.2

Question 13.
0.7 + 0.02 =
Answer:
0.72

Question 14.
0.07 + 0.2 =
Answer:
0.27

Question 15.
0.7 + 2 =
Answer:
2.7

Question 16.
7 + 0.02 =
Answer:
7.02

Question 17.
6 + 0.3 =
Answer:
6.3

Question 18.
0.6 + 0.03 =
Answer:
0.63

Question 19.
0.06 + 0.3 =
Answer:
0.36

Question 20.
0.6 + 3 =
Answer:
3.6

Question 21.
6 + 0.03 =
Answer:
6.03

Question 22.
0.6 + 0.3 =
Answer:
0.9

Question 23.
4.5 + 3.1 =
Answer:
7.6

Question 24.
4.5 + 0.31 =
Answer:
4.81

Question 25.
4.5 + 0.031 =
Answer:
4.531

Question 26.
0.45 + 0.031 =
Answer:
0.481

Question 27.
0.045 + 0.031 =
Answer:
0.076

Question 28.
12 + 0.36 =
Answer:
12.36

Question 29.
1.2 + 3.6 =
Answer:
4.8

Question 30.
1.2 + 0.36 =
Answer:
1.56

Question 31.
1.2 + 0.036 =
Answer:
1.236

Question 32.
0.12 + 0.036 =
Answer:
0.156

Question 33.
0.012 + 0.036 =
Answer:
0.048

Question 34.
0.7 + 3 =
Answer:
3.7

Question 35.
0.7 + 0.3 =
Answer:
1

Question 36.
0.07 + 0.03 =
Answer:
0.1

Question 37.
0.007 + 0.003 =
Answer:
0.01

Question 38.
5 + 0.5 =
Answer:
5.5

Question 39.
0.5 + 0.5 =
Answer:
1

Question 40.
0.05 + 0.05 =
Answer:
0.1

Question 41.
0.005 + 0.005 =
Answer:
0.01

Question 42.
0.11 + 19 =
Answer:
19.11

Question 43.
1.1 + 1.9 =
Answer:
3

Question 44.
0.11 + 0.19 =
Answer:
0.3

Addition of Decimals II – Round 2

Directions: Evaluate each expression.

Question 1.
7.4 + 3 =
Answer:
10.4

Question 2.
7.4 + 0.3 =
Answer:
7.7

Question 3.
7.4 + 0.03 =
Answer:
7.43

Question 4.
7.4 + 0.003 =
Answer:
7.403

Question 5.
7.4 + 0.0003 =
Answer:
7.4003

Question 6.
6 + 2.2 =
Answer:
8.2

Question 7.
0.6 + 2.2 =
Answer:
2.8

Question 8.
0.06 + 2.2 =
Answer:
2.26

Question 9.
0.006 + 2.2 =
Answer:
2.206

Question 10.
0.0006 + 2.2 =
Answer:
2.2006

Question 11.
0.6 + 2.2 =
Answer:
2.8

Question 12.
7 + 0.8 =
Answer:
7.8

Question 13.
0.7 + 0.08 =
Answer:
0.78

Question 14.
0.07 + 0.8 =
Answer:
0.87

Question 15.
0.7 + 8 =
Answer:
8.7

Question 16.
7 + 0.08 =
Answer:
7.08

Question 17.
0.5 + 0.4 =
Answer:
0.9

Question 18.
0.5 + 0.04 =
Answer:
0.54

Question 19.
0.05 + 0.4 =
Answer:
0.45

Question 20.
0.5 + 4 =
Answer:
4.5

Question 21.
5 + 0.04 =
Answer:
5.04

Question 22.
5 + 0.4 =
Answer:
5.4

Question 23.
3.6 + 2.3 =
Answer:
5.9

Question 24.
3.6 + 0.23 =
Answer:
3.83

Question 25.
3.6 + 0.023 =
Answer:
3.623

Question 26.
0.36 + 0.023 =
Answer:
0.383

Question 27.
0.036 + 0.023 =
Answer:
0.059

Question 28.
0.13 + 56 =
Answer:
56.13

Question 29.
1.3 + 5.6 =
Answer:
6.9

Question 30.
1.3 + 0.56 =
Answer:
1.86

Question 31.
1.3 + 0.056 =
Answer:
1.356

Question 32.
0.13 + 0.056 =
Answer:
0.186

Question 33.
0.013 + 0.056 =
Answer:
0.069

Question 34.
2 + 0.8 =
Answer:
2.8

Question 35.
0.2 + 0.8 =
Answer:
1

Question 36.
0.02 + 0.08 =
Answer:
0.1

Question 37.
0.002 + 0.008 =
Answer:
0.01

Question 38.
0.16 + 14 =
Answer:
14.16

Question 39.
1.6 + 1.4 =
Answer:
3

Question 40.
0.16 + 0.14 =
Answer:
0.3

Question 41.
0.016 + 0.014 =
Answer:
0.03

Question 42.
15 + 0.15 =
Answer:
15.15

Question 43.
1.5 + 1.5 =
Answer:
3

Question 44.
0.15 + 0.15 =
Answer:
0.3

Engage NY Eureka Math 6th Grade Module 3 Lesson 3 Answer Key

Eureka Math Grade 6 Module 3 Lesson 3 Example Answer Key

Example 1.
A Look at Sea Level
The picture below shows three different people participating in activities at three different elevations. With a partner, discuss what you see. What do you think the word elevation means in this situation?

Answer:
Pose questions to the class, and define elevation. Students gain additional practice with elevation by completing Exercise 1 independently.
Possible discussion questions:
→ Looking at the picture, if you were to draw a vertical number line to model elevation, which person’s elevation do you think would be at zero? Explain.
Sea level should represent an elevation of zero. So, the person sailing would be at zero because he is sailing on the surface of the water, which is neither above nor below the surface. On a number line, zero is the point or number separating positive and negative numbers.

→ On the same vertical number line, which person’s elevation would be represented above zero?
The elevation of the person hiking would be above zero because she is moving higher above the water.
On a vertical number line, this is represented by a positive value above zero because she is above the surface.

→ On the same vertical number line, which person’s elevation do you think would be below zero?
The elevation of the person scuba diving would be below zero because he is swimming below the surface of the water. On a vertical number line, this is represented by a negative value below zero because he is below the surface.

→ What does zero represent in this situation?
Zero represents the top of the water (the water’s surface).

→ In this example, which numbers correspond to elevations above sea level?
Above sea level means to be above zero, which are positive numbers.

→ In this example, which numbers correspond to elevations below sea level?
Below sea level means to be below zero, which are negative numbers.

→ On a number line, what does it mean to be at sea level?
To be at zero means to be at sea level.

→ Elevation is the height of a person, place, or thing above or below a certain reference point. In this case, what is the reference point?
The reference point is sea level.

Eureka Math Grade 6 Module 3 Lesson 3 Exercise Answer Key

Refer back to Example 1. Use the following information to answer the questions.
→ The scuba diver is 30 feet below sea level.
→ The sailor is at sea level.
→ The hiker is 2 miles (10,560 feet) above sea level.

Exercises 1.
Write an integer to represent each situation.
Answer:
Scuba Diver: – 30
Sailor: 0
Hiker: 2 (to represent the elevation in miles) or 10, 560 (to represent the elevation in feet)

Exercises 2.
Use an appropriate scale to graph each of the following situations on the number line to the right. Also, write an Integer to represent both situations.

a. A hiker is 15 feet above sea level.
Answer:
15

b. A diver is 20 feet below sea level.
Answer:
– 20

Exercises 3.
For each statement, there are two related statements: (i) and (ii). Determine which related statement ((i) or (ii)) is expressed correctly, and circle it. Then, correct the other related statement so that both parts, (i) and (ii), are stated correctly.
a. A submarine is submerged 800 feet below sea level.
i. The depth of the submarine is – 800 feet below sea level.
Answer:
The depth of the submarine is 800 feet below sea level.

ii. 800 feet below sea level can be represented by the integer – 800.
Answer:

b. The elevation of a coral reef with respect to sea level is given as – 150 feet.
i. The coral reef is 150 feet below sea level.
Answer:

ii. The depth of the coral reef is – 150 feet below sea level.
Answer:
The depth of the coral reef is 150 feet below sea level.

Eureka Math Grade 6 Module 3 Lesson 3 Problem Set Answer Key

Question 1.
Write an integer to match the following descriptions.
a. A debit of $40
Answer:
– 40

b. A deposit of $225
Answer:
225

c. 14,000 feet above sea level
Answer:
14,000

d. A temperature increase of 40°F
Answer:
40

e. A withdrawal of $225
Answer:
– 225

f. 14,000 feet below sea level
Answer:
– 14,000

For Problems 2 – 4, read each statement about a real-world situation and the two related statements in parts (a) and (b) carefully. Circle the correct way to describe each real-world situation; possible answers include either (a), (b), or both (a) and (b).

Question 2.
A whale is 600 feet below the surface of the ocean.
a. The depth of the whale is 600 feet from the ocean’s surface.
b. The whale is – 600 feet below the surface of the ocean.
Answer:
a. The depth of the whale is 600 feet from the ocean’s surface.

Question 3.
The elevation of the bottom of an iceberg with respect to sea level is given as – 125 feet.
a. The iceberg is 125 feet above sea level.
b. The iceberg is 125 feet below sea level.
Answer:
b. The iceberg is 125 feet below sea level.

Question 4.
Alex’s body temperature decreased by 2°F.
a. Alex’s body temperature dropped 2°F.
b. The integer – 2 represents the change in Alex’s body temperature in degrees Fahrenheit.
Answer:

Question 5.
A credit of $35 and a debit of $40 are applied to your bank account.
a. What is an appropriate scale to graph a credit of $35 and a debit of $40? Explain your reasoning.
Answer:
Answers will vary. I would count by 5’s because both numbers are multiples of 5.

b. What integer represents “a credit of $35” If zero represents the original balance? Explain.
Answer:
35; a credit is greater than zero, and numbers greater than zero are positive numbers.

c. What integer describes “a debit of $40” if zero represents the original balance? Explain.
Answer:
– 40; a debit is less than zero, and numbers less than zero are negative numbers.

d. Based on your scale, describe the location of both integers on the number line.
Answer:
If the scale is multiples of 5, then 35 would be 7 units to the right of (or above) zero, and – 40 would be 8 units to the left of (or below) zero.

e. What does zero represent in this situation?
Answer:
Zero represents no change being made to the account balance. In other words, no amount is either subtracted or added to the account.

Eureka Math Grade 6 Module 3 Lesson 3 Exit Ticket Answer Key

Question 1.
Write a story problem using sea level that includes both integers – 110 and 120.
Answer:
Answers may vary. On the beach, a man’s kite flies at 120 feet above sea level, which is indicated by the water’s surface. In the ocean, a white shark swims at 110 feet below the water’s surface.

Question 2.
What does zero represent in your story problem?
Answer:
Zero represents the water’s surface level, or sea level

Question 3.
Choose and label an appropriate scale to graph both integers on the vertical number line.
Answer:
I chose a scale of 10.

