enVisionmath 2.0: Grade 6, Volume 1

Chapter 2 Integers And Rational Numbers

Section 2.5: Find Distances On The Coordinate Plane

Page 95 Exercise 1 Answer

Plot the points (3,3), (0,0), (−4,−4), (−9,0), (−4,4), (0,0), (3,−3), and (3,3).

When the points are connected in order they make a fish.

You can use the structure of the grid to find a pair of points that are the same distance from the x-axis. The absolute value of the y-coordinate represents the distance of the point from the x-axis. Opposites have the same absolute value so points with opposite y-coordinates will be the same distance from the x-axis.

A pair of points that are the same distance from the x-axis are for example (−4,4) and (−4,−4), because their y-coordinates are opposites.

∣4∣ = ∣−4∣ = 4

When the points are connected in order they make a fish. The points (−4,4) and (−4,−4) are the same distance from the x-axis because their y-coordinates are opposites.

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Page 95 Exercise 1 Answer

To find the total length of the picture graphed in the Solve & Discuss It! problem, we need to look at the point that is farthest to the left and the point that is farthest to the right.

The point farthest to the left is (−9,0) and the points farthest to the right are (3,−3) and (3,−3).

The length of the picture is the horizontal distance between these two points so we can look at the x-coordinates of these points.

For (−9,0), the horizontal distance to the origin is 9 since ∣−9∣ = 9.

For (3,−3) and (3,3), the horizontal distance to the origin is 3 since ∣3∣ = 3.

The total length of the picture is then 9 + 3 = 12 units.

Result

The total length of the picture is the horizontal distance between the point farthest on the left and the points farthest on the right. Since the x-coordinates of these points are −9 and 3, then the distance between the points is ∣−9∣ + ∣3∣ = 12 units.

Page 96 Exercise 1 Answer

The school is located at (−4,2) and the playground is located at (−4,5). Since the playground and the school have the same x-coordinates, we can use their y-coordinates to find the distance between them. They are in the same quadrant, Quadrant II, so to find the distance between them we must calculate the difference between their y-coordinates, which is ∣5∣ − ∣2∣ = 5 − 2 = 3.

Result

The distance from the school to the playground is 3 miles.

Page 97 Exercise 2 Answer

∣−85.5∣ + ∣65.5∣ = 85.5 + 65.5 = 151

The total distance of the Coulter’s return trip is 151 miles.

If we place the Water Park and the Coulter’s Home in a coordinate plane as points they are not in the same quadrants, the Water Park is in Quadrant III and the Coulter’s Home is in Quadrant IV, so we take the absolute value of their x-coordinates, since their y-coordinates are the same, and add them to find the total distance of their return trip.

Result

The total distance of the Coulter’s return trip is 151 miles.

Page 97 Exercise 3 Answer

The coordinates of point A are (5,7) and of point B (5,0) so the distance from point A to point B is 7, as seen in the Example 3. Point D is the same distance from point B as point A, so if we add the absolute values of the y-coordinates of point B and point D we must get the same number as if we added the absolute values point A and point B.

∣7∣ + ∣0∣ = 7 + 0 = 7

Since point D is in Quadrant IV it has a positive x-coordinate and a negative y-coordinate so the y-coordinate of point D is −7, and the x-coordinate is the same as that of points A and B.

Result

The coordinates of point D are (5,−7).

Page 98 Exercise 1 Answer

To find the distance between two points on a coordinate plane we can use absolute value by adding or subtracting the absolute values of their x-coordinates or y-coordinates.

For example, if A(4,−3) and B(4,8), then to the distance between A and B we must add the absolute values of their y-coordinates since their x-coordinates are the same.

∣−3∣ + ∣8∣ = −(−3) + 8 = 3 + 8 = 11

Thus, the distance from A to B is 11.

Another example, if C(2,7) and D(2,5), then to find the distance from C to D we must subtract their y-coordinates.

∣7∣ − ∣5∣ = 7 − 5 = 2

The distance from C to D is 2.

Result

To find the distance between two points on a coordinate plane we can use absolute value by adding or subtracting the absolute values of their x-coordinates or y-coordinates.

Page 98 Exercise 2 Answer

When two points are in the same quadrant we subtract their absolute values and when they are not in the same quadrant we add their absolute values to find the distance between them.

For example, point A(3,4) and B(7,4) are both in the first quadrant. They have the same y-coordinates, however, their x-coordinates differ. To find the distance from A to B we must subtract the absolute values of their x-coordinates.

∣7∣ − ∣3∣ = 7 − 3 = 4

Thus, the distance from A to B is four.

A different example, C(-3,4) and D(-3,-9) are not in the same quadrant, C is in the second quadrant and D in the third. They have the same x-coordinates and different y-coordinates. To find the distance between them we must add the absolute values of their y-coordinates.

∣4∣ + ∣−9∣ = 4 + (−(−9)) = 4 + 9 = 13

Thus, the distance from C to D is 13.

Result

When two points are in the same quadrant we subtract their absolute values and when they are not in the same quadrant we add their absolute values to find the distance between them.

Page 98 Exercise 3 Answer

From Example 1, Li’s house has coordinates of (−4,−3) and Tammy’s house has coordinates of (0,0).

Li’s house and Tammy’s house do not have the same x-coordinates or the same y-coordinates. Therefore, we cannot use absolute values to find the distance between Li’s house and Tammy’s house.

Result

No, we cannot use absolute values to find the distance because they don’t have the same x-coordinates or the same y-coordinates.

Page 98 Exercise 4 Answer

To find the distance from (-5,2) to (-5,6), since they have the same x-coordinates and are in the same quadrant, the second quadrant, we must subtract the absolute values of their y-coordinates.

∣6∣ − ∣2∣ = 6 − 2 = 4

Result

The distance between (−5,2) and (−5,6) is four.

Page 98 Exercise 5 Answer

To find the distance from (4.5, -3.3) to (4.5, 5.5), since they have the same x-coordinates and are not in the same quadrant, (4.5, -3.3) is in the second quadrant and (4.5, 5.5) in the first, so we must add the absolute values of their y-coordinates.

∣−3.3∣ + ∣5.5∣ = −(−3.3) + 5.5 = 3.3 + 5.5 = 8.8

Result

The distance between (4.5,−3.3) and (4.5,5.5) is 8.8.

Page 98 Exercise 6 Answer

To find the distance from (\(5 \frac{1}{2}\), –\(7 \frac{1}{2}\)) to (\(5 \frac{1}{2}\), –\(1 \frac{1}{2}\)), since they have the same x-coordinates and are in the same quadrant, the fourth quadrant, we must subtract the absolute values of their y-coordinates.

Result

The distance between (​\(5 \frac{1}{2}\), –\(7 \frac{1}{2}\)) and (\(5 \frac{1}{2}\), –\(1 \frac{1}{2}\)) is six.

Page 98 Exercise 7 Answer

To find the distance from (-\(2 \frac{1}{4}\),−8) to (\(7 \frac{3}{4}\),−8), since they have the same y-coordinates and are not in the same quadrant, (-\(2 \frac{1}{4}\),−8) is in the third quadrant and (\(7 \frac{3}{4}\),−8) in the fourth, so we must add the absolute values of their x-coordinates.

Result

The distance between (-\(2 \frac{1}{4}\),−8) and (\(7 \frac{3}{4}\),−8) is ten.

Page 98 Exercise 8 Answer

To find the distance from (\(5 \frac{1}{4}\), –\(\frac{1}{4}\)) to (\(5 \frac{1}{4}\), –\(6 \frac{1}{4}\)), since they have the same x-coordinates and are in the same quadrant, the fourth quadrant, we must subtract the absolute values of their y-coordinates.

Result

The distance between (\(5 \frac{1}{4}\), –\(\frac{1}{4}\)) and (\(5 \frac{1}{4}\), –\(6 \frac{1}{4}\)) is three.

Page 98 Exercise 9 Answer

To find the distance from (-\(1 \frac{1}{2}\), –\(6 \frac{1}{2}\)) to (-\(2 \frac{1}{2}\), –\(6 \frac{1}{2}\)), since they have the same y-coordinates and are in the same quadrant, the third quadrant, we must subtract the absolute values of their x-coordinates.

Result

The distance between (-\(1 \frac{1}{2}\), –\(6 \frac{1}{2}\)) and (-\(2 \frac{1}{2}\), –\(6 \frac{1}{2}\)) is one.

Page 99 Exercise 10 Answer

Since their y-coordinates are the same and (−2,8) is in Quadrant II and (7,8) in Quadrant I, to find the distance between (−2,8) and (7,8), we need to calculate the sum of absolute values of their x-coordinates.

∣−2∣ + ∣7∣ = 2 + 7 = 9

Result

The distance between (−2,8) and (7,8) is 9 units.

Page 99 Exercise 11 Answer

Since their x-coordinates are the same and both (−6.1,−8.4) and (−6.1,−4.2) are in Quadrant III, to find the distance between (−6.1,−8.4) and (−6.1,−4.2), we need to calculate the difference of absolute values of their y-coordinates.

∣−8.4∣ − ∣−4.2∣ = 8.4 − 4.2 = 4.2

Result

The distance between (−6.1,−8.4) and (−6.1,−4.2) is 4.2 units.

Page 99 Exercise 12 Answer

Since their y-coordinates are the same and (\(12 \frac{1}{2}\), \(3 \frac{3}{4}\)) is in Quadrant I and (-\(4 \frac{1}{2}\), \(3 \frac{3}{4}\)) in Quadrant II, to find the distance between (\(12 \frac{1}{2}\), \(3 \frac{3}{4}\)) and (-\(4 \frac{1}{2}\), \(3 \frac{3}{4}\)), we need to calculate the sum of absolute values of their x-coordinates.

Result

The distance between (\(12 \frac{1}{2}\), \(3 \frac{3}{4}\)) and (-\(4 \frac{1}{2}\), \(3 \frac{3}{4}\)) is 17 units.

Page 99 Exercise 13 Answer

Since their x-coordinates are the same and both (−5,−3) and (−5,−6) are in Quadrant III, to find the distance between (−5,−3) and (−5,−6), we need to calculate the difference of absolute values of their y-coordinates.

∣−6∣ − ∣−3∣ = 6 − 3 = 3

Result

The distance between (−5,−3) and (−5,−6) is 3 units.

Page 99 Exercise 14 Answer

Since their y-coordinates are the same and (−5.4,4.7) is in Quadrant II and (0.6,4.7) in Quadrant I, to find the distance between (−5.4,4.7) and (0.6,4.7), we need to calculate the sum of absolute values of their x-coordinates.

∣0.6∣ + ∣−5.4∣ = 0.6 + 5.4 = 6

Result

The distance between (−5.4,4.7) and (0.6,4.7) is 6 units.

Page 99 Exercise 15 Answer

Since their x-coordinates are the same and both (\(7 \frac{1}{2}\), –\(5 \frac{3}{4}\)) and (\(7 \frac{1}{2}\), –\(1 \frac{1}{4}\)) is in Quadrant IV, to find the distance between (\(7 \frac{1}{2}\), –\(5 \frac{3}{4}\)) and (\(7 \frac{1}{2}\), –\(1 \frac{1}{4}\)), we need to calculate the difference of absolute values of their y-coordinates.

Result

The distance between (\(7 \frac{1}{2}\), –\(5 \frac{3}{4}\)) and (\(7 \frac{1}{2}\), –\(1 \frac{1}{4}\)) is \(4 \frac{1}{2}\) units.

Page 99 Exercise 16 Answer

First, we need to find the coordinates which represent the location of roller coaster 1 and the swings. Follow the grid lines directly to the x-axis to find the x-coordinates. Follow the grid lines directly to the y-axis to find the y-coordinates. The coordinates of roller coaster 1 are (−6,7) and the coordinates of the swings are (1,7). Since (−6,7) is in Quadrant II and (1,7) is Quadrant I and their y-coordinates are the same, to find the distance from roller coaster 1 to the swings the sum of absolute values of their x-coordinate.

∣1∣ + ∣−6∣ = 1 + 6 = 7

Result

The distance from roller coaster 1 to the swings is 7 units.

Page 99 Exercise 17 Answer

First, we need to find the coordinates of the Ferris wheel and the roller coaster 3. Follow the grid lines directly to the x-axis to find the x-coordinates. Follow the grid lines directly to the y-axis to find the y-coordinates. The coordinates of the Ferris wheel (-6,2) and the coordinatess of the roller coaster 3 (-6,-8).

Since (-6,2) is in Quadrant II and (-6,-8) is in Quadrant III and their x-coordinates are the same, to find the distance between (-6,2) and (-6,-8) we need to calculate the sum of their y-coordinates.

∣2∣ + ∣−8∣ = 2 + 8 = 10

Result

The distance between the Ferris wheel and the roller coaster 3 is 10 units.

Page 99 Exercise 18 Answer

The total distance from roller coaster 2 to roller coaster 3 and then to the water slide is the sum of the distance from roller coaster 2 to roller coaster 3 and from roller coaster 3 to the water slide. First we must find their coordinates. Follow the grid line directly to the x-axis to find the x-coordinates. Follow the grid line directly to the y-axis to find the y-coordinates.

The coordinates of roller coaster 2 are (-6,-3), of roller coaster 3 are (-6,-8), and of the water slide are (8, -8). Both (-6,-3) and (-6,-8) are in Quadrant III and (8, -8) is in Quadrant IV.

To find the distance between roller coaster 2 and roller coaster 3, since their x-coordinates and they are in the same quadrant, we need to calculate the difference between their y-coordinates.

∣−8∣ − ∣−3∣ = 8 − 3 = 5

The distance between roller coaster 2 and roller coaster 3 is 5 units.

To find the distance between roller coaster 3 and the water slide, since their y-coordinates and (-6,-8) is in Quadrant III and (8, -8) is in Quadrant IV, we need to calculate the sum between their x-coordinates.

∣−6∣ + ∣8∣ = 6 + 8 = 14

The distance between roller coaster 3 and the water slide is 14 units.

The total distance from roller coaster 2 to roller coaster 3 and then to the water slide is the sum

5 + 14 = 19.

Result

The total distance from roller coaster 2 to roller coaster 3 and then to the water slide is 19 units.

Page 99 Exercise 19 Answer

Since distance is defined as the positive length between two points, the distance between two points is always the same no matter which point is chosen as the starting point.

This means the distance between the merry-go-round and the water slide is the same as the distance between the water slide and merry-go-round since the two distances represent the same positive length between the two locations.

Result

The distances are the same since distance between two points is always the same no matter which point is chosen as the starting point.

Page 100 Exercise 20 Answer

H is graphed at (n,−6) and J at (n,−3). Since H and J have the same x-coordinate and are in the same quadrant, to find the distance between them we must caculate the difference of the absolute values of their y-coordinates.

∣−6∣ − ∣−3∣ = 6 − 3 = 3

Result

The distance between H and J is 3 units.

Page 100 Exercise 21 Answer

From the graph we can see that points H and G have the same y-coordinates, which are both equal to -6. Since point G is on the y-axis, its x-coordinate is zero. So, point G is graphed at (0,-6).

The distance between point H and point G is equal to the distance of point H to point J which is 3 units as calculated in the Exercise 20 on page 100.

To find the value of n, we must calcualte the following:

∣n∣ − ∣0∣ = 3

n − 0 = 3

n = 3

Result

n = 3

Page 100 Exercise 22 Answer

Since a, b, and c are all negative numbers, both points (a,b) and (a,c) are in Quadrant III. Their x-coordinates are the same so to find the distance between (a,b) and (a,c) we need to calculate the difference of the absolute values of their y-coordinates.

We don’t know whether b is greater than c or if c is greater than b. Since the distance is always represented by a positive number, we will take the absolute value of the difference.

∣∣b∣ − ∣c∣∣ = ∣b−c∣

Suppose b is greater then c, then the distance is b − c. However if c is greater than b, then the distance is c − b.

Result

If b is greater than c, then the distance is b − c and if c is greater than b, then the distance is c − b.

Page 100 Exercise 23 Answer

To answer the question, we must find the distance between points (-1,8) and (85,8). When placed in a coordinate plane, the epicenter (-1,8) is in Quadrant II and the farthest location reported to have felt the earthquake (85,8) is in Quadrant I. Since their y-coordinates are the same, to find the distance between (-1,8) and (85,8) we need to calculate the sum of their x-coordinates.

∣−1∣ + ∣85∣ = 1 + 85 = 86

Result

The earthquake was felt 86 miles from the epicenter.

Page 100 Exercise 24 Answer

To find the dimensions of the rectangular piece of land we need to find the distance between points A and B, and the distance between points A and D. From the graph we can determine the coordinates of points A, B, and D, which are in order (0,0), (0, \(4 \frac{1}{2}\)), and (\(5 \frac{1}{2}\),0).

Both points A and B are on the y-axis so their x-coordinates are zero. The distance between them is the difference of absolute values of their y-corodinates.

The distance between points A and B is \(4 \frac{1}{2}\) units.

Since each unit represents 1,000 feet, the distance between points A and B is 4,500 feet.

Both points A and D are on the x-axis so their y-coordinates are zero. The distance between them is the difference of absolute values of their x-corodinates.

The distance between points A and D is 5,500 feet.

Result

The width of the rectangular piece of land is 4,500 feet and the length is 5,500 feet.

Page 100 Exercise 25 Answer

To find two ordered pairs that are 4.5 units apart, we can choose from the ones which have either the same x-coordinates or the same y-coordinates, and from those which are in the same quadrant or those that are not. For example, (1,1) and (1,5.5) are both in Quadrant I, have the same x-coordinates, and are 4.5 units apart.

∣5.5∣ − ∣1∣ = 5.5 − 1 = 4.5

Points (3.5,−1) and (−1,−1) have the same y-quadrant, but point (3.5,−1) is in Quadrant IV and point (−1,−1) is in Quadrant III, so the distance among them is the sum of absolute values of their x-coordinates.

∣3.5∣ + ∣−1∣ = 3.5 + 1 = 4.5

Result

(1,1) and (1,5.5), (3.5,−1) and (−1,−1)

Chapter 2 Integers And Rational Numbers

Vocabulary Review

Page 107 Exercise 1 Answer

A point on a coordinate plane is represented by an ordered pair.

Coordinate plane is made up of two perprendicular number lines who meet at the origin, the x-axis and the y-axis. Each point on a coordinate plane has a unique ordered pair which describes its location.

For example, let’s take point A whose x-coordinate is 3 and y-coordinate is −4.

A(3,−4)

Result

Ordered pair.

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Page 107 Exercise 2 Answer

The opposite of a positive integer is a negative integer.

For example, the opposite number of 3 is −3, and the opposite number of −18 is 18.

Result

Opposite.

Page 107 Exercise 3 Answer

A rational number is any number that can be written as the quotient of two integers.

For example, a fraction \(\frac{3}{4}\) is a rational number because it can be written as a quotient 3 ÷ 4, or a mixed number \(7 \frac{1}{4}\), which can be written as \(\frac{29}{4}\) and further as 29 ÷ 4, or a decimal such as 0.25, which can be written as a fraction \(\frac{1}{4}\) and further as 1 ÷ 4.

Result

Rational number.

Page 108 Exercise 1 Answer

We need to plot \(\frac{3}{4}\) on the given number line.

The number line is divided into tenths so it would be easier to graph if we converted \(\frac{3}{4}\) to a decimal:

The location of \(\frac{3}{4}\) is then halfway between 0.7 and 0.8, which are the 7th and 8th tick marks to the right of 0 on the number line:

Result

Plot a point halfway between the 7th and 8th tick marks to the right of 0 on the number line.

Page 108 Exercise 2 Answer

We need to plot –\(\frac{2}{5}\) on the given number line.

The number line is divided into tenths so it would be easier to graph if we converted –\(\frac{2}{5}\) to a decimal:

\(-\frac{2}{5}=-\frac{4}{10}=-0.4[\latex]

The location of –[latex]\frac{2}{5}\) is then −0.4, which is the 4th tick mark to the left of 0 on the number line:

Result

Plot a point at −0.4, which is the 4th tick mark to the left of 0 on the number line.

