Schindler

Chapter 1 Use Postitive Rational Numbers

Section 1.1: Fluently Add, Subtract, and Multiply Decimals

Page 7 Exercise 1 Answer

Maxine is connecting 4 cardboard tubes together vertically and each tube is 0.28 meter in length. The combined measure of the connected tubes is 4 times the length of the tubes:

Result

The combined measure of the connected tubes is 1.12 meter in length.

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Page 7 Exercise 1 Answer

Maxine made another windmill model by connecting 4 cardboard tubes that are each 2.8 meters long.

The combined measure of this model is 11.2 meters.

In the other, similar problem, where Maxine made a windmill model by connecting 4 cardboard tubes, but the length of tubes was 0.28, the combined measure was 1.12. Notice that in this problem the decimal point was moved for one place. The length of tubes was 2.8 and the result was 11.2, the decimal point in the result also moved for one place.

For example, if we were to look at a similar problem, the only difference being that Maxine used tubes 28 meters long, we can assume that the answer would be 112. The decimal point was moved for one place in one of the factors, thus it is moved for one place in the result.

Result

The combined measure of this model is 11.2 meters. Compared to the other similar problem, the decimal point was moved one place for both the factor and the product.

Page 8 Exercise 1 Answer

Kim and Martin swam 50 meters. Martin took 0.47 seconds longer than Kim. Kim’s time was 50.9 seconds.

Martin’s time was 51.37 seconds.

In Martin finished the reace 0.267 seconds after Kim, the solution would be to add 0.267 to 50.9. Adding 0.267 to 50.9 is different from adding 0.26 to 50.9 for a tiny difference between 0.267 and 0.26

So, the result of 0.267 + 50.9 would be a little greater than 0.26 + 50.9, precisely it would be 0.007 points greater.

Result

Martin’s time was 51.37 seconds. Adding 0.267 to 50.9 is 0.007 greater than adding 0.26 to 50.9 since 0.267 is 0.007 greater than 0.26.

Page 9 Exercise 2 Answer

We know Amy finished the race in 20.7 seconds and Katie finish the race 0.13 seconds before Amy. To find Katie’s time in the race, we need to subtract 20.7 and 0.13.

To subtract, we first need to line up the decimals at their decimal points. Since 0.13 has two decimals places, we need to use a zero as a place holder for the 20.7.

We can’t subtract 3 from 0 se we will need to regroup in order to subtract the hundredths places:

Katie’s time in the race was then 20.57 seconds.

To use estimation to check that out answer is resonable, we can estimate the difference by rounding 0.13 to the same decimal place as 20.7. Rounding 0.13 to the nearest tenth gives 0.1 so 20.7 – 0.13 ≈ 20.7 – 0.1 = 20.6. Since 20.57 is close to the estimate of 20.6, then the answer is reasonable

Result

20.57 seconds

Page 9 Exercise 3 Answer

We multiplied the numbers using long multiplication method.

0.42 x 0.2 = 0.086

We can see that the first factor, 0.43, has two decimal places and the second factor, 0.2, has only one decimal place.

The result, 0.086, must have three decimal places because that is the sum of number of decimal places of the first and the second factor.

Result

0.43 has 2 decimal places, 0.2 has 1 decimal place, and the product has 3 decimal places.\

Page 10 Exercise 1 Answer

To add decimals, line up place values and add. Regroup as needed.

For example, add numbers 47.34 and 13.12.

To subtract decimals, line up place values and subtract. Regroup as needed.

For example, subtract 14.08 from 34.78.

To multiply decimals, multiply as you would with whole numbers. Then use the number of decimal places in the factors to place the decimal point in the product.

For example, multiply 7.98 and 21.09.

Result

To add and subtract decimals, line up place values and add or subtract. Regroup as needed. To multiply as you would with whole numbers. Then use the number of decimal places in the factors to place the decimal point in the product.

Page 10 Exercise 2 Answer

Adding or subtracting decimals is similar to adding and subtracting whole numbers in that both involve adding the like place values. However, when adding or subtracting decimals we must be careful to line up place values before adding or subtracting and use zeros as placeholders if necessary.

For example,

Result

Adding or subtracting decimals is similar to adding and subtracting whole numbers in that both involve adding the like place values. However, when adding or subtracting decimals we must be careful to line up place values before adding or subtracting and use zeros as placeholders if necessary.

Page 10 Exercise 3 Answer

If a decimal product has final zeros to the right of the decimal point, there is no need to write them.

For example, multiply 2.05 and 15.2.

2.05 x 15.2 = 31.160

The result is 31.160, but it is equal to write 31.16.

Result

If a decimal product has final zeros to the right of the decimal point, there is no need to write them.

Page 10 Exercise 4 Answer

Diego says that the product of 0.51 × 2.427 will have five decimal places.

Diego is right. One of the factors has two decimal points and the other three. Since, one ends with 1 and the other with 7, the last decimal place won’t be a zero, thus the product will have all five decimal places.

Let’s check.

0.51 x 2.427 = 1.23777

Diego was right, the product has five decimal places.

Result

Diego was right because the product is 1.23777, which has five decimal places.

Page 10 Exercise 5 Answer

We can add numbers using the long addition method.

Result

8.6

Page 10 Exercise 6 Answer

To subtract two decimals, line up the decimals at their decimal places and then subtract each place value. Regroup if needed:

Result

1.06

Page 10 Exercise 7 Answer

To subtract two decimals, line up the decimals at their decimal places and then subtract each place value. Write additional zeros as placeholders and regroup if needed:

Result

5.35

Page 10 Exercise 8 Answer

Subtract numbers using the long subtraction method. We use subscripts for carrying numbers.

9.62 – 0.3 = 9.32

Result

9.32

Page 10 Exercise 9 Answer

We can add numbers using the long addition method.

Result

10.276

Page 10 Exercise 10 Answer

To subtract two decimals, line up the decimals at their decimal places and then subtract each place value. Write additional zeros as place holders and regroup if needed:

Result

8.892

Page 10 Exercise 11 Answer

We need to place the decimal point in the correct location in the product 4 x 0.94 = 476

4 has 0 decimal places and 0.94 has 2 decimal places so the product must have 0 + 2 = 2 decimal places. The placement of the decimal point is then between the 3 and 7 to get 4 x 0.94 = 3.76

Result

4 x 0.94 = 3.76

Page 10 Exercise 12 Answer

We need to place the decimal point in the correct location in the product 5 x 0.487 = 2435.

5 has 0 decimal places and 0.487 has 3 decimal places so the product must have 0 + 3 = 3 decimal places. The placement of the decimal point is then between the 2 and 4 to get 5 x 0.487 = 2.435.

Result

5 × 0.487 = 2.435

Page 10 Exercise 13 Answer

We need to place the decimal point in the correct location in the product 3.4 × 6.8 = 2312

3.4 has 1 decimal place and 6.8 has 1 decimal place so the product must have 1 + 1 = 2 decimal places. The placement of the decimal point is then between the 3 and 1 to get 3.4 × 6.8 = 23.12

Result

3.4 × 6.8 = 23.12

Page 10 Exercise 14 Answer

We need to place the decimal point in the correct location in the product 3.9 × 0.08 = 312

3.9 has 1 decimal place and 0.08 has 2 decimal places so the product must have 1 + 2 = 3 decimal places. The placement of the decimal point is then to the left of the 3 to get 3.9 × 0.08 = 0.312

Result

3.9 x 0.08 = 0.312

Page 10 Exercise 15 Answer

We need to place the decimal point in the correct location in the product 0.9 × 0.22 = 198.

0.9 has 1 decimal place and 0.22 has 2 decimal places so the product must have 1 + 2 = 3 decimal places. The placement of the decimal point is then to the left of the 1 to get 0.9 × 0.22 = 0.198

Result

0.9 × 0.22 = 0.198

Page 10 Exercise 16 Answer

We need to place the decimal point in the correct location in the product 9 x 1.2 = 108.

9 has 0 decimal places and 1.2 has 1 decimal place so the product must have 0 + 1 = 1 decimal place. The placement of the decimal point is then between the 0 and 8 to get 9 × 1.2 = 10.8.

Result

9 × 1.2 = 10.8

Page 10 Exercise 17 Answer

Multiply numbers using long multiplication method.

5.3 x 2.7 = 14.31

Result

14.31

Page 10 Exercise 18 Answer

Multiply numbers using long multiplication method.

8 x 4.09 = 32.72

Result

32.72

Page 10 Exercise 21 Answer

To subtract two decimals, line up the decimals at their decimal places and then subtract each place value. Write additional zeros as placeholders and regroup if needed:

Result

18.21

Page 11 Exercise 22 Answer

We can add numbers using the long addition method.

Result

6.985

Page 11 Exercise 23 Answer

To subtract two decimals, line up the decimals at their decimal places and then subtract each place value. Write additional zeros as placeholders and regroup if needed:

Result

4.62

Page 11 Exercise 24 Answer

We can add numbers using the long addition method.

