HMH Middle School Grade 7 Practice Fluency Workbook 1st Edition Chapter 1 Adding and Subtracting Integers
HMH Grade 7 Practice Fluency Workbook Chapter 1 Exercise 1.4 Solutions Page 7 Problem 1 Answer
It is given that Owen starts with the bait 2 feet below the surface of the water and reels out the bait 19 feet, then reels it back in 7 feet.
It is required to find the final position of the bait relative to the surface of the water.
To do so write an expression to represent the situation and find the absolute value for the first two numbers and then subtract the smaller absolute value from the higher absolute value and the result will take the sign of the higher absolute value.
Repeat the process for the result obtained and the third number.
As a result, the final position of the bait relative to the surface of the water can be determined.
Represent the situation as an expression and find the absolute value for the first two numbers.
The expression can be represented as −2+19−7.
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The final position of the bait is 10 feet above the water.
HMH Middle School Grade 7 Practice Fluency Workbook 1st Edition Chapter 1 Page 7 Problem 2 Answer
It is given that Rita earned 45 points on a test, lost 8 points, earned 53 points, and then lost 6 points.
It is required to find Rita’s final score on the test.
To do so write an expression to represent the situation and find the absolute value for the first two numbers and then subtract the smaller absolute value from the higher absolute value and the result will take the sign of the higher absolute value.
Repeat the process for the result obtained and the third number.
Again repeat the process for the result obtained and the fourth number.
As a result, Rita’s final score on the test can be determined.
Represent the situation as an expression and find the absolute value for the first two numbers.
The expression for the given situation can be represented as 45−8+53−6.
The absolute values 45+(−8) are,∣45∣=45 and,∣−8∣=8
Subtract the number with a lower absolute value from the number with a higher absolute value and the answer takes the sign of the higher absolute value.
45−8=37
Therefore the expression 45+(−8) becomes, 45−8=37
Find the absolute values for the numbers in 37+53 and add their absolute values and result takes the sign of the higher absolute values.
∣37∣=37 and,∣53∣=53
Add the absolute values and the answer takes the sign of the higher absolute value.
37+53=90
Find the absolute values for the numbers in 90−6.
The absolute values 90+(−6) are,∣90∣=90 and,∣−6∣=6
Subtract the number with a lower absolute value from the number with a higher absolute value and the answer takes the sign of the higher absolute value.
90−6=84
The expression 90+(−6) becomes,90−6=84
The value of the expression45−8+53−6 is 84.
Rita’s final score on the test is 84.
HMH Middle School Grade 7 Practice Fluency Workbook 1st Edition Chapter 1 Page 7 Problem 3 Answer
Given the expression −7+12+15.
It is required to find the value of the expression.
To do so, find the absolute value for the first two numbers and then subtract the smaller absolute value from the higher absolute value and the result will take the sign of the higher absolute value.
Repeat the process for the result obtained and the third number in the expression.
As a result the value of the expression can be determined.
Find the absolute value for the first two numbers in the expression.
The absolute values are,∣−7∣=7 and,∣12∣=12
Subtract the number with a lower absolute value from the number with a higher absolute value and the answer takes the sign of the higher absolute value.
12−7=5
Therefore the expression −7+12 becomes, −7+12=5
Find the absolute values for the numbers 5 and 15,∣5∣=5 and,∣15∣=15
Add these absolute values and the answer takes the sign of the higher absolute value.
15+5=20
The value of the of the expression is 20.
HMH Middle School Grade 7 Practice Fluency Workbook 1st Edition Chapter 1 Page 7 Problem 4 Answer
Given the expression −5−9−13.
It is required to find the value of the expression.
To do so, find the absolute value for the first two numbers and then add the absolute values of the numbers and the answer takes the same sign as the numbers.
Repeat the process for the result obtained and the third number in the expression.
As a result the value of the expression can be determined.
Find the absolute value for the first two numbers in the expression.
The absolute values of −5+(−9) are,∣−5∣=5 and,∣−9∣=9.
Add these absolute values and the result takes same sign as the numbers.
5+9=14
The expression −5+(−9) becomes,−5−9=−14
Find the absolute values for the numbers in the expression −14−13.
