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Go Math! Grade 6 Exercise 5.1: Operations with Decimals Solutions

November 18, 2024August 18, 2023 by Alekhya

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 5 Operations with Decimals

Page 25 Problem 1 Answer

The given expression is 585÷13.

We are asked to estimate the quotient by rounding off the dividend and the divisor to the largest possible value.

The answer can be found by rounding off the dividend and the divisor to the nearest value.

Here, 13 is the divisor and 585 dividend.

That would be 10&600 respectively. Now, find the value of 600/10

This will give you the required answer.

The expression is 585÷13. Here 585 is closer to 600

Rounding off the value 585, we get 600.

The value 13 is closer to ten. Rounding of 13, we have:10.

Thus, we have the quotient of 600÷10 as 60

The estimated quotient value of 585÷13 by rounding the dividend and the divisor to the largest place value is 60

Page 25 Problem 2 Answer

The given expression is 2,756÷53.

We are asked to find the quotient. The answer can be found by rounding off the dividend and the divisor to the nearest value.

Here, is the divisor and dividend are 53&2756 respectively.

When rounding off, we have 50&3000 Now, find the value of the quotient. This will give you the required answer.

The given expression is: 2,756÷53. The value 2756 is closer to 3000

Rounding the value 2756 to the largest place value, we have: 3000.

Similarly, 53 is closest to 50. Thus, the quotient value is 3000/50=60.

The estimated value of the quotient of 2,756÷53 by rounding the dividend and the divisor to the largest place value is 60

$Go Math! Practice Fluency Workbook Grade 6 Chapter 5 Operations with Decimals Exercise 5.1 Answer Key$

Go Math Grade 6 Exercise 5.1 Operations With Decimals Solutions

Go Math! Grade 6 Exercise 5.1: Operations with Decimals Solutions Page 25 Problem 3 Answer

The given expression is 22,528÷98.

We are asked to find the value of the quotient when the dividend and the divisor is rounded to the largest value.

The answer can be found by rounding off the dividend and the divisor to the nearest value.

Here, is the divisor and dividend are98&22528 respectively.

When rounding off, we have 100&20,000 Now, find the value of the quotient. This will give you the required answer.

The expression 22,528÷98. The value 22528 is closest to 20,000.

Similarly, the value 98 is closest to 100. Thus, the quotient value is 20,000/100 = 200.

The estimated value of the quotient of the expression 22,528÷98 by rounding the dividend and the divisor to the largest place value is 200.

Page 25 Problem 4 Answer

The given expression is 7,790÷210.

We are asked to find the value of the quotient when the dividend and the divisor is rounded to the largest value.

The answer can be found by rounding off the dividend and the divisor to the nearest value.

Here, is the divisor and dividend are 210&7790 respectively.

When rounding off, we have 200&8000.

Now, find the value of the quotient. This will give you the required answer.

The given expression is 7,790÷210

The value 7790 is closest to 8000.

Similarly, the value 210 is closest to 200. So, the quotient value of 8000/200 =40.

The estimated quotient of the expression 7,790÷210 by rounding the dividend and the divisor to the largest place value is 40.

Go Math! Grade 6 Exercise 5.1: Operations with Decimals Solutions Page 25 Problem 5 Answer

The given expression is 17,658÷360.

We are asked to find the value of the quotient when the dividend and the divisor is rounded to the largest value.

The answer can be found by rounding off the dividend and the divisor to the nearest value. Here, is the divisor and dividend are 360&17658respectively.

When rounding off, we have 400&20,000 Now, find the value of the quotient.

This will give you the required answer.

The expression is 17,658÷360. The value 17658 is closer to 20,000 than 10,000.

Similarly, the value 360 is closest to 400. Thus, the quotient value is: 20,000/400=50.

The estimated quotient value of 17,658÷360 by rounding the dividend and the divisor to the largest place value is 50.

Page 25 Problem 6 Answer

The given expression is 916÷320.

We are asked to find the value of the quotient when the dividend and the divisor is rounded to the largest value.

The answer can be found by rounding off the dividend and the divisor to the nearest value.Here, is the divisor and dividend are320&916 respectively.

When rounding off, we have 300&900 Now, find the value of the quotient.This will give you the required answer.

The expression is 916÷320.

The value 916 is closer to 900 than 1000. Similarly, the value 320 is closer to 300 than 400. Thus, the quotient value of 900/300 =3.

The estimated quotient value of the expression is916÷320 by rounding the dividend and the divisor to the largest place value is 3

Go Math Grade 6 Exercise 5.1 Operations With Decimals Answers

Go Math! Grade 6 Exercise 5.1: Operations with Decimals Solutions Page 25 Problem 7 Answer

The given expression is 1334÷29. We are asked to find the quotient of 1334÷29 by long division method.The answer can be found by writing the number as dq+r.

Then find the quotient for the value 1334 when divided by 29. Perform the operation by long division.

The expression is 1334÷29. 1334 can be written as 29×46+0.

That is, 1334 is divisible by 29. The quotient obtained is: 46

The quotient of 1334÷29 using long division method is 46:

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 5 Operations with Decimals 7$

Page 25 Problem 8 Answer

The given expression is 20884÷92.

We are asked to find the quotient of the expression by long division method. The answer can be found by writing the number as dq+r.

Then find the quotient for the value 20884 when divided by 92. Perform the operation by long division.

The expression is 20884÷92. Write the expression 20884 in the form dq+r.

Thus, we have: 227×92+0. That is the quotient value is q=227.

The quotient value of 20884÷92 using long division method is:

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 5 Operations with Decimals 8$

Page 25 Problem 9 Answer

The given expression is 18175÷25. We are asked to find the value of the quotient 18175÷25 using the long division method.

The answer can be found by found by writing the expression in the form dq+r.

Find the value of q when 18175 is divided by 25. Represent this by the long division method.

The expression is 18175÷25. It can be written as 727×25+0.

Hence, the remainder is 0 .So, the quotient value of 18175÷25 is 727.

The value of the quotient for the expression 18175÷25 using the long division method is 727.

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 5 Operations with Decimals 9$

Go Math! Grade 6 Exercise 5.1: Operations with Decimals Solutions Page 25 Problem 10 Answer

The given expression is 2902÷18. We are asked to find the quotient and the remainder value by the method of long division.

The answer can be found by writing the number as dq+r, where q,r are the quotient and remainder.

Find the quotient value of 2902 when divided by 18. Write the answer in the long division form.

The expression is 2902÷18. It can be written as 18×161+4. Here, the remainder is 4.

That is, the number is not perfectly divisible by 18. The quotient is: 161

The quotient and the remainder value of 2902÷18by long division method is 161 and 4 respectively.:

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 5 Operations with Decimals 10$

Page 25 Problem 11 Answer

The given expression is 34680÷64.We are asked to find the remainder and the quotient of the expression.

The answer can be found by writing the number in dq+r form, where r,q are the remainder and the quotient respectively.

Find the quotient value of 34680 when divided by 64. Similarly, find the remainder. This will give you the required answer.

The expression is 34680÷64. The value 34680 can be written as 34680=64×541+56.

From the above expansion, we have: q=541 and The remainder is r=56.

The remainder and the quotient of the expression 34680÷64 are 56 and 541.

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 5 Operations with Decimals 11$

Solutions For Go Math Grade 6 Exercise 5.1 Operations With Decimals

Go Math! Grade 6 Exercise 5.1: Operations with Decimals Solutions Page 25 Problem 12 Answer

The given expression is 52245÷215. We are asked to find the remainder and the quotient of the expression.

The answer can be found by writing the value as dq+r, where r,q are the remainder and the quotient respectively.

From the expansion, write down the remainder and the quotient value.Reflect the value in the long division form.

The expression is: 52245÷215. Writing the expression 52245 as dq+r, we have:243×215+0.

Here, r=0 and The value of quotient is q=243. The remainder and the quotient of the expression 52245÷215 using the long division method are 0&243respectively.

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 5 Operations with Decimals 12$

Page 25 Problem 13 Answer

Given that there were 4050 students from 15 different school district. We are asked to find the average number of students in the museum.

The answer can be found by using the definition of average.Find the quotient of 4050/15.

This will give you the required answer. The number of students are 4050. Number of school district are: 15.

Thus, the average number of students from each school is: 4050/15

15•270=4050

∴4050/15=270.

Hence, there were an average of 270 students from each school district.

The average number of students from each school district are 270.

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 5 Operations with Decimals 13$

Page 25 Problem 14 Answer

Given that, The Appalachian Trail is 2175 miles long and a hiker averages twelve miles each day.

We are asked to find how long will it take her to hike the length of the trail.

The answer can be found using the definition of average.Find the quotient value of 2175/12.

This would give the number of days. The total distance is 2175 miles.

Each day the hiker completes a distance of 12 miles.

Thus, the number of days taken for him to complete the journey is: 2175/12 = 181.25

That is, it would take 181 days and six hours. It would take 181.25 days for the hiker to hike the length of the trail.

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 5 Operations with Decimals 14$

Go Math Grade 6 Operations With Decimals Exercise 5.1 Key

Go Math! Grade 6 Exercise 5.1: Operations with Decimals Solutions Page 26 Exercise 1 Answer

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 5 Operations with Decimals e1$

We are asked to fill these up and hence find the number of markers available for each table.The answer can be found by the process of division.

Divide the number 47/11 and find the quotient and the remainder, Bring down the remainder and the other remaining number.

Find the quotient when divided by eleven. Find the quotient. This will give you the required answer.

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 5 Operations with Decimals e2$

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 5 Operations with Decimals e3$

She will put on 43 markers on each table:

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 5 Operations with Decimals e4$

Go Math Answer Key

Go Math! Grade 6 Exercise 4.4: Operations with Fractions Solutions

November 18, 2024August 18, 2023 by Alekhya

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 4 Operations with Fractions

Page 23 Problem 1 Answer

Given the values of leftover apple and pumpkin pies.

To do: Find how much more apple pie than pumpkin pie is left.

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 4 Operations with Fractions 1$

1/2 apple pie is left than pumpkin pie.

Go Math! Grade 6 Exercise 4.4: Operations with Fractions Solutions Page 23 Problem 2 Answer

Given about the inches of angelfish.

To do: Find how much it was grown.

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 4 Operations with Fractions 2$

Angelfish has grown 5/6 inches.

$Go Math! Practice Fluency Workbook Grade 6 Chapter 4 Operations with Fractions Exercise 4.4 Answer Key$

Go Math Grade 6 Exercise 4.4 Operations With Fractions Solutions

Go Math! Grade 6 Exercise 4.4: Operations with Fractions Solutions Page 23 Problem 3 Answer

Given about the wrapping paper for a birthday present.

To do: Find how many pieces of 6 square-foot paper are needed to wrap 3 of these presents.