To plot –\(\frac{2}{5}\) divide the first unit to the left of zero into five equal parts and mark the second from the left or find −0,4 on the number line since –\(\frac{2}{5}\) = −0,4.

Result

Open to see the solution.

Page 108 Exercise 3 Answer

To plot 0.5 divide the first unit to the right of zero into half, since 0.5 = \(\frac{1}{2}\).

Result

Open to see the solution.

The given number line is divided into tenths so to plot 0.5 on the number line, plot a point halfway between 0 and 1, which is the 5th tick mark to the right of 0:

Result

Plot a point halfway between 0 and 1.

Page 108 Exercise 4 Answer

We need to compare 0.25 and \(\frac{1}{4}\)

Two rational numbers are easier to compare if they are both fractions or both decimals.

Converting 0.25 to a fraction gives:
​
\(0.25=\frac{25}{100}=\frac{1}{4}\)

Therefore, 0.25 = \(\frac{1}{4}\).

Result

0.25 = \(\frac{1}{4}\)

Fraction \(\frac{1}{4}\) can be written as a quotient 1 ÷ 4 which than gives 0.25. Thus

0.25 = \(\frac{1}{4}\).

Result

0.25 = \(\frac{1}{4}\)

Page 108 Exercise 5 Answer

We need to compare \(1 \frac{5}{8}\) and 1.6.

Two rational numbers are easier to compare if they are both fractions or both decimals.

Dividing 5 and 8 gives:

We can then rewrite \(1 \frac{5}{8}\) as the decimal 1.625.

Since 1.625 > 1.6, then \(1 \frac{5}{8}\) > 1.6

Result

\(1 \frac{5}{8}\) < 1.6

Fraction \(1 \frac{5}{8}\) is equal to \(\frac{13}{8}\), which can be further written as a quotient 13 ÷ 8 which is equal to 1.625. Thus \(1 \frac{5}{8}\) > 1.6.

Result

\(1 \frac{5}{8}\) > 1.6.

Page 108 Exercise 6 Answer

We need to compare 3.65 and \(3 \frac{3}{4}\).

Two rational numbers are easier to compare if they are both fractions or both decimals.

Converting \(3 \frac{3}{4}\) to a decimal gives:

Since 3.65 < 3.75, then 3.65 < \(3 \frac{3}{4}\)

Result

3.65 < \(3 \frac{3}{4}\)
​
Fraction \(3 \frac{3}{4}\) is equal to \(\frac{15}{4}\) which can be written as a quotient 15 ÷ 4 which is equal to 3.75. Thus,

3.65 < \(3 \frac{3}{4}\).

Result

3.65 < \(3 \frac{3}{4}\)
​
Page 108 Exercise 7 Answer

To compare two fractions we must rewrite them with the same denominator. To do that first we need to find a common multiple of 3 and 4.

Thus,

Result

Page 108 Exercise 1 Answer

The integer for point A is 3 since A is three units to the right from zero. The opposite number is −3.

Result

The integer is 3 and its opposite is −3.

Page 108 Exercise 2 Answer

The integer for point B is −1 since point B is one unit left from zero. The opposite number is 1.

Result

The integer is −1 and its opposite is 1.

Page 108 Exercise 3 Answer

The integer for point C is 6 since point C is six units right from zero. Its opposite number is −6.

Result

The integer is 6 and its opposite is −6.

Page 108 Exercise 4 Answer

The integer for point D is −7 since it is seven units left from zero. Its opposite number is 7.

Result

The integer is −7 and its opposite number is 7.

Page 108 Exercise 5 Answer

The integer for point E is −5 since it is five units left from zero. Its opposite number is 5.

Result

The integer is −5 and its opposite number is 5.

Page 108 Exercise 6 Answer

The integer for point F is 1 since it is one unit right from zero. Its opposite number is −1.

Result

The integer is 1 and its opposite is −1.

Page 109 Exercise 1 Answer

Absolute value of a number is either positive or equal to zero. It is equal to zero if and only if the number is zero. The absolute value of a number represents the distance of that number from zero on the number line.

−9 is nine units left from zero on the number line, thus

∣−9∣ = 9.

Result

The absolute value of −9 is 9.

Page 109 Exercise 2 Answer

Absolute value of a number is either positive or equal to zero. It is equal to zero if and only if the number is zero. The absolute value of a number represents the distance of that number from zero on the number line.

−2 is two units left from zero on the number line, thus

∣−2∣ = 2.

Result

The absolute value of −2 is 2.

Page 109 Exercise 3 Answer

Absolute value of a number is either positive or equal to zero. It is equal to zero if and only if the number is zero. The absolute value of a number represents the distance of that number from zero on the number line.

4 is four units right from zero on the number line, thus

∣4∣ = 4.

Result

The absolute value of 4 is 4.

Page 109 Exercise 4 Answer

Absolute value of a number is either positive or equal to zero. It is equal to zero if and only if the number is zero. The absolute value of a number represents the distance of that number from zero on the number line.

Result

The solution is −10.

Page 109 Exercise 5 Answer

First, find the absolute values.

∣−3∣ = 3

∣−2∣ = 2

∣10∣ = 10

Ordered from least to greatest:

2, 3, 10 → ∣−2∣, ∣−3∣, ∣10∣

Result

∣−2∣, ∣−3∣, ∣10∣

Page 109 Exercise 6 Answer

First, find the absolute values.

∣−7∣ = 7

∣0∣ = 0

∣−5∣ = 5

Ordered from least to greatest:

0, 5, 7 → ∣0∣, ∣−5∣, ∣−7∣

Result

∣0∣, ∣−5∣, ∣−7∣

Page 109 Exercise 7 Answer

First, find the absolute values.

∣−18.5∣ = 18.5

∣18∣ = 18

∣−12.5∣ = 12.5

Ordered from least to greatest:

12.5, 18, 18.5 → ∣−12.5∣, ∣18∣, ∣−18.5∣

Result

∣−12.5∣, ∣18∣, ∣−18.5∣

Page 109 Exercise 8 Answer

First, find the absolute values.

∣26∣ = 26

∣−20∣ = 20

∣−24.5∣ = 24.5

Ordered from least to greatest:

20, 24.5, 26 → ∣−20∣, ∣−24.5∣, ∣26∣

Result

∣−20∣, ∣−24.5∣, ∣26∣

Page 109 Exercise 1 Answer

To give the ordered pair for point U follow the grid lines directly from the point to the y-axis to find the y-coordinate, which is 2.5. Since the point is in the y-axis, its x-coordinate is zero.

Result

U(0,2.5)

Page 109 Exercise 2 Answer

To give the ordered pair for point V follow the grid lines directly from the point to the x -axis to find the x-coordinate, which is −2. Follow the grid lines directly from the point to the y – axis to find the y-coordinate, 1.5.

Result

V(−2,1.5)

Page 109 Exercise 3 Answer

To give the ordered pair for point W follow the grid lines directly from the point to the x -axis to find the x-coordinate, which is −4. Follow the grid lines directly from the point to the y -axis to find the y-coordinate, −1.

Result

W(−4,−1)

Page 109 Exercise 4 Answer

To give the ordered pair for point X follow the grid lines directly from the point to the x -axis to find the x-coordinate, which is 2.5. Since the point is in the x-axis, its y-coordinate is zero.

Result

X(2.5,0)

Page 109 Exercise 5 Answer

To give the ordered pair for point Y follow the grid lines directly from the point to the x-axis to find the x-coordinate, which is 2. Follow the grid lines directly from the point to the y-axis to find the y-coordinate, −1.5.

Result

Y(2,−1.5)

Page 109 Exercise 6 Answer

To give the ordered pair for point Z follow the grid lines directly from the point $\textbf{to the x-axis to find the x-coordinate, which is −1.5. Follow the grid lines directly from the point to the y-axis to find the y-coordinate, −3.

Result

Z(−1.5,−3)

Page 110 Exercise 1 Answer

To find the length of side BC first find the ordered pairs which represent them. The ordered pair for point B is B(−1,2), and the ordered pair for point C is C(−1,1). The points are in the same quadrant, Quadrant II and have the same x-coordinates. To find the distance from B to C, subtract the absolute values of their y-coordinates.

∣2∣ − ∣1∣ = 2 − 1 = 1 unit

Result

The length of side BC is 1 unit.

Page 110 Exercise 2 Answer

To find the length of side CD first find the ordered pairs which represent them. The ordered pair for point D is C(−1,1), and the ordered pair for point D is D(1,1). The points are not in the same quadrant and have the same y-coordinates. To find the distance from C to D, add the absolute values of their x-coordinates.

∣−1∣ + ∣1∣ = 1 + 1 = 2 units

Result

The length of side CD is 2 units.

Page 110 Exercise 3 Answer

To find the length of side DE first find the ordered pairs which represent them. The ordered pair for point D is D(1,1), and the ordered pair for point E is E(1,−2). The points are not in the same quadrant and have the same x-coordinates. To find the distance from D to E, add the absolute values of their y-coordinates.

∣1∣ + ∣−2∣ = 1 + 2 = 3 units

Result

The length of side DE is 3 units.

Page 110 Exercise 4 Answer

To find the length of side EF first find the ordered pairs which represent them. The ordered pair for point E is E(1,−2), and the ordered pair for point F is F(−3,−2). The points are not in the same quadrant and have the same y-coordinates. To find the distance from E to F, add the absolute values of their x-coordinates.

∣−3∣ + ∣1∣ = 3 + 1 = 4 units

Result

The length of side EF is 4 units.

Page 110 Exercise 5 Answer

To find the length of side FA first find the ordered pairs which represent them. The ordered pair for point F is F(−3,−2), and the ordered pair for point A is A(−3,2). The points are not in the same quadrant and have the same x-coordinates. To find the distance from F to A, add the absolute values of their y-coordinates.

∣−2∣ + ∣2∣ = 2 + 2 = 4 units

Result

The length of side FA is 4 units.

Page 110 Exercise 6 Answer

The perimeter of polygon ABCDEF is the sum of the lengths of all of its sides.

AB + BC + CD + DE + EF + FA = 2 + 1 + 2 + 3 + 4 + 4 = 16 units

Result

The perimeter of ABCDEF is 16 units.

Page 110 Exercise 7 Answer

To draw a polygon QRST first plot the points Q(−4,−1), R(−4,5), S(2,5), and (2,−1). To do that, find the x-coordinates on the x-axis and find the y-coordinates on the y-axis. Follow the grid lines from each axis for each of the points to where they meet and mark that spot, these are the vertices of the polygon. Connect the adjacent vertices to draw a polygon QRST.

Result

Plot the points Q(−4,−1), R(−4,5), S(2,5), and (2,−1) and connect the adjacent vertices to draw a polygon QRST.

Page 110 Exercise 8 Answer

Polygon QRST is a square if all of its sides are of equal length.

QR = ∣5∣ + ∣−1∣ = 5 + 1 = 6

RS = ∣−4∣ + ∣2∣ = 4 + 2 = 6

ST = ∣5∣ + ∣−1∣ = 5 + 1 = 6

TQ = ∣−4∣ + ∣2∣ = 4 + 2 = 6

All sides of the polygon are of equal length, so the polygon is a square.

Result

Polygon QRST has sides that are of equal length so the polygon is a square.

Chapter 3 Numeric And Algebraic Expressions

Section 3.0: Review What You Know

Page 115 Exercise 1 Answer

A formula is a rule that uses symbols to relate two or more quantities.

Result

Formula.

Page 115 Exercise 2 Answer

The number 12 is a composite number because it has more than two factors.

Result

Composite number.

Page 115 Exercise 3 Answer

A numerical expression is a mathematical phrase that includes numbers and at least one operation.

Result

Numerical expression.

Read And Learn More: enVisionmath 2.0 Grade 6 Volume 1 Solutions

Page 115 Exercise 4 Answer

Formula used to find the perimeter P is

P = 2l + 2w,

where l is length and w is width.

In this exercise l is 13cm and w is 13cm.

P = 2 ⋅ 13 + 2 ⋅ 13 = 26 + 26 = 52cm Perimeter of the square is 52cm.

Formula used to find the area A is A = lw, where l is the length and w is the width,and as before l = 13cm and w = 13 A = lw = 13 ⋅ 13 = 169 cm² Area of the square is 160cm².

Result

P = 52 cm

A = 169 cm²

Page 115 Exercise 5 Answer

Formula used to find the perimeter P is

P = 2l + 2w,

where l is length and w is width.

In this exercise l is 5 in and w is 21 in. P = 2 ⋅ 5 + 2 ⋅ 21 = 10 + 42 = 52 in Perimeter of the square is 52in.

Formula used to find the area A is A = lw, where l is the length and w is the width, and as before l = 5 in and w = 21 in. A = lw = 5 ⋅ 21 = 105 in² Area of the square is 105 in².

Result

P = 52 in

A = 105 in²

Page 115 Exercise 6 Answer

Formula used to find the perimeter P is

P = 2l + 2w,

where l is length and w is width.

In this exercise l is 9m and w is 15m. P = 2 ⋅ 9 + 2 ⋅ 15 = 18 + 30 = 48 m Perimeter of the square is 48m.

Formula used to find the area A is A = lw, where l is the length and w is the width, and as before l = 9m and w = 15m. A = lw = 9 ⋅ 15 = 135 m² Area of the square is 135m².

Result

P = 48 m

A = 135 m²

Page 115 Exercise 7 Answer

Multiple is the product of any quantity and an integer.

The first multiple of any number is that number times one, so the first multiple of any number is the number itself. The second multiple of any number is that number times two. […] The n-th multiple of any number is that number times n.

The first five multiples of 8 are:

8 . 1 = 8,

8 . 2 = 16,

8 . 3 = 24,

8 . 4 = 32,

8 . 5 = 40.

Result

The first five multiples of 8 are 8,16,24,32,40.

Page 115 Exercise 8 Answer

Multiple is the product of any quantity and an integer.

The first multiple of any number is that number times one, so the first multiple of any number is the number itself. The second multiple of any number is that number times two. […] The n-th multiple of any number is that number times n.

The first five multiples of 9 are:

9 . 1 = 9,

9 . 2 = 18,

9 . 3 = 27,

9 . 4 = 36,

9 . 5 = 45.

Result

The first five multiples of 9 are 9,18,27,36,45.

Page 115 Exercise 9 Answer

Multiple is the product of any quantity and an integer.

The first multiple of any number is that number times one, so the first multiple of any number is the number itself. The second multiple of any number is that number times two. […] The n-th multiple of any number is that number times n.

The first five multiples of 9 are:

10 . 1 = 10,

10 . 2 = 20,

10 . 3 = 30,

10 . 4 = 40,

10 . 5 = 50.

Result

The first five multiples of 10 are 10,20,30,40,50.

Page 115 Exercise 10 Answer

Multiple is the product of any quantity and an integer.

The first multiple of any number is that number times one, so the first multiple of any number is the number itself. The second multiple of any number is that number times two. […] The n-th multiple of any number is that number times n.

The first five multiples of 6 are:

6 . 1 = 6,

6 . 2 = 12,

6 . 3 = 18,

6 . 4 = 24,

6 . 5 = 30.

Result

The first five multiples of 6 are 6,12,18,24,30.

Page 115 Exercise 11 Answer

Multiple is the product of any quantity and an integer.

The first multiple of any number is that number times one, so the first multiple of any number is the number itself. The second multiple of any number is that number times two. […] The n-th multiple of any number is that number times n.

The first five multiples of 4 are:

4 . 1 = 4,

4 . 2 = 8,

4 . 3 = 12,

4 . 4 = 16,

4 . 5 = 20.

Result

The first five multiples of 4 are 4,8,12,16,20.

Page 115 Exercise 12 Answer

Multiple is the product of any quantity and an integer.

The first multiple of any number is that number times one, so the first multiple of any number is the number itself. The second multiple of any number is that number times two. […] The n-th multiple of any number is that number times n.

The first five multiples of 3 are:

3 . 1 = 3,

3 . 2 = 6,

3 . 3 = 9,

3 . 4 = 12,

3 . 5 = 15.

Result

The first five multiples of 3 are 3,6,9,12,15.

Page 115 Exercise 13 Answer

To find the factors of a number, you need to find all the pairs of numbers that multiply to the number.

To find the factors of 12, we need to find all the pairs of numbers that multiply to 12:

Since 12 ÷ 1 = 12, then 1 × 12 = 12 so 1 and 12 are factors.

Since 12 ÷ 2 = 6, then 2 × 6 = 12 so 2 and 6 are factors.

Since 12 ÷ 3 = 4, then 3 × 4 = 12 so 3 and 4 are factors.

There are no more pairs of numbers that multiply to 12 so the factors of 12 are 1, 2, 3, 4, 6, and 12.

To find the factors of 15, we need to find all the pairs of numbers that multiply to 15:

Since 15 ÷ 1 = 15, then 1 × 15 = 15 so 1 and 15 are factors.

Since 15 ÷ 3 = 5, then 3 × 5 = 15 so 3 and 5 are factors.

There are no more pairs of numbers that multiply to 15 so the factors of 15 are 1, 3, 5, and 15.

Result

To find the factors of a number, you need to find all the pairs of numbers that multiply to the number. The factors of 12 are then 1, 2, 3, 4, 6, and 12 and the factors of 15 are 1, 3, 5, and 15.

To find the factors start dividing by prime numbers. Start with the least prime number which divides 12 and 15. When the remainder is no longer divisble, try the next prime number.

Repeat until the remaining number is a prime itself.

12 ÷ 2 = 6

6 ÷ 2 = 3

Since three is a prime number it is the last factor.

Twelve can be written as a product of prime numbers as following:

12 = 2 × 2 × 3.

15 is not divisible by two so instead start with three.

15 ÷ 3 = 5

Since five is a prime number it is the last factor.

Fifteen can be written as a product of prime numbers as following:

15 = 3 × 5.

Result

Open to see the solution.

Page 115 Exercise 14 Answer

Difference, sum, quotient, and product are all results of arithmetic operations. Difference is the result of subtraction, sum is the result of addition, quotient is the result of division, and product is the result of multiplication.

Result

Difference, sum, quotient, and products are all results of arithmetic operations.

Chapter 3 Numeric And Algebraic Expressions

Section 3.1: Understand And Represent Exponents

Page 117 Exercise 1 Answer

Folding the sheet in half gives 2 sections.

Folding the sheet in half a second time then gives 4 sections.

Folding in half for the third time gives 8 sections.

Folding in half for the fourth time gives 16 sections.

Folding in half for the fifth time gives 32 sections.

Note that 2 × 2 = 4, 4 × 2 = 8, 8 × 2 = 16, and 16 × 2 = 32. The number of sections then doubled each time we folded it.

Result

2, 4, 8, 16, and then 32 sections

The number of sections doubled each time we folded it.

Read And Learn More: enVisionmath 2.0 Grade 6 Volume 1 Solutions

Page 117 Exercise 1 Answer

From the Solve & Discuss It! exercise, we know after the 5th fold there are 32 sections and that the number of sections doubles each time we fold the paper.

For the 6th fold, there will then be

32 × 2 = 64 sections.

For the 7th fold, there will be

64 × 2 = 128 sections.

Result

6th fold: 64 sections

7th fold: 128 sections

Page 118 Exercise 1 Answer

2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2⁹ = 512

After 3 hours there are 512 bacteria cells.

Since after 3 hours there are 2⁹ bacteria cells, which can be written as 2³ × 3. That means that after one hour there are 2³ bacteria cells. Thus, after two hours there are 2⁶ bacteria cells.

Result

2⁹ = 512

Page 119 Exercise 2 Answer

The base is \(\frac{1}{2}\). The exponent is 3. We need to rewrite the power as repeated multiplication and then evaluate.

Result
​
\(\frac{1}{27}\)

Page 119 Exercise 3 Answer

The evaluation for Rafael’s expression:

1.8 × 10⁴ = 1.8 × 10,000 = 18,000.

Result

18,000

Page 120 Exercise 1 Answer

A power consists of a base and an exponent. Powers can be evaluated using repeated multiplication, just as repeated multiplication can be represented using exponents.

For example,

7⁴ = 7 × 7 × 7 × 7 = 2401.

Result

A power consists of a base and an exponent. Powers can be evaluated using repeated multiplication, just as repeated multiplication can be represented using exponents.