The absolute values for −14+(−13) are,∣−14∣=14 and,∣−13∣=13.
Add these absolute values and answer takes the same sign as the numbers.
14+13=27
The expression −14+(−13) becomes,−14−13=−27
The value of the expression is −27.
HMH Middle School Grade 7 Practice Fluency Workbook 1st Edition Chapter 1 Page 7 Problem 5 Answer
The given expression is −21−17+25+65.
It is required to find the value of the given expression.
To find the value of the given expression, first combine the terms with same sign.
Then add the terms in the parenthesis. Transfer the subtraction into addition by changing the sign of the first number.
Therefore, this will become the addition of two integers with different signs.
To solve that, determine the absolute values of both numbers.
Then subtract the number with a lower absolute value from the number with the higher absolute value.
Finally, assign the sign of the number with the higher absolute value to the answer.
Combine the terms with same sign.
Then add the terms in the parenthesis.
−21−17+25+65=−(21+17)+25+65
−21−17+25+65=−38+90
Transfer the subtraction into addition by changing the sign of the first number and determine the absolute values of both numbers.
−21−17+25+65=(−38)+90
∣−38∣=38 and ∣90∣=90
Subtract the number with a lower absolute value from the number with the higher absolute value and assign the sign of the number with the higher absolute value to the answer.
Here, 90>38.
90−38=52
The larger number 90 has positive sign and therefore,
(−38)+90=52
−21−17+25+65=52
The value of the given expression is 52.
HMH Middle School Grade 7 Practice Fluency Workbook 1st Edition Chapter 1 Page 7 Problem 6 Answer
The given expression is 12+19+5−2.
It is required to find the value of the given expression.
To find the value of the given expression, first, combine the terms with same sign.
Then add the terms in the parenthesis.
Transfer the subtraction into addition by changing the sign of the second number.
Therefore, this will become the addition of two integers with different signs.
To solve that, determine the absolute values of both numbers.
Then subtract the number with a lower absolute value from the number with the higher absolute value.
Finally, assign the sign of the number with the higher absolute value to the answer.
Combine the terms with same sign. Then add the terms in the parenthesis.
12+19+5−2=(12+19+5)−2
12+19+5−2=36−2
Transfer the subtraction into addition by changing the sign of the second number and determine the absolute values of both numbers.
12+19+5−2=36+(−2)
∣36∣=36 and ∣−2∣=2
Subtract the number with a lower absolute value from the number with the higher absolute value and assign the sign of the number with the higher absolute value to the answer.
Here, 36>2.
36−2=34
The larger number 36 has positive sign and therefore,
36+(−2)=34
12+19+5−2=34
The value of the given expression is 34.
HMH Middle School Grade 7 Practice Fluency Workbook 1st Edition Chapter 1 Page 7 Problem 7 Answer
The given expression is 31−4+6 ◯ −17+22−5.
It is required to compare the expressions and write <,> or =.
To find the result, solve the expressions on both the sides.
First, combine the terms with same sign.
Then add the terms in the parenthesis.
Transfer the subtraction into addition by changing the sign of the first number.
Therefore, this will become the addition of two integers with different signs.
To solve that, determine the absolute values of both numbers.
Then subtract the number with a lower absolute value from the number with the higher absolute value.
Finally, assign the sign of the number with the higher absolute value to the answer.
Combine the terms with same sign. Then add the terms in the parenthesis.
31−4+6 ◯ −17+22−5
(31+6)−4 ◯ −(17+5)+22
37−4 ◯ −22+22
Transfer the subtraction into addition by changing the sign of the first number and determine the absolute values of both numbers in the left hand side.
37+(−4) ◯ (−22)+22
∣37∣=37 and ∣−4∣=4
Subtract the number with a lower absolute value from the number with the higher absolute value and assign the sign of the number with the higher absolute value to the answer.
Here, 37>4.
37−4=33
The larger number 37 has positive sign and therefore,37+(−4)=33 (1)
Transfer the subtraction into addition by changing the sign of the first number and determine the absolute values of both numbers in the right hand side.
37+(−4) ◯ (−22)+22
∣−22∣=22 and ∣22∣=22
Subtract the number with a lower absolute value from the number with the higher absolute value and assign the sign of the number with the higher absolute value to the answer.