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 4 Operations with Fractions 3$

We need 2 pieces of 6 square-foot paper are needed to wrap 3 of these presents.

Go Math! Grade 6 Exercise 4.4: Operations with Fractions Solutions Page 23 Problem 4 Answer

Given a bicycle was rode 5×1/2 miles today and 6×1/4 miles yesterday

To do: find the difference in length between the two rides.

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 4 Operations with Fractions 4$

The difference is 3/4 miles. and the fraction of the longer side is 3/25.

Page 23 Problem 5 Answer

Given a survey by the state health department

To do: Find what fraction of the total pounds of fruit and vegetables do the pounds of fruits represent

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 4 Operations with Fractions 5$

1005/2669 is the fraction of the total pounds of fruit and vegetables do the pounds of fruits represent

Go Math! Grade 6 Exercise 4.4: Operations with Fractions Solutions Page 23 Problem 6 Answer

Given the values of leftover apple and pumpkin pies.

To do: if tom ate 1/2 of leftovers, find how much pie in all did he eat.

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 4 Operations with Fractions 6$

Tom ate 3×1/12 pies.

Page 23 Problem 7 Answer

Given about the inches of angelfish.

To do: find how much has the angelfish grown in feet.

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 4 Operations with Fractions 7$

Angelfish has grown 5/72 feet.

Go Math Grade 6 Exercise 4.4 Operations With Fractions Answers

Go Math! Grade 6 Exercise 4.4: Operations with Fractions Solutions Page 24 Exercise 1 Answer

Given the amount of cheese, that deli was ordered.

To do: Find how much cheese was left for Wednesday.

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 4 Operations with Fractions e1$

For Wednesday 1×1/8 wheels of cheese were left.

Solutions For Go Math Grade 6 Exercise 4.4 Operations With Fractions

Go Math Answer Key

Go Math! Grade 6 Exercise 3.3: Rational Numbers Solutions

November 18, 2024August 17, 2023 by Alekhya

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 3 Rational Numbers

Page 15 Problem 1 Answer

Given:- 3/8

To Find:- To write each fraction as a decimal and to round to the nearest hundredth if necessary The given fraction is 3/8.

It can be converted into a decimal by dividing the numerator 3 by the denominator 8 using the long division method. The decimal value of 3/8 would be 0.375

The decimal value of the fraction 3/8 is0.375

Page 15 Problem 2 Answer

Given:- 7/5

To Find:- To write each fraction as a decimal and to round to the nearest hundredth if necessary The given fraction is 7/5.

Here, we can notice that the denominator is 5 , so we can easily convert the denominator in terms of 10 by multiplying both the numerator and denominator by 2.

Then, the fraction 7/5 would become 14/10. Therefore, the decimal value of the fraction 7/5 would be1.4

The decimal value of the fraction 7/5 is 1.4

Go Math! Grade 6 Exercise 3.3: Rational Numbers Solutions Page 15 Problem 3 Answer

Given:- 21/7

To Find:- To write each fraction as a decimal and to round to the nearest hundredth if necessary The given fraction is 21/7.

Here, we can notice that the denominator is7 which cannot be easily converted in terms of 10,100,1000….

So, we divide the numerator by the denominator to get 3.0 or 3 as the numerator 21 is fully divisible by 7.

The fraction 21/7 can be written as 3.0 or 3

Go Math Grade 6 Exercise 3.3 Rational Numbers Solutions

$Go Math! Practice Fluency Workbook Grade 6 Chapter 3 Rational Numbers Exercise 3.3 Answer Key$

Page 15 Problem 4 Answer

Given:- 5/3

To Find:- To write each fraction as a decimal and to round to the nearest hundredth if necessary The given fraction is 5/3.

Here, we can notice that the denominator is 3 which cannot be easily converted in terms of 10,100,1000….

So, we divide the numerator by the denominator to get 1.666which can be rounded to the nearest hundredth as 1.67.

The fraction 5/3 can be written in decimal form as 1.67

Page 15 Problem 5 Answer

Given:- 0.55

To Find:- To write each decimal as a fraction or mixed fraction in simplest form The given decimal is 0.55.

Since there are two digits after the decimal point, we can write 0.55 as 55/100.

Now, we can reduce the 55/100 in its simplest form by dividing both the numerator and denominator by 5 to get 11/20.

The decimal 0.55 can be written in the fractional form 11/20.

Go Math! Grade 6 Exercise 3.3: Rational Numbers Solutions Page 15 Problem 6 Answer

Given:- 10.6

To Find:- To write each decimal as a fraction or mixed fraction in simplest form The given decimal is 10.6.

Since there is one digit after the decimal point, we can write 10.6 as 106/10

This can be further simplified by dividing the numerator and denominator by 2 to get 53/5.

The decimal 10.6 can be written in the fractional form as 53/5.

Page 15 Problem 7 Answer

Given:- −7.08

To Find:- To write each decimal as a fraction or mixed fraction in simplest form The given decimal is 7.08.

Since there are two digits after the decimal point, we can write 7.08 as 708/100

This can be further reduced by dividing the numerator and denominator by 4 to get 177/25.

The decimal−7.08can be written in the fractional form−177/25

Page 15 Problem 8 Answer

Given:- 0.5,0.05,5/8

To Find:- To write the numbers in order from least to greatest

To write the given numbers in order from the least to the greatest, we should first change all the numbers into similar terms such as all the numbers into fractions or all the numbers into decimals.

Here, we can convert 5/8 into a decimal by dividing the numerator by the denominator to get 0.625.

So, the numbers would be 0.5,0.05,0.625.

So, the order from the least to the greatest will be 0.05,0.5,0.625.

Rewriting 0.625 as a fraction and then writing the numbers in order from the least to the greatest will be 0.05,0.5,5/8.

The order of the numbers from the least to the greatest will be0.05,0.5,5/8.

Go Math Grade 6 Exercise 3.3 Rational Numbers Answers

Page 15 Problem 9 Answer

Given:- 1.3,1×1/3 ,1.34

To Find:- To write the numbers in order from least to greatest

To write the given numbers in order from the least to the greatest, we should first change all the numbers into similar terms such as all the numbers into fractions or all the numbers into decimals.

Here, we can rewrite 1×1/3 as 4/3 which can be converted into a decimal by dividing the numerator by the denominator to get 1.33.

So, the numbers would be 1.3,1.33,1.34 in order from the least to the greatest.

The order of the numbers after rewriting 1.33 as a mixed fraction would be 1.3,1×1/3,1.34.

The order of the numbers from the least to the greatest would be 1.3,1×1/3,1.34.

Go Math! Grade 6 Exercise 3.3: Rational Numbers Solutions Page 15 Problem 10 Answer

Given:- 2.07,2×7/10 10 ,2.67,−2.67

To Find:- To write the numbers in order from least to greatest

To write the given numbers in order from the least to the greatest, we should first change all the numbers into similar terms such as all the numbers into fractions or all the numbers into decimals.

Here, we can rewrite 2×7/10 as 2×7/10 which can be converted to decimal as 2.7.

So, the numbers would be2.07,2.7,2.67,−2.67.

Arranging them in the order from the least to the greatest will be −2.67,2.07,2.67,2.7.

Rearranging the numbers in the order from the least to the greatest after rewriting 2.7 as 2×7/10 will be −2.67,2.07,2.67,2×7/10.

The order of the numbers from the least to the greatest would be −2.67,2.07,2.67,2×7/10.

Page 15 Problem 11 Answer

Given:- Out of 45 times at bat, Raul got 19 hits.
To Find:- Raul’s batting average as a decimal It is given that out of 45 times while batting, Raul got 19 hits. So, the average can be represented as 19/45.

Converting it to a decimal, the numerator should be divided by the denominator to get0.4222.

Therefore, Raul’s batting average is0.422.

Raul’s batting average in decimal is0.422

Page 15 Problem 12 Answer

Given:- Karen’s batting average was 0.444. She was at bat 45 times

To Find:- To determine how many hits did Karen get

We know that the batting average is defined as the ratio of the number of hits to the number of times at the bat.

Substituting the given values in the formula of batting average, we get,

⇒ $\text { Batting average }=\frac{\text { number of hits }}{\text { number of times at bat }}$

⇒ $0.444=\frac{\text { number of hits }}{45}$

Multiplying both sides of the equation by 45, we get, the number of hits=0.444×45 =19.98 which can be rounded to the nearest whole number as20

The number of hits Karen got was 20.

Page 15 Problem 13 Answer

Given:- To have batting averages over 0.500 how many hits in 45 times at bat would Raul and Karen need?

To Find:- To determine the number of hits Raul and Karen would need to have batting averages over 0.500

We know that the batting average is defined as the ratio of the number of hits to the number of times at the bat. Substituting the given values in the formula of batting average, we get,

⇒ $Batting average= \frac{\text { number of hits }}{\text { number of times at bat }}$

⇒ $ 0.500=\frac{\text { number of hits }}{45}$

Multiplying both sides of the equation by 45, we get,

the number of hits=0.500×45 =22.5 which can be rounded to the nearest whole number23

The number of hits Raul and Karen would need to have a batting average above 0.500 would be 23.

Solutions For Go Math Grade 6 Exercise 3.3 Rational Numbers

Go Math! Grade 6 Exercise 3.3: Rational Numbers Solutions Page 15 Problem 14 Answer

we have to Solve the given question It is given that a car travels 65 miles per hour, but then travels 3/5 of this speed when going through construction. First, we need to write this fraction as a decimal.

When converting a fraction to a decimal: If the denominator is a factor of 10,100,1000,…, multiply the numerator and denominator by the same number to get a denominator that is a power of 10.

Then write the equivalent fraction as a decimal. If the denominator is not a factor of 10,100,1000,…, then divide the numerator by the denominator to find the decimal.

For3/5, the denominator of 5 is a factor of 10 so we can write an equivalent fraction to find the decimal:

⇒ $\frac{3}{5}=\frac{3 \times 2}{5 \times 2}=\frac{6}{10}$

⇒ $\frac{6}{10} in words is ” six tenths”.$

Since tenths mean one decimal place, then 6/10 =0.6

To find the speed, we can either multiply 65 and 3/5 or multiply 65 and 0.6. It is easier to multiply 65 and 3/5 since 5 is a factor of 65.

From the above step, we get a fraction as a decimal is 0.6 and 3/5 of the given speed of 39 miles per hour.

⇒ Multiplying then gives:$65 \times \frac{3}{5}=65 \times \frac{1}{5} \times 3=\frac{65}{5} \times 3=13 \times 3=39 miles per hour$

Page 15 Problem 15 Answer

Given: A city’s sales tax is 0.07. Write this decimal as a fraction and tell how many cents of tax are on each dollar. 0.07 in words is ‘ ‘seven hundredths” so we can write it as a fraction with 7 as the numerator and 100 as the denominator.

We then get 0.07=7/100.