Result

27.185

Page 11 Exercise 25 Answer

To subtract two decimals, line up the decimals at their decimal places and then subtract each place value. Write additional zeros as placeholders and regroup if needed:

Result

0.051

Page 11 Exercise 26 Answer

We can add numbers using the long addition method.

Result

23.8

Page 11 Exercise 27 Answer

We can add numbers using long addition method.

Result

26.89

Page 11 Exercise 28 Answer

Multiply numbers using long multiplication method.

7 x 0.5 = 3.5

Result

3.5

Page 11 Exercise 29 Answer

Multiply numbers using long multiplication method.

12 x 0.08

Result

0.96

Page 11 Exercise 30 Answer

Multiply numbers using long multiplication method.

24 x 0.17 = 4.08

Result

4.08

Page 11 Exercise 31 Answer

Multiply numbers using long multiplication method.

0.4 x 0.17 = 0.068

Result

0.068

Page 11 Exercise 32 Answer

Multiply numbers using long multiplication method.

1.9 x 0.46 = 0.874

Result

0.874

Page 11 Exercise 33 Answer

Multiply numbers using long multiplication method.

3.42 x 5.15 = 17.613

Result

17.613

Page 11 Exercise 34 Answer

Write an equation that illustrates the following:

A number with two decimal places multiplied by a number with one decimal place. The product has only two nonzero digits.

Since one of the numbers has two decimal places and the other one, the product will have three, so the last digit in the product must be a zero.

We’ll try with 0.12 and 1.5, hoping that 2 × 5 will give 10 “in the right” place.

0.12 x 1.5 = 0.180

The product is 0.180, but the last digit right of the decimal point is zero so the number can be written as 0.18.

The solution is the following equation.

0.12 × 1.5 = 0.18

Result

0.12 x 1.5 = 0.18

Page 11 Exercise 35 Answer

It is given that the entire shampoo bottle holds 6.35 ounces and that 1.078 ounces in the bottle is vanilla oil. We need to find how much of the bottle is NOT vanilla oil so we need to subtract 6.35 and 1.078.

To subtract two decimals, line up the decimals at their decimal places and then subtract each place value. Write additional zeros as placeholders and regroup if needed:

Therefore, 5.272 ounces of the bottle is not vanilla oil.

Result

5.272 ounces

Page 11 Exercise 36 Answer

The fastest speed a table tennis ball has been hit is about 13.07 times as fast as the speed for the fastest swimming which is 5.35 as seen in the graph.

Multiply numbers using long multiplication method.

13.07 x 5.35 = 69.9245

The fastest speed a table tennis ball has been hit is 69.9245 miles per hour.

Result

69.9245 mph

Page 11 Exercise 37 Answer

The graph on the right shows Fastest Sporting Speeds – fastest swimming time is 5.35 miles per hour, fastest running is 27.9 mph, fastest rowing is 13.99 mph, fastest luge is 95.69 mph, and fastest thrown baseball 106 mph.

How fast would 1.5 times the fastest rowing speed be?

The number of decimal places in the answer will be the sum of numbers of decimal places of 1.5 and of the time of the fastest rowing speed. The fastest rowing speed is 13.99 mph, so the sum is equal to 3. The number of decimal places in the answer be 3.

1.5 x 13.99 = 20.985

1.5 times the fastest rowing speed is 20.985.

Result

20.985

Page 11 Exercise 38 Answer

The graph on the right shows Fastest Sporting Speeds – fastest swimming time is 5.35 miles per hour, fastest running is 27.9 mph, fastest rowing is 13.99 mph, fastest luge is 95.69 mph, and fastest thrown baseball 106 mph.

Which activity has a recorded speed about 7 times as fast as the fastest rowing speed?

7 x 13,99 = 97.93

Number closest to 97.93, out Fastest Sporting Speeds, is 95.69, so the answer is fastest luge.

Result

Fastest Luge.

Page 12 Exercise 39 Answer

Matthew bought a jersey, a pennant, and a hat, paid with a 50 bill and some money he borrowed from his friend, and got $6.01 in change from the cashier. We need to find how much he borrowed from his friend to pay for all the items.

The bill shows the following – the price of the jersey $39.99, the pennant $ 10.25, and the had $13.75.

Matthew paid $ 63.99 for the jersey, the pennant and the hat.

Matthew gave the cashier exactly the amount of the bill plus the change she has given him back.

Matthew gave the cashier $70. His friend let him borrow the difference between the money he gave the cashier and the 50 bill.

Matthew borrowed $20 from his friend.

Result

Matthew borrowed $20 from his friend.

Page 12 Exercise 40 Answer

It is given that Anna’s running time was 23.1 seconds and another runner’s time was 5.86 seconds faster. To fin the other runner’s time, we need to subtract 23.1 and 5.68.

To subtract two decimals, line up the decimals at their decimal places and then subtract each place value. Write additional zeros as placeholders and regroup if needed:

The other runner’s time was then 17.24 seconds.

Result

17.24 seconds

Page 12 Exercise 41 Answer

The product of 0.25 × 0.4 has only one decimal place.

The first factor has two decimals places and the other has one, so the product should have three decimal places. However, when 25 and 4 are multiplied they give 100 which mean that the two last digits in the product will be zeros, hence there is no need to write them, and the product has only one decimal place.

0.25 x 0.4

Result

The last two digits in the product will be zeros so there is no need to write them.

Page 12 Exercise 42 Answer

The wings of some hummingbird’s beat 52 times per second when hovering. If a hummingbird hovers for 35.5 seconds, its wings beat 35.5 × 52.

35.5 x 52 = 1846

Result

1846

Page 12 Exercise 43 Answer

The students at Walden Middle School are selling tins of popcorn to raise money for new uniforms. One tin costs $9.25. They sold 42 tins in the first week.

9.25 x 42 = 388.5

Result

The students made $388.5 in the first week.

Page 12 Exercise 44a Answer

There are four trails in Joshua Tree National Park, they are: Lost Horse Mine which is 6.4 kilometers long, Lost Palms Oasis 11.6, Mastodon Peak 4.8, and Skull Rock 2.7 kilometers.

What is the combined length of the Lost Horse Mine trail and the Mastodon Peak trail?

To find the answer we need to calculate the sum of the length of the Lost Horst Mine trail and the Mastodon Peak trail

Result

The combined length of the Lost Horse Mine trail and the Mastodon Peak trail is 11.2 kilometers.

Page 12 Exercise 44b Answer

From the table, we know the Lost Palms Oasis trail is 11.6 km and the Skull Rock trail is 2.7 km. To find how much longer the Lost Palms Oasis trail is than the Skull Rock trail, we need to subtract 11.6 and 2.7.

To subtract two decimals, line up the decimals at their decimal places and then subtract each place value. Regroup if needed:

The Lost Palms Oasis trail is then 8.9 km longer than the Skull Rock Trail.

Result

8.9 km

Chapter 1 Algebra: Understand Numerical And Algebraic Expressions

Homework And Practice 5

Page 35 Exercise 1 Answer

6a + 4 Evaluate

= \(6\left(\frac{1}{3}\right)+4\) Substitute a = \(\frac{1}{3}\)

= 2 + 4 Multiply

= 6 Add

Result

6

Page 35 Exercise 2 Answer

\(5 a-\frac{2}{3}\) Evaluate

= \(5\left(\frac{1}{3}\right)-\frac{2}{3}\) Substitute a = \(\frac{1}{3}\)

= \(\frac{5}{3}\) – \(\frac{2}{3}\) Multiply

= \(\frac{5-2}{3}\) Write over a common denominator

= \(\frac{3}{3}\) Simplify

= 1 Divide

Result

1

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Page 35 Exercise 3 Answer

5d ÷ c + 2 Evaluate

= 5(10) ÷ (5) + 2 Substitute c = 5 and d = 10

= 50 ÷ 5 + 2 Evaluate the parentheses

= 10 + 2 Divide

= 12 Add

Result

12

Page 35 Exercise 4 Answer

\(b^2\) – 9a Evaluate

= \((9)^2-9\left(\frac{1}{3}\right)\) Substitute b = 9 and a = \(\frac{1}{3}\)

= 81 − 3 Evaluate the parentheses

= 78 Subtract

Result

78

Page 35 Exercise 5 Answer

12a + c − b Evaluate

= \(12\left(\frac{1}{3}\right)+(5)-(9)\) Substitute a = \(\frac{1}{3}\), c = 5 and b = 9

= 4 + 5 − 9 Evaluate the parentheses

= 9 − 9 Add

= 0 Subtract

Result

0

Page 35 Exercise 6 Answer

\(\frac{1}{2} d+c^2-b\) Evaluate

= \(\frac{1}{2}(10)+(5)^2-(9)\) Substitute d = 10, c = 5 and b = 9

= 5 + 25 − 9 Evaluate the parentheses

= 30 − 9 Add

= 21 Subtract

Result

21

Page 35 Exercise 7 Answer

\(d^2\) ÷ 2c − b + 3a Evaluate

\((10)^2 \div 2(5)-(9)+3\left(\frac{1}{3}\right)\) Substitute d = 10, c = 5, b = 9 and a = \(\frac{1}{3}\)