Question 4.
Graph and label both points on the vertical number line.

Answer:

Engage NY Eureka Math 6th Grade Module 3 Lesson 4 Answer Key

Eureka Math Grade 6 Module 3 Lesson 4 Example Answer Key

Example 1:
Every Number Has an Opposite
Locate the number 8 and its opposite on the number line. Explain how they are related to zero.

Answer:
→ First, start at zero, and move 8 units to the right to locate positive 8. So, the opposite of 8 must be 8 units to the left of zero. What number is 8 units to the left of zero?
– 8

→ 8 and – 8 are the same distance from zero. Since both numbers are the same distance from zero but on opposite sides of zero on the number line, they are opposites.

Example 2:
A Real-World Example

Maria decides to take a walk along Central Avenue to purchase a book at the bookstore. On her way, she passes the Furry Friends Pet Shop and goes in to look for a new leash for her dog. Furry Friends Pet Shop Is seven blocks west of the bookstore. She leaves Furry Friends Pet Shop and walks toward the bookstore to look at some books. After she leaves the bookstore, she heads east for seven blocks and stops at Ray’s Pet Shop to see If she can find a new leash at a better price. Which location, if any, Is the farthest from Maria while she is at the bookstore?

Determine an appropriate scale, and model the situation on the number line below.
Answer:
Answers will vary.

Explain your answer. What does zero represent in the situation?
Answer:
The pet stores are the same distance from Maria, who is at the bookstore. They are each 7 blocks away but in opposite directions. In this example, zero represents the bookstore.

Eureka Math Grade 6 Module 3 Lesson 4 Exercise Answer Key

Exercise 1:
Walk the Number Line

Each nonzero integer has an opposite, denoted – a; – a and a are opposites if they are on opposite sides of zero and the same distance from zero on the number line.
Answer:

Exercises 2.
Locate and label the opposites of the numbers on the number line.
a. 9
b. – 2
c. 4
d. – 7

Answer:

Exercises 3.
Write the integer that represents the opposite of each situation. Explain what zero means in each situation.
a. 100 feet above sea level
Answer:
– 100; zero represents sea level.

b. 32°C below zero
Answer:
32; zero represents 0 degrees Celsius.

c. A withdrawal of $25
Answer:
25; zero represents no change, where no withdrawal or deposit is made.

Read each situation carefully, and answer the questions.

Exercises 4.
On a number line, locate and label a credit of $15 and a debit for the same amount from a bank account. What does zero represent in this situation?
Answer:
Zero represents no change in the balance.

Exercises 5.
On a number line, locate and label 20°C below zero and 20°C above zero. What does zero represent In this situation?
Answer:
Zero represents 0°C.

Exercises 6.
A proton represents a positive charge. Write an integer to represent 5 protons. An electron represents a negative charge. Write an integer to represent 3 electrons.
Answer:
5 protons: 5
3 electrons: – 3

Eureka Math Grade 6 Module 3 Lesson 4 Problem Set Answer Key

Question 1.
Find the opposite of each number, and describe its location on the number line.
a. – 5
Answer:
The opposite of – 5 is 5, which is 5 units to the right of (or above) 0.

b. 10
Answer:
The opposite of 10 is – 10, which is 10 units to the left of(or below) 0.

c. – 3
Answer:
The opposite of – 3 is 3, which is 3 units to the right of (or above) 0.

d. 15
Answer:
The opposite of 15 is – 15, which is 15 units to the left of(or below) 0.

Question 2.
Write the opposite of each number, and label the points on the number line.
a. Point A: the opposite of 9
Answer:
– 9

b. Point B: the opposite of – 4
Answer:
4

c. Point C: the opposite of – 7
Answer:
7

d. Point D: the opposite of 0
Answer:
0

e. Point E: the opposite of 2
Answer:
– 2

Question 3.
Study the first example. Write the integer that represents the opposite of each real-world situation. In words, write the meaning of the opposite.
a. An atom’s positive charge of 7
Answer:
– 7, an atom’s negative charge of 7

b. A deposit of $25
Answer:
– 25, a withdrawal of $25

c. 3,500 feet below sea level
Answer:
3, 500, 3,500 feet above sea level

d. A rise of 45°C
Answer:
– 45, a decrease of 45°C

e. A loss of 13 pounds
Answer:
13, a gain of 13 pounds

Question 4.
On a number line, locate and label a credit of $38 and a debit for the same amount from a bank account. What does zero represent ¡n this situation?
Answer:
Zero represents no change in the balance.

Question 5.
On a number line, locate and label 40°C below zero and 40°C above zero. What does zero represent in this situation?
Answer:
Zero represents 0°C.

Eureka Math Grade 6 Module 3 Lesson 4 Exit Ticket Answer Key

In a recent survey, a magazine reported that the preferred room temperature in the summer is 68°F. A wall thermostat, like the ones shown below, tells a room’s temperature in degrees Fahrenheit.

a. Which bedroom is warmer than the recommended room temperature?
Answer:
The upstairs bedroom is warmer than the recommended room temperature.

b. Which bedroom is cooler than the recommended room temperature?
Answer:
The downstairs bedroom is cooler than the recommended room temperature.

c. Sarah notices that her room’s temperature is 4°F above the recommended temperature, and the downstairs bedroom’s temperature is 4°F below the recommended temperature. She graphs 72 and 64 on a vertical number line and determines they are opposites. Is Sarah correct? Explain.
Answer:
No. Both temperatures are positive numbers and not the same distance from 0, so they cannot be opposites. Both numbers have to be the same distance from zero, but one has to be above zero, and the other has to be below zero in order to be opposites.

d. After determining the relationship between the temperatures, Sarah now decides to represent 72°F as 4 and 64°F as – 4 and graphs them on a vertical number line. Graph 4 and – 4 on the vertical number line on the right. Explain what zero represents in this situation.
Answer:
Zero represents the recommended room temperature of 68°F. Zero could also represent not being above or below the recommended temperature.

Engage NY Eureka Math 6th Grade Module 3 Lesson 5 Answer Key

Eureka Math Grade 6 Module 3 Lesson 5 Opening Exercise Answer Key

a. Locate the number – 2 and its opposite on the number line below.

Answer:

b. Write an integer that represents each of the following.
i. 90 feet below sea level
Answer:
– 90

ii. $100 of debt
Answer:
– 100

iii. 2°C above zero
Answer:
2

c. Joe is at the Ice cream shop, and his house is 10 blocks north of the shop. The park is 10 blocks south of the ice cream shop. When he is at the ice cream shop, is Joe closer to the park or his house? How could the number zero be used in this situation? Explain.
Answer:
He is the same distance from his house and the park because both are located 10 blocks away from the ice cream shop but in opposite directions. In this situation, zero represents the locution of the ice cream shop.

Eureka Math Grade 6 Module 3 Lesson 5 Example Answer Key

Example 1:
The Opposite of an Opposite of a Number
What is the opposite of the opposite of 8? How can we illustrate this number on a number line?

a. What number is 8 unIts to the right of 0?
Answer:
8

b. How can you illustrate locating the opposite of 8 on this number line?
Answer:
We can illustrate the opposite of 8 on the number line by counting 8 units to the left of zero rather than to the right of zero.

c. What is the opposite of 8?
Answer:
– 8

d. Use the same process to locate the opposite of – 8. What is the opposite of – 8? _________

Answer:

e. The opposite of an opposite of a number is
Answer:
the original number

Eureka Math Grade 6 Module 3 Lesson 5 Exercise Answer Key

Complete the table using the cards in your group.

Answer:

Question 1.
Write the opposite of the opposite of – 10 as an equation.
Answer:
The opposite of – 10: – (- 10) = 10; the opposite of 10: – (10) = – 10. Therefore,(- ( – (- 10))) = – 10.

Question 2.
In general, the opposite of the opposite of a number is the
Answer:
original number.

Question 3.
Provide a real-world example of this rule. Show your work.
Answer:
Answers will vary. The opposite of the opposite of 100 feet below sea level is 100 feet below sea level.
– 100 is 100 feet below sea level.
– (- 100) = 100, the opposite of – 100
– (100) = – 100, the opposite of 100

Eureka Math Grade 6 Module 3 Lesson 5 Problem Set Answer Key

Question 1.
Read each description carefully, and write an equation that represents the description.
a. The opposite of negative seven
Answer:
– (- 7) = 7

b. The opposite of the opposite of twenty-five
Answer:
– (- (25)) = – 25

c. The opposite of fifteen
Answer:
– (15) = – 15

d. The opposite of negative thirty-six
Answer:
– (- 36) = – 36

Question 2.
Jose graphed the opposite of the opposite of 3 on the number line. First, he graphed point P on the number line 3 units to the right of zero. Next, he graphed the opposite of P on the number line 3 units to the left of zero and labeled it K. Finally, he graphed the opposite of K and labeled it Q.

a. Is his diagram correct? Explain. If the diagram Is not correct, explain his error, and correctly locate and label point Q.
Answer:
Yes, his diagram is correct. It shows that point P is 3 because It is 3 units to the right of zero. The opposite of 3 is – 3, which is point K (3 units to the left of zero). The opposite of – 3 is 3, so point Q is 3 units to the right of zero.

b. Write the relationship between the points:
P and K
Answer:
They are opposites.

K and Q
Answer:
They are opposites.

P and Q
Answer:
They are the same.

Question 3.
Read each real-world description. Write the integer that represents the opposite of the opposite. Show your work to support your answer.
a. A temperature rise of 15 degrees Fahrenheit
Answer:
– 15 is the opposite of 15 (fall in temperature).
15 is the opposite of – 15 (rise in temperature).
– (- (15)) = 15

b. A gain of 55 yards
Answer:
– 55 is the opposite of 55 (loss of yards).
55 is the opposite of – 55 (gain of yards).
– (- (55)) = 55

c. A loss of 10 pounds
Answer:
10 is the opposite of – 10 (gain of pounds).
– 10 is the opposite of 10 (loss of pounds).
– (- (- 10)) = – 10

d. A withdrawal of $2,000
Answer:
2,000 is the opposite of – 2,000 (deposit).
– 2,000 is the opposite of 2,000 (withdrawal).
– (- (- 2,000)) = – 2,000

Question 4.
Write the integer that represents the statement. Locate and label each point on the number line below.
Answer:
a. The opposite of a gain of 6
Answer:
– 6

b. The opposite of a deposit of $10
Answer:
– 10

c. The opposite of the opposite of 0
Answer:
0

d. The opposite of the opposite of 4
Answer:
4

e. The opposite of the opposite of a loss of 5
Answer:
– 5

Eureka Math Grade 6 Module 3 Lesson 5 Exit Ticket Answer Key

Question 1.
Jane completes several example problems that ask her to the find the opposite of the opposite of a number, and for each example, the result is a positive number. Jane concludes that when she takes the opposite of the opposite of any number, the result will always be positive. Is Jane correct? Why or why not?
Answer:
She is not correct. The opposite of the opposite of a number is the original number. So, if Jane starts with a negative number, she will end with a negative number.