Since there are 100 cents in a dollar, then the fraction means there are 7 cents of tax in each dollar. the tax is equal to 7 cents

Page 15 Problem 16 Answer

It is given that Norm has 373 sheets of paper left in a ream and each ream of paper initially has 500 sheets of paper. The portion of a ream that Norm has written as a fraction is then 373/500.

To find; A ream of paper contains 500 sheets of paper. Norm has 373 sheets of paper left from a dream. Express the portion of a ream Norm has as a fraction and as a decimal. _______________
When converting a fraction to a decimal:- If the denominator is a factor of 10,100,1000,…, multiply the numerator and denominator by the same number to get a denominator that is a power of 10. Then write the equivalent fraction as a decimal.

If the denominator is not a factor of 10,100,1000,…, then divide the numerator by the denominator to find the decimal.

⇒ $For \frac{373}{500},$ the denominator of 500 is a factor of 1000 so we can write an equivalent fraction to find the decimal:

⇒ $\frac{373}{500}=\frac{373 \times 2}{500 \times 2}=\frac{746}{1000}$

⇒ $\frac{746}{1000}$ in words is “seven hundred forty-six thousandths”.

Since thousandths mean three decimal places, then$ \frac{746}{1000}=0.746.$

So, the portion written as a decimal number is 0.746

Go Math Grade 6 Rational Numbers Exercise 3.3 Key

Go Math! Grade 6 Exercise 3.3: Rational Numbers Solutions Page 16 Exercise 1 Answer

we have to write each decimal as a fraction or mixed number.0.61 in words is ” sixty-one hundredths” so we can write it as a fraction with 61 as the numerator and 100 as the denominator.

We then get 0.61=61/100

From the above step, we will get the answer 61/100

Page 16 Exercise 2 Answer

we have to write each decimal as a fraction or mixed number.3.43 in words is ‘ three and forty-three hundredths” so we can write it as a mixed number with 43 as the numerator and 100 as the denominator.

We then get

⇒ $3.43=3 \frac{43}{100}$

From the above step, we will get the answer 3×43/100

Page 16 Exercise 3 Answer

we have to write each decimal as a fraction or mixed number. 0.009 in words is “nine thousandths” so we can write it as a fraction with 9 as the numerator and 1000 as the denominator.

We then get$0.009=\frac{9}{1000}$

From the above step, we will get the answer 9/1000

Page 16 Exercise 4 Answer

we have to write each decimal as a fraction or mixed number.4.7 in words is ‘ ‘four and seven-tenths” so we can write it as a mixed number with 7 as the numerator and 10 as the denominator.

We then get $4.7=4 \frac{7}{10}$

From the above step, we will get the answer4x7/10

Page 16 Exercise 5 Answer

we have to write each decimal as a fraction or mixed number.1.5 in words is ” one and five tenths” so we can write it as a mixed number with 5 as the numerator and 10 as the denominator.

We then get $1.5=1 \frac{5}{10}$

Since 5 and 10 have a GCF of 5, we can reduce the fraction to:

⇒ $1 \frac{5}{10}=1 \frac{5 \div 5}{10 \div 5}=11 \frac{1}{2}$

From the above step, we will get the answer11x1/2

Go Math! Grade 6 Exercise 3.3: Rational Numbers Solutions Page 16 Exercise 6 Answer

we have to write each decimal as a fraction or mixed number. 0.13 in words is ” thirteen hundredths” so we can write it as a fraction with 13 as the numerator and 100 as the denominator.

We then get $0.13=\frac{13}{100}$

From the above step, we will get the answer13/100

Detailed Solutions For Go Math Grade 6 Exercise 3.3 Rational Numbers

Page 16 Exercise 7 Answer

we have to write each decimal as a fraction or mixed number 5.0002 in words is ” five and two thousandths” so we can write it as a mixed number with 2 as the numerator and 1000 as the denominator.

We then get $5.002=5 \frac{2}{1000} .$

Since 5 and 10 have a GCF of 2, we can reduce the fraction to:

⇒ $5 \frac{2}{1000}=5 \frac{2 \div 2}{1000 \div 2}=5 \frac{1}{500}$

From the above step, we will get the answer5x1/500

Page 16 Exercise 8 Answer

we have to write each decimal as a fraction or mixed number.

0.021 in words is ‘ ‘twenty-one thousandths” so we can write it as a fraction with 21 as the numerator and 1000 as the denominator.

We then get $0.021=\frac{21}{1000}$

From the above step, we will get the answer 21/1000

Practice Problems For Go Math Grade 6 Exercise 3.3 Rational Numbers

Go Math Answer Key

Go Math! Grade 6 Exercise 1.1 Solutions

November 18, 2024August 10, 2023 by Alekhya

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 1 Integers Exercise 1.1 Answer Key

Page 1 Problem 1 Answer

Here given that depositing of $85 in a bank account.
Now we have to represent this condition with positive or negative number.

So here given that deposit of money in bank ,we know that if you deposit some money in bank then our saving amount in bank account increasing so we can represent it by positive number.

Depositing of $ 85 in a bank account =+85

Page 1 Problem 2, Answer

Here given that Riding an elevator down 3 floor.
Now we have to represent this situation with Positive or negative number.

If we take riding of an elevator to up as positive then obviously the riding of an elevator down is negative.

So Riding an elevator down 3 floor is −3

Page 1 Problem 3, Answer

Here given that the foundation of a house sinking 5 inches.
Now we have to present this situation by positive or negative numbers.

The foundation of a house sinking means a decrease of the height of the foundation.

So we can represent the foundation of a house sinking 5 inches =−5

The foundation of a house sinking 5 inches =−5

$Go Math! Practice Fluency Workbook Grade 6 Chapter 1 Integers Exercise 1.1 Answer Key$

Go Math Practice Fluency Workbook Grade 6 Chapter 1 Integers Exercise 1.1 Answer Key

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 1 Integers Exercise 1.1 Answer Key Page 1 Problem 4, Answer

Here given that the temperature of 98° above zero —–
Now we have to represent it by positive or negative number

The temperature above zero degree is assumed as positive so

The temperature of 98∘ above zero =+98°

So the temperature of 98° above zero is +98°

Page 1 Problem 5, Answer

Here given the number line

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1: Integers problem 5

We have to graph the −2 and its opposite number in number line.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1: Integersproblem 5 answer

The graph for −2 and its opposite number +2 is

Problem 6 Page 1 Answer

Here given the number +3

Now we have to write the opposite number of it and have to graph both.

So the opposite number of +3 is −3 And the graph is

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1Integers problem 6$

Page 1 Problem 7, Answer

Here given that the number −5

Now we have to write the opposite number of the given number and have to draw the graph for both of them.

So the opposite number of −5 is +5 And the graph is

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 7$

Go Math Grade 6 Practice Fluency Workbook Exercise 1.1 Integers solutions

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 1 Integers Exercise 1.1 Answer Key Page 1 Problem 8, Answer

Here given that the number +1

Now we have to write the opposite number of it and have to draw the graph.

So the opposite number of +1 is −1 And the graph is

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 8$
Page 1 Problem 9, Answer

Here given that the average temperature in Fairbanks, Alaska, in February is 4°F below zero.
Now we have to write this temperature as an integer.

We know that the temperature below zero degree is represented by negative number.

So the average temperature of Fairbanks, Alaska, in February is 4°F below zero can be represented as −4∘F

So the temperature can be represented as −4° F.

Page 1 Problem 10, Answer

Here given that the average temperature in Fairbanks, Alaska, in November is 2° F above zero.
Now we have to write this temperature as an integer.

So the temperature above zero degree can be represented by positive number.

So the average temperature in Fairbanks, Alaska, in November is 2°F above zero is +2°F

The average temperature in Fairbanks, Alaska, in November is 2° F above zero is +2° F

Page 1 Problem 11, Answer

Here given that the highest point in the state of Louisiana is Driskill Mountain. It rises 535 feet above sea level.
Now we have to represent this elevation as an integer.

We know that the elevation of a mountain is represented by positive number

So the elevation of Driskill Mountain is +535feet

The elevation of Driskill Mountain is +535feet

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 1 Integers Exercise 1.1 Answer Key Page 1 Problem 12, Answer

Here given that the lowest point in the state of Louisiana is New Orleans. The city’s elevation is 8 feet below sea level.
Now we have to represent the elevation of New Orleans as an integer.

If we take the elevation above sea as positive number than the city’s elevation below the sea level is represented by negative integer.

Now the elevation of New Orleans below the sea level is −8feet

The elevation of New Orleans is −8feet

Page 1 Problem 13, Answer

Here given that Death Valley, California, has the lowest elevation in the United States. Its elevation is 282 feet below sea level. Mount McKinley, Alaska, has the highest elevation in the United States. Its elevation is 20,320 feet above sea level

Now we have to write the elevation of two locations United States as an integer.

If we take the elevation above sea level as positive then the elevation below the sea level must be negative.

The elevation of Mount McKinley is −282 feet.

The elevation of is Alaska is +20320 feet

So The height of the two location of united states are as follows

The elevation of Mount McKinley is −282feet and The elevation of Alaska is +20320feet

Page 1 Problem 14, Answer

No there are no integers between zero and one . Because there are no whole number between them

No there are no integers between zero and one . Because there are no whole number between them.

Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1: Integers

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 1 Integers Exercise 1.1 Answer Key Page 2 Exercise 1 Answer

Here given that an increase of 3 points
Now we have to represent this by positive or negative numbers.

So an increase of 3 points is represented by +3

Page 2 Exercise 2, Answer

Here given that spending $10

Now we have to express this by positive or negative numbers.

So the spending of $10 can be represented by $(−10)

Page 2 Exercise 3, Answer

Here given that earning of $25

Now we have to express it by a positive or negative number.

So the earnings of $25 can be represented by $+25

Solutions for Go Math Grade 6 Practice Fluency Workbook Chapter 1 Exercise 1.1

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 1 Integers Exercise 1.1 Answer Key Page 2 Exercise 4, Answer

Here given that a loss of 5 yards.

Now we have to represent it by positive or negative number.

So the loss of 5 yards is −5yards

Page 2 Exercise 5, Answer:

Here given the integer −1 ,now we have to write it’s opposite number and have to place them in number line

So the opposite number of −1 is +1

And the number line is

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 5 page 2$

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 1 Integers Exercise 1.1 Answer Key Page 2 Exercise 6, Answer

Here given the number 9

Now we have to write it’s opposite number and have to present them in the number line

So the opposite number of 9 is −9

And the number line is

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 6 page 2$
Page 2 Exercise 7, Answer

Given: 6 Write each integer and its opposite. Then graph them on the number line.

The graph of the given integer 6 and its opposite is

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 7 page 2$

Go Math Grade 6 Integers Exercise 1.3 Key

Page 2 Exercise 8, Answer

Given: -5. Write each integer and its opposite. Then graph them on the number line.