= 100 ÷ 10 − 9 + 1 Evaluate the parentheses

= 10 − 9 + 1 Divide

= 1 + 1 Subtract

= 2 Add

Result

2

Page 35 Exercise 8 Answer

3c + \(b^2\) ÷ 27a – d Evaluate

= \(3(5)+(9)^2 \div 27\left(\frac{1}{3}\right)-(10)\) Substitute c = 5, b = 9, a = \(\frac{1}{3}\) and d = 10

= 15 + 81 ÷ 9 − 10 Evaluate the parentheses

= 15 + 9 − 10 Divide

= 24 − 10 Add

= 14 Subtract

Result

14

Page 35 Exercise 9 Answer

28 − \(c^3\) + 6

28 − \(c^3\) + 6

= 28 − \((1)^3\) + 6 Substitute c = 1

= 28 − 1 + 6 Evaluate the power

= 27 + 6 Subtract

= 33 Add

28 − \(c^3\) + 6

= 28 − \((2)^3\) + 6 Substitute c = 2

= 28 − 8 + 6 Evaluate the power

= 20 + 6 Subtract

= 26 Add

28 − \(c^3\) + 6

= 28 − \((3)^3\) + 6 Substitute c = 3

= 28 − 27 + 6 Evaluate the power

= 1 + 6 Subtract

= 7 Add

Result

33 ; 26; 7

Page 35 Exercise 10 Answer

\(\frac{d}{7}\) − 3 + 10
​
\(\frac{d}{7}\) − 3 + 10

= \(\frac{28}{7}\) − 3 + 10 Substitute d = 28

= 4 − 3 + 10 Divide

= 1 + 10 Subtract

= 11 Add

\(\frac{d}{7}\) − 3 + 10

= \(\frac{49}{7}\) − 3 + 10 Substitute d = 49

= 7 − 3 + 10 Divide

= 4 + 10 Subtract

= 14 Add

\(\frac{d}{7}\) − 3 + 10

= \(\frac{63}{7}\) − 3 + 10 Substitute d = 63

= 9 − 3 + 10 Divide

= 6 + 10 Subtract

= 16 Add

Result

11 ; 14 ; 16

Page 36 Exercise 11 Answer

a) Expression for the amount she earned for h hours sitting one dog and two days sitting the cats

Cost per hour for one Dog × Number of Hours + Cost per day for two Cats × Number of Days

= 7 × h + 15 × 2

= 7h + 30

b) Evaluate the expression for h = 2

7h + 30

= 7(2) + 30 Substitute h = 2

= 14 + 30 Multiply

= $44

Result

a) 7h + 30

b) $44

Page 36 Exercise 12 Answer

One Dog:

Cost per day = $20

Cost for one hour = $7

Cost for two hours = $7 × 2 = $14

Cost for three hours = $7 × 3 = $21

Thus, we can purchase for 2 hours before it would be cheaper than to pay for one day.

Result

2 hours

Page 36 Exercise 13 Answer

65 + 6x Evaluate

= 65 + 6(4) Substitute x = 4

= 65 + 24 Multiply

= 89 Add

Result

It will cost $89 if he surfs for 4 hours.

Page 36 Exercise 14 Answer

Let the number of lessons be x Assume a variable

12x = 20 + 8x Equate both the expressions

12x − 8x = 20 + 8x − 8x Subtract 8x from both sides

4x = 20 Simplify

\(\frac{4x}{4}\) = \(\frac{20}{4}\) Divide each side by 4

x = 5 Simplify

Result

Thus, it will take 5 lessons for Janet′s class to cost the same amount as Rita′s class.

Page 36 Exercise 15 Answer

3g ÷ \(h^2\) + k − n Evaluate

= 3(12) ÷ \((3)^2\) + (10) − (1) Substitute g = 12, h = 3, k = 10 and n = 1

= 36 ÷ 9 + 10 − 1 Evaluate the parentheses

= 4 + 10 − 1 Divide

= 14 – 1 Add

= 13 Subtract

Result

D) 13

Page 36 Exercise 16 Answer

\(\frac{1}{2} x+y^2-4 z \div t\) Evaluate

= \(\frac{1}{2}(10)+(4)^2-4(5) \div(2)\) Substitute x = 10, y = 4, z = 5, and t = 2

= 5 + 16 − 20 ÷ 2 Evaluate the parentheses

= 5 + 16 − 10 Divide

= 21 − 10 Add

= 11 Subtract

Result

C) 11

Chapter 1 Algebra: Understand Numerical And Algebraic Expressions

Guided Practice 5

Page 39 Exercise 1 Answer

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

Using, Commutative Property of Addition: a + b = b + a

Result

Commutative Property of Addition

The equivalent expression is \(\frac{1}{2}\) + y

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

Page 39 Exercise 2 Answer

Let us substitute the value of z to find \(z^3\) and 3z are equivalent or not.

For algebraic expression to be equivalent they must have the same value for any number substitute for the same variable.

The expression \(z^3\) and 3z are not equivalent when z = 1 and z = 2

So, \(z^3\) and 3z are not equivalent expressions.

Result

\(z^3\) and 3z are not equivalent expressions.

Page 39 Exercise 3 Answer

2(r + 3) Evaluate

= 2(r) + 2(3) Using Distributive Property

= 2r + 6 Multiply

Result

2r + 6

Page 39 Exercise 4 Answer

6(4s − 1) Evaluate

= 6(4s) − 6(1) Using Distributive Property

= (6 ⋅ 4)s − 6(1) Using Associative Property of Multiplication

= 24s − 6 Multiply

Result

24s − 6

Page 39 Exercise 5 Answer

8t + 2 Evaluate

= 2(4t) + 2(1) Using Distributive Property

= 2(4t + 1) 2 is a common factor

Result

2(4t + 1)

Page 39 Exercise 6 Answer

3(m + 3) Evaluate

= 3(m) + 3(3) Using Distributive Property

= 3m + 9 Multiply

Result

3m + 9

Page 39 Exercise 7 Answer

20n − 4m Evaluate

= 4(5n) − 4(m) Using Distributive Property

= 4(5n − 4) 4 is the common factor

Result

4(5n − 4)

Page 39 Exercise 8 Answer

\(4\left(3 p+2 \frac{1}{2}\right)\) Evaluate

= \(4(3 p)+4\left(2 \frac{1}{2}\right)\) Using Distributive Property

= \((4 \cdot 3) p+4\left(\frac{5}{2}\right)\) Using Associative Property of Multiplication

= 12p + 10 Multiply

Result

12p + 10

Page 39 Exercise 9 Answer

3(x − 6) Evaluate

= 3(x) − 3(6) Using Distributive Property

= 3x − 18 Multiply

Result

3(x − 6) and 3x − 18 are equivalent expressions

Page 39 Exercise 10 Answer

2x + 10 Evaluate

= 2(x) + 2(5) Using Distributive Property

= 2(x + 5) 2 is the common factor

Result

2x + 10 and 2(x + 5) are equivalent expression.

Page 39 Exercise 11 Answer

\(8\left(2 y+\frac{1}{4}\right)\) Evaluate

= \(8(2 y)+8\left(\frac{1}{4}\right)\) Using Distributive Property

= 16y + 2 Multiply

Result

16y + 2 are equivalent expression

Page 39 Exercise 12 Answer

5.7 + (3z + 0.3) Equivalent

= 5.7 + 3z + 0.3 Open parentheses

= 5.7 + 0.3 + 3z Group like terms

= 3z + 6 Combine like terms

= 3(z) + 3(2) Using Distributive Property

= 3(z + 2) 3 is the common factor

Result

3(z + 2) is the equivalent expression

Page 39 Exercise 13 Answer

5w − 15 Equivalent

= 5(w) − 5(3) Using Distributive Property

= 5(w − 5) 5 is the common factor

Result

5(w − 5) is the equivalent expression

Page 39 Exercise 14 Answer

2x + 4y Equivalent

= 2(x) + 2(2y) Using Distributive Property

= 2(x + 2y) 2 is the common factor

Result

2(x + 2y) is the equivalent expression

Page 39 Exercise 15 Answer

10(\(y^2\) + 2.45) Equivalent

= \(10\left(y^2\right)+10(2.45)\) Using Distributive Property

10\(y^2\) + 24.5 Multiply

Result

10\(y^2\) + 24.5 is the equivalent expression

Page 39 Exercise 16 Answer

\(\frac{3}{4} \cdot\left(z^3 \cdot 4\right)\) Equivalent

= \(\frac{3}{4} \cdot z^3 \cdot 4\) Open Parentheses

= 3\(z^3\) Multiply

Result

3\(z^3\) is the equivalent expression

Page 40 Exercise 17 Answer

Length = 5 and Width = 2x − 1

Algebraic expression for the area of rectangle:

Length × Width = 5 ⋅ (2x − 1)

Result

5 ⋅ (2x − 1)