Question 2.
To support your answer from the previous question, create an example, written as an equation. Illustrate your example on the number line below.
Answer:
If Jane starts with – 7, the opposite of the opposite of – 7 is written as – (- (- 7)) = – 7 or the opposite of – 7:
– (- 7) = 7; the opposite of 7: – (7) = – 7.

Answer:

Engage NY Eureka Math 6th Grade Module 3 Lesson 6 Answer Key

Eureka Math Grade 6 Module 3 Lesson 6 Example Answer Key

Example 1.
If b is a nonzero whole number, then the unit fraction \(\frac{1}{b}\) is located on the number line by dividing the segment between 0 and 1 into b segments of equal length. One of the b segments has 0 as its left end point; the right end point of this segment corresponds to the unit fraction \(\frac{1}{b}\).
Answer:
→ Since the number is a rational number, a number that can be represented as a fraction, determine how the number line should be scaled.¹
→ First, divide the number line into two halves to represent positive and negative numbers.
Have students complete this task on their student pages.
→ Next, divide the right half of the number line segment between 0 and 1 into ten segments of equal length; each segment has a length of \(\frac{1}{10}\).
Students divide their number lines into ten equal segments as shown. Check for accuracy.

There are 10 equal segments. Each segment has a length of \(\frac{1}{10}\). The first segment has 0 as its left end point, and the right end point corresponds to \(\frac{1}{10}\).

→ Have four students each stand in a square floor tile forming a straight line facing the class. Give each student a number to tie around his neck: 0, \(\frac{1}{10}\), \(\frac{2}{10}\), or \(\frac{3}{10}\). (Use index cards or construction paper.)
→ Ask a fifth student to assist by giving one end of a ball of string to the person at 0. This person holds one end of the string and passes the rest to the person to the left. (So the class sees it moving to the right.)
→ As the string gets passed down the line, each person announces her number, \(\frac{1}{10}, \frac{2}{10}, \frac{3}{10}\) stopping at \(\frac{3}{10}\).
→ The assistant cuts the string at \(\frac{3}{10}\) and gives that end of the string to the person holding , making one segment of length \(\frac{3}{10}\).
→ Have students turn over their numbers to reveal their opposites and rearrange themselves to represent the opposite of \(\frac{3}{10}\) using the same process. This time, students pass the string to the right. (So the class sees it moving to the left.)

The fraction \(\frac{a}{b}\) is located on the number line by joining a segments of length \(\frac{1}{b}\) so that (1) the left end point of the first segment is 0, and (2) the right end point of each segment is the left end point of the next segment. The right end point of the last segment corresponds to the fraction \(\frac{a}{b}\).
Answer:
→ To locate the number \(\frac{a}{b}\) on a number line, students should divide the interval between zero and 1 into b equal parts. Starting at 0, move along the number line a number of times.

→ There are ten equal segments. Each segment has a length of \(\frac{1}{10}\). The first segment has a 0 as its left end point, and the right end point of the third segment corresponds to \(\frac{3}{10}\). The point is located at \(\frac{3}{10}\).
→ The opposite of \(\frac{3}{10}\) is located the same distance from zero as \(\frac{3}{10}\) but in the opposite direction or to the left.
Using your knowledge of opposites, what rational number represents the opposite of \(\frac{3}{10}\)?
– \(\frac{3}{10}\)
→ To locate the opposite of \(\frac{3}{10}\) on the number line, divide the interval between zero and – 1 into ten equal segments. Starting at zero, how far would we move to locate the opposite of \(\frac{3}{10}\) and in what direction?
We would move 3 units to the left of zero because that is the same distance but opposite direction we
moved to plot the point \(\frac{3}{10}\).

→ There are ten equal segments. Each segment has a length of \(\frac{1}{10}\). Three consecutive segments, starting at 0 and moving to the left, would have a total length of \(\frac{3}{10}\). The point is located at –\(\frac{3}{10}\).
\(\frac{3}{10}\)
→ Counting three consecutive segments of length of \(\frac{1}{10}\) from 0 moving to the left and taking the end point of the last segment corresponds to the number –\(\frac{3}{10}\) . Therefore, the opposite of \(\frac{3}{10}\) is –\(\frac{3}{10}\).

Locate and graph the number \(\frac{3}{10}\) and its opposite on a number line.

Answer:

Example 2.
Rational Numbers and the Real World
The water level of a lake rose 1.25 feet after it rained. Answer the following questions using the number line below.

a. Write a rational number to represent the situation.
Answer:
1. 25 or 1\(\frac{1}{4}\)

b. What two integers is 1. 25 between on a number line?
Answer:
1 and 2

c. Write the length of each segment on the number line as a decimal and a fraction.
Answer:
0. 25 and \(\frac{1}{4}\)

d. What will be the water level after It rained? Graph the point on the number line.
Answer:
1.25 feet above the original lake level

e. After two weeks have passed, the water level of the lake Is now the opposite of the water level when It rained. What will be the new water level? Graph the point on the number line. Explain how you determined your answer.
Answer:
The water level would be 1.25 feet below the original lake level. If the water level was 1.25, the opposite of 1.25 is – 1.25.

f. State a rational number that is not an integer whose value is less than 1. 25, and describe Its location between two consecutive integers on the number line.
Answer:
Answers will vary. A rational number whose value is less than 1.25 is 0.75. It would be located between 0 and 1 on a number line.

Eureka Math Grade 6 Module 3 Lesson 6 Exercise Answer Key

Exercise 1.
Use what you know about the point –\(\frac{7}{4}\) and its opposite to graph both points on the number line below. The fraction – \(\frac{7}{4}\) is located between which two consecutive integers? Explain your reasoning.

Answer:

On the number line, each segment will have an equal length of \(\frac{1}{4}\). The fraction is located between – 1 and – 2.

Explanation:
\(\frac{7}{4}\) is the opposite of –\(\frac{7}{4}\). It is the same distance from zero but on the opposite side of zero. Since \(-\frac{7}{4}\) is to the left of zero, \(\frac{7}{4}\)is to the right of zero. The original fraction is located between – 2 (or –\(\frac{8}{4}\) and – 1 (or – \(\frac{4}{4}\)).

Exercise 2.
Our Story Problem
Answer:
Answers will vary.
Melissa and Samantha weigh the same amount. Melissa gained 5.5 pounds last month, while Samantha lost the same amount Melissa gained.

→ Our Scale: 1
→ Our Units: Pounds
→ Description: On the number line, zero represents Melissa and Samantha’s original weight. The point – 5.5 represents the change in Samantha’s weight. The amount lost is 5.5 pounds.
→ Other Information: A rational number to the left of 5.5 is 4. 5. A rational number to the right of 5.5 is 5.75.

Eureka Math Grade 6 Module 3 Lesson 6 Problem Set Answer Key

Question 1.
In the space provided, write the opposite of each number.
a. \(\frac{10}{7}\)
Answer:
\(-\frac{10}{7}\)

b. \(-\frac{5}{3}\)
Answer:
\(\frac{5}{3}\)

c. 3.82
Answer:
– 3. 82

d. -6\(-\frac{1}{2}\)
Answer:
6\(\frac{1}{2}\)

Question 2.
Choose a non-integer between O and 1. Label it point A and its opposite point B on the number line. Write values below the points.
Answer:
(Answers may vary)

a. To draw a scale that would include both points, what could be the length of each segment?
Answer:
Answers may vary. \(\frac{1}{3}\)

b. In words, create a real-world situation that could represent the number line diagram.
Answer:
Answers may vary. Starting at home, I ran \(\frac{1}{3}\) mile. My brother ran \(\frac{1}{3}\) mile from home in the opposite direction.

Question 3.
Choose a value for point P that is between – 6 and – 7.
Answer:
Answers may vary. – \(\frac{13}{2}\), – 6.25, – 6.8

a. What is the opposite of point P?
Answer:
Answers may vary. \(\frac{13}{2}\), 6.25, 6.8

b. Use the value from part (a), and describe its location on the number line in relation to zero.
Answer:
\(\frac{13}{2}\) is the same distance as –\(\frac{13}{2}\) from zero but to the right. \(\frac{13}{2}\) is 6\(\frac{1}{2}\) units to the right of (or above) zero.

c. Find the opposite of the opposite of point P. Show your work, and explain your reasoning.
Answer:
The opposite of an opposite of the number ¡s the original number. If point P is located at – \(\frac{13}{2}\), then the
opposite of the opposite of point P is located at –\(\frac{13}{2}\) . The opposite of – \(\frac{13}{2}\) is \(\frac{13}{2}\). The opposite of \(\frac{13}{2}\) is –\(\frac{13}{2}\)
– (- \(\frac{13}{2}\)) = \(\frac{13}{2}\) and – (- (- \(\frac{13}{2}\))) = – \(\frac{13}{2}\)

Question 4.
Locate and label each point on the number line. Use the diagram to answer the questions.
Jill lives one block north of the pizza shop.
Janette’s house is \(\frac{1}{3}\) block past Jill’s house.
Jeffrey and Olivia are in the park \(\frac{4}{3}\) blocks south of the pizza shop.
Jenny’s Jazzy Jewelry Shop is located halfway between the pizza shop and the park.

a. Describe an appropriate scale to show all the points in this situation.
Answer:
An appropriate scale would be \(\frac{1}{3}\) because the numbers given in the example all have denominators of 3. I would divide the number line into equal segments of \(\frac{1}{3}\).

b. What number represents the location of Jenny’s Jazzy Jewelry Shop? Explain your reasoning.
Answer:
The number is –\(\frac{2}{3}\). I got my answer by finding the park first. It is 4 units below 0. Since the jewelry shop is halfway between the pizza shop and the park, half of 4 is 2. Then, I moved 2 units down on the number line since the shop is south of the pizza shop before the park.

Eureka Math Grade 6 Module 3 Lesson 6 Exit Ticket Answer Key

Use the number line diagram below to answer the following questions.