Page 120 Exercise 2 Answer

n the expression 4⁵, 4 is used five times as a factor.

Written as repeated multiplication 4⁵ = 4 × 4 × 4 × 4 × 4.

Result

4 is used five times as a factor.

4⁵ = 4 × 4 × 4 × 4 × 4

Page 120 Exercise 3 Answer

One to any power always equals one,

since 1n = 1 × 1 × … × 1 (n times).

Thus, 1 to any power equals 18.

Result

1 to any power equals 18.

Page 120 Exercise 4 Answer

Any nonzero number to a power of 0 equals 1 so 100 = 1:

2.5 × 100 = 2.5 × 1 = 2.5

Result

2.5

Page 120 Exercise 5 Answer

For \(\left(\frac{1}{2}\right)^3\), the base is \(\frac{1}{2}\), the number which we multiply three times by itself, since three is the exponent.

\(\left(\frac{1}{2}\right)^3\) can be written as repeated multiplicaton as \(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\)

Result
​
\(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\)

Page 120 Exercise 6 Answer

To write 81 as a repeated multiplication of 3s one must multiply 3 with itself four times.

3 ⋅ 3 = 9

3 ⋅ 3 ⋅ 3 = 27

3 ⋅ 3 ⋅ 3 ⋅ 3 = 81

Thus, 81 can be written as 3 ⋅ 3 ⋅ 3 ⋅ 3 or as 3⁴

Result

81 = 3 ⋅ 3 ⋅ 3 ⋅ 3 = 3⁴

Page 120 Exercise 7 Answer

To write 125 as a repeated multiplication of 5s one must multiply 5 three times.

5 ⋅ 5 = 25

5 ⋅ 5 ⋅ 5 = 25 ⋅ 5 = 125

Thus, 125 can be written as 5 ⋅ 5 ⋅ 5, or as 5³.

Result

125 = 5 ⋅ 5 ⋅ 5 = 5³

Page 120 Exercise 8 Answer

The expression 0.75 × 0.75 × 0.75 × 0.75 × 0.75 can be written as a power with an exponent of 5 since 0.75 is used as a factor 5 times.

0.75 × 0.75 × 0.75 × 0.75 × 0.75 = 0.755

Result

0.75⁵

Page 120 Exercise 9 Answer

The expression \(\frac{3}{8}\) x \(\frac{3}{8}\) x \(\frac{3}{8}\) can be written as a power with an 3 since \(\frac{3}{8}\) is used as a factor 3 times.

Result

Page 120 Exercise 10 Answer

To evaluate \(\left(\frac{1}{6}\right)^2\), we can rewrite the expression as repeated multiplication and then multiply. Since the base is \(\frac{1}{6}\) and the exponent is 2, we will use \(\frac{1}{6}\) as a factor two times:

Result

Page 120 Exercise 11 Answer

Any nonzero number to the power of zero equals one.

450 = 1

Result

1

Page 120 Exercise 12 Answer

To evaluate 0.15, we can rewrite the expression as repeated multiplication and then multiply. Since the base is 0.1 and the exponent is 5, we will use 0.1 as a factor five times:

Result

0.00001

0.1⁵ = 0.1 ⋅ 0.1 ⋅ 0.1 ⋅ 0.1 ⋅ 0.1 = 0.00001

Result

0.00001

Page 120 Exercise 13 Answer

To evaluate 73 , we can rewrite the expression as repeated multiplication and then multiply. Since the base is 7 and the exponent is 3, we will use 7 as a factor three times:

7³ = 7 × 7 × 7 = 49 × 7 = 343

Result

343

7³ = 7 ⋅ 7 ⋅ 7 = 49 ⋅ 7 = 343

Result

343

Page 120 Exercise 14 Answer

4.5 ⋅ 10⁴ = 4.5 ⋅ 10 ⋅ 10 ⋅ 10 ⋅ 10 = 4.5 ⋅ 10000 = 45000

Result

45000

Page 120 Exercise 15 Answer

0.6 ⋅ 10⁶ = 0.6 ⋅ 10 ⋅ 10 ⋅ 10 ⋅ 10 ⋅ 10 ⋅ 10 = 0.6 ⋅ 1000000 = 600000

Result

600000

Page 120 Exercise 16 Answer

Any nonzero number to the power of zero equals one, thus 100 = 1.

3.4 ⋅ 100 = 3.4 ⋅ 1 = 3.4

Result

3.4

Page 121 Exercise 17 Answer

Repeated multiplication can be written as a power where the exponent is the number of times the base is used as a factor.

For 9 × 9 × 9 × 9, the base of 9 is used as a factor 4 times. The exponent for the expression is then 4.

Result

4

Page 121 Exercise 18 Answer

The expression is 1.29 so the exponent is 9.

Result

9

Page 121 Exercise 19 Answer

Repeated multiplication can be written as a power where the exponent is the number of times the base is used as a factor.

For \(\frac{1}{6} \times \frac{1}{6} \times \frac{1}{6}\), the base of \(\frac{1}{6}\) is used as a factor 3 times. The exponent for the expression is then 3.

Result

3

Page 121 Exercise 20 Answer

Any number raised to the power of one equals the number itself, hence there is no need to write the exponent. In cases such as this, where the exponent is not written it is one.

7 = 7¹

Result

1

Page 121 Exercise 21 Answer

The expression 8³ can be written as 8 × 8 × 8, which is then equal to 64 × 8 which gives us 512.

Result

8 × 8 × 8 = 512

Page 121 Exercise 22 Answer

To evaluate \(\left(\frac{1}{5}\right)^4\), we can rewrite the expression as repeated multiplication and then multiply. Since the base is \(\frac{1}{5}\) and the exponent is 4, we will use \(\frac{1}{5}\) as a factor four times:

Result

Page 121 Exercise 23 Answer

To evaluate 0.6², we can rewrite the expression as repeated multiplication and then multiply. Since the base is 0.6 and the exponent is 2, we will use 0.6 as a factor two times:

0.6² = 0.6 × 0.6 = 0.36
​
Result

0.36

Page 121 Exercise 24 Answer

To evaluate \(\left(\frac{1}{4}\right)^2\), we can rewrite the expression as repeated multiplication and then multiply. Since the base is \(\frac{1}{4}\) and the exponent is 2, we will use \(\frac{1}{4}\) as a factor two times:

Result

Page 121 Exercise 25 Answer

Any nonzero number to the power of zero equals one. Thus, 580 = 1.

Result

1

Page 121 Exercise 26 Answer

To evaluate 6.2 × 10³, we can rewrite the expression 103 as repeated multiplication and then multiply. Since the base is 10 and the exponent is 3, we will use 10 as a factor three times:

6.2 × 10³ = 6.2 × 10 × 10 × 10 = 6.2 × 1000 = 6200

Result

6200

6.2 ⋅ 10³ = 6.2 ⋅ 1000 = 6200

Result

6200

Page 121 Exercise 27 Answer

A company rents two storage unites which are both cube-shaped. The volume of a cube is given by the formula V = s³, where s is the side length.

The first unit is 8 ft tall, since it is cube shaped all of its sides are 8 ft. The volume of the first unit is V₁ = 8³ = 512 ft. The second unit is 6.5 ft tall, since it is also cube shaped all of its sides are 6.5 ft, and the volume is

6.5³ = 274.625 ft.

The difference in the volume of the two storage units is

512 − 274.625 = 237.375.

A common mistake may be to first find the difference between the lengths of sides and then raise that to the power of three. For example, (8−6.5)³ = 1.53 = 3.375, which is obviously not equal to 237.375.

Result

237.375 ft

Page 121 Exercise 28 Answer

Jia is tiling a floor with side length 12 feet and she wants the tiles to be squares with side length 2 feet. How many tiles does Jia need to cover the entire floor?

The area of the floor which needs to be covered is A₁ = 12² = 144 feet. One square tile with side length 2 feet covers the area of A₂ = 2² = 4. To cover all 144 feet of the floor Jia needs 144 ÷ 4 = 36 tiles.

Result

Jia needs 36 tiles to cover the entire floor.

Page 121 Exercise 29 Answer

A marine biologist studies the population of seals in a research area. He found out that there is 3.27 × 102 seals in the area, which is exactly 327 seals since:

3.27 × 10² = 3.27 × 10 × 10 = 3.27 × 100 = 327

Result

There is 327 seals in the research area.

Page 121 Exercise 29 Answer

A marine biologist studies the population of seals in a research area. He found out that there is 3.27 × 10² seals in the area, which is exactly 327 seals since:

3.27 × 10² = 3.27 × 10 × 10 = 3.27 × 100 = 327

Result

There is 327 seals in the research area.

Page 122 Exercise 30 Answer

Zach invested $50 and tripled his money in two years, which means after two years he has $50 × 3 = $150.

Kayla also invested $50 and after two years the amount was equal to 503 dollars, so Kayla had 50³ = 50 × 50 × 50 = 2500 × 50 = $175,000.

Result

After two years, Kayla had more money.

Page 122 Exercise 31 Answer

Malik read that the land area of Alaska is about 5.7 ⋅ 10⁵ square miles, which is equal to 570000 square miles.

5.7 ⋅ 10⁵ = 5.7 ⋅ 100000 = 570000

Result

The land area of Alaska is about 570000 square miles.

Page 122 Exercise 32 Answer

Any nonzero number to the power of zero equals one, thus 100 = 1 and 1 × 1.00 = 1 × 1 = 1.

One to any power gives one because raising a number to the power of n means multiplying that number n times with itself, and 1 ⋅ 1 = 1. Thus, 1⁴ = 1.

All three expressions are equal to 1, thus 100 = 14 = 1 × 1.00.

Result

All three expressions are equal to 1.

Page 122 Exercise 33 Answer

If 0.33 = n, what is n?

0.33 = 0.3 ⋅ 0.3 ⋅ 0.3 = 0.09 ⋅ 0.3 = 0.027 ⟹ n = 0.027

Result

n = 0.027

Page 122 Exercise 34 Answer

The same digits are used for the expressions 2⁵ and 5², however in the expression 2⁵ two is the base and five is the exponent and in the expression 5² five is the base and two is the exponent.

Values of expressions which contain exponents can be compared without calculating their actual value if either bases or exponents are the same, for example: 2³ is less than 2⁶, and 5³ is less than 6³ , Since neither is the case for 2⁵ and 5² , they have different bases and different exponents, we must first calculate their vales and then compare them.

2⁵ = 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 = 32

5² = 5 ⋅ 5 = 25

Now, we can see that 2⁵ is greater than 5².

Result

2⁵ is greater than 5² since 2⁵ = 32 and 5² = 25.

Page 122 Exercise 35 Answer

Kristen rewrote the expression 80,000 × 25 using exponents as (8 × 10³) × 52. We need to determine if she was right.

80,000 × 25

80,000 can be written as 8 ⋅ 10,000, which can further be written using exponents as 8 × 10⁴. Kristen has the wrong exponent for the power of 10 and she didn’t notice that 8 can also be written as 8 = 2³.

25 can be written as 52.

So, the correct answer is:

(2³ × 10⁴) × 52.

Result

The correct response is (2³ × 10⁴) × 52.

Page 122 Exercise 36 Answer

Rewriting 1,000,000 as repeated multiplication gives:

1,000,000 = 10 × 100,000

= 10 × 10 × 10,000

= 10 × 10 × 10 × 1,000

= 10 × 10 × 10 × 10 × 100

= 10 × 10 × 10 × 10 × 10 × 10
​
Since 1,000,000 can be rewritten as repeated multiplication where 10 is used as a factor 6 times, then it can be written as a power where the base is 10 and the exponent is 6.

Result

Since 1,000,000 = 10 × 10 × 10 × 10 × 10 × 10, the expression can be written as a power where the base is 10 and the exponent is 6.

Page 122 Exercise 37 Answer

Isabella saved 2 nickles today and she doubles the number of nickles she saves each day. The number of nickels she has saved each day is then:

First day: 2

Second day: 2 ⋅ 2 = 4

Third day: 2 ⋅ 2 ⋅ 2 = 2 ⋅ 4 = 8

Forth day: 2 ⋅ 2 ⋅ 2 ⋅ 2 = 2 ⋅ 8 = 16

Fifth day: 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 = 2 ⋅ 16 = 32

Sixth day: 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 = 2 ⋅ 32 = 64

Seventh day: 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 = 2 ⋅ 64 = 128

Eighth day: 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 = 2 ⋅ 128 = 256

Ninth day: 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 = 2 ⋅ 256 = 512

Result

It will take Isabella nine days to save more than 500 nickles.

Page 122 Exercise 38 Answer

We are given the four expressions 2¹⁰, 5 × 5 × 5 × 5, 4⁵, and 4 × 4 × 4 × 4 × 4 and need to determine which one is NOT equal to 1,024.

Note that 1,024 is an even number so it must have even factors. The expression 5 × 5 × 5 × 5 does not have even factors since 5 is an odd number. This means that 5 × 5 × 5 × 5 ≠ 1,024. Therefore, B. 5 × 5 × 5 × 5 is not equal to 1,024.

We can verify our answer by showing the other three expressions are equal.

4⁵ has a base of 4 and an exponent of 5 so it can be written as repeated multiplication where the base of 4 is used as a factor 5 times. Therefore, 4⁵ = 4 × 4 × 4 × 4 × 4.

2¹⁰ has a base of 2 and an exponent of 10 so it can be written as repeated multiplication where the base of 2 is used as a factor 10 times. Therefore, 2¹⁰ = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2. Multiplying each pair of 2s then gives 2¹⁰ = 4 × 4 × 4 × 4 × 4.

The other three expressions are all equal to each other so they must all equal 1,024.

Result

B. 5 × 5 × 5 × 5

Which expression is NOT equal to 1024?

We know that the expression A 210 is equal to 1024. Expressions C and D are equal, 45 = 4 ⋅ 4 ⋅ 4 ⋅ 4 ⋅ 4.

Futher more, expressions C and D are equal to expression A, thus equal to 1024.

Expression B 5 ⋅ 5 ⋅ 5 ⋅ 5 is equal to 54 which is equal to 625, and not equal to 1024.

The answer to the question which expression is not equal to 1024 is expression B 5 ⋅ 5 ⋅ 5 ⋅ 5.

Result

Expression B 5 ⋅ 5 ⋅ 5 ⋅ 5 is not equal to 1024.

Page 122 Exercise 39 Answer

Which expression is equal to \(\frac{1}{36}\)?

Expression A:

Expression B:

Expression C:

Expression D:

Result

The expression C is equal to \(\frac{1}{36}\).

Chapter 3 Numeric And Algebraic Expressions

Section 3.2: Find Greatest Common Factor And Least Common Multiple

Page 123 Exercise 1 Answer

Since Mark sets the table every two days and dries the dishes every three days, he will do both chores on the days which are multiples of both 2 and 3. The least common multiple of 2 and 3 is 6.

All multiples of 6 are also multiples of 2 and 3, so Mark will do both chores on days that are multiples of 6.

Result

He will do both chores on days which are multiples of both 2 and 3 so Mark will do both chores on days that are multiples of 6.

Page 123 Exercise 1 Answer

We know that Mark sets the dinner table ever 2 days and he sets the table of Day 2. He will then also set the table on Day 4, Day 6, Day 8, Day 10, ……

We know that Mark dries the dishes every 3 days and that he dries the dishes on Day 3. He also then dries on Day 6, Day 9, Day 12, ….

The table can then be completed as:

From the table, he will perform both chores on Day 6.

Read And Learn More: enVisionmath 2.0 Grade 6 Volume 1 Solutions

Result

Day 6

If every two days Marks sets the table and every 3 days he dries the dishes, he would do both chores on the day which is the least common multiple of 2 and 3.

Since both 2 and 3 are prime numbers, than their least common mulitple is their product.

2 × 3 = 6

Result

Mark would first perform both chores on the same day on Day 6.

Page 124 Exercise 1 Answer

To find the prime factorization of 56 divide the number by primes, start with two. When the remainder is no longer divisible by two, try with the next prime.

56 = 2 × 28 = 2 × 2 × 14 = 2 × 2 × 2 × 7

or

56 = 23 × 7

A composite number is by definition a number with more than two factors. Since the number in question is greater than 2 and it has 2 as a factor, it must have more than 2 factors since it at least has 1, 2, and the number itself as factors. It must then be a composite number.

Result

The prime factorization of 56 is 2 × 2 × 2 × 7 or 2³ × 7.

Page 125 Exercise 2 Answer

Keesha can add an equal number of beads to each bag since there are 6 bags, 24 beads, and 24 is divisible by 6.

24 = 4 × 6

She adds 4 beads to each bag.

Result

Keesha can add the beads and have no supplies left over.

Page 125 Exercise 3 Answer

First, we find the GCF of 12 and 36.

12 = 2 × 6 = 2 × 2 × 3

36 = 2 × 18 = 2 × 2 × 9 = 2 × 2 × 3 × 3

The common factors are 2, 2, and 3, so the GCF is

2 × 2 × 3 = 12.

Next we write each number as a product using the GCF as a factor.

12 + 36 = 12 + 12 × 3 = 12(1 + 3) = 12(4) = 48

Result

The sum of 12 and 36 is 48.

Page 126 Exercise 4 Answer

To answer the question we need to find the least common multiple of 12 and 10.

12 = 2 × 6 = 2 × 2 × 3

10 = 2 × 5

The LCM of 12 and 10 is the product 2 × 2 × 3 × 5 or 60.

Since bottled water is sold in cases of 12 bottles and juice in boxes of 10, write 60 as a product of two number so that one of them is 12, and write 60 as s product of two number so that one of them is 10.

60 = 12 × 5

60 = 10 × 6

Result

Grant should buy 5 cases of bottled water and 6 boxes of juice.

Page 127 Exercise 1 Answer

Whole numbers greater than 1 are either prime or composite numbers. A composite number can be written as a product of its prime factors, called its prime factorization.

To find the prime factorization of a number, write the number as products of its factors. We start with the least prime factor and continue using prime factors until all factors are prime numbers.

For example, to find the prime factorization of 630 we first divide the number by 2 beacuse it is the least prime number which divides 630.

630 = 2 ⋅ 315

Than, we are further looking for the prime factorization of 315.

2 doesn’t divide 315 so we are looking at the next least prime number which is 3.

3 divides 315 so we use 3.

315 = 3 ⋅ 105

3 divides 105 so we use 3 again.

105 = 3 ⋅ 35

3 doesn’t divide 35 so we look at the next least prime number which is 5. 5 divides 35 so we use 5.

35 = 5 ⋅ 7

Since both factors in 5⋅7 are prime numbers we stop here.

630 = 2 ⋅ 315 = 2 ⋅ 3 ⋅ 105 = 2 ⋅ 3 ⋅ 3 ⋅ 35 = 2 ⋅ 3 ⋅ 3 ⋅ 5 ⋅ 7

The prime factorization of 630 is 2 ⋅ 3 ⋅ 3 ⋅ 5 ⋅ 7.

To find the greatest common factor and the least common multiple of two numbers first we need to find their prime factorizations.

For example, let’s find the greatest common factor and least common multiple of 42 and 60.

First, we find their prime factorizations.

42 = 2 ⋅ 21 = 2 ⋅ 3 ⋅ 7

60 = 2 ⋅ 30 = 2 ⋅ 2 ⋅ 15 = 2 ⋅ 2 ⋅ 3 ⋅ 5

Their greatest common factor is found by multiplying all their common prime factors which are, in this case, 2 and 3. The greatest common factor of 42 and 60 is 6. Their least common multiple is found by listing the greatest number of times each factor appeared in either prime factorization. We then multiply these factors to find the least common multiple.

22 ⋅ 3 ⋅ 5 ⋅ 7

The least common multiple of 42 and 60 is 420.

Result

To find the prime factorization of a number, write the number as products of its factors until all factors are prime numbers. To find the GCF and the LCM of two numbers, first we need to find their prime factorizations. Their GCF is found by multiplying all their common prime factors. Their LCM is found by listing the greatest number of times each factor appeared in either prime factorization.

Page 127 Exercise 2 Answer

In Example 1, we saw that there were two ways to find the prime factorization of a number.