Page 40 Exercise 18 Answer

5(2x − 1) Evaluate

= 5(2x) − 5(1) Using Distributive Property

= 10x − 5 Multiply

Result

10x − 5 is the equivalent expression

Page 40 Exercise 19 Answer

10x − 5 Evaluate

= \(10\left(5 \frac{1}{2}\right)-5\) Substitute x = \(5 \frac{1}{2}\)

= \(10 \cdot \frac{11}{2}-5\) Open parentheses

= 55 − 5 Multiply

= 50 Subtract

Result

Area = 50 square units

Page 40 Exercise 20 Answer

Number of magnifying glasses . Cost of each magnifying glass + Number of safety glasses ⋅ Cost of each safety glass Given

7 ⋅ 1.25 + 7 ⋅ 3.75 Evaluate

= 8.75 + 26.25 Multiply

= $35 Add

Result

The total cost is $35

Page 40 Exercise 21 Answer

5y − 20 Evaluate

= 5(y) − 5(4) Using Distributive Property

= 5(y − 4) 5 is the common factor

Result

5 is the common factor in the expression 5y − 20

Page 40 Exercise 22 Answer

2(2n − 1) Evaluate

= 2(2n) − 2(1) Using Distributive Property

= 4n − 2 Multiply

Result

Yes, Chrish is correct that the expression 4n − 2 and 2(2n − 1) are equivalent

Page 40 Exercise 23 Answer

(f ⋅ \(g^2\)) + 5 − (\(g^2\) ⋅ f) Evaluate

= f\(g^2\) + 5 − f\(g^2\) Multiply

= 5 Simplify

Result

5 is the only one term and is equivalent to the expression (f ⋅ \(g^2\)) + 5 − (\(g^2\) ⋅ f)

Page 40 Exercise 24 Answer

8.5 + (2s + 0.5)

= (8.5 + 2s) + 0.5 Using Associative Property of Addition

8.5 + (2s + 0.5)

= (8.5 + 0.5) + 2s Using Associative Property of Addition

8.5 + (2s + 0.5)

= (8.5 + 0.5) + 2s Using Associative Property of Addition

= 9 + 2s Add

= 2(4.5) + 2(s) Using Distributive Property

= 2(4.5 + s) Using Associative Property of Addition

8.5 + (2s + 0.5) is equivalent to the following:

→ (8.5 + 2s) + 0.5

→ (8.5 + 0.5) + 2s

→ 2(4.5 + s)

Result

(8.5 + 2s) + 0.5

(8.5 + 0.5) + 2s

2(4.5 + s)

Page 40 Exercise 25 Answer

5n + 20

= 5(n) + 5(4) Using Distributive Property

= 5(n + 4)

15 + 5n + 5

= 20 + 5n Add

= 5(4) + 5(n) Using Distributive Property

= 5(n + 4) 5 is the common factor

5(n + 3) + 5

= 5(n) + 5(3) + 5 Using Distributive Property

= 5n + 15 + 5 Multiply

= 5n + 20 Add

= 5(n) + 5(4) Using Distributive Property

= 5(n + 4) 5 is the common factor

Result

5n + 20 is equivalent to the following:

→ 5n + 20

→ 15 + 5n + 5

→ 5(n + 3) + 5

Chapter 1 Algebra: Understand Numerical And Algebraic Expressions

Homework And Practice 6

Page 41 Exercise 1 Answer

Question 1: Solve The Expression 5(m − 2)

Solution:

Given

5(m − 2)

= 5(m) − 5(2) Using Distributive Property

= 5m − 10 Multiply

Result

5m − 10

Page 41 Exercise 2 Answer

Question 2: Solve The Expression 24x + 18y

Solution:

24x + 18y Given

= 6(4x) + 6(3y) Using Distributive Property

= 6(4x + 3y) 6 is the common factor

Result

6(4x + 3y)

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

Question 3: Solve The Expression \(2\left(9 p-\frac{1}{2}\right)\)

Solution:

Page 41 Exercise 3 Answer

\(2\left(9 p-\frac{1}{2}\right)\) Given

= \(2(9 p)-2\left(\frac{1}{2}\right)\) Using Distributive Property

= 18p − 1 Multiply

Result

18p − 1

Page 41 Exercise 4 Answer

Question 4: Solve The Expression 8(2x − 3)

Solution:

8(2x − 3) Given

= 8(2x) − 8(3) using Distributive Property

= 16x − 24 Multiply

Result

8(2x − 3) and 16x − 24 are equivalent

Page 41 Exercise 5 Answer

Question 5: Solve The Expression 5(3x − 9)

Solution:

5(3x − 9) Given

= 5(3x) − 5(9) Using Distributive Property

= 15x − 45 Multiply

Result

5(3x − 9) and 15x − 45 are equivalent

Page 41 Exercise 6 Answer

Question 6: Solve The Expression 6(2x + 9)

Solution:

6(2x + 9) Given

= 6(2x) + 6(9) Using Distributive Property

= 12x + 54 Multiply

Result

6(2x + 9) and 12x + 54 are equivalent

Page 41 Exercise 7 Answer

Question 7: Solve The Expression 3(6x − 7)

Solution:

3(6x − 7) Given

= 3(6x) − 3(7) Using Distributive Property

= 18x − 21 Multiply

Result

18x − 21

Page 41 Exercise 8 Answer

Question 8: Solve The Expression 4(9x − 2)

Solution:

4(9x − 2) Given

= 4(9x) − 4(2) Using Distributive Property

= 36x − 8 Multiply

Result

36x − 8

Page 41 Exercise 9 Answer

Question 9: Solve The Expression 6(8x + 1)

Solution:

6(8x + 1) Given

= 6(8x) + 6(1) Using Distributive Property

= 48x + 6 Multiply

Result

48x + 6

Page 41 Exercise 10 Answer

Question 10: Solve The Expression 35x + 30

Solution:

35x + 30 Given

= 5(7x) + 5(6) Using Distributive Property

= 5(7x + 6) Multiply

Result

5(7x + 6)

Page 41 Exercise 11 Answer

Question 11: Solve The Expression 4(x + 7)

Solution:

4(x + 7) Given

= 4(x) + 4(7) Using Distributive Property

= 4x + 28 Multiply

Result

4x + 28

Page 41 Exercise 12 Answer

Question 12: Solve The Expression 5x − 15y

Solution:

5x − 15y Given

= 5(x) − 5(3y) Using Distributive Property

= 5(x − 3y) 5 is the common factor

Result

5(x − 3y)

Page 41 Exercise 13 Answer

Question 13: Solve The Expression \(6\left(3 y-\frac{1}{2}\right)\)

Solution:

\(6\left(3 y-\frac{1}{2}\right)\) Given

= \(6(3 y)-6\left(\frac{1}{2}\right)\) Using Distributive Property

= 18y − 3 Multiply

Result

18y − 3

Page 41 Exercise 14 Answer

Question 14: Solve The Expression 1.6 + (2z + 0.4)

Solution:

1.6 + (2z + 0.4) Given

= (1.6 + 2z) + 0.4 Using Associative Property of Addition

Result

(1.6 + 2z) + 0.4

Page 41 Exercise 15 Answer

Question 15: Solve The Expression 8w − 16

Solution:

8w − 16 Given

= 8(w) − 8(2) Using Distributive Property

= 8(w − 2) 8 is the common factor

Result

8(w – 2)

Page 41 Exercise 16 Answer

Question 16: Solve The Expression 2.2x + 2.2

Solution:

2.2x + 2.2 Given

= 2.2(x) + 2.2(1) Using Distributive Property

= 2.2(x + 1) 2.2 is the common factor

Result

2.2(x + 1)

Page 41 Exercise 17 Answer

Question 17: Solve The Expression \(100\left(z^2-5.38\right)\)

Solution:

\(100\left(z^2-5.38\right)\) Given

= 100\((z)^2\) – 100(5.38) Using Distributive Property

100\(z^2\) − 538 Multiply

Result

100\(z^2\) − 538

Page 41 Exercise 18 Answer

Question 18: Solve The Expression \(8 \cdot\left(y^3 \cdot \frac{3}{4}\right)\)

Solution:

\(8 \cdot\left(y^3 \cdot \frac{3}{4}\right)\) Evaluate

= \(8 \cdot \frac{3 y^3}{4}\) Evaluate inside parentheses

= 6\(y^3\)Multiply

Result

6\(y^3\)

Page 42 Exercise 19 Answer

Cost of 1 Pencil Pack = $1.50

Cost of 1 Notebook = $2

Cost of 1 Marker = $2.50

Number of Pencil Packs ordered = 5

Number of Notebooks ordered = n

Number of Markers ordered = 5 sets

Algebraic Expression for the total cost of Ms. Thomas′s ordered:

Number of Pencil pack order × Cost of each pencil pack + Number of Notebooks order × Cost of each Notebook + Number of Markers order × Cost of each Marker

= 5 × 1.50 + n × 2 + 5 × 2.50

Result

5 × 1.50 + n × 2 + 5 × 2.50

Page 42 Exercise 20 Answer

5 × 1.50 + n × 2 + 5 × 2.50 Algebraic Expression

= 7.50 + 2n + 12.50 Multiply

= 7.50 + 12.50 + 2n Commutative Property of Addition

= 20 + 2n Add

= 2(10) + 2(n) Distributive Property

= 2(10 + n) 2 is the common factor

Result

2(10 + n)

Page 42 Exercise 21 Answer

Question 21: Solve The Expression 6(2x + 9)

Solution:

2(10 + n) Evaluate

= 2(10 + 20) Substitute n = 20

= 2(30) Add

= $60 Multiply

Result

The total cost of Ms. Thomas′s Order if she ordered 20 Notebooks is $60

Page 42 Exercise 22 Answer

Question 4: Solve The Expression 6(2x + 9)

Solution:

2l + 2w Evaluate

= 2(l) + 2(w) Using Distributive Property

=2(l + w) 2 is the common factor

Result

2l + 2w and 2(l + w) are equivalent expressions.