Question 1.
What is the length of each segment on the number line?
Answer:
\(\frac{1}{12}\)

Question 2.
What number does point K represent?
Answer:
\(\frac{8}{12}\) or \(\frac{2}{3}\)

Question 3.
What is the opposite of point K?
Answer:
\(-\frac{8}{12}\) or \(-\frac{2}{3}\)

Question 4.
Locate the opposite of point K on the number line, and label it point L.
Answer:

Question 5.
In the diagram above, zero represents the location of Martin Luther King Middle School. Point K represents the library, which is located to the east of the middle school. In words, create a real-world situation that could represent point L, and describe its location in relation to 0 and point K.
Answer:
Answers may vary. Point L is \(\frac{8}{12}\) units to the left of 0, so it is a negative number. Point L represents the recreation center, which is located \(\frac{8}{12}\) mile west of Martin Luther King Middle School. This means that the recreation center and library are the same distance from the middle school but in opposite directions because the opposite of \(\frac{8}{12}\) is \(-\frac{8}{12}\)

Eureka Math Grade 6 Module 3 Lesson 6 Opening Exercise Answer Key

a. Write the decimal equivalent of each fraction.
i. \(\frac{1}{2}\)
Answer:
0.5

ii. \(\frac{4}{5}\)
Answer:
0.8

iii. 6\(\frac{7}{10}\)
Answer:
6.70

b. Write the fraction equivalent of each decimal.

i. 0.42
Answer:
\(\frac{42}{100}=\frac{21}{50}\)

ii. 3.75
Answer:
\(3 \frac{75}{100}=3 \frac{3}{4}\)

iii. 36.90
Answer:
\(36 \frac{90}{100}=36 \frac{9}{10}\)

Engage NY Eureka Math 6th Grade Module 3 Lesson 7 Answer Key

Eureka Math Grade 6 Module 3 Lesson 7 Example Answer Key

Example 1.
The record low temperatures for a town in Maine are – 20°F for January and – 19°F for February. Order the numbers from least to greatest. Explain how you arrived at the order.
Answer:
→ Read: January: – 20 and February: – 19
→ Draw: Draw a number line model.
→ Write: Since – 20 is farthest below zero and – 19 is above – 20 on the vertical number line, – 20 is less than – 19.
→ Answer: – 20, – 19

Example 2.
Henry, Janon, and Clark are playing a card game. The object of the game Is to finish with the most points. The scores at the end of the game are Henry: – 7, Janon: 0, and Clark: – 5. Who won the game? Who came in last place? Use a number line model, and explain how you arrived at your answer.
Answer:
→ Read: Henry: – 7, fanon: O, and Clark: —5
→ Draw:

→ Explain: – 7, – 5, 0
→ Janon won the game, and Henry came in last place. I ordered the numbers on a number line, and – 7 is farthest to the left. That means – 7 is the smallest of the three numbers, so Henry came in last place. Next on the number line is – 5, which is to the right of – 7 but to the left of 0. Farthest to the right is 0; therefore, 0 is the greatest of the three numbers. This means Janon won the game.

Eureka Math Grade 6 Module 3 Lesson 7 Exercise Answer Key

Exercise 1.
a. Graph the number 7 and its opposite on the number line. Graph the number 5 and its opposite on the number line.

Answer:

b. Where does 7 lie in relation to 5 on the number line?
Answer:
On the number line, 7 is 2 units to the right of 5.

c. Where does the opposite of 7 lie on the number line in relation to the opposite of 5?
Answer:
On the number line, – 7 is 2 units to the left of – 5.

d. I am thinking of two numbers. The first number lies to the right of the second number on a number line. What can you say about the location of their opposites? (If needed, refer to your number line diagram.)
Answer:
On the number line, the opposite of the second number must lie to the right of the opposite of the first number. If we call the first number f and the second number s, then – f and – s will have the opposite order off and s because – f and – s are opposites off and s, so they lie on the opposite side of zero.

For each problem, order the rational numbers from least to greatest by first reading the problem, then drawing a number line diagram, and finally, explaining your answer.

Exercise 2.
Jon’s time for running the mile in gym class is 9.2 minutes. Jacky’s time is 9. 18 minutes. Who ran the mile in less
time?
9.18, 9.2
I drew a number line and graphed 9.2 and 9. 18; 9.2 is to the right of 9. 18. So, 9. 18 is less than 9.2, which means Jacky ran the mile in less time than Jon.

Exercise 3.
Mrs. Rodriguez is a teacher at Westbury Middle School. She gIves bonus points on tests for outstanding written answers and deducts points for answers that are not written correctly. She uses rational numbers to represent the points. She wrote the following on students’ papers: Student A: – 2 points, Student B: – 2. 5 points. Did Student A or Student B perform worse on the test?
Answer:
– 2.5, – 2
I drew a number line, and – 2 and – 2.5 are both to the left of zero, but – 2.5 is to the left of – 2. So, – 2.5 is less than – 2. That means Student B did worse than Student A.

Exercise 4.
A carp is swimming approximately 8\(\frac{1}{4}\) feet beneath the water’s surface, and a sunfish is swimming approximately 3\(\frac{1}{2}\) feet beneath the water’s surface. Which fish is swimming farther beneath the water’s surface?
Answer:
– 8\(\frac{1}{4}\), – 3\(\frac{1}{2}\)
I drew a vertical number line, and – 8\(\frac{1}{4}\) as farther below zero than – 3\(\frac{1}{2}\). So, – 8\(\frac{1}{4}\) as less than – 3\(\frac{1}{2}\) , which means the carp is swimming farther beneath the water’s surface.

For each problem, order the rational numbers from least to greatest by first reading the problem, then drawing a number line diagram, and finally, explaining your answer.

Exercise 5.
Henry, Janon, and Clark are playing another round of the card game. Their scores this time are as follows: Clark: – 1, Janon: – 2, and Henry: – 4. Who won? Who came in last place?
Answer:
– 4, – 2, – 1
Clark won the game, and Henry came in last place. I ordered the numbers on a number line, and —4 is farthest to the left. That means – 4 is the smallest of the three numbers, so Henry lost. Next on the number line is – 2, which is to the right of – 4 and to the left of – 1. Farthest to the right is – 1, which is the greatest of the three negative numbers, so Clark won the game.

Exercise 6.
Represent each of the following elevations using a rational number. Then, order the numbers from least to greatest.
Cayuga Lake 122                                            meters above sea level
Mount Marcy 1, 629                                       meters above sea level
New York Stock Exchange Vault 15.24           meters below sea level
Answer:
– 15.24; 122; 1,629
I drew a number line, and – 15.24 is the only number to the left of zero, so it is the least (because as you move to the right, the numbers increase). Next on the number line is 122, which is to the right of zero. Last on the number line is 1,629, which is to the right of 122, so 1,629 meters is the greatest elevation.

Closing: What is the Value of Each Number, and Which is Larger?
Use your teacher’s verbal clues and this number line to determine which number is larger.

Answer:

Eureka Math Grade 6 Module 3 Lesson 7 Problem Set Answer Key

Question 1.
In the table below, list each set of rational numbers in order from least to greatest. Then, list their opposites. Finally, list the opposites in order from least to greatest. The first example has been completed for you.

Answer:

Question 2.
For each row, what pattern do you notice between the numbers in the second and fourth columns? Why is this so?
Answer:
For each row, the numbers in the second and fourth columns are opposites, and their order is opposite. This is because on the number line, as you move to the right, numbers increase. But as you move to the left, the numbers decrease. So, when comparing 5 and 10, 10 is to the right of 5; therefore, 10 is greater than 5. However, – 10 is to the left of – 5; therefore, – 10 is less than – 5.

Eureka Math Grade 6 Module 3 Lesson 7 Exit Ticket Answer Key

Question 1.
In math class, Christina and Brett are debating the relationship between two rational numbers. Read their claims below, and then write an explanation of who is correct. Use a number line model to support your answer.

Christina’s Claim: “I know that 3 is greater than 2\(\frac{1}{2}\). So, – 3 must be greater than – 2\(\frac{1}{2}\).”

Brett’s Claim: “Yes, 3 is greater than 2\(\frac{1}{2}\), but when you look at their opposites, their order will be opposite. So, that means – 2\(\frac{1}{2}\) is greater than – 3.”
Answer:

Brett is correct. I graphed the numbers on the number line, and – 3 is to the left of – 2\(\frac{1}{2}\). The numbers increase as you move to the right, so – 2\(\frac{1}{2}\) is greater than – 3.

Engage NY Eureka Math 6th Grade Module 3 Lesson 8 Answer Key

Eureka Math Grade 6 Module 3 Lesson 8 Example Answer Key

Example 1.
Ordering Rational Numbers from Least to Greatest
Sam has $10.00 in the bank. He owes his friend Hank $2.25. He owes his sister $1.75. Consider the three rational numbers related to this story of Sam’s money. Write and order them from least to greatest.
Answer:
– 2.25, – 1.75, 10.00

Example 2.
Ordering Rational Numbers from Greatest to Least
Jason is entering college and has opened a checking account, which he will use for college expenses. His parents gave him $200. 00 to deposit into the account. Jason wrote a check for $85.00 to pay for his calculus book and a check for $25.34 to pay for miscellaneous school supplies. Write the three rational numbers related to the balance in Jason’s checking account in order from greatest to least.
Answer:
200.00, – 25.34, – 85.00

Eureka Math Grade 6 Module 3 Lesson 8 Exercise Answer Key

Exercises 2 – 4
For each problem, list the rational numbers that relate to each situation. Then, order them from least to greatest, and explain how you made your determination.

Exercise 2.
During their most recent visit to the optometrist (eye doctor), Kadijsha and her sister, Beth, had their vision tested. Kadijsha’s vision in her left eye was – 1.50, and her vision in her right eye was the opposite number. Beth’s vision was -1. 00 in her left eye and +0. 25 in her right eye.
Answer:
– 1. 50, – 1.00, 0.25, 1.50
The opposite of – 1.50 is 1.50, and 1.5015 farthest right on the number line, so if is the greatest. – 1.50 is the same distance from zero but on the other side, so it is the least number. – 1.00 is to the right of – 1.50, so it is greater than – 1.50, and 0.25 is to the right of – 1.00, so it is greater than – 1.00. Finally, 1.50 is the greatest.

Exercise 3.
There are three pieces of mail in Ms. Thomas’s mailbox: a bill from the phone company for $38. 12, a bill from the electric company for $67. 55, and a tax refund check for $25.89. (A bill is money that you owe, and a tax refund check is money that you receive.)
Answer:
– 67.55, – 38. 12, 25.89
The change in Ms. Thomas’s money is represented by – 38.12 due to the phone bill, and – 67.55 represents the change in her money due to the electric bill. Since – 67.55 is farthest to the left on the number line, it is the least. Since – 38.12 is to the right of – 67.55, It comes next. The check she has to deposit for $25.89 can be represented by 25.89, which Is to the right of – 38.12, and so it is the greatest number.

Exercise 4.
Monica, Jack, and Destiny measured their arm lengths for an experiment in science class. They compared their arm lengths to a standard length of 22 inches. The listing below shows, in inches, how each student’s arm length compares to 22 inches.
Monica: – \(\frac{1}{8}\)
Jack: 1\(\frac{3}{4}\)
Destiny: – \(\frac{1}{2}\)
Answer:
I ordered the numbers on a number line, and –\(\frac{1}{2}\) was farthest to the left. To the right of that was –\(\frac{1}{8}\). Lastly, 1\(\frac{3}{4}\) is to the right of –\(\frac{1}{8}\) so 1\(\frac{3}{4}\) is the greatest.

Exercise 5 – 6
For each problem, list the rational numbers that relate to each situation in order from greatest to least. Explain how you arrived at the order.

Exercise 5.
The following are the current monthly bills that Mr. McGraw must pay:
$122. 00 Cable and Internet
$73.45 Gas and Electric
$45.00 Cell Phone
Answer:
– 45.00, – 73.45, – 122.00
Because Mr. McGraw owes the money, I represented the amount of each bill as a negative number. Ordering them from greatest to least means I have to move from right to left on a number line. Since – 45.00 is farthest right, it is the greatest. To the left of that is – 73.45, and to the left of that is – 122.00, which means – 122.00 is the least.