The first way is to write its factors as a product by starting with the least prime factor and continuing until all the factors are prime. The second way is to make a factor tree by writing the number as the product of two factors and then writing each composite factor as a product of two factors until all of the factors are prime factors.

Result

The first way is to write its factors as a product by starting with the least prime factor and continuing until all the factors are prime. The second way is to make a factor tree by writing the number as the product of two factors and then writing each composite factor as a product of two factors until all of the factors are prime factors.

Page 127 Exercise 3 Answer

The greatest common factor of two numbers is the greatest positive number that is a factor of both numbers. Prime numbers, by definition, have only two factors, one and themselves. When looking for the GCF of two prime numbers we are looking for the greatest number which is a factor of both of them. The only factor they have in common is 1 so the GCF must be 1.

Result

The only common factor of two prime number is 1 so the GCF must be 1.

Page 127 Exercise 4 Answer

In Example 4, we had to find the LCM of 6 and 8 since the juice bottles came 6 per pack and the applesauce came 8 per pack. We found that the LCM was 24.

If the juice bottles come in packages of 3 instead of 6, then we need to find the LCM of 3 and 8. The prime factorization of the two numbers is:

3 = 3

8 = 2 × 2 × 2

The factor of 3 appears 1 time in the factorizations and the factor of 2 appears 3 times. The LCM is then:

3 × 2 × 2 × 2 = 24

This is the same LCM we found in Example 4 so the LCM did not change.

Result

The LCM did not change.

Page 127 Exercise 5 Answer

You can’t find the LCM of any two numbers by simply multiplying them together. This works if and only if those two numbers don’t have any common factors in their prime factorization.

For example, we can find the LCM of 5 and 23 by simply mulitplying them since they are both prime, however we can’t do the same with any number, for example we can’t find the LCM of 12 and 48 by multiplying them because they have common prime factors.

12 = 2 × 6 = 2 × 2 × 3,

48 = 2 × 24 = 2 × 2 × 12 = 2 × 2 × 2 × 6 = 2 × 2 × 2 × 2 × 3.

Result

You can’t find the LCM of any two numbers by simply multiplying them together. This works if and only if those two numbers don’t have any common factors in their prime factorization. For example, we can find the LCM of 5 and 23 by simply multiplying them since they are prime, however, we can’t do the same when finding the LCM of 12 and 48.

Page 127 Exercise 6 Answer

To find the prime factorization of 33 we write 33 as a product of prime numbers. We start with the least prime number 2, but since 2 doesn’t divide 33 we can’t use it in the product. We try with the next least prime number which is 3. 3 divides 33 so we can use it.

33 = 3 ⋅ 11

Since 11 is also a prime number we are done. The solution is 3 ⋅ 11.

Result

3 ⋅ 11

Page 127 Exercise 7 Answer

To find the prime factorization of 32 we write 32 as a product of prime numbers. We start with the least prime number 2, since it divides 32 we can use it in the product.

32 = 2 ⋅ 16 = 2 ⋅ 2 ⋅ 8 = 2 ⋅ 2 ⋅ 2 ⋅ 4 = 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2

Prime factorization of 32 is 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2.

Result

2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2

Page 127 Exercise 8 Answer

19 is a prime number.

Result

Prime.

Page 127 Exercise 9 Answer

To find the greatest common factor we first find prime factorizations of both numbers.

18 = 2 ⋅ 9 = 2 ⋅ 3 ⋅ 3 = 2 ⋅ 3 ⋅ 3

36 = 2 ⋅ 18 = 2 ⋅ 2 ⋅ 9 = 2 ⋅ 2 ⋅ 3 ⋅ 3 = 2 ⋅ 2 ⋅ 3 ⋅ 3

The GCF is the product of prime numbers which appear in both prime factorizations.

2 ⋅ 3 ⋅ 3 = 18

Result

18

Page 127 Exercise 10 Answer

To find the greatest common factor we first find prime factorizations of both numbers.

22 = 2 ⋅ 11 = 2 ⋅ 11

55 = 5 ⋅ 11 = 5 ⋅ 11

The GCF is the product of prime numbers which appear in both prime factorizations, but in this case it is only one number 11, which is then the GCF of 22 and 55.

Result

11

Page 127 Exercise 11 Answer

To find the greatest common factor we first find prime factorizations of both numbers.

100 = 2 ⋅ 50 = 2 ⋅ 2 ⋅ 25 = 2 ⋅ 2 ⋅ 5 ⋅ 5 = 2 ⋅ 2 ⋅ 5 ⋅ 5

48 = 2 ⋅ 24 = 2 ⋅ 2 ⋅ 12 = 2 ⋅ 2 ⋅ 2 ⋅ 6 = 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 3 = 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 3

The GCF is the product of prime numbers which appear in both prime factorizations.

2 ⋅ 2 = 4

Result

4

Page 127 Exercise 12 Answer

Both numbers, 2 and 5 are prime, so their least common multiple is simply their product.

2 ⋅ 5 = 10

Result

10

Page 127 Exercise 13 Answer

To find the least common multiple of 8 and 12 we first need to find their prime factorizations.

8 = 2 ⋅ 4 = 2 ⋅ 2 ⋅ 2

12 = 2 ⋅ 6 = 2 ⋅ 2 ⋅ 3

We then list the greatest number of times each factor appeared in either prime factorization.

2,2,2,3

Now, we multiply these factors as the last step.

2 ⋅ 2 ⋅ 2 ⋅ 3 = 24

Result

24

Page 127 Exercise 14 Answer

To find the least common multiple of 8 and 12 we first need to find their prime factorizations.

8 = 2 ⋅ 4 = 2 ⋅ 2 ⋅ 2

10 = 2 ⋅ 5

We then list the greatest number of times each factor appeared in either prime factorization.

2,2,2,5

Now, we multiply these factors as the last step.

2 ⋅ 2 ⋅ 2 ⋅ 5 = 40

Result

40

Page 128 Exercise 15 Answer

To find the prime factorization of 27 start dividing by prime numbers, start with 3.

27 = 3 × 9 = 3 × 3 × 3

Result

The prime factorization of 27 is 3 × 3 × 3.

Page 128 Exercise 16 Answer

To find the prime factorization of 30 start dividing by primes, start with 2.

30 = 2 × 15 = 2 × 3 × 5

Result

The prime factorization of 30 is 2 × 3 × 5.

Page 128 Exercise 17 Answer

Prime factorization of 26 is found by writing 26 as a product of prime numbers. When looking for prime factors, we start with the least prime number, which is 2. Since 2 divides 26, we can write 26 = 2 ⋅ 13 . 13 is a prime number so we stop here, because we have found the prime factors whose product give 26.

Result

Prime factorization of 26 is 2 ⋅ 13.

Page 128 Exercise 18 Answer

Number 47 is prime.

Result

Prime.

Page 128 Exercise 19 Answer

To find the greatest common factor of 21 and 49, we must first find prime factorizations of 21 and 49.

21 = 3 ⋅ 7

49 = 7 ⋅ 7

The greatest common factor is the product of prime numbers which are common to both 21 and 49. In this case, the GCF is 7.

Result

The GCF is 7.

Page 128 Exercise 20 Answer

To find the greatest common factor of 8 and 52, we must first find prime factorizations of 8 and 52.

8 = 2 ⋅ 2 ⋅ 2

52 = 2 ⋅ 26 = 2 ⋅ 2 ⋅ 13

The greatest common factor is the product of prime numbers which are common to both 8 and 52. In this case, the GCF is

2 ⋅ 2 = 4.

Result

The GCF is 4.

Page 128 Exercise 21 Answer

To find the greatest common factor of 32 and 81, we must first find prime factorizations of 32 and 81.

32 = 2 ⋅ 16 = 2 ⋅ 2 ⋅ 8 = 2 ⋅ 2 ⋅ 2 ⋅ 4 = 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2

81 = 3 ⋅ 27 = 3 ⋅ 3 ⋅ 9 = 3 ⋅ 3 ⋅ 3 ⋅ 3

The greatest common factor is the product of prime numbers which are common to both 32 and 81, however, in the case of 32 and 81, they have no common prime factors. The only number which divides both 32 and 81 is 1 so the GCF is 1.

Result

The GCF is 1.

Page 128 Exercise 22 Answer

Using the GCF and the Distributive Property we need to find the sum 30+66.

First, we find prime factorizations of 30 and 66.

30 = 2 ⋅ 15 = 2 ⋅ 3 ⋅ 5

66 = 2 ⋅ 33 = 2 ⋅ 3 ⋅ 11

Then, we write the sum 30+60 as 2 ⋅ 3 ⋅ 5 + 2 ⋅ 3 ⋅ 11. The GCF is 2 ⋅ 3 = 6, the number which divides both 30 and 66. We can extract it and write a product of the GCF and the sum to simplify calculations.

30 + 60 = 2 ⋅ 3 ⋅ 5 + 2 ⋅ 3 ⋅ 11 = 2 ⋅ 3 ⋅ (5 + 11) = 6 ⋅ 16 = 96

Result

The sum is 96.

Page 128 Exercise 23 Answer

Using the GCF and the Distributive Property we need to find the sum 34 + 51.

First, we find prime factorizations of 34 and 51.

34 = 2 ⋅ 17

51 = 3 ⋅ 17

Then, we write the sum 34 + 51 as 2 ⋅ 17 + 3 ⋅ 17. The GCF is 17, the number which divides both 34 and 51. We can extract it and write a product of the GCF and the sum to simplify calculations.

34 + 51 = 2 ⋅ 17 + 3 ⋅ 17 = 17 ⋅ (2 + 3) = 17 ⋅ 5 = 85

Result

The sum is 85.

Page 128 Exercise 24 Answer

Using the GCF and the Distributive Property we need to find the sum 15 + 36.

First, we find prime factorizations of 15 and 36.

15 = 3 ⋅ 5

36 = 2 ⋅ 18 = 2 ⋅ 2 ⋅ 9 = 2 ⋅ 2 ⋅ 3 ⋅ 3

Then, we write the sum 15 + 36 as 3 ⋅ 5 + 2 ⋅ 2 ⋅ 3 ⋅ 3. The GCF is 3, the number which divides both 15 and 36. We can extract it and write a product of the GCF and the sum to simplify calculations.

15 + 36 = 3 ⋅ 5 + 2 ⋅ 2 ⋅ 3 ⋅ 3 = 3 ⋅ (5 + 2 ⋅ 2 ⋅ 3) = 3 ⋅ (5 + 12) = 3 ⋅ 17 = 51

Result

The sum is 51.

Page 128 Exercise 25 Answer

To find the least common multiple we must first find prime factorizations of 12 and 11.

12 = 2 ⋅ 6 = 2 ⋅ 2 ⋅ 3

11 is a prime number.

They don’t have any common factors, so to find the LCM we simply multiply 12 and 11.

12 ⋅ 11 = 132

Result

The LCM of 12 and 11 is 132.

Page 128 Exercise 26 Answer

To find the least common multiple of 4 and 12 we must first find prime factorizations of 4 and 11.

4 = 2 ⋅ 2

12 = 2 ⋅ 6 = 2 ⋅ 2 ⋅ 3

We can see that 12 is a multiple of 4. Since, every number is its least multiple, the least multiple of 12 is 12.

The LCM of 4 and 12 is 12.

Result

The LCM of 4 and 12 is 12.

Page 128 Exercise 27 Answer

To find the least common multiple of 5 and 8 we must first find prime factorizations of 5 and 8.

5 is a prime number.

8 = 2 ⋅ 4 = 2 ⋅ 2 ⋅ 2

Since 5 is a prime number and doesn’t divide 8, then to find the LCM we simply multiply 5 and 8.

5 ⋅ 8 = 40

Result

The LCM of 5 and 8 is 40.

Page 128 Exercise 28 Answer

Both Gabrielle and John got the same answer of 2 × 2 × 2 × 2 × 2 × 2. Neither student made a mistake in their work. We can verify that their answer is correct by multiplying it out and showing it equals 64:

2 × 2 × 2 × 2 × 2 × 2 = 4 × 2 × 2 × 2 × 2

= 8 × 2 × 2 × 2

= 16 × 2 × 2

= 32 × 2

= 64

Result

Neither student made a mistake in their work.

Page 128 Exercise 29 Answer

To celebrate its grand opening, a store is giving customers gift certificates. Every 8th customer gets a $50 gift certificate, and every 6th customer gets a $10 gift certificate. A $50 gift certificate will get 8th, 16th, 24th, … customer. We notice that these are multiples of 8. The same for a $10 gift certificate which will get 6th, 12th, 18th, … customer, and these are the multiples of 6. When looking for the first person who will get two gift certificates we are actually looking for the least common multiple of 8 and 6.

First, we find prime factorizations of 8 and 6.

8 = 2 ⋅ 4 = 2 ⋅ 2 ⋅ 2 = 2³

6 = 2 ⋅ 3

The least common multiple is the product of multiplying the highest power of each prime number when looking at prime factorizations together. The LCM of 8 and 6 is 2³ ⋅ 3 = 8 ⋅ 3 = 24.

The first customer to get two gift certificates is the 24th customer who comes to the store.

Result

The 24th customer is the first customer that will recieve two gift certificates.

Page 129 Exercise 3a Answer

The Venn diagram shows the factors of 24 and 40. The left circle contains factors of 24, which are 1,2,3,4,6,8,12 and 24. The right circle contains factors of 40, which are 1,2,4,5,8,10,20 and 40. Factors in the region which is common to both circles, their intersection, contains common factors – the numbers which divide both 24 and 40. They are 1,2,4 and 8.

Result

The left circle contains the factors of 24, the right circle contains the factors of 40, and the intersection of the circles contains the factors common to 24 and 40.

Page 129 Exercise 30b Answer

The Venn diagram shows the factors of 24 and 40 where the intersection of circles are the common factors. The intersection contains the factors: 1,2,4,8. Since 8 is the greatest number, it is the greatest common factor of 24 and 40.

Result

8 since the greatest number in the intersection is 8.

Page 129 Exercise 31 Answer

If we have 50 blueberry scones and 75 cranberry scones, how many identical bags can we make? Each bag should have an equal number of blueberry scones and an equal number of cranberry scones. What is the greatest number of bags we can fill?

First, we find the prime factorization of 50 and 75.

50 = 2 ⋅ 25 = 2 ⋅ 5 ⋅ 5

75 = 3 ⋅ 25 = 3 ⋅ 5 ⋅ 5

If we put 2 blueberry scones and 3 cranberry scones in each bag, we can make 25 identical bags. However, if we put 1 blueberry scones and 1 cranberry scones in each bag, we can fill 50 bags.

Result

50 bags if we put 1 blueberry scone and 1 cranberry scone in each bag.

Page 129 Exercise 32 Answer

The prime factorization of A and B are shown. Find the value n that needs to be listed as a prime factor of B so that the greatest common factor of A and B is 9.

Prime factorization of A: 3 ⋅ 3 ⋅ 3.

Prime factorization of B: 2 ⋅ 2 ⋅ 3 ⋅ n.

The greatest common factor of A and B is 9, which means that 9 is the greatest number which divides both A and B. 9 already divides A, but it doesn’t divide B. If we put n = 3, then 9 will divide B.

Also, when looking for the greatest common factor, we usually first find the prime factorization, which is here given, and than we take numbers which are common to both numbers. In this case what is common to A and B is 3 and possibly n, it depends on what we decide n is.

Prime factorization of A: 3 ⋅ 3 ⋅ 3.

Prime factorization of B: 2 ⋅ 2 ⋅ 3 ⋅ n.

Since the prime factorization of A contains only number 3, if we want n to be common to both, we must put n = 3.

Result

n = 3

Page 129 Exercise 33 Answer

Gena has 28 trading cards, Sam has 91 trading cards, and Tiffany has 49 trading cards. Using the GCF and the Distributive Property we have to find the total number of trading cards they all have together.

First, we find prime factorizations of 28,91 and 49.

28 = 2 ⋅ 14 = 2 ⋅ 2 ⋅ 7

91 = 7 ⋅ 13

49 = 7 ⋅ 7

28 + 91 + 49 = 2 ⋅ 2 ⋅ 7 + 7 ⋅ 13 + 7 ⋅ 7 = 7 ⋅ (2 ⋅ 2 + 13 + 7) = 7 ⋅ 24 = 168

Gena, Sam and Tiffany together have 168 trading cards.

Result

Gena, Sam and Tiffany together have 168 trading cards.

Page 129 Exercise 34 Answer

One of the species emerges every 13 years, and the other every 17 years.

Both 13 and 17 are prime numbers and the GCF of two prime numbers is always 1, so the GCF of the years is 1.

Result

The GCF of the years is 1.

Page 129 Exercise 35 Answer

People are waiting in line for a theater premiere. Every 5th person in line will receive a free theater ticket. Every 6th person will receive a gift card for $40.

The first person to receive both a free theater ticket and a gift card is the one whose number in the line is the least common multiple of 5 and 6.

The prime factorization of 5 is 5 and the prime factorization of 6 is 2 × 3. The LCM of 5 and 6 is then 5 × 2 × 3 = 30.

The 30th person will receive both prizes.

Result

The 30th person will recieve both prizes.

Page 129 Exercise 36 Answer

Group A plants trees in clusters of 3, Group B plants the trees in clusters of 10, and both groups plant the same number of trees. We need to find the least number of clusters that Group B plants.

Since both groups planted the same number of trees, the number of planted trees must be the lowest common multiple of 3 and 10. The prime factorization of 3 is 3 and the prime factorization of 10 is 10 = 2 × 5. The LCM of 3 and 10 is then 3 × 2 × 5 = 30.

Since Group B plants trees in clusters of 10 they must plant at 3 clusters to plant 30 trees. So, Group B had to plant at least 3 clusters.

Result

The least number of clusters that Group B plants is 3.

Page 130 Exercise 37 Answer

Since all three numbers are prime the LCM of each two is their product.

2 × 3 = 6

3 × 7 = 21

2 × 7 = 14

6 = F

21 = U

14 = N

Result

The word is FUN.

Page 130 Exercise 38 Answer

Rami has swimming lessons every 3 days, so the days on which he has swimming lessons are multiples of 3. He has guitar lessons every 8 days, so the days in which he has guitar lessons are multiples of 8. The days on which he has both lessons are multiples of both 3 and 8. Since 3 is a prime number and the prime factorization of 8 is 2 ⋅ 2 ⋅ 2, the LCM of 3 and 8 is their product 3 ⋅ 8 = 24.

Rami has both lessons every 24 days. So if he had both lessons on the first day of the month, he will have both lessons on the same day again in 24 days.

Result

Rami will have both lessons on the same day again in 24 days.

Page 130 Exercise 39 Answer

A number is between 58 and 68. It has prime factors of 2, 3 and 5. Let’s call that number n.

58 < n < 68

n is one of the following numbers 59,60,61,62,63,64,65,66,67. Since we now its prime factors are

5 we know it is divisible by 2,3 and 5, and is not a prime number.

59,61 and 67 are prime numbers, which leaves us 60,61,62,63,64,65,66.

However, most of these numbers do not have 2 as a factor since they are odd and only even number have 2 as a factor. That leaves us 60,62,64 and 66.

Furthermore, n must be divisible by 5, and the only number which is still on the list of potential candidates which is divisible by 5 is 60. Now, we must check if 60 is divisible by 3. Since, 60 = 3 ⋅ 20, 3 divides 60 so n = 60.

Result

The number is 60.

Page 130 Exercise 40 Answer

The shuttle departs from Dickson Hall every 12 minutes and from Lot B every 10 minutes. They first both depart from their location at 9:10 A.M, to find the next time when they both depart for campus at the same time we need to find the LCM of 10 and 12 and add that to 9:10.

12 = 2 × 6 = 2 × 2 × 3

10 = 2 × 5

The LCM of 12 and 10 is the product

2 × 2 × 3 × 5 = 60.

60 minutes is equal to one hour, so both shuttles will depart for campus at the same time every hour. Since the first time they departed at the same time is 9:10 A.M., the next is 10:10 A.M.

Result

The next time both shuttle will depart for campus at the same time is 10:10 A.M.