Page 42 Exercise 23 Answer

Question 4: Solve The Expression 6(2x + 9)

Solution:

It is easier to use 2(l + w) than to use 2l + 2w because →

2l + 2w: We need to multiply length with 2 and then width with 2 and then add to get the solution.

2(l + w): We need to add length and width and then Multiply it with 2 to get the solution.

Example: Let length = 5 units and Width = 3 units

2l + 2w = 2 ⋅ 5 + 2 ⋅ 3 = 10 + 2 ⋅ 3 = 10 + 6 = 16 units

2(l + w) = 2(5 + 3) = 2(8) = 16 units

Result

Because it is less time consuming than the expression 2l+2w

Page 42 Exercise 24 Answer

Question 4: Solve The Expression 6(2x + 9)

Solution:

\(4 \frac{1}{2}+\left(3 t+1 \frac{1}{2}\right)\)
​
= \(\left(4 \frac{1}{2}+3 t\right)+1 \frac{1}{2}\) Using Associative Property of Addition

= \(\left(4 \frac{1}{2}+1 \frac{1}{2}\right)+3 t\) Using Associative Property of Addition

= \(\left(4 \frac{1}{2}+1 \frac{1}{2}\right)+3 t\) Using Associative Property of Addition

= 6 + 3t Adding

= \(\left(4 \frac{1}{2}+1 \frac{1}{2}\right)+3 t\) Using Associative Property of Addition

= 6 + 3t Adding

= 3(2) + 3(t) Using Distributive Property

= 3(2 + t) 2 is the common factor

Result

6 + 3t

3(2 + t)

Page 42 Exercise 25 Answer

8(x − 3)

= 8(x) − 8(3) Using Distributive Property

= 8x − 24

8(x − 24)

= 8(x) − 8(24) Using Distributive Property

= 8x − 192

9(x − 3) − (x – 3)

= 9(x) − 9(3) − x + 3 Using Distributive Property

= 9x − 27 − x + 3 Multiply

= 9x − x − 27 + 3 Using Commutative Property of Addition

= 8x − 24

(5 + 3)x − 24

= x(5) + x(3) − 24 Using Distributive Property

= 5x + 3x − 24 Multiply

= 8x − 24

Result

8x − 24 is equivalent to the following:

8(x − 3)

9(x − 3) − (x − 3)

(5 + 3)x − 24

Chapter 1 Algebra: Understand Numerical And Algebraic Expressions

Guided Practice 6

Page 45 Exercise 1 Answer

The expression 2y – y can be written as y because 2y and y are like terms so we can simplify it by using Properties of Operations.

2y − y

= 2y − 1y Using Identity Property of Multiplication

= (2 − 1)y Using Distributive Property

= y

Result

Because 2y and y are like terms so we can simplify the expression

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

Page 45 Exercise 2 Answer

\(\frac{1}{2}\)x + \(\frac{1}{2}\)x and x are equivalent because if we simplify the expression \(\frac{1}{2}\)x + \(\frac{1}{2}\)x we will get x

\(\frac{1}{2}\)x + \(\frac{1}{2}\)x

= (\(\frac{1}{2}\) + \(\frac{1}{2}\))x Using Distrtibutive Property

= x

Result

Because the expression \(\frac{1}{2}\)x + \(\frac{1}{2}\)x when simplify will become x

Page 45 Exercise 3 Answer

No, 4\(z^2\) − \(z^2\) and 4 are not equivalent expressions.

4\(z^2\) − \(z^2\)

= (4 − 1)\(z^2\) Using Distributive Property

= 3\(z^2\)

Result

No, 4\(z^2\) − \(z^2\) and 4 are not equivalent expressions.

Page 45 Exercise 4 Answer

x + x + x + x Given

= 1x + 1x + 1x + 1x Use Identity Property of Multiplication

= (1 + 1 + 1 + 1)x Use Distributive Property

= 4x Simplify

Result

4x

Page 45 Exercise 5 Answer

4y − y Given

= 4y − 1y Use Identity Property of Multiplication

= (4 − 1)y Use Distributive Property

= 3y Simplify

Result

3y

Page 45 Exercise 6 Answer

3x + 8 + 2x Given

= 3x + 2x + 8 Using Commutative Property of Addition

= 5x + 8 Combine like terms

Result

5x + 8

Page 45 Exercise 7 Answer

7y − 4.5 − 6y Given

= 7y − 6y − 4.5 Group like terms

= y − 4.5 Combine Like terms

Result

y − 4.5

Page 45 Exercise 8 Answer

4x + 2 − \(\frac{1}{2} x\) Given

= 4x – \(\frac{1}{2} x\) + 2 Group like terms

= \(\left(4-\frac{1}{2}\right) x+2\) Using Distributive Property

= \(3 \frac{1}{2}\)x + 2 Simplify

Result

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

Page 45 Exercise 9 Answer

3 + 3y − 1 + y Given

= 3y + y + 3 − 1 Group like terms

= 3y + 1y + 3 − 1 Using Identity Property of Multiplication

= (3 + 1)y + 3 − 1 Using Distributive Property

= 4y + 2 Simplify

Result

4y + 2

Page 45 Exercise 10 Answer

x + 6x Given

= 1x + 6x Using Identity Property of Multiplication

= (1 + 6)x Using Distributive Property

Step 4

= 7x Simplify

Result

7x

Page 45 Exercise 11 Answer

9y − 3y Given

= (9 − 3)y Using Distributive Property

= 6y Simplify

Result

6y

Page 45 Exercise 12 Answer

2z + \(\frac{1}{4}\) + 2z Given

= 2z + 2z + \(\frac{1}{4}\) Group like terms

= (2 + 2)z + \(\frac{1}{4}\) Using Distributive Property

= 4z + \(\frac{1}{4}\) Simplify

Result

4z + \(\frac{1}{4}\)

Page 45 Exercise 13 Answer

5 + 3w + 3 − w Given

= 5 + 3 + 3w − w Group Like terms

= 5 + 3 + 3w − 1w Using Identity Property of Multiplication

= 5 + 3 + (3 − 1)w Using Distributive Property

= 8 + 2w Simplify

Result

8 + 2w

Page 45 Exercise 14 Answer

5w − 5w Given

= (5 − 5)w Using Distributive Property

= 0 Simplify

Result

0

Page 45 Exercise 15 Answer

2x + 5 + 3x + 6 Given

= 2x + 3x + 5 + 6 Group Like Terms

= (2 + 3)x + 5 + 6 Using Distributive Property

= 5x + 11 Simplify

Result

5x + 11

Page 45 Exercise 16 Answer

10\(y^2\) + 2\(y^2\) Given

= (10 + 2)\(y^2\) Using Distributive Property

= 12\(y^2\) Simplify

Result

12\(y^2\)

Page 45 Exercise 17 Answer

\(\frac{3}{4} z^3+4-\frac{1}{4} z^3\) Given

= \(\frac{3}{4} z^3-\frac{1}{4} z^3+4\) Group Like terms

= \(\left(\frac{3}{4}-\frac{1}{4}\right) z^3+4\) Using Distributive Property

= \(\left(\frac{3-1}{4}\right) z^3+4\) LCM is 4

= \(\left(\frac{2}{4}\right) z^3+4\) Subtract

= \(\frac{1}{2} z^3+4\) Simplify

Result

Page 45 Exercise 18 Answer

3.4m + 2.4m Given

= (3.4 + 2.4)m Using Distributive Property

= 5.8m Simplify

Result

5.8m

Page 45 Exercise 19 Answer

4.2n + 5 − 3.2n Given

= 4.2n − 3.2n + 5 Group Like Terms

= (4.2 − 3.2)n + 5 Using Distributive Property

= n + 5 Simplify

Result

n + 5

Page 45 Exercise 20 Answer

5\(p^2\) – 5 – 2\(p^2\) Given

= 5\(p^2\) – 2\(p^2\) – 5 Group like terms

= (5 – 2)\(p^2\) – 5 Using Distributive Property

= \(3p^2\) − 5 Simplify

Result

\(3p^2\) − 5

Page 45 Exercise 21 Answer

\(q^5+q^5+q^5\) Given

= \(1 q^5+1 q^5+1 q^5\) Using Identity Property of Multiplication

= (1 + 1 + 1)\(q^5\) Using Distributive Property

= \(3q^5\) Simplify

Result

Page 45 Exercise 22 Answer

\(3 x+\frac{1}{4}+2 y+\frac{1}{4}+7 x-y\) Given

= \(3 x+7 x+2 y-y+\frac{1}{4}+\frac{1}{4}\) Group like terms

= \((3+7) x+(2-1) y+\frac{1+1}{4}\) Using Distributive Property

= 10x + y + \(\frac{2}{4}\) Simplify

= 10x + y + \(\frac{1}{2}\) Simplify

Result

10x + y + \(\frac{1}{2}\)