Exercise 6.
\(-\frac{1}{3}\), 0, \(-\frac{1}{5}\), \(\frac{1}{8}\)
Answer:
\(\frac{1}{8}\), 0, \(-\frac{1}{5}\), \(-\frac{1}{3}\)
I graphed them on the number line. Since I needed to order them from greatest to least, I moved from right to left to record the order. Farthest to the right is \(\frac{1}{8}\) so that is the greatest value. To the left of that number is 0. To the left of 0 is –\(\frac{1}{5}\), and the farthest left is –\(\frac{1}{3}\) so that is the least.

Eureka Math Grade 6 Module 3 Lesson 8 Problem Set Answer Key

Question 1.
a. In the table below, list each set of rational numbers from greatest to least. Then, in the appropriate column, state which number was farthest right and which number was farthest left on the number line.

Answer:

b. For each row, describe the relationship between the number in Column 3 and its order in Column 2. Why is this?
Answer:
The number in Column 3 is the first number listed in Column 2. Since it is farthest right on the number line, it will be the greatest; therefore, it comes first when ordering the numbers from greatest to least.

c. For each row, describe the relationship between the number in Column 4 and Its order in Column 2. Why is this?
Answer:
The number in Column 4 is the last number listed in Column 2. Since it is farthest left on the number line, it will be the smallest; therefore, it comes last when ordering the numbers from greatest to least.

Question 2.
If two rational numbers, a and b, are ordered such that a is less than b, then what must be true about the order for their opposites: – a and – b?
Answer:
The order will be reversed for the opposites, which means – a is greater than – b.

Question 3.
Read each statement, and then write a statement relating the opposites of each of the given numbers:
a. 7 is greater than 6.
Answer:
– 7 is less than – 6.

b. 39.2 is greater than 30.
Answer:
– 39.2 is less than – 30.

c. –\(\frac{1}{5}\) is less than \(\frac{1}{3}\).
Answer:
\(\frac{1}{5}\) is greater than –\(\frac{1}{3}\)

Question 4.
Order the following from least to greatest: – 8, – 19, 0, \(\frac{1}{2}\), \(\frac{1}{4}\).
Answer:
– 19, – 8, 0, \(\frac{1}{4}\), \(\frac{1}{2}\)

Question 5.
Order the following from greatest to least: – 12, 12, – 19, 1\(\frac{1}{2}\), 5.
Answer:
12, 5, 1\(\frac{1}{2}\), – 12, – 19

Eureka Math Grade 6 Module 3 Lesson 8 Exit Ticket Answer Key

Question 1.
Order the following set of rational numbers from least to greatest, and explain how you determined the order.
– 3, 0, –\(\frac{1}{2}\), 1, – 3\(\frac{1}{3}\), 6, 5, – 1, \(\frac{21}{5}\), 4
Answer:

– 3, 0, –\(\frac{1}{2}\), 1, – 3\(-\frac{1}{2}\), 0, 1, 4, \(\frac{21}{5}\), 5, 6
I drew a number line and started at zero. I located the positive numbers to the right and their opposites (the negative numbers) to the left of zero. The positive integers listed in order from left to right are 1, 4, 5, 6. And since \(\frac{21}{5}\) is equal to 4\(\frac{1}{5}\), I know that it is \(\frac{1}{5}\) more than 4 but less than 5. Therefore, I arrived at 0, 1, 4, \(\frac{21}{5}\), 5, 6. Next, I ordered the negative numbers. Since – 1 and – 3 are the opposites of 1 and 3, they are 1 unit and 3 units from zero but to the left of zero. And – 3\(\frac{1}{3}\) is even farther left, since it is 3\(\frac{1}{3}\) units to the left of zero. The smallest number is farthest to the left, so l arrived at the following order: – 3\(\frac{1}{3}\) , – 3, – 1, –\(\frac{1}{2}\) , 0, 1, 4, \(\frac{21}{5}\), 5, 6.

Engage NY Eureka Math 6th Grade Module 3 Lesson 9 Answer Key

Eureka Math Grade 6 Module 3 Lesson 9 Example Answer Key

Example 1.
Interpreting Number Line Models to Compare Numbers

Answer:
Answers may vary. Every August, the Boy Scouts go on an 8-day 40-mile hike. At the halfway point (20 miles into the hike), there is a check-in station for Scouts to check in and register. Thomas and Evan are Scouts in 2 different hiking groups. By Wednesday morning, Evan’s group has 10 miles to go before it reaches the check-in station, and Thomas’s group is 5 miles beyond the station. Zero on the number line represents the check-in station.

Eureka Math Grade 6 Module 3 Lesson 9 Exercise Answer Key

Exercise 1.
Create a real-world situation that relates to the points shown in the number line model. Be sure to describe the relationship between the values of the two points and how it relates to their order on the number line.

Answer:
Answers will vary.
Alvin lives in Canada and is recording the outside temperature each night before he goes to bed. On Monday night, he recorded a temperature of 0 degrees Celsius. On Tuesday night, he recorded a temperature of – 1 degree Celsius. Tuesday night’s temperature was colder than Monday night’s temperature. – 1 is less than 0, so the associated point is below 0 on a vertical number line.

For each problem, determine if you agree or disagree with the representation. Then, defend your stance by citing specific details in your writing.

Exercise 2.
FelIcia needs to write a story problem that relates to the order In which the numbers – 6 and – 10 are represented on a number line. She writes the following:

“During a recent football game, our team lost yards on two consecutive downs. We lost 6\(\frac{1}{2}\) yards on the first down. During the second down, our quarterback was sacked for an additional 10-yard loss. On the number line, I represented this situation by first locating -6\(\frac{1}{2}\). I located the point by moving 6\(\frac{1}{2}\) units to the left of zero. Then, I graphed the second point by moving 10 units to the left of 0.”
Answer:
Agree. – 10 is less than – 6\(\frac{1}{2}\) since – 10 is to the left of – 6\(\frac{1}{2}\) on the number line. Since both numbers are negative, they indicate the team lost yards on both football plays, but they lost more yards on the second play.

Exercise 3.
Manuel looks at a number line diagram that has the points –\(\frac{3}{4}\) and –\(\frac{1}{2}\) graphed. He writes the following related story:
“I borrowed 50 cents from my friend, Lester. I borrowed 75 cents from my friend, Calvin. I owe Lester less than I owe Calvin.”
Answer:
Agree. – \(\frac{3}{4}\) is equivalent to – 0.75 and – \(\frac{1}{2}\) is equivalent to – 0.50. – 0.50 and – 0.75 both show that he owes money. But – 0.50 is farther to the right on a number line, so Manuel does not owe Lester as much as he owes Calvin.

Exercise 4.
Henry located 2\(\frac{1}{4}\) and 2. 1 on a number line. He wrote the following related story:
“In gym class, both Jerry and I ran for 20 minutes. Jerry ran 2\(\frac{1}{4}\) miles, and 1 ran 2. 1 miles. I ran a farther distance.”
Answer:
Disagree. 2\(\frac{1}{4}\) is greater than 2.1 since 2\(\frac{1}{4}\) is equivalent to 2.25. On the number line, the point associated with 2.25 is to the right of 2. 1. Jerry ran a farther distance.

Exercise 5.
Sam looked at two points that were graphed on a vertical number line. He saw the points – 2 and 1. 5. He wrote the following description:
“I am looking at a vertical number line that shows the location of two specific points. The first point is a negative number, so it is below zero. The second point is a positive number, so it is above zero. The negative number is – 2.
The positive number is \(\frac{1}{2}\) unit more than the negative number.”
Answer:
Disagree. Sam was right when he said the negative number is below zero and the positive number is above zero. But 1.5 is 1\(\frac{3}{4}\) units above zero, and – 2 is 2 units below zero. So, altogether, that means the positive number is 3\(\frac{1}{2}\) units more than – 2.

Exercise 6.
Claire draws a vertical number line diagram and graphs two points: – 10 and 10. She writes the following related story:

“These two locations represent different elevations. One location is 10 feet above sea level, and one location is 10 feet below sea level. On a number line, 10 feet above sea level is represented by graphing a point at 10, and 10 feet below sea level is represented by graphing a point at – 10.”
Answer:
Agree. Zero in this case represents sea level Both locations are 10 feet from zero but in opposite directions, so they are graphed on the number line at 10 and – 10.

Exercise 7.
Mrs. Kimble, the sixth-grade math teacher, asked the class to describe the relationship between two points on the number line, 7.45 and 7. 5, and to create a real-world scenario. Jackson writes the following story:

“Two friends, Jackie and Jennie, each brought money to the fair. Jackie brought more than Jennie. jackie brought $7.45, and Jennie brought $7. 50. Since 7.45 has more digits than 7.5, It would come after 7.5 on the number line, or to the right, so it is a greater value.”
Answer:
Disagree. Jackson is wrong by saying that 7.45 is to the right of 7.5 on the number line. 7. 5 is the same as 7.50, and it is greater than 7.45. When I count by hundredths starting at 7.45, I would say 7.46, 7.47, 7.48, 7.49, and then 7.50. So, 7. 50 is greater than 7.45, and the associated point falls to the right of the point associated with 7.45 on the number line.

Exercise 8.
Justine graphs the points associated with the following numbers on a vertical number line: – 1\(\frac{1}{4}\) – 1\(\frac{1}{2}\), and 1. She then writes the following real-world scenario:
“The nurse measured the height of three sixth-grade students and compared their heights to the height of a typical sixth grader. Two of the students’ heights are below the typical height, and one is above the typical height. The point whose coordinate is 1 represents the student who has a height that is 1 inch above the typical height. Give this information, Justine determined that the student represented by the point associated with – 1\(\frac{1}{4}\) is the shortest of the three students.”
Answer:
Disagree. Justine was wrong when she said the point – 1\(\frac{1}{4}\) represents the shortest of the three students. If zero stands for no change from the typical height, then the point associated with – 1\(\frac{1}{2}\) is farther below zero than the point associated with – 1\(\frac{1}{4}\). The greatest value is positive 1. Positive 1 represents the tallest person. The shortest person is represented by – 1\(\frac{1}{2}\).

Eureka Math Grade 6 Module 3 Lesson 9 Problem Set Answer Key

Write a story related to the points shown in each graph. Be sure to include a statement relating the numbers graphed on the number line to their order.

Question 1.

Answer:
Answers will vary. Marcy earned no bonus points on her first math quiz. She earned 4 bonus points on her second math quiz. Zero represents earning no bonus points, and 4 represents earning 4 bonus points. Zero is graphed to the left of 4 on the number line. Zero is less than 4.

Question 2.

Answer:
Answers will vary. My uncle’s investment lost $200 in May. In June, the investment gained $150. The situation is represented by the points – 200 and 150 on the vertical number line. Negative 200 is below zero, and 150 is above zero. – 200 is less than 150.

Question 3.

Answer:
Answers will vary. I gave my sister $1.50 last week. This week, I gave her $0.50. The points – 1.50 and – 0.50 represent the change to my money supply. We know that – 1.50 is to the left of – 0. 50 on the number line; therefore, – 0. 50 is greater than – 1.50.