The graph of the given integer -5 and its opposite 5 is

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 8 page 2$

Go Math Grade 6 Integers Exercise 1.1 key from Practice Fluency Workbook

Go Math Answer Key

Go Math! Grade 6 Chapter 1 Integers Exercise 1.2 Answers

November 18, 2024August 9, 2023 by Alekhya

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 1 Integers Exercise 1.2 Answer Key

Page 3 Problem 1, Answer

Given: 10? -2 Use the number line to compare each pair of integers. Write < or >.ExplanationThe number line is

From the number line, 10 is to the right of -2. So, 10>−2.

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 1 page 3$

Thus, 10>−2.

Page 3 Problem 2, Answer

Given: 0? 3.

Use the number line to compare each pair of integers. Write < or >.ExplanationThe number line is

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 2$

From the number line, 0 is to the left of 3.So, 0<3.

Page 3 Problem 3, Answer

Given: −5?0.

Use the number line to compare each pair of integers. Write < or >.ExplanationThe number line is
From the number line, -5 is to the left of 0.So, −5<0.

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 3$

Thus, −5<0.

Go Math Grade 6 Chapter 1 Integers Exercise 1.2 Answers

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 1 Integers Exercise 1.2 Answer Key Page 3 Problem 4, Answer

Given: −7?6.

$Go Math! Practice Fluency Workbook Grade 6 Chapter 1 Integers Exercise 1.2 Answer Key$

Use the number line to compare each pair of integers. Write < or >.ExplanationThe number line is

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 4$

From the number line, -7 is to the left of 6.So, −7<6

Thus, −7<6.

Page 3 Problem 5 , Answer

Given: −6?−9.

Use the number line to compare each pair of integers. Write < or >.ExplanationThe number line is

From the number line, -6 is to the right of -9. So, −6>−9.

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 5$

Thus, −6>−9.

Page 3 Problem 6, Answer

Given: −8?−10.

Use the number line to compare each pair of integers. Write < or >.ExplanationThe number line is

From the number line, -8 is to the right of -10. So, −8>−10.

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 6$

Thus, −8>−10.

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 1 Integers Exercise 1.2 Answer Key Page 3 Problem 7, Answer

Given: 5,−2,6.

Order the integers in each set from least to greatest.Use the number line to order the integers.

List all the numbers in the order in which they appear from left to right. So, the integers in order from least to greatest are −2,5,6.

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 7$

Thus, the Order of Integers from least to greatest is −2,5,6.

Page 3 Problem 8, Answer

Given: 0,9,-3. Order the integers in each set from least to greatest.

List all the numbers in the order in which they appear from left to right. So, the integers in order from least to greatest are −3,0,9.

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 8$

Thus, the Order of Integers from least to greatest is −3,0,9.

Page 3 Problem 9, Answer

Given: -1,6,1.Order the integers in each set from least to greatest.Use the number line to order the integers.

List all the numbers in the order in which they appear from left to right. So, the integers in order from least to greatest are −1,1,6.

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 9$
Thus, the Order of Integers from least to greatest is −1,1,6.

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 1 Integers Exercise 1.2 Answer Key Page 3 Problem 10 , Answer

Given: -1,1,0.Order the integers in each set from greatest to least.

List all the numbers in the order in which they appear from right to left. So, the integers in order from greatest to least are 1,0,−1.

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 10$

Thus, the Order of Integers from greatest to least is 1,0,−1.

Page 3 Problem 11, Answer

Given: −12,2,1.

Order the integers in each set from greatest to least.

List all the numbers in the order in which they appear from right to left. So, the integers in order from greatest to least are 2,1,−12.

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 11$

Go Math Grade 6 Exercise 1.2 Integers Solutions

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 1 Integers Exercise 1.2 Answer Key Page 3 Problem 12, Answer

Given: −10,−12,−11.

Order the integers in each set from greatest to least.

List all the numbers in the order in which they appear from right to left. So, the integers in order from greatest to least are −10,−11,−12.

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 12$

Thus, the Order of Integers from greatest to least is −10,−11,−12.

Page 3 Problem 13, Answer

Given: 205,−20,−5,50.

Order the integers in each set from greatest to least.

List all the numbers in the order in which they appear from right to left.So, the integers in order from greatest to least are 205,50,−5,−20.

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 13$

Thus, the Order of Integers from greatest to least is 205,50,−5,−20.

Page 3 Problem, 14 Answer

Given: -78, -89, 78, 9.

Order the integers in each set from greatest to least.

List all the numbers in the order in which they appear from right to left. So, the integers in order from greatest to least are 78,9,−78,−89

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 14$
Thus, the Order of Integers from greatest to least is 78,9,−78,−89.

Page 3 Problem 15, Answer

Given: -55, -2, -60, 0.Order the integers in each set from greatest to least.

List all the numbers in the order in which they appear from right to left. So, the integers in order from greatest to least are 0,−2,−55,−60.

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 15$

Thus, the Order of Integers from greatest to least is 0,−2,−55,−60.

Page 3 Problem 16, Answer

Given: 28, 8, -8, 0.

Order the integers in each set from greatest to least.

List all the numbers in the order in which they appear from right to left. So, the integers in order from greatest to least are 28,8,0,−8.

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 16$

Thus, the Order of Integers from greatest to least is 28,8,0,−8

Page 3 Problem 17, Answer

Given: 37, -37, -38, 38.

Order the integers in each set from greatest to least.

List all the numbers in the order in which they appear from right to left.So, the integers in order from greatest to least are 38,37,−37,−38.

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 17$
Thus, the Order of Integers from greatest to least is 38,37,−37,−38.

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 1 Integers Exercise 1.2 Answer Key Page 3 Problem 18, Answer

Given: -111, -1, 1, 11.

Order the integers in each set from greatest to least.

List all the numbers in the order in which they appear from right to left. So, the integers in order from greatest to least are11,1,−1,−111.

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 18$

Thus, the Order of Integers from greatest to least is 11,1,−1,−111

Solutions For Go Math Grade 6 Chapter 1 Exercise 1.2 Integers

Page 3 Problem 19, Answer

Given: Four friends went scuba diving today. Ali dove 70 feet, Tim went down 50 feet, Carl dove 65 feet, and Brenda reached 48 feet below sea level. Write the 4 friends’ names in order from the person whose depth was closest to the surface to the person whose depth was the farthest from the surface.

Here Ali dove 70 feet, Tim went down 50 feet, Carl dove 65 feet, and Brenda reached 48 feet below sea level. We know that 48<50<65<70.

So, from the given, the order from the person whose depth was closest to the surface to the person whose depth was the farthest from the surface is
Brenda, Tim, Carl, Ali.

The order from the person whose depth was closest to the surface to the person whose depth was the farthest from the surface is Brenda, Tim, Carl, Ali.

Page 3 Problem 20, Answer

Given: The temperatures on Monday and Tuesday were opposites.
The temperature on Wednesday was neither positive nor negative.
The temperature dropped below zero on Monday. To find: Write the 3 days in order from the highest to the lowest temperature.

We know that Monday and Tuesday were opposites and Monday dropped below zero. so, Tuesday had the highest temperature. Wednesday wasn’t positive or negative so it would be exactly zero. Monday dropped below zero so it had the lowest temperature. So, the order from the highest to the lowest temperature is Tuesday, Wednesday, Monday.

The 3 days in order from the highest to the lowest temperature are Tuesday, Wednesday, Monday.

Page 4 Exercise 1 Answer

Given: 1? −4. Use the number line to compare each pair of integers. Write < or >.

The number line is From the number line, 1 is to the right of -4. So, 1>−4.

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers exercise 1$

Thus, 1>−4.

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 1 Integers Exercise 1.2 Answer Key Page 4 Exercise 2 Answer

Given: −5? −2.

Use the number line to compare each pair of integers. Write < or >.ExplanationThe number line is
From the number line, -5 is to the left of -2. So, −5<−2.

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers exercise 2$

Thus, −5<−2.

Page 4 Exercise 3 Answer

Given: −3?2.

Use the number line to compare each pair of integers. Write < or >.ExplanationThe number line is
From the number line, -3 is to the left of 2. So, −3<2.

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers exercise 3$

Thus, −3<2.

Page 4 Exercise 4 Answer

Given: −1? −4.

Use the number line to compare each pair of integers. Write < or >.ExplanationThe number line is

From the number line, -1 is to the right of -4.So, −1>−4.

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers exercise 4$
Thus, −1>−4.

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 1 Integers Exercise 1.2 Answer Key Page 4 Exercise 5 Answer

Given: 5? 0.

Use the number line to compare each pair of integers. Write < or >.ExplanationThe number line is
From the number line, 5 is to the right of 0. So, 5>0

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers exercise 5$

Thus, 5>0.

Page 4 Exercise 6 Answer

Given: −2?3.

Use the number line to compare each pair of integers. Write < or >.Explanation The number line is From the number line, -2 is to the left of 3. So, −2<3

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers exercise 6$

Thus, −2<3

Page 4 Exercise 7 Answer

Given: −2,−5,1.

Order the integers from least to greatest.

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers exercise 7$

List all the numbers in the order in which they appear from left to right. So, the integers in order from least to greatest are −5,−2,−1.

Thus, the Order of Integers from least to greatest is −5,−2,−1.

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 1 Integers Exercise 1.2 Answer Key Page 4 Exercise 8 Answer

Given:0,−5,5
Order the integers from least to greatest.

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers exercise 8$

List all the numbers in the order in which they appear from left to right.So, the integers in order from least to greatest are−5,0,5.

Thus, the Order of Integers from least to greatest is−5,0,5.

Page 4 Exercise 9 Answer

Given:−5,2,−3
Order the integers from least to greatest. The given integers are−5,2,−3and the number line is

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers exercise 9$

List all the numbers in the order in which they appear from left to right. So, the integers in order from least to greatest are -5,−3,2.

Hence the integers in order from least to greatest are−5,−3,2.

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 1 Integers Exercise 1.2 Answer Key Page 4 Exercise 10 Answer

Given:3,−1,−4
Order the integers from least to greatest. The given integers are3,−1,−4 and the number line is

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers exercise 10$
List all the numbers in the order in which they appear from left to right.

So, the integers in order from least to greatest are−4,−1,3.Hence the integers in order from least to greatest are−4,−1,3

Page 4 Exercise 11 Answer

Given:3,−5,0
Order the integers from least to greatest.

The given integers are3,−5,0, and the number line is

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers exercise 11$

List all the numbers in the order in which they appear from left to right.
So, the integers in order from least to greatest are−5,0,3

Hence the integers in order from least to greatest are−5,0,3

Page 4 Exercise 12 Answer

Given:−2,−4,1.

Order the integers from least to greatest.