Page 45 Exercise 23 Answer

1.5\(z^2\) + 4.5 + 6z − 0.3 − 3z + \(z^2\) Given

= 1.5\(z^2\) + \(z^2\) + 6z − 3z + 4.5 − 0.3 Group like terms

= 1.5\(z^2\) + 1z + 6z − 3z + 4.5 − 0.3 Using Identity Property of Multiplication

= (1.5 + 1)\(z^2\) + (6 − 3)z + 4.5 − 0.3 Using Distributive Property

= 2.5\(z^2\) + 3z + 4.2 Simplify

Result

2.5\(z^2\) + 3z + 4.2

Page 46 Exercise 24 Answer

Length = 2y + 1; Width = y

Algebraic Expression:

Perimeter = 2(Length + width) = 2(2y + 1 + y)

Result

2(2y + 1 + y)

Page 46 Exercise 25 Answer

2(2y + 1 + y) Given

= 2(2y) + 2(1) + 2(y) Using Distributive Property

= 4y + 2 + 2y Multiply

= 4y + 2y + 2 Group Like Terms

= 6y + 2 Combine Like Terms

Result

6y + 2

Page 46 Exercise 26 Answer

6y + 2

= \(6\left(2 \frac{1}{2}\right)+2\) Substitute y = \(2 \frac{1}{2}\)

= 6 . \(\frac{5}{2}\) + 2 Multiply

= 15 + 2 Simplify

= 17 Add

Result

The perimeter of the rectangle is 17 units

Page 46 Exercise 27 Answer

\(\frac{1}{2}\)(2x + 7) Given

= \(\frac{1}{2}(2 x)+\frac{1}{2}(7)\) Using Distributive Property

= \(\frac{2x}{2}\) + \(\frac{7}{2}\) Multiply

= x + \(3 \frac{1}{2}\) Simplify

Result

Rodney use Distributive Property to rewrote \(\frac{1}{2}\)(2x + 7) as x + \(3 \frac{1}{2}\)

Page 46 Exercise 28 Answer

n > \(n^2\) Given

= (0.5) > \((0.5)^2\) Substitute n = 0.5

= 0.5 > 0.25 Simplify

Result

n > \(n^2\) is true for 0 < n < 1

Page 46 Exercise 29 Answer

4x − 3x + 2 Given

= (4 – 3)x + 2 Using Distributive Property

= x + 2 Simplify

Result

Yes, Thea is Correct

Page 46 Exercise 30 Answer

\(\frac{a}{3}+\frac{a}{3}+\frac{a}{3}\) Given

= \(\left(\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\right) a\) Using Distributive Property

= \(\frac{3}{3} a\) Add

= a Simplify

Result

a

Page 46 Exercise 31 Answer

2x + 7 + 6x − x

= 2x + 6x − x + 7 Group like terms

= (2 + 6)x − x + 7 Using Distributive Property

= 8x − x + 7 Simplify

= (8x − 1)x + 7 Using Distributive Property

= 7x + 7

Result

Equivalent to: 2x + 7 + 6x − x

7 + 7x and 7x + 7

Not Equivalent to: 2x + 7 + 6x − x

2x + 13 and 14x

Chapter 1 Algebra: Understand Numerical And Algebraic Expressions

Homework And Practice 7

Page 47 Exercise 1 Answer

n + n + n Given

= 1n + 1n + 1n Using Identity Property of Multiplication

= (1 + 1 + 1)n Using Distributive Property

= 3n Simplify

Result

3n

Page 47 Exercise 2 Answer

3n + 6 − n − 4 Given

= 3n − n + 6 − 4 Using Commutative Property of Addition

= 3n − 1n + 6 − 4 Using Identity Property of Multiplication

= (3 − 1)n + 6 − 4 Using Distributive Property

= 2n + 2 Simplify

Result

2n + 2

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

Page 47 Exercise 3 Answer

\(1 \frac{1}{2} z^2+3 \frac{1}{2}+5 z-3+6 z-\frac{1}{2} z^2\) Given

= \(1 \frac{1}{2} z^2-\frac{1}{2} z^2+5 z+6 z+3 \frac{1}{2}-3\) Using Commutative Property of Addition

= \(\left(1 \frac{1}{2}-\frac{1}{2}\right) z^2+(5+6) z+\left(3 \frac{1}{2}-3\right)\) Using Distributive Property

= \(z^2+11 z+\frac{1}{2}\) Simplify

Result

Page 47 Exercise 4 Answer

4y + 9y Given

= (4 + 9)y Using Distributive Property

= 13y Simplify

Result

13y

Page 47 Exercise 5 Answer

3z + \(\frac{3}{4}\) – 2z Given

= 3z – 2z + \(\frac{3}{4}\) Using Commutative Property of Addition

= (3 – 2)z + \(\frac{3}{4}\) Using Distributive Property

= z + \(\frac{3}{4}\) Simplify

Result

z + \(\frac{3}{4}\)

Page 47 Exercise 6 Answer

25 + 5w − 10 + w Given

= 5w + w + 25 − 10 Using Commutative Property of Addition

= 5w + 1w + 25 − 10 Using Identity Property of Multiplication

= (5 + 1)w + 25 − 10 Using Distributive Property

= 6w + 15 Simplify

Result

6w + 15

Page 47 Exercise 7 Answer

7.7w − 4.6w Given

= (7.7 − 4.6)w Using Distributive Property

= 3.1w Simplify

Result

3.1w

Page 47 Exercise 8 Answer

\(\frac{1}{2} x+\frac{1}{2}+\frac{1}{2} x+\frac{1}{2}\) Given

= \(\frac{1}{2} x+\frac{1}{2} x+\frac{1}{2}+\frac{1}{2}\) Using Commutative Property of Addition

= \(\left(\frac{1}{2}+\frac{1}{2}\right) x+\left(\frac{1}{2}+\frac{1}{2}\right)\) Using Distributive Property

= x + 1 Simplify

Result

x + 1

Page 47 Exercise 9 Answer

12\(y^2\) – 6\(y^2\) Given

= (12 – 6)\(y^2\) Using Distributive Property

= \(6y^2\)Simplify

Result

Page 47 Exercise 10 Answer

\(3 z^3+2 \frac{1}{4}-z^3\) Given

= \(3 z^3-z^3+2 \frac{1}{4}\) Using Commutative Property of Addition

= \(3 z^3-1 z^3+2 \frac{1}{4}\) Using Identity Property of Multiplication

= \((3-1) z^3+2 \frac{1}{4}\) Using Distributive Property

= \(2 z^3+2 \frac{1}{4}\) Simplify

Result
​
\(2 z^3+2 \frac{1}{4}\)

Page 47 Exercise 11 Answer

6.6m + 3m Given

=(6.6 + 3)m Using Distributive Property

= 9.6m Simplify

Result

9.6m

Page 47 Exercise 12 Answer

100n − 1 − 25n Given

= 100n − 25n − 1 Using Commutative Property of Addition

= (100 − 25)n − 1 Using Distributive Property

= 75n − 1 Simplify

Result

75n − 1

Page 47 Exercise 13 Answer

5x + \(\frac{1}{2}\) + 3y + \(\frac{1}{4}\) + 2x − 2y Given

= \(5 x+2 x+3 y-2 y+\frac{1}{2}+\frac{1}{4}\) Using Commutative Property of Addition

= \((5+2) x+(3-2) y+\left(\frac{1}{2}+\frac{1}{4}\right)\) Using Distributive Property

= \(7 x+y+\frac{2+1}{4}\) Simplify

= 7x + y + \(\frac{3}{4}\) Simplify

Result

7x + y + \(\frac{3}{4}\)

Page 47 Exercise 14 Answer

\(p^2\)+ 2.3 + \(3p^2\) Given

= \(p^2\) + \(3p^2\) + 2.3 Using Commutative Property of Addition

= \(1p^2\) + \(3p^2\) + 2.3 Using Identity Property of Multiplication

= (1 + 3)\(p^2\) + 2.3 Using Distributive Property

= \(4p^2\) + 2.3 Simplify

Result

\(4p^2\) + 2.3

Page 47 Exercise 15 Answer

\(z^4+z^4+z^4+z^4\) Given

= \(1 z^4+1 z^4+1 z^4+1 z^4\) Using Identity Property of Multiplication

= \((1+1+1+1) z^4\) Using Distributive Property

= \(4z^4\) Simplify

Result

Page 48 Exercise 16 Answer

Cost of Small drink = $1.10

Cost of Medium drink = $1.25

Cost of Large drink = $1.50

Number of drinks ordered by Casey′s Family :