Question 4.

Answer:
Answers will vary. A fish is swimming 7 feet below the water’s surface. A turtle is swimming 2 feet below the water’s surface. We know that – 7 is to the left of – 2 on the number line. This means – 7 is less than – 2.

Question 5.

Answer:
Answers will vary. I spent $8 on a CD last month. I earned $5 in allowance last month. – 8 and 5 represent the changes to my money last month. – 8 is to the left of 5 on a number line. – 8 is 3 units farther away from zero than 5, which means that I spent $3 more on the CD than I made in allowance.

Question 6.

Answer:
Answers will vary. Skip, Mark, and Angelo were standing in line in gym class. Skip was the third person behind
Mark. Angelo was the first person ahead of Mark. If Mark represents zero on the number line, then Skip is associated with the point at -3, and Angelo is associated with the point at 1. 1 is 1 unit to the right of zero, and -3 is 3 units to the left of zero. -3 is less than 1.

Question 7.

Answer:
Answers will vary. I rode my bike \(\frac{3}{5}\) miles on Saturday and \(\frac{4}{5}\) miles on Sunday. On a vertical number line, \(\frac{3}{5}\) and \(\frac{4}{5}\) are both associated with points above zero, but \(\frac{4}{5}\) is above \(\frac{3}{5}\). This means that \(\frac{4}{5}\) is greater than \(\frac{3}{5}\).

Eureka Math Grade 6 Module 3 Lesson 9 Exit Ticket Answer Key

Question 1.
Interpret the number line diagram shown below, and write a statement about the temperature for Tuesday compared to Monday at 11:00 p.m.

Answer:
At 11:00p.m. on Monday, the temperature was about 40 degrees Fahrenheit, but at 11:00 p.m. on Tuesday, it was – 10 degrees Fahrenheit. Tuesday’s temperature of – 10 degrees is below zero, but 40 degrees is above zero. It was much warmer on Monday at 11:00 p.m. than on Tuesday at that time.

Question 2.
If the temperature at 11:00 p.m. on Wednesday is warmer than Tuesday’s temperature but still below zero, what is a possible value for the temperature at 11:00 p.m. Wednesday?
Answer:
Answers will vary but must be between 0 and – 10. A possible temperature for Wednesday at 11:00 p.m. is – 3 degrees Fahrenheit because – 3 is less than zero and greater than – 10.

Engage NY Eureka Math 5th Grade Module 6 End of Module Assessment Answer Key

Eureka Math Grade 5 Module 6 End of Module Assessment Task Answer Key

Question 1.
Follow the directions.
a. Draw a ray that starts at point L at (1\(\frac{1}{2}\), 3) and includes point K at (5, 3). Label points K and L.
b. Give the coordinates of three other points on the ray.
c. Draw a second ray with the same initial point and containing point M with coordinates (3\(\frac{1}{4}\), 4\(\frac{1}{2}\)). Label point M.

Answer:
a.

Explanation :
Point L and K are marked and a ray is drawn passing from these points .

b. The 3 points on the ray are  ( 2, 3) , (4, 3) , (3\(\frac{1}{2}\), 3).
c. The initial point is L (1\(\frac{1}{2}\), 3) and M (3\(\frac{1}{4}\), 4\(\frac{1}{2}\)) are plotted and a ray is drawn through these points .

Question 2.
David draws a line segment from point Q (\(\frac{1}{4}\), \(\frac{7}{8}\)) to point R (\(\frac{5}{8}\), \(\frac{1}{2}\)). He then draws a line perpendicular to the first segment that intersects segment \(\overline{Q R}\) and includes point S (\(\frac{3}{4}\), 1).

a. Draw \(\overline{Q R}\), and label the endpoints on the grid.
b. Draw the perpendicular line, and label point S.
c. Name another point that lies on the perpendicular line whose x-coordinate is between 1 and 1\(\frac{1}{2}\).
Answer:

Explanation :
The Points Q and R are plotted then Point S ia marked then a straight line from point S is drawn and intersect the \(\overline{Q R}\) . Then Point between 1 and 1\(\frac{1}{2}\) is marked at (1\(\frac{1}{4}\), 1\(\frac{1}{2}\) ).

Question 3.
Complete the table for the rule multiply by 2 and then add 2 for the values of x from 0 to 4. Then, use the coordinate plane to answer the questions.

a. Which line shows the rule in the table?
b. Give the coordinates for the intersection of lines b and c.
c. Draw a line on the graph such that any point on the line has a y-coordinate of 2. Label your line as e.
d. Which coordinate is 2 for any point on line c?
e. Write a rule that tells how to find the y-coordinate when the x-coordinate is given for the points on line b.
f. Kim and Lacy want to draw a line on the coordinate plane that is parallel to line a. Kim uses the rule multiply by 4 and add 2 to generate her y-coordinates. Lacy uses the rule multiply by 2 and add 4 to generate her y-coordinates. Which girl’s line will be parallel to line a? Without graphing the lines, explain how you know.
Answer:
Rule : multiply by 2 and then add 2
y = 2x + 2

x= 0 ; y = 2(0) + 2 = 2
x= 1 ; y = 2(1) + 2 = 4
x=2 ; y = 2(2) + 2 = 6
x=3 ; y = 2(3) + 2 = 8
x= 4; y = 2(4) + 2 = 10
a. Line a follows the rule .
b. The coordinates for the intersection of lines b and c is (2, 6)
c. For all x-coordinates the y coordinate is 2 line is drawn and labeled as line e. 
d. The x-coordinate is 2 for all y-coordinates on line c .
e.
The coordinates are (2, 6) , (3, 7) , ( 4, 8 )
The Rule for y-coordinate when the x-coordinate is given for the points on line b is x-coordinate and add 4 .
f. Lacy’s line rule will make a line parallel to line a .The rule for line ???? is multiply ???? by 2, and then add 2. The rule for Lacy’s line is multiply ????- coordinate by 2, and then add 4. Lacy’s line is parallel because the steepness of the line is the same. (That is, the multiplication part of the rule is the same.) The adding part of the rule will make the ????- coordinates two more than those in line ????.)

Question 4.
An airplane is descending into an airport. When its altitude is 5 miles, it is 275 miles from the airport. When its altitude is 4 miles, it is 200 miles from the airport. At 3 miles, it is 125 miles from the airport.

a. If the pilot follows the same pattern, what will the plane’s altitude be at 50 miles from the airport?
b. For the plane to land at the airport, the altitude will need to be 0, and the distance from the airport will need to be 0. Should the pilot continue this pattern? Why or why not?
Answer:
a.
For 50miles the attitude will be 2 miles as we can notice from the above graph .
b.
The pilot should not follow this pattern . He will be way past the airport when his altitude will be 0 .

Engage NY Eureka Math 5th Grade Module 6 Mid Module Assessment Answer Key

Eureka Math Grade 5 Module 6 Mid Module Assessment Task Answer Key

Question 1.
Give the coordinates of each point.

A ________________
B ________________
C ________________
D ________________
E ________________
Answer:
A ( 3, 4)
B (4, 2)
C (\(\frac{1}{2}\) , \(\frac{1}{4}\))
D ( 1, 2\(\frac{1}{2}\))
E (1\(\frac{3}{4}\), 4\(\frac{1}{4}\))
Explanation :
The respective x – coordinate and y- coordinates are written of the given points .

Question 2.
Plot each point in the coordinate plane above, and label each point with F, G, or H.
F (0, 4)
G (2, 1)
H (4\(\frac{3}{4}\), 3\(\frac{3}{4}\))
Answer:

Explanation :
The Given Points F, G and H are marked and is shown in the above graph .

Question 3.
a. Give coordinates for any three points that are on the same vertical line. Include at least one point that has a mixed number as a coordinate.
b. Give coordinates for any three points that are on the same horizontal line. Include at least one point that has a fraction as a coordinate.
Answer:
a. The 3 points are P ( 3, 2 ), Q ( 3, 3 ) and R (3, 3\(\frac{1}{4}\))
Explanation :
To form a Vertical line on the graph that mean x-coordinates will be same for all y-coordinates. then a vertical line is formed that is parallel to y-axis .
The points are Marked in the above graph .

b. The 3 points are A ( 1, 2 ), B ( 2\(\frac{1}{2}\), 2 ) and C (3\(\frac{1}{4}\), 2)

Explanation :
To form a Horizontal line on the graph that mean y-coordinates will be same for all x-coordinates. then a vertical line is formed that is parallel to x-axis .
The points are Marked in the above graph .

Question 4.
Garrett and Jeffrey are planning a treasure hunt. They decide to place a treasure at a point that is a distance of 5 units from the x-axis and 3 units from the y-axis. Jeffrey places a treasure at point J, and Garrett places one at point G. Who put the treasure in the right place? Explain how you know.

Answer:
Jeffrey puts the treasure in the right place
Explanation :
The treasure is at at a point that is a distance of 5 units from the x-axis and 3 units from the y-axis at a point ( 5, 3)
Jeffrey marks it correct where as the Garrett marks at a point that is a distance of 3 units from the x-axis and 5 units from the y-axis at a point ( 3, 5)

Question 5.
a. Find the y-coordinates by following the rules given for each table.
Table A: Multiply by \(\frac{1}{2}\).

Table B: Multiply by \(\frac{1}{4}\).

b. Graph and label the coordinate pairs from Table A. Connect the points, and label the line a. Graph and label the coordinate pairs from Table B. Connect the points, and label the line b.

c. Describe the relationship between the y-coordinates in Table A and Table B that have the same x-coordinate.
Answer:
a.
Table A: Multiply by \(\frac{1}{2}\).
y = \(\frac{1}{2}\)(x) .

Table B: Multiply by \(\frac{1}{4}\).
y=\(\frac{1}{4}\)(x)

Explanation :By following the given rule the y-coordinates are calculated and written in the above tabular columns .
b.

Explanation :
The points of tabular Column A are plotted and written as line a .
The points of tabular Column B are plotted and written as line b
All points are plotted and shown in above graph .
c.
The y coordinates in Tabular Column A are half of the y coordinates in the Tabular Column B

Question 6.
a. Use the graph to give the coordinate pairs of the points marked on the line.

b. Using this rule, generate three more points that would be on this line but lie beyond the portion of the coordinate plane that is pictured.
______ _______ ________
Answer:
a.
The coordinate points that are marked on the above graph are .

b.
The Rule used is the y-coordinate is double the x-coordinate and 2 added.
y = 2x + 2
The 3 points that lie on the above line beyond the given line in the above graph are .
x = 6 ; y = 2 (6) + 2 = 12 + 2 = 14 then (6, 14)
x= 8 ; y = 2 (8) + 2 = 16 + 2 = 18 then (8, 18)
x= 10; y = 2(10) + 2 = 20 + 2 = 22 then (10, 22 )
(6, 14), (8, 18) and (10, 22 ) are 3 points .