The given integers are−2,−4,1 and the number line is

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers exercise 12$

list all the numbers in the order in which they appear from left to right. So, the integers in order from least to greatest are−4,−2,1

Hence the integers in order from least to greatest are−4,−2,1.

Go Math Grade 6 Integers Exercise 1.2 Key

Go Math Answer Key

Go Math! Grade 6 Chapter 1 Integers Exercise 1.3 Answers

November 18, 2024August 8, 2023 by Alekhya

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 1 Integers

Page 5 Problem 1, Answer

Given the number 6.

We need to graph the number on the number line

The number6 on the number line is shown below: $Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 1 page 5$

Page 5 Problem 2, Answer

Given the number 3.

We need to graph the number on the number line.

The number 3 on the number line is shown below: $Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 2$

Page 5 Problem 3, Answer

Given the number−3.

We need to graph the number on the number line.

The number−3 on the number line is shown below:

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 3$

Go Math Grade 6 Chapter 1 Integers Exercise 1.3 Answers

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 1 Integers Page 5 Problem 4, Answer

$Go Math! Practice Fluency Workbook Grade 6 Chapter 1 Integers Exercise 1.3 Answer Key$

Given the number 5.

We need to graph the number on the number line.

The number on the number line is shown below:

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 4$
Page 5 Problem 5, Answer

Given the number∣−6∣ .

We need to use the number line to find each absolute value.

The absolute value of∣−6∣ is 6 as shown below:

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 5$

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 1 Integers Page 5 Problem 6, Answer

Given the number|3|.

We need to use the number line to find each absolute value.

The absolute value of|3|is 3 as shown below:

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 6$ Page 5 Problem 7, Answer

Given the number|8| .

We need to use the number line to find each absolute value.

The absolute value of|8| is 8 as shown below: $Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 7$

Page 5 Problem 8, Answer

Given the number|6|.

We need to use the number line to find each absolute value.

The absolute value of |6| is 6 as shown below: $Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 8$

Page 5 Problem 9, Answer

Given the number∣−3∣.

We need to use the number line to find each absolute value.

The absolute value of∣−3∣ is 3 as shown below: $Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 9$

Go Math Grade 6 Exercise 1.3 Integers Solutions

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 1 Integers Page 5 Problem 10, Answer

Given the number|5|.

We need to use the number line to find each absolute value.

The absolute value of|5| is5 as shown below: $Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 10$

Page 5 Problem 11, Answer

Given the numbers6 and −6.

We need to find each absolute value and tell what do we notice.

The absolute values of 6 and −6 is shown below: $Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 11$

Page 5 Problem 12, Answer

Given the numbers 6 and −6 or 3 and −3.

We need to find each absolute value and tell what do we notice.

The absolute values are: We call∣−3∣ and ∣3∣ ∣−6∣ and ∣6∣ as equal numbers. $Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1$

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 1 Integers Page 5 Problem 13, Answer

Given the table:

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers table 1$

Write a negative integer to show the amount spent on each purchase on monday.

Hence the negative integer to show the amount spent on each purchase on Monday is−20.

Page 5 Problem 14, Answer

Given the table:

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 14$

Write a negative integer to show the amount spent on each purchase on Tuesday.

Hence the negative integer to show the amount spent on each purchase on Tuesday is−6.

Page 5 Problem 15, Answer

Given the table:

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 15$

Write a negative integer to show the amount spent on each purchase on Friday.

Hence the negative integer to show the amount spent on each purchase on Friday is−8.

Solutions For Go Math Grade 6 Chapter 1 Exercise 1.3 Integers

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 1 Integers Page 5 Problem 16, Answer

Given the table:

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 16$

Find the absolute value of each transaction on Monday.

The absolute value of the transaction on Monday is20.

Page 5 Problem 17, Answer

The given table is,

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 17$

We have to find the absolute value of each transaction on Tuesday.

We will take the modulus of the transaction.

The absolute value of the transaction on Tuesday is 6

Page 5 Problem 18, Answer

The given table is,

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers problem 18$

We have to find the absolute value of each transaction on Wedneday.

We will take the modulus of the transaction.

The absolute value of a payment of Wednesday is15

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 1 Integers Page 5 Problem 19, Answer

The given table is,

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 19$

We have to tell on which day did Andrea spend the most on her card.

We will observe the table.

On observing the table carefully we observed that Andrea spent most on her card on Monday

Page 5 Problem 20, Answer

We have to show that, |3+10|=|3|+|10|.

We will find the value of L.H.S. and R.H.S.

We will check the values are the same or not.

Hence we showed |3+10|=|3|+|10|

Page 5 Problem 21, Answer

The given statement is How many different integers can have the same absolute value? ________ Give an example.

We have to fill in the blanks.

We will see the cases of the modulus.

The revised statement is, How many different integers can have the same absolute value? 2

Give an example. −3 and 3

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 1 Integers Page 6 Exercise 1, Answer

We have to match the absolute value of 15

The absolute value of 15 is ⇒∣15∣ ⇒15

The row C is the correct match.

Page 6 Exercise 2,Answer

We have to match the negative integer.

The negative integer is −15

The correct match is d

Page 6 Exercise 3, Answer

We have to match opposite of −7

The opposite of −7 is 7

The correct match is b

Go Math Grade 6 Integers Exercise 1.3 Key

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 1 Integers Page 6 Exercise 4,Answer

We have to match opposite of 7

The opposite of 7 is −7

The correct match is a

Page 6 Exercise5,Answer

We have to match ∣−15∣ The value of ∣−15∣ is 15

The correct match is c

Page 6 Exercise6,Answer

We have to find the value of ∣−3∣

We will use the concept of modulus.

The value of ∣−3∣ is 3

Page 6 Exercise7,Answer

We have to find the value of |5|

We will use the concept of modulus.

The value of |5| is 5

Page 6 Exercise 8, Answer

We have to find the value of ∣−7∣

We will use the concept of modulus.

The value of ∣−7∣ is 7

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 1 Integers Page 6 Exercise 9, Answer

We have to find the value of |6|

We will use the concept of modulus.

The value of |6| is 6

Page 6 Exercise 10,Answer

We have to find the value of |0|

We will use the concept of modulus.

The value of |0| is 0

Page 6 Exercise 11,Answer

We have to find the value of ∣−2∣

We will use the concept of modulus.

The value of ∣−2∣ is 2

Page 6 Exercise 12,Answer

We have to find the value of∣−10∣

We will use the concept of modulus.

The value of ∣−10∣ is 10

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 1 Integers Page 6 Exercise 13,Answer

We have to find the value of |−3/4|

We will use the concept of modulus.

The value of |−3/4| is 3/4

Page 6 Exercise 14,Answer

We have to find the value of|0.8|

We will use the concept of modulus.

The value of |0.8| is 0.8

Page 6 Exercise15, Answer

The given conditions are, Abby has been absent from class. How would I explain to her what absolute value is? Use the number line and an example in our explanation.

The absolute value of any number is its distance from 0 on the number line . Since the distance is always positive or 0 ,absolute value is positive or 0.

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 1 Integers exercise 15$

Detailed Answers For Go Math Grade 6 Chapter 1 Exercise 1.3

Go Math Answer Key

Go Math! Grade 6 Chapter 2 Factors and Multiples Exercise 2.1 Answers

November 18, 2024August 7, 2023 by Alekhya

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 2 Factors And Multiples

Page 7 Problem 1 Answer

A number is given to us 5.

We have to find all the factors of the given number.

The factors of the given number are: => 5, 1

Page 7 Problem 2 Answer

A number is given to us 15.

We have to find all the factors of the given number.

The factors of the given number are: 15,5,3,1

Page 7 Problem 3 Answer

A number is given to us 60.

We have to find all the factors of the given number.

The factors of the given number are: 60,30,20,15,12,10,6,5,4,3,2,1

$Go Math! Practice Fluency Workbook Grade 6 Chapter 2 Factors and Multiples Exercise 2.1 Answer Key$

Page 7 Problem 4 Answer

A number is given to us 6.

We have to find all the factors of the given number.

The factors of the given number are: 6,3,2,1

Go Math Grade 6 Chapter 2 Factors And Multiples Exercise 2.1 Answers

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 2 Factors And Multiples Page 7 Problem 5 Answer

A number is given to us 12.

We have to find all the factors of the given number.

The factors of the given number are: 12,6,4,3,2,1

Page 7 Problem 6 Answer

A number is given to us 36.

We have to find all the factors of the given number.

The factors of the given number are: 36,18,12,9,6,4,3,2,1

Page 7 Problem 7 Answer

Numbers are given to us 6,9.

We have to find the GCD of the given numbers.

The GCD of the given numbers is:3

Page 7 Problem 8 Answer

Numbers are given to us 4,8.

We have to find the GCF of the given numbers.

The factors of the given numbers are: 4=2×2×1, 8=2×2×2×1

So the GCF of the given numbers is: 2×2×1=4

The GCF of the given numbers is: 4

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 2 Factors And Multiples Page 7 Problem 9 Answer

Numbers are given to us 8,12.

We have to find the GCF of the given numbers.

The factors of the given numbers are:

8=2×2×2×1
12=3×2×2×1

So the GCF of the given numbers is:

2×2×1 ⇒4

The GCF of the given numbers is: 4

Page 7 Problem 10 Answer

Numbers are given to us 6,15.

We have to find the GCF of the given numbers.

The factors of the given numbers are:

6=3×2×1
15=5×3×1

So the GCF of the given numbers is: 3

The GCF of the given numbers is: 3

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 2 Factors And Multiples Page 7 Problem 11 Answer

Numbers are given to us10,15.

We have to find the GCF of the given numbers.

The factors of the given numbers are:

10=5×2×1
15=5×3×1

So the GCF of the given numbers is: 5×1=5

The GCF of the given numbers is: 5

Page 7 Problem 12 Answer

Numbers are given to us 9,12.

We have to find the GCF of the given numbers.

The factors of the given numbers are:

9=3×3×1
12=3×2×2×1

So the GCF of the given numbers is 3×1=3

The GCF of the given numbers is: 3

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 2 Factors And Multiples Page 7 Problem 13 Answer

Sum of numbers if given to us44+40.

We have to rewrite the sum as the product of GCF and a new sum.

The factors of the numbers are:

44=11×2×2×1
40=5×2×2×2×1

So the GCF is: 2×2×1=4

Now rewriting the sum: 44+40=4×(11+10)

The given sum can also be written as: 44+40=4×(11+10)

Page 7 Problem 14 Answer

Sum of numbers if given to us15+81 .

We have to rewrite the sum as the product of GCF and a new sum.

The factors of the numbers are:

15=5×3×1
81=3×3×3×3×1

So the GCF is: 3×1=3

Now rewriting the sum: 15+81=3×(5+27)

The given sum can also be written as: 15+81=3×(5+27)

Go Math Grade 6 Exercise 2.1 Factors And Multiples Solutions

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 2 Factors And Multiples Page 7 Problem 15 Answer

Sum of numbers if given to us13,52.