Small = 1 and Medium = m

Number of drinks ordered by Anika′s Family :

Medium = m and Large = 1

Algebraic Expression for Total cost of both orders:

Casey′s Family order + Anika′s Family order

= 1 × 1.10 + m × 1.25 + m × 1.25 + 1 × 1.50

= 1.10 + 1.25m + 1.25m + 1.50

Result

1.10 + 1.25m + 1.25m + 1.50

Page 48 Exercise 17 Answer

1.10 + 1.25m + 1.25m + 1.50 Given

= 1.25m + 1.25m + 1.50 + 1.10 Using Commutative Property of Addition

= (1.25 + 1.25)m + 1.50 + 1.10 Using Distributive Property

= 2.50m + 2.60 Simplify

Result

2.50m + 2.60

Page 48 Exercise 18 Answer

2.50m + 2.60 Given

= 2.50(3) + 2.60 Substitute m = 3

= 7.50 + 2.60 Multiply

= $10.10 Add

Result

The total cost of both order is $10.10

Page 48 Exercise 19 Answer

\(\frac{1}{2} y\) . 5 Given

= 5.\(\frac{1}{2} y\) Using Commutative Property of Multiplication

Result

Jan use Commutative Property of Multiplication

Page 48 Exercise 20 Answer

\(n^3\) < \(n^2\) Given

= \((0.5)^3\) < \((0.5)^2\) Substitute n = 0.5

= 0.125 < 0.25 Simplify

Result

\(n^3\) < \(n^2\) is true for 0 < n < 1

Page 48 Exercise 21 Answer

6x − x + 5 Given

= (6 − 1)x + 5 Using Distributive Properties

= 5x + 5 Simplify

Result

6x − x + 5 and 6 + 5 are not equivalent expressions.

Page 48 Exercise 22 Answer

\(\frac{b}{2}\) + \(\frac{b}{2}\) Given

= \(\left(\frac{1}{2}+\frac{1}{2}\right) b\) Using Distributive Property

= b Simplify

Result

b

Page 48 Exercise 23 Answer

\(\frac{1}{2} x+4 \frac{1}{2}+\frac{1}{2} x-\frac{1}{2}\)
​
= \(\frac{1}{2} x+\frac{1}{2} x+4 \frac{1}{2}-\frac{1}{2}\) Using Commutative Property of Addition

= \(\left(\frac{1}{2}+\frac{1}{2}\right) x+\left(4 \frac{1}{2}-\frac{1}{2}\right)\) Using Distributive Property

= x + 4

Result

Equivalent to: \(\frac{1}{2} x+4 \frac{1}{2}+\frac{1}{2} x+\frac{1}{2}\)

x + 4

Not Equivalent to: \(\frac{1}{2} x+4 \frac{1}{2}+\frac{1}{2} x+\frac{1}{2}\)

\(\frac{1}{2} x\) + 4

x + \(4 \frac{1}{2}\)

x − 4

Chapter 1 Algebra: Understand Numerical And Algebraic Expressions

Guided Practice 7

Page 57 Exercise 1 Answer

Formulas are useful because it generalize a relationship between two or more quantitites.

Result

Because it generalize a relationship between quantities

Page 57 Exercise 2 Answer

It is important to define each variable used in formulas because a formula uses symbols to relate between quantities.

Result

Because formula uses symbols to relate between quantities

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

Page 57 Exercise 3 Answer

Step 1 → Identify the values in the formulas

d = distance

r = rate = 400 miles per hour

t = time = 5 hours

Step 2 → Substitute the values in the formula and evaluate

d = rt

d = 400 × 5

d = 2000

The plane will travel 2000 miles in 5 hours.

Result

2000 miles

Page 57 Exercise 4 Answer

Step 1 → Identify the values in the formulas

d = distance = 510 miles

r = rate = 68 miles per hour

t = time

Step 2 → Substitute the values in the formula and evaluate

d = rt

510 = (68)t Substitute the value

510 = 68t Simplify

\(\frac{510}{68}=\frac{68 t}{68}\) Divide both sides by 68

t = 7.5

It will take 7.5 hours for a car travelling at a rate of 68 miles per hour to go 510 miles.

Result

7.5 hours

Page 57 Exercise 5 Answer

Step 1 → Identify the values in the formulas

V = volume

s = length = width = height = 8 cm

Step 2 → Substitute the values in the formula and evaluate

V = \(s^3\)

V = \((8)^3\) Substitute the value

V = 512

The Volume of the cube is 512\(cm^3\)

Result

512\(cm^3\)

Page 57 Exercise 6 Answer

Step 1 → Identify the values in the formulas

A = Total surface area

s = length = width = height = 8 cm

Step 2 → Substitute the values in the formula and evaluate

A = \(6s^2\)

A = \(6(8)^2\) Substitute the value

A = 6 × 64 Simplify

A = 384\(cm^2\)

The Total surface area of the cube is 384\(cm^2\)

Result

384 \(cm^2\)

Page 57 Exercise 7 Answer

Step 1 → Identify the values in the formulas

g = number of gallons of gasoline used

m = miles per gallon = 16

d = distance traveled = 296

Step 2 → Substitute the values in the formula and evaluate

g = \(\frac{d}{m}\)

g = \(\frac{296}{16}\) Substitute the value

g = 18.5

Myra will need 18.5 gallons of gasoline to travel 296 miles.

Result

18.5 gallons

Page 58 Exercise 8 Answer

Step 1 → Identify the values in the formulas

F = Fahrenheit

C = Celsius = 26°

Step 2 → Substitute the values in the formula and evaluate

F = (C × 1.8) + 32

F = (26 × 1.8) + 32 Substitute the value

F = 46.8 + 32 Simplify

F = 78.8°

The temperature in Fahrenheit is 78.8°

Result

78.8°

Page 58 Exercise 9 Answer

Step 1 → Identify the values in the formulas

F = Fahrenheit = 45°F

C = Celsius = 13°C

Step 2 → Substitute the values in the formula and evaluate

F = (C × 1.8) + 32

45 = (13 × 1.8) + 32 Substitute the value

45 = 23.4 + 32 Simplify

45 ≠ 55.4

Since Left hand side is not equal to Right hand side.

So, the thermometer cannot show as both 45°F and 13°C

Result

NO

Page 58 Exercise 10 Answer

Step 1 → Identify the values in the formulas

A = Average grade

X = test score = 78

Y = test score = 90

Z = test score = 81

Step 2 → Substitute the values in the formula and evaluate

A = \(\frac{X+Y+Z}{3}\)
​
A = \(\frac{78+90+81}{3}\) Substitute the value

A = \(\frac{249}{3}\) Simplify

A = 83

Jules average test grade is 83

Result

83

Page 58 Exercise 11 Answer

Step 1 → Identify the values in the formulas

d = density

m = mass of the object = 65 grams

v = volume = 8 cubic meters

Step 2 → Substitute the values in the formula and evaluate

d = \(\frac{m}{v}\)
​
d = \(\frac{65}{8}\) Substitute the value

d = 8.125 gram per cubic meter

Density of the object is 8.125 gram per cubic meter

Result

8.125 gram per cubic meter

Page 58 Exercise 12 Answer

\(7\left(3^2+5\right)-\left(\frac{81}{9}\right)\) Evaluate

= \(7(9+5)-\left(\frac{81}{9}\right)\) \(3^2\) = 9

= 7(14) − 9 Evaluate inside the parentheses

= 98 − 9 Multiply

= 89 Subtract

Result

89

Page 58 Exercise 13 Answer

Step 1→Identify the values in the formulas

I = Interest = $696

p = Principal loan amount = $5800

r = rate = 4% = 0.04

t = time

Step 2 → Substitute the values in the formula and evaluate

I = prt

696 = 5800 ⋅ 0.04 ⋅ t Substitute the value

696 = 232t Simplify

\(\frac{696}{232}=\frac{232 t}{232}\) Divide both sides by 232

t = 3

Janie will take 3 years to pay off the loan

Result

3 years

Page 58 Exercise 14 Answer

Option 1 → 4.50 for each hour worked.

Option 2 → Final payment of $350

Step 1 → Identify the values in the formulas

p = total payment after 15 days

h = hours worked each day=6 hours

Step 2 → Substitute the values in the formula and evaluate

p = 15 × 4.50h

p = 15 × 4.50(6) Substitute the value

p = 15 × 27 Simplify

p = 405

Jeremiah should choose Option 1 as that he will get more payment than Option 2.

Result

Option 1

Chapter 1 Algebra: Understand Numerical And Algebraic Expressions

Homework And Practice 8

Page 59 Exercise 1 Answer

Step 1 → Identify the values in the formulas

P = Perimeter of the rectangle

l = length of the rectangle = 12 ft

w = width of the rectangle = 8 ft

Step 2 → Substitute the values in the formula and evaluate

P = 2l + 2w

P = 2(12) + 2(8) Substitute the value

P = 24 + 16 Simplify

P = 40 ft

The perimeter of the rectangle LMNP is 40 feet.