Engage NY Eureka Math 5th Grade Module 6 Lesson 34 Answer Key

Eureka Math Grade 5 Module 6 Lesson 34 Problem Set Answer Key

Use the chart below to evaluate your friend’s two boxes and lid. Measure and record the dimensions, and calculate the box volumes. Then, assess suitability, and suggest improvements in the adjacent columns.

Answer:

Eureka Math Grade 5 Module 6 Lesson 34 Reflection Answer Key

What are you most looking forward to learning about in Grade 6 or in math in your future?
Answer:
Necessary computational skills. This is by far among the very important basic math skills you should be able to learn and understand. Everyday situations require you to be knowledgeable in computations of whole numbers or fractions, decimals and this should be done without any calculator.

Engage NY Eureka Math 5th Grade Module 6 Lesson 33 Answer Key

Eureka Math Grade 5 Module 6 Lesson 33 Sprint Answer Key

A
Divide Decimals

Question 1.
1 ÷ 1 =
Answer:
1 ÷ 1 = 1

Question 2.
1 ÷ 0.1 =
Answer:
1 ÷ 0.1 = 10
Let us multiply the 0.1 by 10, which shifts the decimal point out of the way:
0.1 × 10 = 1
But we must also do it to the 1:
1× 10 = 10
So, 1 ÷ 0.1 has become 10 ÷ 1 (they are both 10 times larger):
10 ÷ 1 = 10
And So, the answer is:
1 ÷ 0.1 = 10

Question 3.
2 ÷ 0.1 =
Answer:
2 ÷ 0.1 = 20

Question 4.
7 ÷ 0.1 =
Answer:
7 ÷ 0.1 = 70

Question 5.
1 ÷ 0.1 =
Answer:
1 ÷ 0.1 = 10

Question 6.
10 ÷ 0.1 =
Answer:
10 ÷ 0.1 = 100
Explanation :
Let us multiply the 0.1 by 10, which shifts the decimal point out of the way:
0.1 × 10 = 1
But we must also do it to the 1:
10× 10 = 100
So, 10 ÷ 0.1 has become 100 ÷ 1 (they are both 10 times larger):
100 ÷ 1 = 100
And So, the answer is:
10 ÷ 0.1 = 100

Question 7.
20 ÷ 0.1 =
Answer:
20 ÷ 0.1 = 200

Question 8.
60 ÷ 0.1 =
Answer:
60 ÷ 0.1 = 600

Question 9.
1 ÷ 1 =
Answer:
1 ÷ 1 = 1

Question 10.
1 ÷ 0.1 =
Answer:
1 ÷ 0.1 = 10

Question 11.
10 ÷ 0.1 =
Answer:
10 ÷ 0.1 = 100

Question 12.
100 ÷ 0.1 =
Answer:
100 ÷ 0.1 = 1000
Explanation :
Let us multiply the 0.1 by 10, which shifts the decimal point out of the way:
0.1 × 10 = 1
But we must also do it to the 1:
100× 10 = 1000
So, 100 ÷ 0.1 has become 1000 ÷ 1 (they are both 10 times larger):
1000 ÷ 1 = 100
And So, the answer is:
100 ÷ 0.1 = 1000

Question 13.
200 ÷ 0.1 =
Answer:
200 ÷ 0.1 = 2000

Question 14.
800 ÷ 0.1 =
Answer:
800 ÷ 0.1 = 8000

Question 15.
1 ÷ 0.1 =
Answer:
1 ÷ 0.1 = 10

Question 16.
1 ÷ 0.01 =
Answer:
1 ÷ 0.01 = 100
Explanation :
Let us multiply the 0.01 by 100, which shifts the decimal point out of the way:
0.01 × 100 = 1
But we must also do it to the 1:
1× 100 = 100
So, 1 ÷ 0.01 has become 100 ÷ 1 (they are both 10 times larger):
100 ÷ 1 = 100
And So, the answer is:
1 ÷ 0.01 = 100

Question 17.
2 ÷ 0.01 =
Answer:
2 ÷ 0.01 = 200

Question 18.
9 ÷ 0.01 =
Answer:
9 ÷ 0.01 = 900

Question 19.
5 ÷ 0.01 =
Answer:
5 ÷ 0.01 = 500

Question 20.
50 ÷ 0.01 =
Answer:
50 ÷ 0.01 = 5000
Explanation :
Let us multiply the 0.01 by 100, which shifts the decimal point out of the way:
0.01 × 100 = 1
But we must also do it to the 1:
50× 100 = 5000
So, 50 ÷ 0.01 has become 5000 ÷ 1 (they are both 10 times larger):
5000 ÷ 1 = 5000
And So, the answer is:
50 ÷ 0.01 = 5000

Question 21.
60 ÷ 0.01 =
Answer:
60 ÷ 0.01 = 6000

Question 22.
20 ÷ 0.01 =
Answer:
20 ÷ 0.01 = 2000

Question 23.
5 ÷ 0.1 =
Answer:
5 ÷ 0.1 = 50

Question 24.
0.5 ÷ 0.1 =
Answer:
0.5 ÷ 0.1 = 5
Explanation :
Let us multiply the 0.1 by 10, which shifts the decimal point out of the way:
0.1 × 10 = 1
But we must also do it to the 1:
0.5 × 10 = 5
So, 0.5 ÷ 0.1has become 5 ÷ 1 (they are both 10 times larger):
5 ÷ 1= 5
And So, the answer is:
0.5 ÷ 0.1 = 5

Question 25.
0.05 ÷ 0.1 =
Answer:
0.05 ÷ 0.1 = 0.5
Explanation :
Let us multiply the 0.1 by 10, which shifts the decimal point out of the way:
0.1 × 10 = 1
But we must also do it to the 1:
0.05 × 10 = 0.5
So, 0.05 ÷ 0.1has become 0.5 ÷ 1 (they are both 10 times larger):
0.5 ÷ 1= 0.5
And So, the answer is:
0.05 ÷ 0.1 = 0.5

Question 26.
0.08 ÷ 0.1 =
Answer:
0.08 ÷ 0.1 = 0.8

Question 27.
4 ÷ 0.01 =
Answer:
4 ÷ 0.01 = 400

Question 28.
40 ÷ 0.01 =
Answer:
40 ÷ 0.01 = 4000

Question 29.
47 ÷ 0.01 =
Answer:
47 ÷ 0.01 = 4700

Question 30.
59 ÷ 0.01 =
Answer:
59 ÷ 0.01 = 5900

Question 31.
3 ÷ 0.1 =
Answer:
3 ÷ 0.1 = 30

Question 32.
30 ÷ 0.1 =
Answer:
30 ÷ 0.1 = 300

Question 33.
32 ÷ 0.1 =
Answer:
32 ÷ 0.1 = 320

Question 34.
32.5 ÷ 0.1 =
Answer:
32.5 ÷ 0.1 = 325
Explanation :
Let us multiply the 0.1 by 10, which shifts the decimal point out of the way:
0.1 × 10 = 1
But we must also do it to the 1:
32.5 × 10 = 325
So, 32.5 ÷ 0.1 has become 325 ÷ 1 (they are both 10 times larger):
325 ÷ 1 = 325
And So, the answer is:
32.5 ÷ 0.1 = 325