We have to rewrite the sum as the product of GCF and a new sum.

The factors of the numbers are:

13=13×1
52=13×2×2×1

So the GCF is:
13×1=13

Now rewriting the sum: 13+52=13×(1+4)

The given sum can also be written as: 13+52=13×(1+4)

Page 7 Problem 16 Answer

Sum of numbers if given to us is 64,28.

We have to rewrite the sum as the product of GCF and a new sum.

The factors of the numbers are:

64=2×2×2×2×2×2×1
28=7×2×2×1

So the GCF is: 2×2×1=4

Now rewriting the sum: 64+28=4×(16+7)

The given sum can also be written as: 64+28=4×(16+7)

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 2 Factors And Multiples Page 7 Problem 17 Answer

The data about beads in necklaces are given to us. We have to find now many necklaces that can be made from the given number of beads and how many beads are there in every necklace.

The GCF of the number of beads is:

24=3×2×2×2×1
30=5×3×2×1
⇒3×2×1=6

So the number of necklaces is six.

Now the number of individual beads are:

24/6 =4 Jade beads
30/6 =5 Teak beads

The total number of necklaces that can be made from the beads is 6.

There will be4 Jade beads and 5 teak beads in each of the necklace.

Page 7 Problem 18 Answer

The data about a marine-life store is given to us.We have to find the greatest number of tanks that can be set up, that contain equal number of fishes.

The GCF of the number of fishes is:

12=3×2×2×1
24=3×2×2×2×1
30=5×3×2×1
⇒3×2×1=6

A total of6 tanks can be set up in the marine-life store.

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 2 Factors And Multiples Page 8 Exercise 1 Answer

Numbers are given to us 32,48.

We have to find the GCF of the given numbers.

The factors of the given numbers are:

32=2×2×2×2×2×1
48=3×2×2×2×2×1

So the GCF of the given numbers is:

2×2×2×2×1=16

The GCF of the given numbers is: 16

Page 8 Exercise 2 Answer

Numbers are given to us 18,36.

We have to find the GCF of the given numbers.

The factors of the given numbers are:

18=3×3×2×1
36=3×3×2×2×1

So the GCF of the given numbers is:

3×3×2×1=18

The GCF of the given numbers is:18

Solutions For Go Math Grade 6 Chapter 2 Exercise 2.1 Factors And Multiples

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 2 Factors And Multiples Page 8 Exercise 3 Answer

Numbers are given to us 28,56,84.

We have to find the GCF of the given numbers.

The factors of the given numbers are:

28=7×2×2×1
56=7×2×2×2×1
84=7×3×2×2×1

So the GCF of the given numbers is: 7×2×2×1=28

The GCF of the given numbers is: 28

Page 8 Exercise 4 Answer

Numbers are given to us 30,45,75.

We have to find the GCF of the given numbers.

The factors of the given numbers are:

30=5×3×2×1
45=5×3×3×1
75=5×5×3×1

So the GCF of the given numbers is: 5×3×1=15

The GCF of the given numbers is: 15

Page 8 Exercise 5 Answer
Sum of numbers if given to us is 9,15.

We have to rewrite the sum as the product of GCF and a new sum.

The factors of the numbers are:

9=3×3×1
15=5×3×1

So the GCF is 3×1=3

Now rewriting the sum: 9+15=3×(3+5)

The given sum can also be written as: 9+15=3×(3+5)

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 2 Factors And Multiples Page 8 Exercise 6 Answer

Sum of numbers if given to us100+350.

We have to rewrite the sum as the product of GCF and a new sum.

The factors of the numbers are:

100=5×5×2×2×1
350=7×5×5×2×1

So the GCF is: 5×5×2×1=50

Now rewriting the sum: 100+350=50×(2+7)

The given sum can also be written as: 100+350=50×(2+7)

Page 8 Exercise 7 Answer

Sum of numbers if given to us12+18+21.

We have to rewrite the sum as the product of GCF and a new sum.

The factors of the numbers are:

12=3×2×2×1
18=3×3×2×1
21=7×3×1

So the GCF is 3×1=3

Now rewriting the sum: 12+18+21=3×(4+6+7)

The given sum can also be written as: 12+18+21=3×(4+6+7)

Go Math Grade 6 Factors And Multiples Exercise 2.1 Key

Go Math Answer Key

Go Math! Practice Fluency Workbook Grade 6 Chapter 2 Factors and Multiples Exercise 2.2 Answer Key

November 18, 2024August 6, 2023 by Alekhya

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 2 Factors and Multiples

Page 9 Problem 1 Answer

Given :- number 3

Find to the question first three multiple of 3

Multiples of 3 are the product obtained when an integer is multiplied by 3

First three multiples are:- 3,6,9

First three multiples of 3 are 3,6,9.

Page 9 Problem 2 Answer

Given:- number is 7

Find to the question first three multiples of 7

Multiples of 7 are the products obtained when an integer is multiplied by 7

First three multiples are:- 7,14,21

First three multiple of 7 are 7,14,21

Go Math! Practice Fluency Workbook Grade 6 Chapter 2 Factors and Multiples Exercise 2.2 Page 9 Problem 3 Answer

Given:- number is 12

Find to the question first three multiples of 12

Multiples of 12 are the products obtained when an integer is multiplied by 12

We need to multiply 12 by 1,2,3

First three multiples are:- 12,24,36

First three multiples of 12 are 12,24,and 36.

$Go Math! Practice Fluency Workbook Grade 6 Chapter 2 Factors and Multiples Exercise 2.2 Answer Key$

Go Math Practice Fluency Workbook Grade 6 Chapter 2 Factors And Multiples Exercise 2.2 Answer Key

Page 9 Problem 4 Answer

Given:- number is 200

Find to the question first three multiples of 200

Multiples of 200 are the products of obtained when an integer is multiplied by 200

We nee to multiply 200 by 1,2,3

First three multiples are :- 200,400,600

First three multiples of 200 are 200,400,600.

Page 9 Problem 5 Answer

Given :- number is 2 & 3

Find to the question least common multiple of 2 & 3

Multiples of 2 are 2,4,6,8,… etc.

Multiples of 3 are 3,6,9,12… etc.

Here we can see the least multiple of 2 is 2 and 3 is 3

Least common multiple of 2 & 3 is 6

LCM of 2 &3 is =6

Least Common multiple of 2 & 3 is 6 and the LCM of 2 & 3 = 6

Go Math! Practice Fluency Workbook Grade 6 Chapter 2 Factors and Multiples Exercise 2.2 Page 9 Problem 6 Answer

Given :- number is 4 & 5

Find to the question least common multiple of 4 & 5

Multiples of 4 is 4,8,12,16,20….etc.

Multiples of 5 are 5,10,15,20… etc.

Here we can see the least common multiple of 4 & 5 is 20

LCM of 4 & 5= 20

Least common multiple of 4 & 5 is 20 LCM of 4 & 5=20

Page 9 Problem 7 Answer

Given :- number is 6 & 7

Find to the question Least common multiple of 6 & 7

Multiples of 6 are 6,12,18,24,30,36,42…etc.

Multiples of 7 are 7,14,21,28,35,42…etc.

Here we can see Least common multiple of 6 & 7 is 42.

Least common multiple of 6 & 7 = 42

Page 9 Problem 8 Answer

Given :- number is 2,3 & 4

Find to the question least common multiple of 2,3 & 4

Multiple of 2 are 2,4,6,8,10,12,14,…etc.

Multiple of 3 are 3,6,9,12,15,…etc.

Multiples of 4 are 4,8,12,16…etc.

Here we can see Least common factor of 2,3 & 4 is 12

Least common factor of 2,3 & 4 is 12 .

Go Math! Practice Fluency Workbook Grade 6 Chapter 2 Factors and Multiples Exercise 2.2 Page 9 Problem 9 Answer

Given :- number is 5,6 & 7

Find to the question the least common factor of 5,6 & 7

Multiples of 5 are 5,10,15,20,25,30,35…etc.

Multiples of 6 are 6,12,18,24,30,…etc

Multiples of 7 are 7,14,21,28,…etc.

Here we can see the least common factor of 5,6,&7 is 210

The least common factor of 5,6 & 7 is 210 .

Page 9 Problem 10 Answer

Given :- number is 8,9 & 10

Find to the question the least common factor of 8,9& 10

Multiples of 8 are 8,16,24,32…etc.

Multiples of 9 are 9,18,27,36…etc.

Multiples of 10 are 10,20,30,40…etc.

Here we can see the least common multiple of 8,9 & 10 is 360

The lease common multiple of 8,9 & 10 is 360 .

Page 9 Problem 11 Answer

Given :- 60 peoples are invited to a party .there are 24 cups in a package and 18 napkins in a package.

Find :- least number of packages of cups and napkins that can be bought if each party guest gets one cup and one napkin . Using multiple and factor concept and proceed .

Given,
Number of cups in a package=24

Number of napkins in a package =18

Let total number of package of cups and napkins respectively x & y

Equation of total cups bought =24×x

Equation of total napkin bought =18×y

Total people invited to the party =60

as party guest gets one cup and one napkin total cups and napkins required :- Total number of cups bought

24×x≥60

x≥60

24 x≥2.5

Number of package should be an integer therefor x=3

Total number of napkins bought

18×y≥60

y≥10

3 y≥3.3

Number of package should be an integer therefor y=3

Least number of packages of cups bought are 3 and napkins is 4

Go Math Grade 6 Practice Fluency Workbook Exercise 2.2 Factors And Multiples Solutions

Go Math! Practice Fluency Workbook Grade 6 Chapter 2 Factors and Multiples Exercise 2.2 Page 9 Problem 12 Answer

Given :- Given 45 members in science club .caps come in package of 3 and the shirts come in package of 5

Find :- what is the least number of packages of caps and shirts . Using multiple and factor concept and proceed .

Given , Caps come in package of 3 and the shirts come in package of 5

Total number of member in science club =45

Let total number of packages of caps and shirts x & y respectively .

Equation of caps ordering =3×x

Equation of shirts ordering =y×5

If club members get one cap and one shirts then required caps and shirts package :- Total number of caps ordered

3×x≥45

x=15

Total number of shirts ordered

5×y≥45

y=9

The least number of caps and shirts 15 & 9 respectively.

We need 15 packages of caps and 9 packages of shirts and that is least number of caps and shirts packages.

Page 9 Problem 13 Answer

Given :- Some hot dogs come in packages of 8 and hot dog buns packages of 7

Find :- why would a backer of hot dog buns package 7 hot dog buns to a package? We use here least common multiple concept .

Given , Hot dogs come in packages of 8 and hot dog buns packages of 7

So we need to find here the LCM of 8 & 7 LCM of 8,7 =56

We need 7 packages of hot dogs and 8 packages of buns.