Result

40 feet

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

Page 59 Exercise 2 Answer

Step 1 → Identify the values in the formulas

A = Area of the rectangle

l = length of the rectangle = 12 ft

w = width of the rectangle = 8 ft

Step 2 → Substitute the values in the formula and evaluate

A = lw

A = (12)(8) Substitute the value

A = 96 \(ft^2\)

The area of the rectangle LMNP is 96 square feet.

Result

96 \(ft^2\)

Page 60 Exercise 3 Answer

Step 1 → Identify the values in the formulas

a = rate of acceleration

f = final speed = 77 meters per second

s = starting speed = 44 meters per second

t = time = 11 seconds

Step 2 → Substitute the values in the formula and evaluate

\(a=\left(\frac{f-s}{t}\right)\)
​
\(a=\left(\frac{77-44}{11}\right)\) Substitute the value

\(a=\left(\frac{33}{11}\right)\) Simplify

a = 3

The acceleration of the racecar is 3 meters per second squared.

Result

3 meters per second squared

Page 60 Exercise 4 Answer

Step 1 → Identify the values in the formulas

a = rate of acceleration

f = final speed = 26.9 meters per second

s = starting speed = 0 meters per second

t = time = 6.5 seconds

Step 2 → Substitute the values in the formula and evaluate

\(a=\left(\frac{f-s}{t}\right)\)
​
\(a=\left(\frac{26.9-0}{6.5}\right)\) Substitute the value

\(a=\left(\frac{26.9}{6.5}\right)\) Simplify

a = 4.13

The rate of acceleration of the car is 4.13 meters per second squared.

I do not support the claim.

Result

I do not support the claim

Page 60 Exercise 5 Answer

Cups of flour requires for one recipe = \(\frac{3}{4}\)

Cups of flour requires for other recipe = \(\frac{1}{2}\)

Total cups of flour requires

= \(\frac{3}{4}\) + \(\frac{1}{2}\)

= \(\frac{3+2}{4}\)

= \(\frac{5}{4}\)

= 1.25

Both the recipes require a total of 1.25 cups of flour and Jenna has 2 cups of flour which is enough to make both recipes.

Result

Yes, 2 cups of flour is enough to make both recipes.

Page 60 Exercise 6 Answer

P = 2l + 2w

P = 2(l) + 2(w) Using Distributive Property

P = 2(l + w) 2 is the common factor

Since both the formula P = 2l + 2w and P = 2(l + w) is equivalent.

So, Jack and Sandy both are correct.

Result

Both the formula are correct as they are equivalent

Page 60 Exercise 7 Answer

Length:

Step 1 → Identify the values in the formulas

c = measures in centimeters

f = measure in feet = 9 ft

Step 2 → Substitute the values in the formula and evaluate

c = 0.3f × 100

c = 0.3(9) × 100 Substitute the value

c = 2.7 × 100 Simplify

c = 270

Width:

Step 1 → Identify the values in the formulas

c = measures in centimeters

f = measure in feet = 5 ft

Step 2 → Substitute the values in the formula and evaluate

c = 0.3f × 100

c = 0.3(5) × 100 Substitute the value

c = 1.5 × 100 Simplify

c = 150

Result

Dimension of rectangle in centimeter:

Length = 270 cm and Width = 150 cm

Chapter 1 Algebra: Understand Numerical And Algebraic Expressions

Guided Practice 8

Page 63 Exercise 1 Answer

3(3x + 2x + x)

Write Equivalent expressions using properties

→ By Combining like terms, I can write this expression:

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

→ By using the Distributive Properties, I can write this expression:

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

Result

3(6x) ; 9x + 6x + 3x

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

Page 63 Exercise 2 Answer

I can substitute the value of x in the expression to find the number of hours Patrick plans to spend training during the summer.

3(3x + 2x + x)

= 3(3x) + 3(2x) + 3(x) Using Distributive Property

= 9x + 6x + 3x Multiply

= 18x Combining Like terms

= 18(9) Substitute x = 9

= 162

Result

Total hours Patrick plans to spend training over 9 weeks is 162 hours

Page 63 Exercise 3 Answer

(25t + 15) + (20t + 10) + 50t

Write Equivalent expressions using properties

→ By using Commutative Property, I can write this expression:

(25t + 15) + (20t + 10) + 50t = (20t + 10) + (25t + 15) + 50t

→ By Combining Like Terms, I can write this expression:

(25t + 15) + (20t + 10) + 50t = 25t + 20t + 50t + 15 + 10 = 95t + 25

Result

(20t + 10) + (25t + 15) + 50t ; 95t + 25

Page 63 Exercise 4 Answer

95t + 25 Evaluate

= 95(5) + 25 Substitute t = 5

= 475 + 25 Multiply

= 500 Add

Result

The Total cost Yolanda will pay is $500

Page 64 Exercise 5 Answer

Alex
​
Overtime Bonus = $10

Hourly wage = $7

Number of days = 2

Number of hours = x

Alex′s earning = Number of days(Overtime Bonus + Hourly wage ⋅ Number of hours) = 2(10 + 7 ⋅ x)

= 2(10 + 7x)

Tyrone

Overtime Bonus = $15

Hourly wage = $8

Number of days = 2

Number of hours = x

Tyrone′s earning = Number of days(Overtime Bonus + Hourly wage ⋅ Number of hours) = 2(15 + 8 ⋅ x)

= 2(15 + 8x)

Result

Alex′s earning: 2(10 + 7x)

Tyrone′s earning: 2(15 + 8x)

Page 64 Exercise 6 Answer

Mrs. Hayes accurately sum each of their wages to get the total amount she paid.

She also multiply their hourly wages with the number of hours, x

Hence, Mrs. Hayes′s expression is accurate

Result

Yes, Mrs. Hayes′s expression is accurate.

Page 64 Exercise 7 Answer

2(10 + 7x) + 2(15 + 8x)

Write Equivalent expressions using properties

→ By using the Distributive Properties, I can write this expression:

2(10 + 7x) + 2(15 + 8x) = 2(10) + 2(7x) + 2(15) + 2(8x) = 20 + 14x + 30 + 16x = 30x + 50

Result

30x + 50

Page 64 Exercise 8 Answer

If each worked 9 hours each day then we can substitute the value of x to find the total amount Mrs. Hayes paid to Alex and Tyrone.

30x+ 50

= 30(9) + 50

= 270 + 50

= $320

Result

Mrs. Hayes paid the total amount $320

Chapter 1 Algebra: Understand Numerical And Algebraic Expressions

Homework And Practice 9

Page 65 Exercise 1 Answer

4(3 ⋅ 8x) + 4(2 ⋅ 7x)

4(3 ⋅ 8x) → This part of expression represent the amount of time he runs for 3 days of the week.

4(2 ⋅ 7x) → This part of expression represent the amount of time he runs for 2 days of the week.

Result

4(3 ⋅ 8x) → For 3 days of the week

4(2 ⋅ 7x) → For 2 days of the week.

Page 65 Exercise 2 Answer

4(3 ⋅ 8x) + 4(2 ⋅ 7x)

Write Equivalent expressions using properties

→ By using Associative Properties of Multiplication, I can write this expression:

4(3 ⋅ 8x) + 4(2 ⋅ 7x) = 3(4 ⋅ 8x) + 2(4 ⋅ 7x)

Result

Associative Property of Multiplication: 3(4 ⋅ 8x) + 2(4 ⋅ 7x)

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

Page 66 Exercise 3 Answer

5(6x + 25x + 10)

The part 6x and 25x represent the time each group member spent surveying volunteers.

Result

6x and 25x

Page 66 Exercise 4 Answer

5(6x + 25x + 10)

Since the total number of group members is 5. So, Josephine multiply the time taken by each member with 5.

She multiply the survey time in January with the number of volunteers.

She also multiply the survey time in May with the number of volunteers.

In her expression she also added the time taken in giving presentation by each member.

Hence, Josephine′s expression is accurate.

Result

Yes, Josephine′s expression is accurate

Page 66 Exercise 5 Answer

5(6x + 25x + 10)

Write Equivalent expressions using properties

→ By Combining like terms, I can write this expression:

5(6x + 25x + 10) = 5((6 + 25)x + 10) = 5(31x + 10)

→ By using the Distributive Properties, I can write this expression:

5(6x + 25x + 10) = 5(6x) + 5(25x) + 5(10) = 30x + 125x + 50

Result

Combining Like terms: 5(31x + 10)

Distributive Property: 30x + 125x + 50

Page 66 Exercise 6 Answer

If each members worked with 20 volunteers each then we can substitute the value of x and then combine the like terms to calculate the total time Josephine′s group spent on the research.

5(31x + 10)

= 5(31(20) + 10)

= 5(620 + 10)

= 5(630)

= 3150

Result

Total time Josephine′s group spent on research project is 3150 minutes