Page 130 Exercise 41 Answer

Mr. Jenkins bought the same number of blue pencils as green pencils. Blue pencils come in packs of 5 so the number of blue pencils he bought is a multiple of 5. Green pencils come in packs of 4 so the number of green pencils he bought is a multiple of 4. He bought the same number of blue and green pencils, which means that the number of pencils he bought for each color is the least common multiple of 4 and 5.

5 is a prime number.

4 = 2 ⋅ 2

So the least common multiple of 5 and 4 is 5 ⋅ 2 ⋅ 2 = 20.

Result

The least number of blue or green pencils Mr. Jenkins could have bought is 20.

Page 130 Exercise 42 Answer

Mr. Jenkins has 48 brushes and 60 tubes of paint to put into cups. The greatest common factor of the number of paintbrushes and the number of tubes of paint is equal to the number of cups Mr. Jenkins has.

To find the greatest common factor of 48 and 60 we first need to find their prime factorizations.

48 = 2 ⋅ 24 = 2 ⋅ 2 ⋅ 12 = 2 ⋅ 2 ⋅ 2 ⋅ 6 = 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 3

60 = 2 ⋅ 30 = 2 ⋅ 2 ⋅ 15 = 2 ⋅ 2 ⋅ 3 ⋅ 5

The common part of prime factorization of 48 and 60 is 2 ⋅ 2 ⋅ 3, thus the greatest common factor is 12.

Result

Mr. Jenkins has 12 cups.

Chapter 3 Numeric And Algebraic Expressions

Section 3.3: Write And Evaluate Numerical Expressions

Page 131 Exercise 1 Answer

The airline company charges $49 for each overwight bag and $75 for each oversized bag. To find the total amount of fees collected for that flight, first, claculate the products 49 x 50 and 75 x 6, Since there were 50 overweight bags and 6 oversized bags, and then find the sum of those products.

49 x 50 + 75 x 6 = 2450 + 450 = 2900

Result

The total amount of fees collected for that night is $2900.

Read And Learn More: enVisionmath 2.0 Grade 6 Volume 1 Solutions

Page 132 Exercise 1 Answer

The Order of Operations is:

Evaluate parentheses and brackets from inside out.

Evaluate any powers.

Multiply and divide in order from left to right.

Add or subtract in order from left to right.

Evaluating the given expression using the order of operations gives:

= \(\frac{1}{8}\left[6^3+8\right]-20\) Evaluate inside the parentheses by dividing.

= \(\frac{1}{8}[{216}+{8}]-20\) Evaluate the power.

= \(\frac{1}{8}\)[224] – 20 Evaluate inside the brackets by adding.

= 28 – 20 Multiply.

= 8 Subtract.

It is important to follow the order of operations because evaluating in a different order will not give the correct value.

Consider the expression 4 + 2 × 3. Order of operations states we need to multiply and then add so the correct value is 4 + 2 × 3 = 4 + 6 = 10. If we added and then multiplied we would get the incorrect value of 4 + 2 × 3 = 6 × 3 = 18.

Result

= \(\frac{1}{8}\left[6^3+8\right]-20\) Evaluate inside the parentheses by dividing.

= \(\frac{1}{8}[{216}+{8}]-20\) Evaluate the power.

= \(\frac{1}{8}\)[224] – 20 Evaluate inside the brackets by adding.

= 28 – 20 Multiply.

= 8 Subtract.

It is important to follow the order of operations because evaluating in a different order will not give the correct value.

Page 133 Exercise 2a Answer

Find the product inside the parentheses.

Find the sum inside the brackets.

Evaluate the power.

Evaluate the power.

Find the product.

= 10.24 – 5

Find the difference.

= 5.24

Result

5.24

Page 133 Exercise 2b Answer

To insert grouping symbols so that the expression has a value of 80 use the trial and error method. Try putting parentheses in some place and calculating the value. If the result isn’t what you need, try putting the parentheses in some other place. Repeat until you find the correct result.

For example:

\((6+12) \times\left(\frac{2}{3}\right)^2 \times 3+7\)
​
= \(18 \times\left(\frac{2}{3}\right)^2 \times 3+7\)

= 18 x \(\frac{4}{9}\) x 3 + 7

= 8 × 3 + 7

= 24 + 7

= 32

≠ 80
​
A second try:

= \(18 \times\left(\frac{2}{3}\right)^2 \times 10\)
​
= 18 x \(\frac{4}{9}\)

= 8 × 10

= 80

Result

Page 134 Exercise 1 Answer

To write a numerical expression, we use the given relationships in the problem to determine what operations are needed in the expression. To evaluate the expression, we then use the order of operations.

Page 134 Exercise 2 Answer

From the Order of Operations, we know that evaluating inside of parentheses and brackets must be done first. Grouping symbols can then change the value of a numerical expression because they change what part of the expression is evaluated first.

We are given the expression

80 ÷ 8 x 5 + 4².

Grouping the first two numbers and then evaluating gives:

(80 ÷ 8) x 5 + 4²

= 10 x 5 + 4² Evaluate inside the parentheses by dividing.

= 10 x 5 + 16 Evaluate the power.

= 50 + 16 Multiply.

= 66 Add.

Grouping the second and third numbers and then evaluating gives:

​80 ÷ (8 x 5) + 4²

= 80 ÷ 40 + 4² Evaluate inside the parentheses by multiplying.

= 80 ÷ 40 + 16 Evaluate the power.

= 2 + 16 Divide.

= 18 Add.

Grouping the last two numbers and then evaluating gives:

​80 ÷ 8 x (5 + 4²)

= 80 ÷ 8 x (5 + 16) Evaluate inside the parentheses by evaluating the power.

= 80 ÷ 8 x 21 Evaluate inside the parentheses by adding.

= 10 x 21 Divide.

= 210 Multiply.

Grouping the last three numbers and then evaluating gives:

​80 ÷ (8 x 5 + 4²)

= 80 ÷ (8 x 5 + 16) Evaluate inside the parentheses by evaluating the power.

= 80 ÷ (40 + 16) Evaluate inside the parentheses by multiplying.

= 80 ÷ 56 Evaluate inside the parentheses by adding.

= \(\frac{10}{7}\) Divide.
​
Result

Grouping symbols can change the value of a numerical expression because they change what part of the expression is evaluated first.

(80 ÷ 8) × 5 + 4² = 66

80 ÷ (8 × 5) + 4² = 18

80 ÷ 8 × (5 + 4²) = 210

80 ÷ (8 × 5 + 4²) = \(\frac{10}{7}\)

Page 134 Exercise 3 Answer

First perform the operations in brackets, when that is done you are left with (7) × (9) ÷ (8). Since there are multiplication and division left, start from left to right.

Thus, the last operation you should perform is division.

Result

Division.

Page 134 Exercise 4 Answer

Charles says that 2 × 3 − 2 is 4, and Seth says that 2 × 3 − 2 is 2.

2 × 3 − 2 … First we multiply the numbers from left to right.

= 6 − 2 … Than subtract the numbers.

= 4

The correct answer is 4, so Charles is correct.

How did Seth find the answer 2 and why is it wrong?

2 × 3 − 2 = 2 × 1 = 2

Seth first subtracted 2 from 3 and than multiplied the result by 2. However, we know that when there is no parenetheses nor brackets we have a specific order of operations. First, evaluate the parentheses and brackets from inside out. Second, evaluate powers. Third, multiply and divide from left to right. Fourth, Add and subtract from left to right. Seth was wrong to have first subtracted 2 from 3 and than multipled the result by 2.

Result

Charles is correct.

Page 134 Exercise 5 Answer

52 + (6.7−3.1) … Evaluate the expression inside the parentheses.

= 52 + 3.6 … Evaluate the power.

= 25 + 3.6 … Add the numbers.

= 28.6

Result

28.6

Page 134 Exercise 6 Answer

(8.2 + 5.3) ÷ 5 … Evalute the expression inside the parentheses.

= 13.5 ÷ 5 … Divide the numbers.

= 2.7

Result

2.7

Page 134 Exercise 7 Answer

(1.5−0.52) ÷ [(3 + 2) × 2] … Add the numbers inside the parentheses.

= (1.5 − 0.52)÷[5 × 2] … Evalute the expression inside the brackets.

= (1.5 − 0.52) ÷ 10 … Evaluate the power.

= (1.5 − 0.25) ÷ 10 … Evaluate the expression inside the parentheses.

= 1.25 ÷ 10 … Divide the numbers.

= 0.125

Result

0.125

Page 134 Exercise 8 Answer

36.8 ÷ [11.5 − (2.5×3)]² … Evalute the expression inside the parentheses.

= 36.8 ÷ [11.5 − 7.5]² … Evalute the expression inside the brackets.

= 36.8 ÷ 4² … Evalute the power.

= 36.8 ÷ 16 … Divide the numbers.

= 2.3

Result

2.3

Page 134 Exercise 9 Answer

The Order of Operations is:

Evaluate parentheses and brackets from inside out.

Evaluate any powers.

Multiply and divide in order from left to right.

Add or subtract in order from left to right.

Using the order of operations then gives:

6 + 4 × 5 ÷ 2 − 8 × 1.5

= 6 + 20 ÷ 2 – 12 Multiply.

= 6 + 10 – 12 Divide.

= 16 – 12 Add.

= 4 Subtract.

Result

4

6 ÷ 4 × 5 ÷ 2 − 8 × 1.5 … Multiply and divide from left to right.

= 1.5 × 5 ÷ 2 − 8 × 1.5 … Multiply and divide from left to right.

= 7.5 ÷ 2 − 8 × 1.5 … Multiply and divide from left to right.

= 3.75 − 8 × 1.5 … Multiply and divide from left to right.

= 3.75 − 12 … Add and subtract from left to right.

= −8.25

Result

−8.25

Page 134 Exercise 10 Answer

The given expression is 12 × 3² + 36 and the target value is 540.

12 × 3² + 36 = 12 × 9 + 36 = 108 + 36 = 144 ≠ 540

12 × (3² + 36) = 12 × (9 + 36) = 12 × 45 = 540

Result

12 × (3² + 36)

Page 134 Exercise 11 Answer

The given expression is 32 ÷ 2³ − 4 and the target value is 8.

32 ÷ 2³ − 4 = 32 ÷ 8 − 4 = 4 − 4 = 0 ≠ 8

32 ÷ (2³ − 4) = 32 ÷ (8 − 4) = 32 ÷ 4 = 8

Result

32 ÷ (2³ − 4)

Page 134 Exercise 12 Answer

The given expression is 2.32 + 9 × 4 ÷ 2 and the target value is 28.58.

2.32 + 9 × 4 ÷ 2 = 5.29 + 9 × 4 ÷ 2 = 5.29 + 45 ÷ 2 = 5.29 + 22.5 = 27.79 ≠ 28.58

(2.32 + 9) × 4 ÷ 2 = (5.29 + 9) × 4 ÷ 2 = 14.29 × 4 ÷ 2 = 57.16 ÷ 2 = 28.58.

Result

(2.32 + 9) × 4 ÷ 2

Page 135 Exercise 13 Answer

42 − (3.1 + 6.4) + 4.5 … Evaluate the expression in the parentheses.

= 42 − 9.5 + 4.5 … Evaluate the power.

=16 − 9.5 + 4.5 … Add and subtract numbers from left to right.

= 6.5 + 4.5 … Add and subtract numbers from left to right.

= 11

Result

11

Page 135 Exercise 14 Answer

\((8.7+3.3) \times\left(\frac{1}{2}\right)^2\) …..Evaluate the expression in the parentheses.

= \(12 \times\left(\frac{1}{2}\right)^2\) …….Evaluate the power.

= 12 x \(\frac{1}{4}\) …..Multiply the numbers.

= 3

Result

3

Page 135 Exercise 15 Answer

157.8 − (32 + 6) × 3 … Evaluate the power.

= 157.8 − (9 + 6) × 3 … Evaluate the expression in the parentheses.

= 157.8 − 15 × 3 … Multiply the numbers from left to right.

= 157.8 − 45 … Subtract the numbers from left to right.

= 112.8

Result

112.8

Page 135 Exercise 16 Answer

4.3 + (8.4 − 5.1) … Evalute the expression in the parentheses.

= 4.3 + 3.3 … Add the numbers.

= 7.6

Result

7.6

Page 135 Exercise 17 Answer

\(1.25 \times 4+3 \times 2 \div\left(\frac{1}{2}\right)^3\) …..Evaluate the power.

= \(1.25 \times 4+3 \times 2 \div \frac{1}{8}\) …. Multiply and divide the number from left to right.

= \(5+3 \times 2 \div \frac{1}{8}\) ….Multiply and divide the number from left to right.

= 5 + 6 ÷ \(\frac{1}{8}\) …Multiply and divide the number from left to right.

= 5 + 6 x \(\frac{8}{1}\) … Multiply and divide the number from left to right.

= 5 + 48 …. Add the numbers.

= 53

Result

53

Page 135 Exercise 18 Answer

[2³ x (152 ÷ 8)] − 52 … Evalute the expression in the parentheses.

= [8 × 19] − 52 … Evalute the power.

= 152 – 52 … Subtract the numbers.

= 100

Result

100

Page 135 Exercise 19 Answer

The given expression is 2 × 9 + 7 and the target value is 32.

(2 × 9) + 7 = 19 + 7 = 26 ≠ 32

2 × (9 + 7) = 2 × 16 = 32

Result

2 × (9 + 7)

Page 135 Exercise 20 Answer

The given expression is \(\frac{1}{3}\) x 21 – 3 and the target value is 6.

\(\frac{1}{3}\) x 21 – 3 = 7 – 3 = 4

\(\frac{1}{3}\) x (21 – 3) = \(\frac{1}{3}\) x 18 = 6

Result

\(\frac{1}{3}\) x (21 – 3)

Page 135 Exercise 21 Answer

The given expression is 2.5 + 5 × 6 − 2 and the target value is 43.

2.5 + 5 × 6 − 2 = 2.5 + 30 − 2 = 30.5 ≠ 43

(2.5 + 5) × 6 − 2 = 7.5 × 6 − 2 = 45 − 2 = 43

Result

(2.5 + 5) × 6 − 2

Page 135 Exercise 22 Answer

Cory bought some baseball equipment, a bat, a glove, and 3 baseballs. A bat costs $69, a glove $75, and a baseball $5.50. He used a coupon for \(\frac{1}{2}\) off the price of the bat and glove. \(\frac{1}{2}\) × (69 + 75) + 3 × 5.50 … Evalute the expression in the parentheses. = ​\(\frac{1}{2}\) × 144 + 3 × 5.50 … Multiply the numbers from left to right. = 72 + 16.50 … Add the numbers. = 88.50$

Result

The total cost of the bat, the glove, and 3 baseballs is $88.50.

Page 135 Exercise 23 Answer

First, calculate the value of the given expression.

5 + (8 − 4) ÷ 2 + 3

Find the difference inside the parenthesis.

= 5 + (4) ÷ 2 + 3

Find the quotient.

= 5 + 2 + 3

Find the sum.

= 7 + 3

Find the sum.

= 10

To write a numerical expression with the same value as the previous one, 10, start with that value. Rewrite it as a result of one of the arithmetic operations.

For example, write 10 as a difference of two numbers.

10 = 20 − 10

Rewrite one of the numbers as a product of two of its factors.

10 = 20 − 10 = 4 × 5 − 10

Rewrite one of the numbers as a power.

10 = 20 − 10 = 4 × 5 − 10 = 2² × 5 − 10

A numerical expression with the same value as the one given in the Exercise:

2² × 5 − 10

For example, write 10 as a product of two of its factors.

10 = 2 × 5

Rewrite one of the factors as a quotient.

10 = 2 × 5 = \(\frac{8}{4}\) × 5

Rewrite one of the numbers as a difference.

10 = 2 × 5 = \(\frac{8}{4}\) × (10 − 5)

A numerical expression with the same value as the one given in the Exercise: \(\frac{8}{4}\) × (10−5).

Result

Possible answers: 2² × 5 − 10 and \(\frac{8}{4}\) × (10−5)

Page 136 Exercise 24 Answer

The first part of the numerical expression to evaluate is the part inside the parentheses or bracket. However, in this example, there are more than one parentheses, so the first part we evaluate is the power since powers come second in the order of operations.

(26 + 2.5) − [(8.3 × 3) + (13 − 0.25)]

=(26 + 2.5) − [(8.3×3) + (1−0.25)]

Next, we evaluate the product.

= (26 + 2.5) − [(8.3 × 3)+(1 − 0.25)]

= (26 + 2.5) − [(24.9)+(1−0.25)]

Find the sum.

= (26 + 2.5) − [(24.9)+(1−0.25)]

= (28.5) − [(24.9)+(1−0.25)]

Find the difference.

= (28.5) − [(24.9)+(1−0.25)]

= (28.5) − [(24.9)+(0.75)]

Find the sum inside the brackets.

= (28.5) − [(24.9)+(0.75)]

= (28.5) − [25.65]

Find the difference.

= (28.5) − [25.65]

= 2.85

Result

The first part of the numerical expression to evaluate is the part inside the parentheses or bracket. However, in this example, there are more than one parentheses, so the first part we evaluate is the power since powers come second in the order of operations.

Page 136 Exercise 25 Answer

Solve the numerical expression and then compare it to Evan’s solution.

0.2² + 12 ÷ (1.5 × 4)

Evalute the power.

= 0.04 + 12 ÷ (1.5 × 4)

Find the product inside the parentheses.

= 0.04 + 12 ÷ (6)

Find the quotient.

= 0.04 + 2

Find the sum.

= 2.04

Evan’s solution is not correct. The solution is 2.04.

Result

Evan’s solution is not correct. The solution is 2.04.

Page 136 Exercise 26 Answer

Since the length of the rectangular drawing is 12 inches, one-third of that length is the product \(\frac{1}{3}\) × 12. Add three to that product.
​
\(\frac{1}{3}\) × 12 + 3

Find the product.

= 4 + 3

Find the sum.

= 7

The width of the rectangular drawing is 7 inches.

To find the perimeter calculate the sum of all sides of the rectangular.

12 + 7 + 12 + 7 = 38

Result

The perimeter of the drawing is 38 square inches.

Page 136 Exercise 27 Answer

Even though the expressions have the same numbers and the same operations, they are not the same since Frederick’s expression has parentheses and brackets in multiple places and Lana’s doesn’t. Frederick has then carried out the operations in a different order than Lana, and thus the results are different but both correct.

[(53.7+37.2) − (33 + 3.8)] − 8.6 = 51.5

Frederick first evaluated the power.

= [(53.7+37.2) − (27+3.8)] − 8.6

He than found the sum in the parentheses.

= [(53.7 + 37.2) − (30.8)] − 8.6

He found the sum in the other parentheses.

= [(90.9) − (30.8)] − 8.6

He found the difference in the brackets.

= [60.1] − 8.6

Finally, he found the difference, the end result.

= 51.5

53.7 + 37.2 − 33 + 3.8 − 8.6 = 59.1

Lana also first evaluated the power.

= 53.7 + 37.2 − 27 + 3.8 − 8.6

However, in her expression she doesn’t have any parentheses or brackets, so next she found the sums and differences – from left to right.

= 90.9 − 27 + 3.8 − 8.6

= 63.9 + 3.8 − 8.6

= 67.7 − 8.6

= 59.1

Result

Even though the expressions have the same numbers and the same operations, they are not the same since Frederick’s expression has parentheses and brackets in multiple places and Lana’s doesn’t. Frederick has then carried out the operations in a different order than Lana, and thus the results are different but both correct.

Page 136 Exercise 28 Answer

Lillian both four hairbrushes at $3.99 each. She had a coupon for $1 off. Her mom paid for half of the remaining cost.

\(\frac{1}{2}\) × [(4×3.99) − 1] … Evalute the expression in the parentheses.

= \(\frac{1}{2}\) × [15.96−1] … Evalute the expression in the brackets.
​
= \(\frac{1}{2}\) × 14.96 … Multiply the numbers.

= 7.48

Result

Lillian paid $7.48 toward the purchase of the hairbrushes.

Page 136 Exercise 29 Answer

In an ecosystem, some animals get energy by eating plants. An elk can eat 20 pounds of plants each day and we need to find how many pounds of plants a herd of 18 elk can eat in one week.