Question 35.
25 ÷ 5 =
Answer:
25 ÷ 5 = 5

Question 36.
2.5 ÷ 0.5 =
Answer:
2.5 ÷ 0.5 = 5

Question 37.
2.5 ÷ 0.05 =
Answer:
2.5 ÷ 0.05 = 500

Question 38.
3.6 ÷ 0.04 =
Answer:
3.6 ÷ 0.04 = 90

Question 39.
32 ÷ 0.08 =
Answer:
32 ÷ 0.08 = 400

Question 40.
56 ÷ 0.7 =
Answer:
56 ÷ 0.7 = 80

Question 41.
77 ÷ 1.1 =
Answer:
77 ÷ 1.1 = 70

Question 42.
4.8 ÷ 0.12 =
Answer:
4.8 ÷ 0.12 = 40

Question 43.
4.84 ÷ 0.4 =
Answer:
4.84 ÷ 0.4 = 12.1

Question 44.
9.63 ÷ 0.03 =
Answer:
9.63 ÷ 0.03 = 321

B
Divide Decimals

Question 1.
10 ÷ 1 =
Answer:
10 ÷ 1 = 10

Question 2.
1 ÷ 0.1 =
Answer:
1 ÷ 0.1 = 10

Question 3.
2 ÷ 0.1 =
Answer:
2 ÷ 0.1 = 20

Question 4.
8 ÷ 0.1 =
Answer:
8 ÷ 0.1 = 80

Question 5.
1 ÷ 0.1 =
Answer:
1 ÷ 0.1 = 10

Question 6.
10 ÷ 0.1 =
Answer:
10 ÷ 0.1 = 100

Question 7.
20 ÷ 0.1 =
Answer:
20 ÷ 0.1 = 200

Question 8.
70 ÷ 0.1 =
Answer:
70 ÷ 0.1 = 700

Question 9.
1 ÷ 1 =
Answer:
1 ÷ 1 = 1

Question 10.
1 ÷ 0.1 =
Answer:
1 ÷ 0.1 = 10

Question 11.
10 ÷ 0.1 =
Answer:
10 ÷ 0.1 = 100

Question 12.
100 ÷ 0.1 =
Answer:
100 ÷ 0.1 = 1000

Question 13.
200 ÷ 0.1 =
Answer:
200 ÷ 0.1 = 2000

Question 14.
900 ÷ 0.1 =
Answer:
900 ÷ 0.1 = 9000

Question 15.
1 ÷ 0.1 =
Answer:
1 ÷ 0.1 = 10

Question 16.
2 ÷ 0.01 =
Answer:
2 ÷ 0.01 = 200

Question 17.
7 ÷ 0.01 =
Answer:
7 ÷ 0.01 = 700

Question 18.
7 ÷ 0.01 =
Answer:
7 ÷ 0.01 = 700

Question 19.
4 ÷ 0.01 =
Answer:
4 ÷ 0.01 = 400

Question 20.
40 ÷ 0.01 =
Answer:
40 ÷ 0.01 = 4000

Question 21.
50 ÷ 0.01 =
Answer:
50 ÷ 0.01 = 5000

Question 22.
80 ÷ 0.01 =
Answer:
80 ÷ 0.01 = 8000

Question 23.
4 ÷ 0.1 =
Answer:
4 ÷ 0.1 = 40

Question 24.
0.4 ÷ 0.1 =
Answer:
0.4 ÷ 0.1 = 4

Question 25.
0.04 ÷ 0.1 =
Answer:
0.04 ÷ 0.1 = 0.4

Question 26.
0.07 ÷ 0.1 =
Answer:
0.07 ÷ 0.1 = 0.7

Question 27.
5 ÷ 0.01 =
Answer:
5 ÷ 0.01 = 0.5

Question 28.
50 ÷ 0.01 =
Answer:
50 ÷ 0.01 =5000

Question 29.
53 ÷ 0.01 =
Answer:
53 ÷ 0.01 = 5300

Question 30.
68 ÷ 0.01 =
Answer:
68 ÷ 0.01 = 6800

Question 31.
2 ÷ 0.1 =
Answer:
2 ÷ 0.1 = 20

Question 32.
20 ÷ 0.1 =
Answer:
20 ÷ 0.1 = 200

Question 33.
23 ÷ 0.1 =
Answer:
23 ÷ 0.1 = 230

Question 34.
23.6 ÷ 0.1 =
Answer:
23.6 ÷ 0.1 = 236

Question 35.
15 ÷ 5 =
Answer:
15 ÷ 5 = 3

Question 36.
1.5 ÷ 0.5 =
Answer:
1.5 ÷ 0.5 = 3

Question 37.
1.5 ÷ 0.05 =
Answer:
1.5 ÷ 0.05 = 30

Question 38.
3.2 ÷ 0.04 =
Answer:
3.2 ÷ 0.04 = 80

Question 39.
28 ÷ 0.07 =
Answer:
28 ÷ 0.07 = 400

Question 40.
42 ÷ 0.6 =
Answer:
42 ÷ 0.6 = 80

Question 41.
88 ÷ 1.1 =
Answer:
88 ÷ 1.1 = 80

Question 42.
3.6 ÷ 0.12 =
Answer:
3.6 ÷ 0.12 = 30

Question 43.
3.63 ÷ 0.3 =
Answer:
3.63 ÷ 0.3 = 12.1

Question 44.
8.44 ÷ 0.04 =
Answer:
8.44 ÷ 0.04 = 2.11

Eureka Math Grade 5 Module 6 Lesson 33 Problem Set Answer Key

Record the dimensions of your boxes and lid below. Explain your reasoning for the dimensions you chose for Box 2 and the lid.
BOX 1 (Can hold Box 2 inside.)
The dimensions of Box 1 are __________ × __________ × __________ .
Its volume is __________ .

BOX 2 (Fits inside of Box 1.)
The dimensions of Box 2 are __________ × __________ × __________ .
Reasoning:

LID (Fits snugly over Box 1 to protect the contents.)
The dimensions of the lid are __________ × __________ × __________ .
Reasoning:
Answer:
BOX 1 (Can hold Box 2 inside.)
The dimensions of Box 1 are 19cms × 13cms  × 4cms .
Its volume is Length × Breadth ×  Height .
v= l × b ×  h
v = 19 ×  13 ×  4 = 19 ×  52 = 988 cubic cms .

BOX 2 (Fits inside of Box 1.)
The dimensions of Box 2 are 9.5cms × 12.5cms × 3.5cms .
Reasoning:
I need a box that is smaller box to be just with the same width and height but its length up to half of the length .

LID (Fits snugly over Box 1 to protect the contents.)
The dimensions of the lid are 19.5 cms × 12.5 cms × 2 cms .
Reasoning:
The length and width need only to be little bit longer than the box of 0.25 cms on each side . The height of the lid needs to be probably needs to be 2cms .

Question 1.
What steps did you take to determine the dimensions of the lid?
Answer:
The dimensions of Box 1 are 19cms × 13cms  × 4cms .
First I, decided to add 0.25 cms to the length and the width so, that it is  slightly bigger than the box .
Then later, I figured the edges of the lid should cover half of the height of the box.
later checked the materials to make sure 21cms ×27cms had enough material …. if done .

Question 2.
Find the volume of Box 2. Then, find the difference in the volumes of Boxes 1 and 2.
Answer:
BOX 1
The dimensions of Box 1 are 19cms × 13cms  × 4cms .
Its volume is Length × Breadth ×  Height .
v= l × b ×  h
v = 19 ×  13 ×  4 = 19 ×  52 = 988 cubic cms .
BOX 2
The dimensions of Box 1 are 9.5cms × 12.5cms  × 3.5cms .
Its volume is Length × Breadth ×  Height .
v= l × b ×  h
V = 9.5cms × 12.5cms  × 3.5cms = 9.5  × 43.75 = 415.625 cubic cms.
The Differences between Box 1 and Box 2 = 988 – 415.625 =572.375 cubic cms.

Question 3.
Imagine Box 3 is created such that each dimension is 1 cm less than that of Box 2. What would the volume of Box 3 be?
Answer:
BOX 2
The dimensions of Box 1 are 9.5cms × 12.5cms  × 3.5cms .
Box 3 : 1 cms less in each dimension
The dimensions of Box 1 are 8.5cms × 11.5cms  × 2.5cms .
Its volume is Length × Breadth ×  Height .
v= l × b ×  h
V = 8.5cms × 11.5cms  × 2.5cms = 8.5  × 28.75 = 244.375 cubic cms.

Eureka Math Grade 5 Module 6 Lesson 33 Reflection Answer Key

Today, you made a box for a special purpose. It shows one way that math is used all the time to create containers. When might there be other opportunities for you to use the math you have learned in elementary school?
Answer:
Math is an important part of learning for children in the early years because it provides vital life skills. They will help children problem solve, measure and develop their own spatial awareness, and teach them how to use and understand shapes.

Eureka Math Grade 5 Module 6 Lesson 33 Homework Answer Key

Question 1.
Find various rectangular boxes at your home. Use a ruler to measure the dimensions of each box to the nearest centimeter. Then, calculate the volume of each box. The first one is partially done for you.

Answer:
Its volume is Length × Breadth ×  Height .
v= l × b ×  h

Question 2.
The dimensions of a small juice box are 11 cm by 4 cm by 7 cm. The super-size juice box has the same height of 11 cm but double the volume. Give two sets of the possible dimensions of the super-size juice box and the volume.
Answer:
The Small Juice box =11 cm by 4 cm by 7 cm.
The Volume of the small juice box = Length × Breadth ×  Height .
v = 11 × 4 × 7 = 11 × 28 = 308 cubic cms.
Height = 11 cm
If height is same volume is double means
volume = 2 × 308 = 616 cubic cms .
The super juice box will have change in length or breadth
As volume is double it is multiplied by 2
616 = 11 × length × width
56 = length × width
56 = 14 × 4 or 8 × 7
Length can be 14 cm then width will be 4 cm
length can be 8 cm then width will be 7 cm .

Engage NY Eureka Math 5th Grade Module 6 Lesson 32 Answer Key

Eureka Math Grade 5 Module 6 Lesson 32 Problem Set Answer Key

Question 1.
Ashley decides to save money, but she wants to build it up over a year. She starts with $1.00 and adds 1 more dollar each week. Complete the table to show how much she will have saved after a year.

Answer:

Explanation :
Amount with which ashley started = $1
Each week she adds $1 more that the previous week
So, Week she has $1 in her account
Week 2 she adds one more dollar than previous week so, that is $2 after adding her account balance is $2 +$1 = $3
Week 3 she adds one more than than previous week so, that is $3 after adding her account balance is $3 + $3 = $ 6  and so on …. till Week 52 the balance amount is calculated .

Question 2.
Carly wants to save money, too, but she has to start with the smaller denomination of quarters. Complete the second chart to show how much she will have saved by the end of the year if she adds a quarter more each week. Try it yourself, if you can and want to!

Answer:

Explanation :
Amount with which Carlie started = $0.25 (Quarters)
Each week she adds $0.25 more that the previous week
So, Week 1 she starts with  $0.25 in her account
Week 2 she adds one more Quarter dollar than previous week so, that is $0.50 after adding her account balance is $0.25 +$0.50 = $0.75
Week 3 she adds one more Quarter dollar than previous week so, that is $0.75 after adding her account balance is $0.75 + $0.75 = $ 1.50  and so on ….  till week 52 amount is calculated .

Question 3.
David decides he wants to save even more money than Ashley did. He does so by adding the next Fibonacci number instead of adding $1.00 each week. Use your calculator to fill in the chart and find out how much money he will have saved by the end of the year. Is this realistic for most people? Explain your answer.

Answer:

Explanation :
By end of the year he will save $86,267,571,271 .
It is not realistic because In Fibonacci number the rule is followed
So we can write the rule:
The Rule is x_n = x_n−1 + x_n−2
where:
x_n is term number “n”
x_n−1 is the previous term (n−1)
x_n−2 is the term before that (n−2)
In Fibonacci rule the previous term and the term before that is added to get next fibonacci number , That means the David should starts with $1 . for week he should add $1 for week1 later, he should add $1 the previous money and before that week in week 2 total is $2,
In week 3 he should add $2 that double of previous then amount becomes $4. and so on till week 52 for every week he should add nearly 2 weeks amount which will be difficulty to add. so this not realistic to follow this rule as Savings are done with smaller amounts not with huge amounts .

Eureka Math Grade 5 Module 6 Lesson 32 Reflection Answer Key

Today, we watched how savings can grow over time, but we did not discuss how the money saved was earned. Have you ever thought about how math skills might help you to earn money? If so, what are some jobs that might require strong math skills? If not, think about it now. How might you make a living using math skills?
Answer:
Money math involves many of the math skills you learn in school, such as addition, subtraction, multiplication, division, fractions, decimals, and percentages. … Handling money can help you learn how to count it. Making purchases in a store can help you become comfortable counting money.
Some jobs that might require strong math skills are in Auditor, Statistician, Actuary, Mathematician, Operations Research Analyst, Math Professor, Banking , Cashier ,

Eureka Math Grade 5 Module 6 Lesson 32 Homework Answer Key

Question 1.
Jonas played with the Fibonacci sequence he learned in class. Complete the table he started.

Answer:

Explanation :
The Fibonacci Sequence can be written as a “Rule”
First, the terms are numbered from 0 onwards like this:
So we can write the rule:
The Rule is x_n = x_n−1 + x_n−2
where:
x_n is term number “n”
x_n−1 is the previous term (n−1)
x_n−2 is the term before that (n−2)

Question 2.
As he looked at the numbers, Jonas realized he could play with them. He took two consecutive numbers in the pattern and multiplied them by themselves and then added them together. He found they made another number in the pattern. For example, (3 × 3) + (2 × 2) = 13, another number in the pattern. Jonas said this was true for any two consecutive Fibonacci numbers. Was Jonas correct? Show your reasoning by giving at least two examples of why he was or was not correct.
Answer:
Yes he is write .

Explanation :
The Fibonacci number of 6 is 8
The Fibonacci number of 7 is  5
Taking two consecutive numbers and squaring and adding them we get,
( 3 × 3)+(5 × 5 ) = 9 + 25 = 34
34 is the Fibonacci number of 9 (4 + 5 = 9) It is true
The Fibonacci number of 6 is 8
The Fibonacci number of 7 is 13
Taking two consecutive numbers and squaring and adding them we get,
( 8 × 8 )+(13 × 13 ) = 64 + 169 =  233
233 is the Fibonacci number of 13 (6 + 7 = 13) It is true

Question 3.
Fibonacci numbers can be found in many places in nature, for example, the number of petals in a daisy, the number of spirals in a pine cone or a pineapple, and even the way branches grow on a tree. Find an example of something natural where you can see a Fibonacci number in action, and sketch it here.
Answer:
In the below figure we see two different pine cone patterns.
In figure 1 The Fibonacci spirals are shown clearly
In Figure 2 The Fibonacci spirals of another pine cone is shown