Page 9 Problem 14 Answer

Find to the question how are the GCF and the LCM alike and different .

GCF is the greatest common factor is the greatest real number shared between two integers .what makes this number a factor is that it is a whole,real number that two integers share- that is , when broken down to their lowest multiples,the largest integer that is shared between the two numbers is their greatest common factor.

And the other hand,the lowest common multiple is the integer shared by two numbers that can be divided by both number.

The biggest difference between the GCF and LCM is that one is based upon what can divide evenly into two numbers,while the other depends on what number shared between two integers can be divided by the two integers (LCM).

Page 10 Exercise 1 Answer

Given:- number is 2,9

Find to the question the least common multiple of 2,9

Multiples of 2 are 2,4,6,8…etc

Multiples of 9 are 9,18,27…etc

Here we can see The least common multiple of 2 & 9 is 18

Least common multiple of 2 & 9 is 18.

Go Math! Practice Fluency Workbook Grade 6 Chapter 2 Factors and Multiples Exercise 2.2 Page 10 Exercise 2 Answer

Given :- number is 4 & 6

Find to the question least common multiple of 4 & 6

Multiples of 4 are 4,8,12…etc

Multiples of 6 are 6,12,18…etc

Here we can see the least common factor of 4 & 6 is 12.

The least common factor of 4 & 6 is 12.

Solutions For Go Math Grade 6 Chapter 2 Exercise 2.2 Factors And Multiples

Page 10 Exercise 3 Answer

Given :- number is 4 & 10

Find to the question the least common multiple of 4 & 10

Multiples of 4 are 4,8,12,16,20…etc.

Multiples of 10 are 10,20,30…etc

Here we can see the least common multiple of 4 & 10 is 20

The least common factor of 4 & 10 is 20.

Page 10 Exercise 4 Answer

Given :- number is 2,5 & 6

Find to the question the least common multiple of 2,5 & 6

Multiples of 2 are 2,4,6,8,10,12,14…etc.

Multiples of 5 are 5,10,15,20,25…etc

Multiples of 6 are 6,12,18,24…etc.

Here we can see the least common multiple of 2,5 & 6 is 30.

The least common multiple of 2,5,& 6 is 30.

Go Math! Practice Fluency Workbook Grade 6 Chapter 2 Factors and Multiples Exercise 2.2 Page 10 Exercise 5 Answer

Given :- number is 3,4 & 9

Find to the question the least common factor of 3,4 & 9

Multiples of 3 are 3,6,9,12…etc

Multiples of 4 are 4,8,12,16…etc.

Multiples of 9 are 9,18,27,36…etc.

Here we can see the least common multiple of 3,4 & 9 is 36.

The least common multiple of 3,4 & 9 is 36.

Page 10 Exercise 6 Answer

Given :- number is 6,8,10 & 12

Find to the question the least common multiple of 6,8,10 &12

Multiples of 6 are 6,12,18,24,30,36…etc

Multiples of 8 are 8,16,24,32,40…etc.

Multiples of 10 are 10,20,30,40,50…etc.

Multiples of 12 are 12,24,36,48,60…etc.

Here we can see the least common factor of 6,8,10,12 is 120

The least common factor of 6,8,10 & 12 is 120.

Go Math! Practice Fluency Workbook Grade 6 Chapter 2 Factors and Multiples Exercise 2.2 Page 10 Exercise 7 Answer

Given:- pads of paper 4 to a box, pencils come 27 to a box, and erasers come 12 to a box.

Find:- What is the least number of kits that can be made with paper, pencils, and erasers Using least common multiple concepts?

Given,
Pads of paper 4 to a box, pencils come 27 to a box and erasers come 12 to a box.

We want to make kits that can be made with paper,pencils, and erasers so Least common multiple of 4,27 & 12 is 108

So we can make 108 kits with paper pencils and eraser.

We can make 108 kits with pencil, eraser, and paper.

Go Math Grade 6 Factors And Multiples Exercise 2.2 Key From Practice Fluency Workbook

Go Math Answer Key

Grade 6 Chapter 3 Rational Numbers Exercise 3.1 Solutions

November 18, 2024August 5, 2023 by Alekhya

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 3 Rational Numbers

Page 11 Problem 1 Answer

We have given the number as 0.3 Asked to write each rational number in the form a/b

To get the required answer we just need to change the given number in the fractional form.To write in the fractional form we must know that the by default denominator of the number is always one.

We have given the number as 0.3

So here we get the fractional form as 0.3/1=3/10

For the given number 0.3 we get the fractional form as3/10

Page 11 Problem 2 Answer

We have given the number as 2 x 7/8

Asked to write each rational number in the form a/b

To get the required answer we just need to change the given number in the fractional form.To write in the fractional form we must know that the by default denominator of the number is always one.

We have given the number as

⇒ $2 \frac{7}{8}$

So here we get the fractional form as

⇒ $2 \frac{7}{8}$

⇒ $=\frac{23}{8}$

For the given number 2×7/8

we get the fractional form as23/8

Page 11 Problem 3 Answer

$Go Math! Practice Fluency Workbook Grade 6 Chapter 3 Rational Numbers Exercise 3.1 Answer Key$

We have given the number as −5 and Asked you to write each rational number in the form

To get the required answer we just need to change the given number in the fractional form. To write in the fractional form we must know that the by default denominator of the number is always one.

We have given the number as −5

So here we get the fractional form as −5/1

For the given number −5

we get the fractional form as−5/1

Page 11 Problem 4 Answer

We have given the number as 16 and Asked to write each rational number in the forma/b

To get the required answer we just need to change the given number in the fractional form. To write in the fractional form we must know that the by default denominator of the number is always one.

We have given the number as 16

So here we get the fractional form as 16/1

For the given number 16

we get the fractional form as16/1

Grade 6 Chapter 3 Rational Numbers Exercise 3.1 Solutions

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 3 Rational Numbers Page 11 Problem 5 Answer

We have given the number as −1×3/4

Asked to write each rational number in the forma/b

To get the required answer we just need to change the given number in the fractional form.To write in the fractional form we must know that the by default denominator of the number is always one.

We have given the number as

⇒ $-1 \frac{3}{4}$

So here we get the fractional form as

⇒ $ -1 \frac{3}{4}$

⇒ $\frac{-7}{4}$

For the given number −1×3/4

we get the fractional form as−7/4

Page 11 Problem 6 Answer

We have given the number as −4.5
Asked to write each rational number in the form a/b

To get the required answer we just need to change the given number in the fractional form. To write in the fractional form we must know that the by default denominator of the number is always one.

We have given the number as

-4.5

So here we get the fractional form as

⇒ $\frac{-4.5}{1}$

⇒ $ \frac{-45}{10}$

⇒ $ \frac{-9}{2}$

For the given number −4.5

we get the fractional form as−9/2

Page 11 Problem 7 Answer

We have given the number as 3
Asked to write each rational number in the form a/b

To get the required answer we just need to change the given number in the fractional form.To write in the fractional form we must know that the by default denominator of the number is always one.

We have given the number as 3

So here we get the fractional form as 3/1

For the given number 3

we get the fractional form as 3/1

Page 11 Problem 8 Answer

We have given the number as 0.11

Asked to write each rational number in the form a/b

To get the required answer we just need to change the given number in the fractional form.To write in the fractional form we must know that the by default denominator of the number is always one.

We have given the number a

0.11

So here we get the fractional form as

⇒ $\frac{0.11}{1} $

⇒ $\frac{11}{100}$

For the given number 0.11

we get the fractional form as11/100

Grade 6 Exercise 3.1 Rational Numbers Answers

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 3 Rational Numbers Page 11 Problem 9 Answer

We have given the number as −13
Asked to place each number in the correct place on the Venn diagram.To get the right place for the number in the diagram we have to analyze the number first. To place the number in the diagram we have to write that number within the circle of that category only.

We have given the number as −13

The nature of the given number is an integer

The given number −13 is an integer and placed in the diagram is given as

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 3 Rational Numbers 9$

Page 11 Problem 10 Answer

We have given the number as 1/6

Asked to place each number in the correct place on the Venn diagram. To get the right place for the number in the diagram we have to analyze the number first.

To place the number in the diagram we have to write that number within the circle of that category only. We have given the number as 1/6

The nature of the given number is a rational number

The given number1/6 is a rational number and placed in the diagram is given

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 3 Rational Numbers 10$

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 3 Rational Numbers Page 11 Problem 11 Answer

We have given the number as 0. Asked to place each number in the correct place on the Venn diagram.

To get the right place for the number in the diagram we have to analyze the number first.

To place the number in the diagram we have to write that number within the circle of that category only.

We have given the number as 0

The nature of the given number is a whole number The given number 0 is a whole number and the place in the diagram is given as

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 3 Rational Numbers 11$

Page 11 Problem 12 Answer

We have given the number as 0.99 Asked to place each number in the correct place on the Venn diagram.

To get the right place for the number in the diagram we have to analyze the number first.

To place the number in the diagram we have to write that number within the circle of that category only.

We have given the number as 0.99

The nature of the given number is rational number The given number 0.99 is a rational number and the place in the diagram is given as

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 3 Rational Numbers 12$

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 3 Rational Numbers Page 11 Problem 13 Answer

We have given the number as −6.7
Asked to place each number in the correct place on the Venn diagram. To get the right place for the number in the diagram we have to analyze the number first.

To place the number in the diagram we have to write that number within the circle of that category only. We have given the number as −6.7

The nature of the given number is a rational number

The given number 6.7 is a rational number and its place in the diagram is given as

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 3 Rational Numbers 13$

Page 11 Problem 14 Answer

We have given the number as 34 Asked to place each number in the correct place on the Venn diagram. To get the right place for the number in the diagram we have to analyze the number first.

To place the number in the diagram we have to write that number within the circle of that category only. We have given the number as 34

The nature of the given number is a whole number

The given number 34 is a whole number and the place in the diagram is given as

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 3 Rational Numbers 14$

Page 11 Problem 15 Answer

We have given the number as −14x/2 and Asked to place each number in the correct place on the Venn diagram.

To get the right place for the number in the diagram we have to analyze the number first.

To place the number in the diagram we have to write that number within the circle of that category only.

We have given the number as −14×1/2

The nature of the given number is a rational number

The given number −14×1/2 is a rational number and placed in the diagram is given

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 3 Rational Numbers 15$

Solutions For Grade 6 Chapter 3 Exercise 3.1 Rational Numbers

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 3 Rational Numbers Page 12 Exercise 1 Answer

We have given the number as−12

So here we get the fractional form as−12/1

For the given number −12

we get the fractional form as −12/1 and the nature of the number as

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 3 Rational Numbers exerciose 1$

Page 12 Exercise 2 Answer

We have given the number as 7.3 Asked to write each rational number in the form a/b, then circle the name of each set to which the number belongs.

o, get the required answer we just need to change the given number in the fractional form.