20 × 18 × 7 … Multiply the numbers from left to right.

= 360 × 7 = 2520

Result

A herd of 18 elk can in one week eat 2520 pounds of plants.

Page 136 Exercise 30 Answer

12.3 × [(2×1.7) + 6.6]

Find the product in the parentheses.

= 12.3 × [(3.4) + 6.6]

Find the sum in the brackets.

= 12.3 × [10]

Find the product.

= 123

24 ÷ [(3.2 × 0.8) + 1.44]

Find the product in the parentheses.

= 24 ÷ [(2.56) + 1.44]

Find the sum in the brackets.

= 24 ÷ [4]

Evaluate the power.

= 16 ÷ [4]

Find the quotient.

= 4

6.2 + (3 × \(\frac{1}{3}\) + 4.8)

Find the product in the parentheses..

= 6.2 + (1 + 4.8)

Find the sum in the parentheses.

= 6.2 + (5.8)

Find the sum.

= 12

[4 x (9.6 ÷ 3)] + 8.2

Find the quotient in the parenetheses.

= [4 x (3.2)] + 8.2

Find the product in the brackets.

= [12.8] + 8.2

Find the sum.

= 21

Result

12.3 × [(2 × 1.7) + 6.6] = 123

24 ÷ [(3.2 × 0.8) + 1.44] = 4

6.2 + (3 × \(\frac{1}{3}\) + 4.8) = 12

[4 × (9.6÷3)] + 8.2 = 21

Page 136 Exercise 31 Answer

[4 × (6.6÷3)] + 18 (Find the quotient in the parenthesis.)

= [4 × (2.2)] + 18 (Find the product in the brackets.)

= [8.8] + 18 (Find the sum.)
​
= 26.8

18.9 × [(2 × 2.7) − 4.6] − 2² (Find the product in the parenthesis.)

= 18.9 × [(5.4) − 4.6] − 2² (Find the difference in the brackets.)

= 18.9 × [0.8] − 2² (Evaluate the power. )

= 18.9 × [0.8] − 4 (Find the product. )

= 15.12 − 4 (Find the difference.)
​
= 11.12

3³ ÷ [(2.6 x 0.7) + 1.18] (Find the product in the parenthesis.)

= 3³ ÷ [(1.82) + 1.18] (Find the sum in the brackets.)

= 3³ ÷ [3] (Evaluate the power. )

= 27 ÷ [3] (Find the quotient. )

= 9
​
6.9 + (2 x 42 – 4.1) (Evaluate the power inside the parenthesis)

= 6.9 + (32 – 4.1) (Find the product in the parenthesis)

= 6.9 + (32 – 4.1) (Find the difference in the parenthesis)

= 6.9 + (27.9) (Find the sum)

= 34.8

Result

[4 x (6.6 ÷ 3)] + 18 = 26.8

18.9 × [(2×2.7) − 4.6] − 2² = 11.12

3³ ÷ (2.6 × 0.7) + 1.18] = 9

6.9 + (2 × 4² − 4.1) = 34.8

Chapter 3 Numeric And Algebraic Expressions

Midpoint

Page 137 Exercise 1 Answer

For the power 4³, the base is 4 and the exponent is 3. The exponent means the base will be a factor 3 times when the power is rewritten as repeated multiplication. That is, 4³ = 4 × 4 × 4.

Page 137 Exercise 2 Answer

The statement “The GCF of 12 and 28 is 4.” is true.

12 = 2 × 6 = 2 × 2 × 3

28 = 2 × 14 = 2 × 2 × 7
​
Common factors of 12 and 28 are 2 and 2, so their product, which equals 4, is the GCF of 12 and 28.

Result

True.

Read And Learn More: enVisionmath 2.0 Grade 6 Volume 1 Solutions

Page 137 Exercise 3 Answer

The statement “The GCF of 18 and 84 is 9.” is false.

18 = 2 × 9 = 2 × 3 × 3

84 = 2 × 42 = 2 × 2 × 21 = 2 × 2 × 3 × 7

Common factors of 18 and 84 are 2 and 3, so their product, which equals 6, is the GCF of 18 and 84.

Result

False.

Page 137 Exercise 4 Answer

The statement “The LCM of 12 and 6 is 6.” is false.

Since 12 is greater than 6, 6 is not the LCM of 6 and 12.

Result

False.

Page 137 Exercise 5 Answer

The statement “The LCM of 4 and 8 is 8.” is true.

Since 8 can be written as a product of 4, 8 is a multiple of 4. Also 8 is the first multiple 8. Thus, the LCM of 4 and 8 is 8.

Result

True.

Page 137 Exercise 6 Answer

0.52 x (20 – 22 x 3) x (\(\frac{2}{5}\) x 25) (Evaluate the power.)

= 0.52 x (20 – 4 x 3) x (\(\frac{2}{5}\) x 25) (Find the product.)

= 0.52 x (20 – 12) x (\(\frac{2}{5}\) x 25) (Find the product.)

= 0.52 x (20 – 12) x (10) (Find the difference.)

= 0.52 ×(8) × (10) (Evaluate the power.)

= 0.25 × (8) × (10) (Find the product.)

= 2 × (10) (Find the product.)
​
= 20

2[(14 – 3) + 43 ÷ 4] – 2.5 x 12 (Find the difference in the parenthesis.)

= 2[(11) + 43 ÷ 4] – 2.5 x 12 (Evaluate the power.)

= 2[(11) + 64 ÷ 4] – 2.5 x 12 (Find the quotient in the brackets.)

= 2[(11) + 16] – 2.5 x 12 (Find the sum in the brackets.)

= 2[27] – 2.5 x 12 (Find the product.)

= 54 – 2.5 x 12 (Find the product.)

= 54 – 30 (Find the difference.)
​
= 24

9 + 12 ÷ \(\frac{2}{3}\) – 33 ÷ 3 (Evaluate the power.)

= 9 + 12 ÷ \(\frac{2}{3}\) (Find the quotient.)

= 9 + 18 − 27 ÷ 3 (Find the quotient.)

= 9 + 18 – 9 (Find the sum.)

= 27 – 9 (Find the difference.)
​
= 18

Result

0.52 x (20 – 22 x 3) x (\(\frac{2}{5}\) x 35) = 20

2[(14−3) + 43 ÷ 4]− 2.5 × 12 = 24

9 + 12 ÷ \(\frac{2}{3}\) ÷ 3 = 18

Page 137 Exercise 7 Answer

First, square the given fraction.

The given fraction \(\left(\frac{2}{3}\right)^2\) is equal to the first option \(\frac{4}{9}\).

The second option the fractions have the same numerators, but the denominators are different, thus the fractions are not equal.

To compare the third option – a product of two fractions, with a given fraction, find the product and than compare.

The fractions have the same denominators, but the numerators are different, thus the fractions are not equal.

To compare the fourth option – a product of a fraction and a whole number, with a given fraction, find the product and than compare.

\(\frac{1}{9}\) x 4 (Rewrite the whole number as a fraction)

= \(\frac{1}{9} \times \frac{4}{1}\)

= \(\frac{1 \times 4}{9 \times 1}\)

= \(\frac{4}{9}\)

The given fraction \(\left(\frac{2}{3}\right)^2\) is equal to the fourth option \(\frac{1}{9}\) x 4.

To compare the fifth option – a product of two fractions, with a given fraction, find the product and than compare.

The given fraction \(\left(\frac{2}{3}\right)^2\) is equal to the fifth option \(\frac{2}{3}\) x \(\frac{2}{3}\).

Result

Expressions \(\frac{2}{3} \times \frac{2}{3}, \frac{1}{9} \times 4, \frac{4}{9}\) are equal to \(\left(\frac{2}{3}\right)^2\).

Page 137 Exercise 8 Answer

Liam bought 2 vintage movie posters which cost $28.50, so the cost of it is the product 2 × 28.50. He bought 2 rock posters which cost $29.75, and 1 rap poster which costs $19.50. The total cost, without the gift card and the \(\frac{1}{2}\) – off coupon, is the sum of the products mentioned before.

Since he applied a $35 gift card to the total purchase, so from the sum mention before we subtract 35. A \(\frac{1}{2}\) – off coupon is applied only to the rap poster so we multiply 1 × 19.50 by \(\frac{1}{2}\).

The numerical expression which then describes the total cost for the posters is:

= [2 × 28.50 + 2 × 29.75 + \(\frac{1}{2}\) (19.50)] − 35

= [2 × 28.50 + 2 × 29.75 + 9.75] − 35

= [57 + 2 × 29.75 + 9.75] − 35

= [57 + 59.5 + 9.75] − 35

= [116.5 + 9.75] − 35

= [126.25] − 35

= 91.25

Result

Liam paid $91.25 for the posters.

Page 137 Exercise 9 Answer

Since Eva volunteers every 3 days and Jin volunteers every 5 days, they will volunteer together on the days which are multiples of both 3 and 5. To find this numbers, first find the least common multiple of 3 and 5. Multiples of the LCM of 3 and 5 are then mulitples of both 3 and 5.

Both 3 and 5 are prime so the LCM is simply their product.

3 x 5 = 15

Since they worked the first day of the month, they will volunteer together on the 15-th day and on the 30-th day.

Result

They will volunteer together on the same day again in 15 days.

Page 138 Exercise 1a Answer

To find the number of gift bags Raoul needs to make, find the greatest common factor of of 72 and 96.

72 = 2 × 36 = 2 × 2 × 18 = 2 × 2 × 2 × 9 = 2 × 2 × 2 × 3 × 3

96 = 2 × 48 = 2 × 2 × 24 = 2 × 2 × 2 × 12 = 2 × 2 × 2 × 2 × 6

= 2 × 2 × 2 × 2 × 2 × 3

The greatest common factor of 72 and 96 is the product of common factors in the prime factorization of 72 and 96.

2 × 2 × 2 × 3 = 24

The GCF of 72 and 96 is 24.

Raoul needs to make 24 gift bags.

To find how many wristbands and how many movie posters go in each gift bag if evenly distributed write 72 and 96 as a product of two numbers whose one factor is 24.

72 = 24 × 3

96 = 24 × 4

In each bag Raoul needs to put 3 wristbands and 4 movie posters.

Result

Raoul needs to make 24 gift bags and in each bag he needs to put 3 wristbands and 4 movie posters.

Page 138 Exercise 1b Answer

To answer the question find the least common multiple of 12 and 10.

2 = 2 × 6 = 2 × 2 × 3

10 = 2 × 5

The LCM of 12 and 10 is the product of the union of factors of 12 and 10.

2 × 2 × 3 × 5 = 60

The LCM of 12 and 10 is 60.

Rewrite 60 as a product of two numbers, one being 12.

60 = 12 × 5
​
Rewrite 60 as a product of two numbers, one being 10.

60 = 10 × 6

To have an equal number of cups and napkins Monique needs to buy 5 packages of cups and 6 packages of napkins.

Result

Monique needs to buy 5 packages of cups and 6 packages of napkins.

Page 138 Exercise 1c Answer

Monique will have 60 cups and 60 napkins in Part B.

Find the value of each expression to check if they are equal to 60.

21 x 30 (A number raised to the power of one is that number.)

= 2 x 30 (Find the product.)
​
= 60

102 x 60 (Evaluate the power.)

= 100 x 60 (Find the product.)
​
= 6000

≠ 60

22 x 15 (Evaluate the power.)

= 4 x 15 (Find the product.)
​
= 60

4600 (Any number raised to the power of 0 is equal to 1.)
​
= 1

≠ 60

40 x 60 (Any number raised to the power of 0 is equal to 1.)

= 1 x 60 (Find the product.)
​
= 60

Result

40 × 60, 22 × 15, and 21 × 30.

Page 138 Exercise 1d Answer

In Part B we calculated that the teachers need to buy 5 packages of cups and 6 packages of napkins. The total cost of cups and napkins is the sum of products 5 × $3.50 and 6 × $4.25.

To answer the question

find the difference between the total cost of supplies and the money teachers have.

(5 x 3.50 + 6 x 4.25) – 25 (Find the products.)

= (17.50 + 25.50) – 25 (Find the sum.)

= 43 – 25 (Find the difference.)
​
= 18

Result

The teachers need $18 more.

Chapter 3 Numeric And Algebraic Expressions

Section 3.4: Write Algebraic Expressions

Page 139 Exercise 1 Answer

The difference between the number of games the Hornets won and the number of games the Lynx won is two. The Lynx always win two more games than the Hornets. Thus, if the Lynx won g games, the Hornets must win two less.

A mathematical expression describing how many games the Hornets won is g − 2.

In this expression the greater number (the number of games the Lynx win) is given as the variable and the lesser number (the number of games the Hornets win) is then expressed in relation to that variable.

In the other expression, given as

n + 2,

where n marks the number of games the Hornets won, and n + 2 the number of games the Lynx won. Here the lesser number is given as the variable and the greater number is then expressed in relation to that variable.

Read And Learn More: enVisionmath 2.0 Grade 6 Volume 1 Solutions

Result

g − 2

For g − 2, the number of games the Lynx won is the variable and the number of games the Hornets won is expressed in relation to that variable. For n + 2, the number of games the Hornets won is the variable and the number of games the Lynx won is expressed in relation to that variable.

Page 139 Exercise 1a Answer

The pairs of numbers are: 3 and 5, 6 and 8, 9 and 11. What those pairs have in common is that when subtracted they all give two as a result.

The pattern seen in the data is then the number of games won by the Lynx is 2 more than the number of games won by the Hornets.

Result

Each pair of number has a difference of 2. The pattern seen in the data is the number of games won by the Lynx is 2 more than the number of games won by the Hornets.

Page 139 Exercise 1b Answer

When the Hornets won 3 games, the Lynx won 5, which can be written as 3 + 2 = 5.

When the Hornets won 6 games, the Lynx won 8, which can be written as 6 + 2 = 8.

When the Hornets won 9 games, the Lynx won 11, which can be written as 9 + 2 = 11.

Result

Page 139 Exercise 1c Answer

In the table we can see that when the Hornets won 3 games, the Lynx won 5, the difference being two. The Hornets 6, the Lynx 8, the difference again is two. The Hornets 9, the Lynx 11, the difference again is two.

Thus, the pattern is: if the Hornets win n games, the Lynx win n + 2 games, since the difference is than (n + 2) – n which than equals two.

Result

If the Hornets win n games, the Lynx win n+2 games.

Page 140 Exercise 1 Answer

The algebraic expressions which represent the total cost of the mystery books Rachel bought are

m × 6.50,m ⋅ 6.50 and m(6.50).

The expressions above are algebraic expressions because each of them contains a variable – a letter which represents an unknown quantity, in this case the letter m which represents the number of mystery book Rachel bought. The expressions also contain an operation – multiplication.

Result

m × 6.50, m ⋅ 6.50, and m(6.50)

The expressions are algebraic expressions because each of them contains a variable and an operation.

Page 141 Exercise 2 Answer

The algebraic expression which represents “8 minus the quantity b divided by 6” is

8 − b ÷ 6.

Result

8 − b ÷ 6

Page 141 Exercise 3 Answer

A term is each part of an expression separated by a plus sign or a minus sign.

The expression r ÷ 9 + 5.5 has two terms. The first term is the quotient r ÷ 9 and the second term is 5.5.

Result

The expression has two terms.

Page 142 Exercise 1 Answer

To write an algebraic expression we can use a variable, which is written as a letter and it represents a quantity that can change, and at least one of the following operations: addition, subtraction, multiplication, and division.

For example, we can use the variable x and the operation addition to write the following algebraic expression:

x + 3,

or we can use the variable y and the operation multiplication to write:

7 × y.

Result

To write an algebraic expression we can use a variable, which is written as a letter and it represents a quantity that can change, and at least one of the following operations: addition, subtraction, multiplication, and division.

Page 142 Exercise 2 Answer

The variable in the given algebraic expression \(\frac{6}{x}\) is x since the letter is x, and the operation is division since fractions mean division.

Result

The variable is x and the operation is division.

Page 142 Exercise 3 Answer

An algebraic expression is a type of math expression that has at least one variable and at least one operation.

In the expression

n is the variable and the operation is addition, so the given expression is an algebraic expression.

Result

The expression has the variable n and the operation addition.

Page 142 Exercise 4 Answer

The expression 2(3+4) can be described as a product of two factors, where one factor is 2 and the second factor is the sum 3 + 4.

Result

Yes, one factor is 2 and the second factor is 3 + 4.

Page 142 Exercise 5 Answer

In the expression 2(3+4) we have two factors, one is 2 and the other is 3 + 4. The other factor, 3 + 4 is the sum of two terms.

Result

3 + 4 is the sum of two terms.

Page 142 Exercise 6 Answer

The given situation is “five less than y”, which can be written as an algebraic expression: y − 5.

y − 5

Page 142 Exercise 7 Answer

The given situation is “six times the quantity two x plus three y”, which can be written as an algebraic expression:6 × (2x + 3y).

Result

6 × (2x + 3y)

Page 142 Exercise 8 Answer

The expression

has three terms.

A term is a part of an expression that is separated by a plus or a minus sign. Thus, the given expression has the following terms: \(\frac{w}{4}\), 12.5, and 7z.

The first term, \(\frac{w}{4}\), is written as a fraction and represents the quotient of w divided by 4.

The second term, 12.5, is a constant numerical value.

The third term, 7z, is a product of two factors. A coefficient, the number that is multiplied by a variable, in this case the variable z, is 7. So, 7 is the coefficient of z.

Result

Three terms since the terms are the parts of the expression that are separated by a plus or a minus sign.

Page 142 Exercise 9 Answer

In the expression

which has three terms of \(\frac{w}{4}\), 12.5, and 7z, the third term of 7z, has a coefficient.

When a term is a product of two factors, a variable and a constant numerical value , a coefficient is the number that is multiplied by a variable. In this case, the variable is z and the coefficent is 7. So, 7 is the coefficient of z.

Result

7z has a coefficient since it has a number multiplied by a variable.

Page 142 Exercise 10 Answer

In the expression

which has three terms of \(\frac{w}{4}\), 12.5, and 7z, the second term of 12.5 is a constant numerical value since it has no variable.

Result

12.5

Page 143 Exercise 11 Answer

The given situation is “12 times a number g”, which can be written as an algebraic expression: 12 × g.

Result

12 × g

Page 143 Exercise 12 Answer

The given situation is “p pennies added to 22 pennies”, which can be written as an algebraic expression: p + 22.

Result

p + 22

Page 143 Exercise 13 Answer

The given situation is “22 divided by a number s”, which can be written as an algebraic expression: 22 ÷ s, which can also be written as \(\frac{22}{s}\).

Result

22 ÷ s or \(\frac{22}{s}\)
​
Page 143 Exercise 14 Answer

The given situation is “\(12 \frac{3}{4}\) less than the product of 7 and a number x”, which can be written as an algebraic expression:

7x − \(12 \frac{3}{4}\)

Result

7x − \(12 \frac{3}{4}\)
​
Page 143 Exercise 15 Answer

The expression

5 − g

has two terms.

A term is a part of an expression that is separated by a plus or a minus sign. Thus, the given expression has the following terms: 5 and g.

The first term is a constant numerical value and the second term is a variable. The operation is subtraction.

Result

The given expression has two terms.

Page 143 Exercise 16 Answer

The expression

3 + \(\frac{1}{2}\)b

has two terms.

A term is a part of an expression that is separated by a plus or a minus sign. Thus, the given expression has the following terms: 3 and ​\(\frac{1}{2}\)b.

The first term is a constant numerical value and the second term is a product of a coefficient \(\frac{a}{b}\) and a variable b.

Result

The given expression has two terms.

Page 143 Exercise 17 Answer

The expression

\(\frac{v}{3}\) + 3⋅ 5

has two terms.

A term is a part of an expression that is separated by a plus or a minus sign. Thus, the given expression has the following terms: \(\frac{v}{3}\) and 3 ⋅ 5.

The first term is a constant numerical value and the second is a product of two numbers.

Result

The given expression has two terms.

Page 143 Exercise 18 Answer

The expression

16.2 − (3 ⋅ 4) + (14 ÷ 2)

has three terms.

A term is a part of an expression that is separated by a plus or a minus sign. Thus, the given expression has the following terms: 16.2, (3 ⋅ 4), and (14 ÷ 2).

The first term is a constant numerical value, the second term is a product of two numbers, and the third term is a quotient of two numbers.

Result

The given expression has three terms.

Page 143 Exercise 19 Answer

In the expression

5.3t – (20 ÷ 4) + 11

there are three terms: 5.3t, (20 ÷ 4), and 11.

The second term, (20 ÷ 4), is a quotient of two numbers, 20 and 4.

Result

The second term (20 ÷ 4) is a quotient of two numbers, 20 and 4.

Page 143 Exercise 20 Answer

In the expression

5.3t – (20 ÷ 4) + 11

there are three terms: 5.3t, (20 ÷ 4), and 11.

The first term, 5.3t, is a product of a variable t, and a constant numerical value, 5.3.

Result

The first term 5.3t is a product of a variable t, and a constant numerical value, 5.3.

Page 143 Exercise 21 Answer

To following expression shows how much longer is the round-trip to San Diego, which is 1,012 miles, than the round-trip to San Jose, which is 236 miles:

1012 − 236.

Result

The expression has two terms.

Page 143 Exercise 22 Answer

A truck driver made 5 round-trips to Los Angeles and some round-trips to San Diego. Let the number of round-trips to San Diego be x.

Each round trip to Los Angeles is 770 miles and each round trip to San Diego is 1,012. The following expression then shows how many miles he drove in all:

5 ⋅ 770 + x ⋅ 1012

The second term, x ⋅ 1012, describes how many miles he drove to San Diego, while x stands for how many trips he made to San Diego.

Result

The expression is 5 ⋅ 770 + x ⋅ 1012. The second term, x ⋅ 1012, describes how many miles he drove to San Diego, while x stands for how many trips he made to San Diego.

Page 144 Exercise 23 Answer

The expression is:

y ÷ 3(4 – 2) + 5.5

The given expression has the following parts: y, 3(4 – 2), and 5.5.

The first part, y, is a variable.

The second part, 3(4 – 2), is a product of the numbers 3 and 4 – 2. This second factor of the product is a difference of 4 and 2.

The third part, 5.5, is a constant numerical value.

Result

Page 144 Exercise 24 Answer

In one year a florist sells f flowers, so in 6 years he sells 6 ⋅ f flowers.

The expression 6 ⋅ f shows how many flowers a float in the parade may use.

Result

6 ⋅ f

Page 144 Exercise 25 Answer

The expression abc is a product of three variables, however it has only one term since a term is a part of an expression that is separated by a plus or a minus sign. Thus, Anthony is wrong.

Page 144 Exercise 26 Answer

In a 5-day period Yuri walked the same number of poodles and the same number of bulldogs each day as he did on Monday, when he walked p puddles and b bulldogs. This means he walked p + b dogs each of the 5 days. The following expression shows how many dogs were walked in this 5-day period:

5 ⋅ (p+b).

Result

5 ⋅ (p+b)

Page 144 Exercise 27 Answer

There are two baskets of apples and each has 12 apples, so we will multiply 2 by 12.

There is no exact number of students, so let’s mark the number of students with n. They equally shared two baskets of apples thus we will divide the product 2 ⋅ 12 by n.

The following expression represents the given situation:

(2⋅12) ÷ n.

Result

(2⋅12) ÷ n where n is the number of students.

Page 144 Exercise 28 Answer

A regular octagon has 8 sides which are all equal length, in this case s. The following two expressions represent the perimeter P of the figure.

P = s + s + s + s + s + s + s + s

P = 8 ⋅ s

The first expression uses addition to represent the perimeter, and the second expression uses multiplication to express the same.

Result

s + s + s + s + s + s + s + s and 8 ⋅ s

Page 144 Exercise 29 Answer

Four more than the product 3 times the number of c cats can be represented by the following expressions:

4 + 3c,

3 ⋅ c + 4,

(3×c) + 4.

The expression (4+3)c doesn’t represent four more than the product 3 times the number of c cats. The sum 4 + 3 is in brackets so it takes priority over multiplication. The result is then 7c which is not equal to 4 + 3c.

Result

(4+3)c

Page 144 Exercise 30 Answer

The phrase “Four less than w divided by 4.”could be represented by the algebraic expression

\(\frac{w}{4}\) – 4.

The other phrases couldn’t be represented by the given algebraic expression.

“The quotient of four and a number w” could be represented by

The difference between a number w and 4″ and Four less than a number w” could be represented by

w − 4.

Result

four less than w divided by 4

Chapter 3 Numeric And Algebraic Expressions

Section 3.6: Generate Equivalent Expressions

Page 155 Exercise 1a Answer

Evaluating 8n + 6 for n = 1 gives:

8n + 6 = 8(1) + 6 Substitute n = 1.

= 8 + 6 Multiply.

= 14 Add.

Evaluating 2(4n + 3) for n = 1 gives:

2(4n + 3) = 2(4.1 + 3) Substitute n = 1.

= 2(4 + 3) Multiply.

= 2 . 7 Add.

= 14 Multiply.

Read And Learn More: enVisionmath 2.0 Grade 6 Volume 1 Solutions

Evaluating 14n for n = 1 gives:

14n = 14(1) Substitute n = 1.

= 14 Multiply.
​
Result

Page 155 Exercise 1b Answer

Evaluating 8n + 6 for n = 2 gives:

= 8(2) + 6 Substitute n = 2.

= 16 + 6 Multiply.

= 22 Add.

Evaluating 2(4n+3) for n = 2 gives:

2(4n+3) = 2(4⋅2+3) Substitute n = 2.

= 2(8 + 3) Multiply.

= 2 . 11 Add.

= 22 Multiply.

Evaluating 14n for n = 2 gives:

= 14(2) Substitute n = 2.

= 28 Multiply.

Result

Page 155 Exercise 1c Answer

Expressions are equivalent if they give the same result when evaluated for different values of the variable.

In parts A and B, we made the following tables:

From the tables, all three expressions had the same result when n = 1 but they did not all have the same result when n = 2. This means the three expressions are not equivalent.

Notice that the first two expressions have the same result through for n = 1 and n = 2. Using the Distributive Property gives 2(4n + 3) = 2(4n) + 2(3) = 8n + 6 so 8n + 6 and 2(4n + 3) are equivalent.

Result

All three expressions are not equivalent. Only 8n+6 and 2(4n+3) are equivalent.

Page 155 Exercise 1 Answer

Two expressions are only equivalent if they have the same result for all values of the variable. Therefore, the expressions only need to have different results one time before you know they are not equivalent.

Page 156 Exercise 1 Answer

To write an expression which is equivalent to 3y − 9 use the Distributive Property in reverse order. To do so, look for a common factor of both terms that is greater than 1. Since y is an unknown, find a common factor of 3 and 9.

A common factor of 3 and 9 is 3, since 9 can be writen as 3 × 3.

3y − 9

= 3(y) − 3(3) (Rewrite both terms as products using a common factor.)

= 3(y−3) (Extract the common factor.)
​
3y − 9 is equivalent to 3(y−3).

Equivalent expressions have the same value even when the value of the variable in the expression changes. Using different properties of operations you can write equivalent expressions in more than one way. Either use the Distributive and Associative Properties or use the Distributive Propert in reverse order.

Result

3(y−3)

Page 157 Exercise 2 Answer

Use different properties of operations to write the expressions in more than one way and than compare them.

10y + 5

= 5(2y) + 5(1) (Look for a common factor of both terms.)

= 5(2y+1) (Use distributive property in reverse.)

Notice that the first expression 10y + 5 can be written differently as the third expression 5(2y+1), thus they are equivalent.

However, the second expression has only one term and can’t be written as either the first or as the third expression.

Result

The first, 10y + 5, and the second, 5(2y+1), expressions are equivalent.

Page 157 Exercise 3 Answer

2(x−3) + 1

= 2(x) − 2(3) + 1 (Use the Distributive Property.)

= 2x – 6 + 1 (Use the Associative Property of Multiplication.)

= 2x − 5 (Find the sum.)

Notice that 2(x−3) + 1 ≠ 2x + 6, so the expressions are not equivalent.

Substitute the variable x with any number to show that the expression are not equivalent. Suppose x = 3.

2(x−3) + 1
​
= 2x − 5 (The expression can be written differently.)

= 2(3) − 5 (Substitute x with 3.)

= 6 − 5 (Find the product.)

= 1 (Find the difference.)
​
2x + 6

= 2(3) + 6 (Substitute x with 3.)

= 6 + 6 (Find the product.)

= 12 (Find the sum.)

Result

The expressions are not equivalent. When x = 3, 2(x−3) + 1 = 1 and 2x + 6 = 12 so they don’t have the same value.

Page 158 Exercise 1 Answer

To identify and write equivalent expressions, we can use the properties of operations, which are the Commutative Properties of Addition and Multiplication, Associative Properties of Addition and Multiplication, and the Distributive Properties across Addition and Subtraction.

Page 158 Exercise 2 Answer

Two possible answers are shown below:

For the expression y + \(\frac{1}{2}\), we could use the Commutative Property of Addition to write the equivalent expression \(\frac{1}{2}\) + y.

For the expression y + \(\frac{1}{2}\), we could use the Distributive Property to write the an equivalent expression that is a product:

Result

Possible answers:

We can use the Commutative Property of Addition to write the equivalent expression \(\frac{1}{2}\) + y.

We can use the Distributive Property to write the equivalent expression \(\frac{1}{2}\)(2y+1).

Page 158 Exercise 3 Answer

The expressions z³ and 3z are not equivalent expressions. The expression 3z means 3 × z while z³ means z × z × z.

You can verify the expressions are not equivalent using substitution. When z = 1, we get:

z³ = 1³ = 1

3z = 3(1) = 3

Since 1 ≠ 3, then z3 and 3z are not equivalent.

Result

No, when z = 1 we get z³ = 1 and 3z = 3 so the expressions are not equivalent.

Page 158 Exercise 4 Answer

When y = 1, we get:

3(y+1) = 3(1+1) = 3(2) = 6

3y + 3 = 3(1) + 3 = 3 + 3 = 6

Both expressions equal 6 when y = 1 so the expressions are equivalent when y = 1.

When y = 2, we get:

3(y+1) = 3(2+1) = 3(3) = 9

3y + 3 = 3(2) + 3 = 6 + 3 = 9

Both expressions equal 9 when y = 2 so the expressions are equivalent when y = 2.

When y = 3, we get:

3(y+1) = 3(3+1) = 3(4) = 12

3y + 3 = 3(3) + 3 = 9 + 3 = 12

Both expressions equal 12 when y = 3 so the expressions are equivalent when y = 3.

Result

The expressions are equivalent for y = 1, y = 2, and y = 3.

Page 158 Exercise 5 Answer

Using the Distributive Property a(b+c) = a(b) + a(c) gives:

3(y+1) = 3(y) + 3(1) = 3y + 3

Therefore, 3(y+1) and 3y + 3 are equivalent for any value of y.

Result

By the Distributive Property 3(y+1) = 3y + 3 so they are equivalent for any value of y.

Page 158 Exercise 6 Answer

2(r+3)

= 2(r) + 2(3) (Use the Distributive Property.)

= 2r + 6 (Multiply.)
​
Result

2r + 6

Page 158 Exercise 7 Answer

6(4s−1)

= 6(4s) − 6(1) (Use the Distributive Property.)

= (6×4)s − 6(1) (Use the Associative Property of Multiplication.)

= 24s − 6 (Multiply.)

Result

24s − 6

Page 158 Exercise 8 Answer

8t + 2

= 2(4t) + 2(1) (Find a common factor of both terms.)

= 2(4t+1) (Use the Distributive Property in reverse.)
​
Result

2(4t+1)

Page 158 Exercise 9 Answer

To complete the table, we need to evaluate the expressions 12x − 6, 3x + 3, and 6(2x−1) for x = 1, 2, and 3.

Evaluating the expressions for x = 1 gives:

12x − 6 = 12(1) − 6 = 12 − 6 = 6

3x + 3 = 3(1) + 3 = 3 + 3 = 6

6(2x−1) = 6(2⋅1−1) = 6(2−1) = 6(1) = 6

Evaluating the expressions for x = 2 gives:

12x − 6 = 12(2) − 6 = 24 − 6 = 18

3x + 3 = 3(2) + 3 = 6 + 3 = 9

6(2x−1) = 6(2⋅2−1) = 6(4−1) = 6(3) = 18

Evaluating the expressions for x = 3 gives:

12x − 6 = 12(3) − 6 = 36 − 6 = 30

3x + 3 = 3(3) + 3 = 9 + 3 = 12

6(2x−1) = 6(2⋅3−1) = 6(6−1) = 6(5) = 30

The completed table is then:

Result

​

Page 158 Exercise 10 Answer

In Exercise 9, the first and the third expression are equivalent.

12x – 6

= 6(2x) – 6(1) (Find a common factor of both terms)

= 6(2x – 1) (Use the Distributive Property in reverse)

12x – 6 = 6(2x – 1)

Result

In Exercise 9, the first and the third expression are equivalent.

Page 159 Exercise 11 Answer

3(m+3)

= 3m + 3(3) (Use the Distributive Property.)

= 3m + 9 (Multiply.)
​
Result

3m + 9

Page 159 Exercise 12 Answer

20n − 4m

= 4(5n) − 4(m) (Find a common factor of both terms.)

= 4(5n – m) (Use the Distributive Property in reverse.)

Result

4(5n−m)

Page 159 Exercise 13 Answer

3(x−6)

= 3(x) − 3(6) (Use the Distributive Property.)

= 3x − 18 (Multiply.)

Result

3x − 18

Page 159 Exercise 14 Answer

2x + 10

= 2(x) + 2(5) (Find a common factor of both terms.)

= 2(x+5) (Use the Distributive Property in reverse.)

Result

2(x+5)

Page 159 Exercise 15 Answer

8(2y + \(\frac{1}{4}\)

= 8(2y) + 8(\(\frac{1}{4}\) (Use the Distributive Property.)

= (8 x 2)y + (8 x \(\frac{1}{4}\))(Use the Associative Property of Multiplication.)

= 16y + 2 (Multiply.)
​
Result

16y + 2

Page 159 Exercise 16 Answer

5.7 + (3z+0.3)

= 5.7 + (0.3+3z) (Use the Commutative Property of Addition.)

= (5.7+0.3) + 3z (Use the Associative Property of Addition.)

= 6 + 3z (Find the sum inside the parentheses.)

= 3(2) + 3(z) (Find a commmon factor of both terms.)

= 3(2+z) (Use the Distributive Property in reverse.)

Result

3(2+z)

Page 159 Exercise 17 Answer

5w − 15

= 5(w) – 5(3) (Find a commmon factor of both terms.)

= 5(w-3) (Use the Distributive Property in reverse.)

Result

5(w−3)

Page 159 Exercise 18 Answer

2x + 4y

= 2(x) + 2(2y) (Find a commmon factor of both terms.)

= 2(x + 2y) (Use the Distributive Property in reverse.)
​
Result

2(x + 2y)

Page 159 Exercise 19 Answer

10(y² + 2.45)

= 10(y²) + 10(24.5) (Use the Distributive Property.)

= 10y² + 24.5 (Multiply.)

Result

10y² + 24.5

Page 159 Exercise 20 Answer

​\(\frac{3}{4} \times\left(z^3 \times 4\right)\)

= \(\frac{3}{4}\) x 4 x z³) (Use the Commutative Property of Multiplication.)

= (\(\frac{3}{4}\) x 4) x z³ (Use the Associative Property of Multiplication.)

= 3 x z³ (Find the product inside the parentheses.)

= 3z³ (Multiply.)

Result

3z³

Page 159 Exercise 21a Answer

To check if the given expression, 5(2x+3), is equivalent to 10x + 15, rewrite the given expression.

5(2x+3) = 5(2x) + 5(3)

= 10x + 15

They are equivalent.

Result

They are equivalent.

Page 159 Exercise 21b Answer

To check if the given expression, 5(2x+3), is equivalent to 5x + 15 + 5x, rewrite both expressions.

5(2x+3) = 5(2x) + 5(3)

= 10x + 15

5x + 15 + 5x = 10x + 15

Result

They are equivalent.

Page 159 Exercise 21c Answer

To check if the given expression, 5(2x+3), is equivalent to 10x + 8, rewrite the given expression.

5(2x+3) = 5(2x) + 5(3)

= 10x + 15 ≠ 10x + 8
​
Result

They are not equivalent.

Page 159 Exercise 22a Answer

To check if the given expression, 4x − 8, is equivalent to 2(2x−6), rewrite one of the expressions.

2(2x−6) = 2(2x) − 2(6)

= 4x − 12 ≠ 4x − 8

Result

They are not equivalent.

Page 159 Exercise 22b Answer

To check if the given expression, 4x − 8, is equivalent to 2(2x−4), rewrite one of the expressions.

2(2x−4) = 2(2x) − 2(4)

= 4x − 8
​
Result

They are equivalent.

Page 159 Exercise 22c Answer

To check if the given expression, 4x − 8, is equivalent to x − 8 + 3x, rewrite one of the expressions.

x − 8 + 3x = 4x − 8

Result

They are equivalent.

Page 159 Exercise 23a Answer

To check if the given expression, 12x − 16, is equivalent to 9.6x − 16 + 2.4x, rewrite one of the expressions.

When 9.6x and 2.4x are added they give 12x, thus the expressions are equivalent.

9.6x − 16 + 2.4x = 12x − 16

Result

They are equivalent.

Page 159 Exercise 23b Answer

To check if the given expression, 12x − 16, is equivalent to 3(3x−5), rewrite one of the expressions.

3(3x−5) = 3(3x) − 3(5)

= 9x − 15 ≠ 12x − 16

Result

They are not equivalent.

Page 159 Exercise 23c Answer

To check if the given expression, 12x − 16, is equivalent to 4(3x−4), rewrite one of the expressions.

4(3x−4) = 4(3x) − 4(4)

= 12x − 16

Result

They are equivalent.

Page 159 Exercise 24a Answer

To check if the given expression, 2(6x + \(\frac{1}{2}\)), is equivalent to 12x + 2, rewrite one of the expressions.

= \(2\left(6 x+\frac{1}{2}\right)=2(6 x)+2\left(\frac{1}{2}\right)\)

= 12x + 1 ≠ 12x + 2
​
Result

They are not equivalent.

Page 159 Exercise 24b Answer

To check if the given expression, 2(6x + \(\frac{1}{2}\)), is equivalent to 12x + 1, rewrite one of the expressions.

= \(2\left(6 x+\frac{1}{2}\right)=2(6 x)+2\left(\frac{1}{2}\right)\)

= 12x + 1

Result

They are equivalent.

Page 159 Exercise 24c Answer

To check if the given expression, 2(6x + \(\frac{1}{2}\)), is equivalent to 6x + \(\frac{1}{2}\) + 6x + \(\frac{1}{2}\), rewrite both expressions.

= \(2\left(6 x+\frac{1}{2}\right)=2(6 x)+2\left(\frac{1}{2}\right)\)

= 12x + 1

= \(6 x+\frac{1}{2}+6 x+\frac{1}{2}=12 x+\frac{1}{2}+\frac{1}{2}\)
​
= 12x + 1

Result

They are equivalent.

Page 159 Exercise 25a Answer

Since Mr. Tonkery bought x soccer balls which cost $15 each, the product x × $15 represents what he paid for them.

Also, he bought 3 baseballs which cost $6 each, so the product3 × $6 represents how much he paid for them.

The sum of this two products represents the total amount Mr. Tonkery paid.

An algebraic expression which represents the situation is:

x × $15 + 3 × $6.
​
Result

x × $15 + 3 × $6

Page 159 Exercise 25b Answer

First, make an expression which describes what each of them bought. Since they all bought the same and there are three of them, multiply the first expression by three.

Each of them bought a baseball, which costs $6, and x pairs of sweat socks, which cost $5 per pair, that is the productx × $5 describes the second part.