To write in the fractional form we must know that the by default denominator of the number is always one.

We have given the number as

7.3

So here we get the fractional form as

⇒ $\frac{7.3}{1}$

⇒ $\frac{73}{10}$

For the given number 7.3 we get the fractional form as73/10 and the nature of the n

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 3 Rational Numbers exerciose 2$

Page 12 Exercise 3 Answer

We have given the number as 0.41 Asked to write each rational number in the form a/b, then circle the name of each set to which the number belongs.

To get the required answer we just need to change the given number in the fractional form.

To write in the fractional form we must know that the by default denominator of the number is always one.

We have given the number as

0.41

So here we get the fractional form as

⇒ $\frac{0.41}{1}$

⇒ $\frac{41}{100}$

For the given number 0.41

we get the fractional form as 41/100 and the nature of the number as

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 3 Rational Numbers exerciose 3$

Page 12 Exercise 4 Answer

We have given the number 6 Asked to write each rational number in the form a/b, then circle the name of each set to which the number belongs.

To get the required answer we just need to change the given number in the fractional form.

To write in the fractional form we must know that the by default denominator of the number is always one.

We have given the number as 6

So here we get the fractional form as 6/1

For the given number 6

we get the fractional form as 6/1 and the nature of the number

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 3 Rational Numbers exerciose 4$

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 3 Rational Numbers Page 12 Exercise 5 Answer

We have given the number as 3×1/2

Asked to write each rational number in the form a/b, then circle the name of each set to which the number belongs.

To get the required answer we just need to change the given number in the fractional form.

To write in the fractional form we must know that the by default denominator of the number is always one.

We have given the number as

⇒ $3 \frac{1}{2}$

So here we get the fractional form as

⇒ $3 \frac{1}{2}$

⇒ $\frac{7}{2}$

For the given number 3×1/2

we get the fractional form as 7/2 and the nature of the number as

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 3 Rational Numbers exerciose 5$

Grade 6 Rational Numbers Exercise 3.1 Key

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Go Math! Grade 6 Exercise 3.2: Rational Numbers Answers

November 18, 2024August 3, 2023 by Alekhya

Go Math! Practice Fluency Workbook Grade 6 California 1st Edition Chapter 3 Rational Numbers

Page 13 Problem 1 Answer

We have given the number as 3.5 Asked to graph each number and its opposite on a number line.

To get the required number we first find the opposite of the given number by multiplying the negative sign with it.

Then mark the numbers that are opposite of each other on the given graph.

We have given the number as 3.5

We get the opposite of the given number as −3.5

For the given number 3.5

we get the opposite of it −3.5 and numbers on the line as

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 3 Rational Numbers 1$

Page 13 Problem 2 Answer

We have given the number as −2.5 and Asked to graph each number and its opposite on a number line.

To get the required number we first find the opposite of the given number by multiplying the negative sign with it.

$Go Math! Practice Fluency Workbook Grade 6 Chapter 3 Rational Numbers Exercise 3.2 Answer Key$

Then mark the numbers that are opposite of each other on the given graph.

We have given the number as 2.5

We get the opposite of the given number as
−(−2.5)
=2.5

For the given number −2.5

we get the opposite number as 2.5 and on the num

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 3 Rational Numbers 2$

Go Math Grade 6 Exercise 3.2 Rational Numbers Answers

Page 13 Problem 3 Answer

We have given the number as 2×1/2 Asked to graph each number and its opposite on a number line.

To get the required number we first find the opposite of the given number by multiplying the negative sign with it.

Then mark the numbers that are opposite of each other on the given graph.

For the given number 2×1/2

we get the opposite number as −2×1/2 and on the number line as

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 3 Rational Numbers 3$

Page 13 Problem 4 Answer

We have given the number as −1×1/2 Asked to graph each number and its opposite on a number line.

To get the required number we first find the opposite of the given number by multiplying the negative sign with it.

Then mark the numbers that are opposite of each other on the given graph.

For the given number −1×1/2, we get the opposite number as 1×1/2 and on the number line a

$Go Math! Practice Fluency Workbook Grade 6, California 1st Edition, Chapter 3 Rational Numbers 4$

Page 13 Problem 5 Answer

We have given the number as 4.25 and Asked to name the opposite of the given number.

To get the required number we first find the opposite of the given number by multiplying the negative sign with it.

Then on solving the sign convention, we will get the required opposite number. We have given the number as 4.25

We get the opposite of the number as −4.25

For the given number 4.25

we get the opposite of it as −4.25

Page 13 Problem 6 Answer

We have given the number as −5×1/4 and Asked to name the opposite of the given number.

To get the required number we first find the opposite of the given number by multiplying the negative sign with it.

Then on solving the sign convention, we will get the required opposite number.

We have given the number as

⇒ $-5 \frac{1}{4}$

We get the opposite of the number as

⇒ $-\left(-5 \frac{1}{4}\right)$

⇒ $5 \frac{1}{4}$

For the given number −5×1/4

we get the opposite of it as 5×1/4

Go Math Grade 6 Exercise 3.2 Rational Numbers Solutions

Page 13 Problem 7 Answer

We have given the number as 1/2 and Asked to name the opposite of the given number.

To get the required number we first find the opposite of the given number by multiplying the negative sign with it.

Then on solving the sign convention, we will get the required opposite number.

For the given number 1/2

we get the opposite of it as −1/2

Page 13 Problem 8 Answer

We have given the number as 2×1/3 Asked to Name the absolute value of the given number.

Here to get the absolute value of the given number we just have to write the numerical part of the number.

We have given the number as

⇒ $2 \frac{1}{3}$

We get the absolute of the number as

⇒ $\left|2 \frac{1}{3}\right|$

⇒ $2 \frac{1}{3}$

For the given number 2×1/3

we get the absolute of it as 2×1/3

Page 13 Problem 9 Answer

We have given the number −3.85 and Asked to Name the absolute value of the given number.

Here to get the absolute value of the given number we just have to write the numerical part of the number.

We have given the number as −3.85

We get the absolute of the number as ∣−3.85∣ =3.85

For the given number −3.85

we get the absolute of it as 3.85

Page 13 Problem 10 Answer

We have given the number as −6.1 and Asked to Name the absolute value of the given number.

Here to get the absolute value of the given number we just have to write the numerical part of the number.

We have given the number as−6.1

We get the absolute of the number as ∣−6.1∣ =6.1

For the given number −6.1

we get the absolute of it as 6.1

Page 13 Problem 11 Answer

Given:- The elevations of checkpoints along a marathon route in a table

To Find:- To determine the opposite values of each checkpoint elevation

The elevation of checkpoints A, B, C.D, and E are 15.6,17.1,5.2,−6.5,−18.5 feet respectively.

The opposite values of the checkpoints A, B, C, D, and E would be −15.6,−17.1,−5.2,6.5,18.5 feet respectively.

The opposite values of the checkpoints A, B, C, D, and E would be 15.6,−17.1,−5.2,6.5,18.5 feet respectively.

Solutions For Go Math Grade 6 Exercise 3.2 Rational Numbers

Page 13 Problem 12 Answer

Given:- The elevations of checkpoints along a marathon route in a table

To Find:- To determine the checkpoint which is closest to the sea level

The elevation of any point is considered from the sea level which would be at 0 feet.

So, the checkpoint that would be closest to the sea level would be the smallest value or the smallest absolute value of the data given in the table.

From the data given in the table, we can notice that the smallest value and the checkpoint that would be closest to the sea level would be 5.2 feet at checkpoint C.

The checkpoint that would be closest to the sea level would be at checkpoint C at 5.2 feet.

Page 13 Problem 13 Answer

Given:- The elevations of checkpoints along a marathon route in a table

To Find:- To determine the checkpoint which is furthest from the sea level

The elevation of any point is considered from the sea level which would be at 0 feet.

So, the checkpoint that would be furthest from the sea level would be the largest value or the largest absolute value of the data given in the table.

From the data given, the absolute values of the checkpoints A, B, C, D, and E are 15.6,17.1,5.2,6.5,18.5.

Therefore, we can conclude from the absolute values of the data given that Checkpoint E at 18.5 feet would be the furthest from the sea level.

The checkpoint that would be furthest from the sea level would be at checkpoint E at 18.5 feet.

Page 14 Exercise 1 Answer

Given:- Are the opposite of −6.5 and the absolute value of −6.5 the same?

To Find:- To determine whether the opposite and the absolute value of the given number −6.5 are the same or not

We know that the opposite value of a negative number would always be positive of the same number. So, the opposite of −6.5 would be 6.5.

We also know that the absolute value of a number is always positive. So, the absolute value of −6.5 would be 6.5.

Therefore, the opposite and the absolute value of the given number−6.5 are the same.

The opposite and the absolute value of the given number−6.5 are the same.

Page 14 Exercise 2 Answer

Given:- Are the opposite of 3×2/5 and the absolute value of 3×2/5 the same?

To Find:- To determine whether the opposite and the absolute value of the given fraction or mixed number are the same We know that the opposite value of a positive number would always be negative of the same number. So, the opposite of 3×2/5 would be−3×2/5.

We also know that the absolute value of a number is always positive. So, the absolute value of 3×2/5 would be 3×2/5.

Therefore, we can clearly say that the opposite and the absolute value of the mixed fraction 3×2/5 are not the same.

The opposite and the absolute value of the mixed fraction 3×2/5 are not the same.

Go Math Grade 6 Rational Numbers Exercise 3.2 Key

Page 14 Exercise 3 Answer

Given:- Write a rational number whose opposite and absolute value are the same

To Find:- To write the example of a rational number whose opposite and absolute value are the same and to given the appropriate explanation We know that the opposite value of a positive number would always be negative of the same number and the opposite value of a negative number would be positive of the same number.

Also, the absolute value of a number is always positive. So, we can say that the opposite value and the absolute value of any number would be positive and the same.

For example, the absolute value and the opposite of the number 5.5 would be 5.5.

The rational number whose opposite and absolute values are the same are−5.5 which would be 5.5

Page 14 Exercise 4 Answer

Given:- Write a rational number whose opposite and absolute values are opposite

To Find:- To write the example of a rational number whose opposite and absolute value are opposite and to give the appropriate explanation

We know that the opposite value of a positive number would always be negative of the same number and the opposite value of a negative number would be positive of the same number.

Also, the absolute value of a number is always positive. So, we can say that a rational number whose opposite and absolute values are opposite would be any positive number.

For example, the opposite value of the rational number 5.5 would be − 5.5. At the same time, the absolute value of the same rational number 5.5 is 5.5.

The rational number whose opposite and absolute values are not the same or opposite is 5.5

Detailed Answers For Go Math Grade 6 Exercise 3.2 Rational Number

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