Each of them paid:

$6 + x × $5.
​
All three of them paid:

3($6 + x × $5).
​
Result

3($6 + x × $5)

Page 159 Exercise 26 Answer

To check if the expression x × $15 + 3 × $6 and 3($6 + x × $5), if the value of x is the same, rewrite the expressions.

x × $15 + 3 × $6 = 15x + 18

3($6 + x × $5) = 18 + 3(x)(5)

= 15x + 18

Result

They are equivalent.

Page 159 Exercise 27 Answer

If soccer balls cost \(2 \frac{1}{2}\) times as much as baseballs, than the following equality must be true:

\(2 \frac{1}{2}\) x $6 = $15

\(\frac{5}{2} \times \frac{6}{1}=15\) = 15

\(\frac{5 \times 6}{2 \times 1}\) = 15

\(\frac{30}{2}\) = 15

15 = 15

The equality is balanced, that is it is true.

Wendy is right.

Result

Yes, \(2 \frac{1}{2}\) × $6 = $15.

Page 160 Exercise 28 Answer

Since the length of the rectangular rug is given as 2(x – 1), its width is 5, and the area is calculated as the product of the width and length, the area is:

A = 5 x 2(x – 1)

= 10(2x – 1)

= 10(2x) – 10(1)

= 10x – 10

Result

10x – 10

Page 160 Exercise 29 Answer

Jamie may not be right, the expressions may be equivalent despite one having a term which is added and the other having a term which is subtracted.

To check, rewrite the expressions using properties of operations.

6x − 2x + 4 = 4x + 4

4(x+1) = 4(x) + 4(1)

= 4x + 4
​
When the expressions are rewritten we can see that they are equivalent.

Result

No, the expression are equivalent.

Page 160 Exercise 30 Answer

Use properties of operations to check if the expressions are equivalent.

4(n+3) − (3+n)

= 4(n) + 4(3) − (3) − (n) (Use the Distributive Property.)

= 4n + 12 – 3 – n (Multiply.)

= 4n – n + 12 – 3 (Use the Commutative Property.)

= (4n – n) + (12 – 3) (Use the Associative Property.)

= 3n + 9 (Subtract.)

Result

The expressions are equivalent.

Page 160 Exercise 31 Answer

Use properties of operations to rewrite 4n − 2 as 2(2n−1).

4n − 2 = 2(2n) − 2(1)

= 2(2n−1)

Chris is right.

Result

Chris is right.

Page 160 Exercise 32 Answer

Using the properties of operations we get:

​(f .g²) + 5 – (g² . f)

= fg² + 5 – g²f Multiply.

= fg² – g²f + 5 Use the Commutative Property of Addition.

= fg² – fg² + 5 Use the Commutative Property of Multiplication.

= 5 Subtract.

Result

5

Page 160 Exercise 33 Answer

5(g+14) = 5(g) + 5(14)

= 5g + 70 ≠ 5g + 14
​
When the expression 5(g+14) is rewritten we can see that it is not equivalent to the expression 5g + 14 because of the Distributive Property across Addition.

Result

No, the expressions are not equivalent.

Page 160 Exercise 34 Answer

(8.5+2s) + 0.5 = 8.5 + (2s+0.5) (Use Associative Prop. of Addition.)

(8.5+0.5) + 2s = 8.5 + (0.5+2s) (Use Associative Prop. of Addition.)

= 8.5 + (2s+0.5) (Use Comm. Prop. of Addition.)

Using the properties of Addition the given expression can be rewritten as:

8.5 + 2s + 0.5 = 8.5 + 0.5 + 2s

= 9 + 2s.

However, even now it is not equivalent to 9 + 2.

2(4.5+s) = 2(4.5) + 2(s)

= 9 + 2s

8.5(2s+0.5) = 8.5(2s) + 8.5(0.5)

= 17s + 4.25 ≠ 9 + 2s

The expressions equivalent to the given expression are: (8.5+2s) + 0.5, (8.5+0.5) + 2s, and 2(4.5+s).

Result

(8.5+2s) + 0.5, (8.5+0.5) + 2s, and 2(4.5+s).

Page 160 Exercise 35 Answer

Using the Distributive Property, the given expression can also be written as

5(n+4) = 5(n) + 5(4) = 5n + 20.

5n + 4 ≠ 5n + 20

5n + 20 = 5(n+4)

15n + 5n + = 5n + 15 + 5 (Use the Commutative Property of Addition.)

= 5n + 20

5(n + 3) + 5 = 5n + 15 + 5(Use the Distributive Property.)
​
= 5n + 20

5n + 54 ≠ 5n + 20

Result

5n + 20, 15 + 5n + 5, and 5(n+3) + 5.

Chapter 3 Numeric And Algebraic Expressions

Section 3.7: Simplify Algebraic Expressions

Page 161 Exercise 1 Answer

Using the properties of operations we get:

x + 5 + 2x + 2

= x + 2x + 5 + 2 Use the Commutative Property of Addition.

= (x+2x) + (5+2) Use the Associative Property of Addition.

= (1+2)x + (5+2) Use the Distributive Property.

= 3x + 7 Add

Result

3x + 7

Read And Learn More: enVisionmath 2.0 Grade 6 Volume 1 Solutions

Page 161 Exercise 1 Answer

Two expressions are equivalent if they have the same value each time a number is substituted for the variable.

Evaluating x + 5 + 2x + 2 and 3x + 7 for x = 1 gives:

x + 5 + 2x + 2 = 1 + 5 + 2(1) + 2

= 1 + 5 + 2 + 2

= 10
​
3x + 7 = 3(1) + 7

= 3 + 7

= 10

Evaluating x + 5 + 2x + 2 and 3x + 7 for x = 2 gives:

x + 5 + 2x + 2 = 2 + 5 + 2(2) + 2

= 2 + 5 + 4 + 2

= 13

3x + 7 = 3(2) + 7

= 6 + 7

= 13

Evaluating x + 5 + 2x + 2 and 3x + 7 for x = 3 gives:

x + 5 + 2x + 2 = 3 + 5 + 2(3) + 2

= 3 + 5 + 6 + 2

= 16
​
3x + 7 = 3(3) + 7

= 9 + 7

= 16
​
The expressions x + 5 + 2x + 2 and 3x + 7 have the same value for all values of x so they are equivalent.

Result

The expressions x + 5 + 2x + 2 and 3x + 7 have the same value for all values of x so they are equivalent.

Page 162 Exercise 1 Answer

Using the properties of operations we get:

4z − z + z − 2z

= 4z – 1z + 1z – 2z Use the Identity Property of Multiplication.

= (4 – 1 + 1 – 2)z Use the Distributive Property.

= 2z Add and subtract.

The simplified expression is then 2z.

Like terms are terms that have the same variable part. The terms in 2x + 4y have different variables so the terms are not like terms. This means the terms cannot be combined to get 6xy. We then know that 2x + 4y is not equivalent to 6xy.

We can verify this by substituting values for x and y. If x = 1 and y = 2, then:

2x + 4y = 2(1) + 4(2)

= 2 + 8

= 10

6xy = 6(1)(2)

= 6(2)

= 12

The two expressions do not have the same value so they are not equivalent.

Result

4z − z + z − 2z

= 4z − 1z + 1z − 2z

= (4−1+1−2)z

= 2z
​
The simplified expression is 2z.

We know 2x + 4y is not equivalent to 6xy since the terms 2x and 4y are not like terms so they can’t be combined. Verifying this using substitution for x = 1 and y = 2 gives 2x + 4y = 10 and 6xy = 12, which are not the same value so the expressions are not equivalent.

Page 163 Exercise 2 Answer

Using the properties of operations we get:

​\(\frac{1}{2} n+n+\frac{1}{2}\)

= \(\frac{1}{2} n+1 n+\frac{1}{2}\) Use the Identity Property of Multiplication.

= \(\left(\frac{1}{2}+1\right) n+\frac{1}{2}\) Use the Distributive Property.

= \(\frac{3}{2} n+\frac{1}{2}\) Add.

= \(1 \frac{1}{2} n+\frac{1}{2}\) Rewrite as a mixed number.
​
Result

Page 163 Exercise 3 Answer

In Example 3, the expression 2.5(20.50x + 5.50x – 10) represented her total earnings where 2.5 was how many times the amount she charged and her expenses increased by.

If the amount she charges and her expenses increase by 3.5 times, then we need to replace the 2.5 in the expression with 3.5. The first expression is then 3.5(20.50x + 5.50x − 10).

Using the properties of operations, an equivalent expression is:

3.5(20.50x + 5.50x − 10)

= 3.5(26x−10) Combine like terms inside the parentheses.

= 3.5(26x) − 3.5(10) Use the Distributive Property.

= 91x − 35 Multiply.

Result

3.5(20.50x + 5.50x − 10) and 91x − 35

Page 164 Exercise 1 Answer

To simplify algebraic expressions, we can use properties of operations to combine like terms to get an equivalent expression with no like terms and no parentheses.

Page 164 Exercise 2 Answer

To combine terms in an algebraic expression, the terms must be like terms, which are terms that have the same variable part.

Page 164 Exercise 3 Answer

Using the Identity Property of Multiplication, 2y − y = 2y − 1y. Using the Distributive Property gives 2y−1y = (2−1)y. Subtracting then gives (2−1)y = y. Therefore, 2y − y can be written as y.

Page 164 Exercise 4 Answer

Using the Distributive Property gives \(\frac{1}{2}\)x + \(\frac{1}{2}\)x = \(\left(\frac{1}{2}+\frac{1}{2}\right) x\). Adding gives \(\left(\frac{1}{2}+\frac{1}{2}\right) x\) = 1x. Using the Identity Property of Multiplication then gives 1x = x so \(\frac{1}{2}\)x + \(\frac{1}{2}\)x and x are equivalent.

Page 164 Exercise 5 Answer

4z² − z² and 4 are not equivalent expressions since 4z² − z² does not equal 4 for all values of z.

For example, when z = 2 we get:

4z² − z² = 4(2)² − 2²

= 4(4) − 4

= 16 − 4

= 12 ≠ 4

Result

4z² − z² and 4 are not equivalent expressions since 4z² − z² does not equal 4 for all values of z.

Page 164 Exercise 6 Answer

Using the properties of operations we get:

x + x + x + x

= 1x + 1x + 1x + 1x Use the Identity Property of Multiplication.

= (1+1+1+1)x Use the Distributive Property.

= 4x Add.
​
Result

4x

Page 164 Exercise 7 Answer

Using the properties of operations we get:

4y − y

= 4y − 1y Use the Identity Property of Multiplication.

= (4−1)y Use the Distributive Property.

= 3y Subtract.

Result

3y

Page 164 Exercise 8 Answer

7y − 4.5 − 6y = 7y − 6y − 4.5

= y − 4.5
​
Result

y − 4.5

Page 164 Exercise 9 Answer

Using the properties of operations we get:

= \(4 x+2-\frac{1}{2} x\) Use the Commutative Property.

= \(\left(4-\frac{1}{2}\right) x+2\) Use the Distributive Property.

= \(\left(3 \frac{2}{2}-\frac{1}{2}\right)+2\) Regroup 4 as a mixed number.

= \(3 \frac{1}{2} x+2\) Subtract.

Result

Page 164 Exercise 10 Answer

Using the properties of operations we get:

3 + 3y − 1 + y

= 3 + 3y − 1 + 1y Use the Identity Property of Multiplication.

= 3y + y + 3 − 1 Use the Commutative Property of Addition.

= (3y+y) + (3−1) Use the Associative Property of Addition.

= (3+1)y + (3−1) Use the Distributive Property.

= 4y + 2 Add and subtract.
​
Result

4y + 2

Page 164 Exercise 11 Answer

Using the properties of operations we get:

x + 6x

= 1x + 6x Use the Identity Property of Multiplication.

= (1+6)x Use the Distributive Property.

= 7x Add.
​
Result

7x

Page 164 Exercise 12 Answer

Using the properties of operations we get:

0.5w + 1.7w − 0.5

= (0.5+1.7)w − 0.5 Use the Distributive Property.

= 2.2w − 0.5 Add.

Result

2.2w − 0.5

Page 164 Exercise 13 Answer

Using the properties of operations we get:

​\(12 \frac{1}{3} b+6 \frac{2}{3}-10 \frac{2}{3} b\)

= \(12 \frac{1}{3} b-10 \frac{2}{3} b+6 \frac{2}{3}\) Use the Commutative Property of Addition.

= \(\left(12 \frac{1}{3}-10 \frac{2}{3}\right) b+6 \frac{2}{3}\) Use the Distributive Property.

= \(\left(11 \frac{4}{3}-10 \frac{2}{3}\right) b+6 \frac{2}{3}\) Regroup.

= \(1 \frac{2}{3} b+6 \frac{2}{3}\) Subtract.
​
Result

\(1 \frac{2}{3} b+6 \frac{2}{3}\)
​
Page 164 Exercise 14 Answer

Using the properties of operations we get:

​\(\frac{3}{4} x+2+3 x-\frac{1}{2}\)

= \(\frac{3}{4} x+3 x+2-\frac{1}{2}\) Use the Commutative Property of Addition.

= \(\left(\frac{3}{4} x+3 x\right)+\left(2-\frac{1}{2}\right)\) Use the Associative Property of Addition.

= \(\left(\frac{3}{4}+3\right) x+\left(2-\frac{1}{2}\right)\) Use the Distributive Property.

= \(\left(\frac{3}{4}+3\right) x+\left(1 \frac{2}{2}-\frac{1}{2}\right)\) Rewrite 2 as a mixed number.

\(3 \frac{3}{4} x+1 \frac{1}{2}\) Add and subtract.
​
Result

\(3 \frac{3}{4} x+1 \frac{1}{2}\)
​
Page 164 Exercise 15 Answer

Using the properties of operations we get:

3.2x + 6.5 − 2.4x − 4.4

= 3.2x − 2.4x + 6.5 − 4.4 Use the Commutative Property of Addition.

= (3.2x−2.4x) + (6.5−4.4) Use the Associative Property of Addition.

= (3.2−2.4)x + (6.5−4.4) Use the Distributive Property.

= 0.8x + 2.1 Subtract.
​
Result

0.8x + 2.1

Page 165 Exercise 16 Answer

2.1x² + 3 − 0.5x² − 1 = (2.1x² − 0.5x²) + (3−1)

= 1.6x² + 2

Result

1.6x² + 2

Page 165 Exercise 17 Answer

= \(3 \frac{2}{3} n+\frac{18}{3}-\frac{2}{3}\)

= \(3 \frac{2}{3} n+\frac{16}{3}\)

= \(3 \frac{2}{3} n+5 \frac{1}{3}\)

Result

Page 165 Exercise 18 Answer

5 + 3w + 3 − w = 3w − w + 5 + 3

= 2w + 8
​
Result

2w + 8

Page 165 Exercise 19 Answer

5w – 5w = 0

Result

0

Page 165 Exercise 20 Answer

2x + 5 + 3x + 6 = 2x + 3x + 5 + 6

= 5x + 11
​
Result

5x + 11

Page 165 Exercise 21 Answer

= \(\frac{2}{4} z^3+4\)

= \(\frac{1}{2} z^3+4\)

Result

Page 165 Exercise 22 Answer

3.4m + 2.4m = 5.8m

Result

5.8m

Page 165 Exercise 23 Answer

4.2n + 5 – 3.2n = 4.2n – 3.2n + 5

= n + 5

Result

n + 5

Page 165 Exercise 24 Answer

q⁵ + q⁵ + q⁵ = 3q⁵

Result

3q⁵

Page 165 Exercise 25 Answer

= 10x + y + \(\frac{2}{4}\)

= 10x + y + \(\frac{1}{2}\)

Result

10x + y + \(\frac{1}{2}\)

Page 165 Exercise 26 Answer

1.5z² + 4.5 + 6z − 0.3 − 3z + z² = 1.5z² + z² + 6z − 3z + 4.5 − 0.3

= 2.5z² + 3z + 4.2
​
Result

2.5z² + 3z + 4.2

Page 165 Exercise 27a Answer

The rooms have hourly rental fees, in order, of $25, $20, and $50.

Since Yolanda rents all the rooms for t hours, multiply each hourly rental fee by t to find the cost for each room. The total cost of the rooms is then $25t + $20t + $50t.

For the sound system in each room Yolanda needs to pay, in order, $15 and $10. There is no charge for the sound system in the third room. The total cost of the sound systems is then $15 + $10.

The total cost is the sum of the total cost of the rooms and the total cost of the sound systems.

The expression which describes the total amount Yolanda needs to pay is:

$25t + $20t + $50t + $15 + $10.
​
Result

$25t + $20t + $50t + $15 + $10

Page 165 Exercise 27b Answer

The expression is $25t + $20t + $50t + $15 + $10. Using the Distributive Property across Addition.

$25t + $20t + $50t + $15 + $10 = ($25+$20+$50)t + $15 + $10

= $95t + $25
​
Result

$95t + $25

Page 166 Exercise 28 Answer

The perimeter of a rectangle is the sum of its four side lengths. From the figure, the rectangle has side lengths of y, 2y + 1, y, and 2y + 1. The perimeter is then:

y + 2y + 1 + y + 2y + 1

= (y+2y+y+2y) + (1+1) Group like terms.

= 6y + 2 Combine like terms.
​
Result

6y + 2

Page 166 Exercise 29 Answer

The expression from Exercise 28 is 6y + 2.

6 and 2 have a greatest common factor of 2 so we can use the Distributive Property to rewrite 6y + 2 as a product of two factors:

6y + 2

= 2(3y) + 2(1) Rewrite each term as a product.

= 2(3y + 1)Use the Distributive Property.
​
Result

2(3y + 1)

Page 166 Exercise 30 Answer

The expressions from Exercise 28 and 29 are 6y + 2 and 2(3y + 1).

In Exercise 29, we used the Distributive Property to rewrite 6y + 2 as 2(3y + 1). Since we used one of the properties of operations to rewrite the expression, then the two expressions must be equivalent.

Result

6y + 2 and 2(3y + 1) are equivalent by the Distributive Property.

Page 166 Exercise 31 Answer

Rodney rewrote \(\frac{1}{2}\)(2x+7) as 3 + \(3 \frac{1}{2}\) so he rewrote a product of two factors as a sum.

To rewrite an expression that is a product of two factors as a sum, you need to use the Distributive Property. We can verify this by using the Distributive Property on \(\frac{1}{2}\)(2x+7), which gives:
​
\(\frac{1}{2}(2 x+7)=\frac{1}{2}(2 x)+\frac{1}{2}(7)=1 x+\frac{7}{2}=x+3 \frac{1}{2}\)

Result

Distributive Property

Page 166 Exercise 32 Answer

I don’t agree with Annie. She rewrote the expression using the Distributive Property so the two expressions must be equivalent but she forgot to combine like terms:

6.5(x+0.5x+1) = 6.5(x) + 6.5(0.5x) + 6.5(1) = 6.5x + 3.25x + 6.5

Simplified expressions must have no like terms so to simplify the expression we need to combine the like terms of 6.5x and 3.25x to get:

6.5x + 3.25x + 6.5 = 9.75x + 6.5

Result

I don’t agree. She used the Distributive Property to write an equivalent expression but she didn’t combine the like terms. The simplified expression should be 9.5x + 6.5.

Page 166 Exercise 33 Answer

Thea is correct. Combining like terms gives 4x − 3x + 2 = x + 2 since 4x − 3x = (4−3)x = 1x = x.

Page 166 Exercise 34 Answer

Using the properties of operations gives:

​= \(\frac{a}{3}+\frac{a}{3}+\frac{a}{3}\)

= \(a\left(\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\right)\) Use the Distributive Property.

= a(\(\frac{3}{3}\)) Add.

= a(1) Simplify.

= a Multiply.

Result

Page 166 Exercise 35 Answer

Simplifying the given expression gives:

2x + 7 + 6x − x

= (2x+6x−x) + 7 Group like terms.

= 7x + 7 Combine like terms.

= 7 + 7x Use the Commutative Property of Addition.

The two expressions that are equivalent to 2x + 7 + 6x − x are then 7x + 7 and 7 + 7x. The two expressions that are NOT equivalent are 2x + 13 and 14x. The table is then completed as:

Result