Here, 22=22.
22−22=0
(−22)+22=0 (2)
From (1) and (2),33>0
The given expression can be written as 31−4+6>−17+22−5.
HMH Middle School Grade 7 Practice Fluency Workbook 1st Edition Chapter 1 Page 7 Problem 8 Answer
It is given that Anna and Maya are competing in a dance tournament where a dancer earns points if a dance move is done correctly and losses points if dance move is done incorrectly.
It is also given that Anna currently has 225 points.
It is also given that before the dance routine ends Anna earns 75 points and losses 30 points.
It is required to find Anna’s final score.
To find Anna’s final score, add the points earned and subtract the points loosed from her current score.
First, combine the terms with same sign. Then add the terms in the parenthesis.
Transfer the subtraction into addition by changing the sign of the first number.
Therefore, this will become the addition of two integers with different signs.
To solve that, determine the absolute values of both numbers.
Then subtract the number with a lower absolute value from the number with the higher absolute value.
Finally, assign the sign of the number with the higher absolute value to the answer.
Add the points earned and subtract the points loosed from Anna’s current score.
P=225+75−30
Combine the terms with same sign. Then add the terms in the parenthesis.
P=(225+75)−30
P=300−30
Transfer the subtraction into addition by changing the sign of the second number and determine the absolute values of both numbers.
P=300+(−30)
∣300∣=300 and ∣−30∣=30
Subtract the number with a lower absolute value from the number with the higher absolute value and assign the sign of the number with the higher absolute value to the answer.
Here, 30>300.
300−30=270
The larger number 300 has positive sign and therefore,
300+(−30)=270
P=270
Anna’s final score is 270.
Adding And Subtracting Integers Exercise 1.4 Chapter 1 Answers HMH Grade 7 Workbook Page 7 Problem 9 Answer
It is given that Anna and Maya are competing in a dance tournament where a dancer earns points if a dance move is done correctly and losses points if dance move is done incorrectly.
It is also given that Anna currently has 225 points.
It is also given that before the dance routine ends Anna earns 75 points and losses 30 points.
It is also given that Maya’s final score is 298.
It is required to find which dancer has greatest final score.
To find Anna’s final score, add the points earned and subtract the points loosed from her current score.
First, combine the terms with same sign.
Then add the terms in the parenthesis.
Transfer the subtraction into addition by changing the sign of the first number.
Therefore, this will become the addition of two integers with different signs.
To solve that, determine the absolute values of both numbers.
Then subtract the number with a lower absolute value from the number with the higher absolute value.
Finally, assign the sign of the number with the higher absolute value to the answer.
Then compare it with Maya’s final score.
Add the points earned and subtract the points loosed from Anna’s current score.
P=225+75−30
Combine the terms with same sign.
Then add the terms in the parenthesis.
P=(225+75)−30
P=300−30
Transfer the subtraction into addition by changing the sign of the second number and determine the absolute values of both numbers.
P=300+(−30)
∣300∣=300 and ∣−30∣=30
Subtract the number with a lower absolute value from the number with the higher absolute value and assign the sign of the number with the higher absolute value to the answer.
Here, 30>300.
300−30=270
The larger number 300 has positive sign and therefore,
300+(−30)=270
P=270
Therefore, Anna’s final score is 270 which less than Maya’s final score 298.
Maya has greatest final score.
Step-By-Step Solutions For Exercise 1.4 Adding And Subtracting Integers HMH Grade 7 Practice Workbook Page 8 Problem 10 Answer
The given expression is 10−19+5.
It is required to regroup the integers in the given expression.
To do so, regroup the expression and keep the integers of same sign together.
10−19+5=(10+5)−19
The given expression can be regrouped as (10+5)−19.
Page 8 Problem 11 Answer
The given expression is 10−19+5.
It is required to add and subtract the integers in the given expression.
To do so, regroup the integers of same sign.
First, add the terms in the parenthesis.
Transfer the subtraction into addition by changing the sign of the first number.
Therefore, this will become the addition of two integers with different signs.
To solve that, determine the absolute values of both numbers.
Then subtract the number with a lower absolute value from the number with the higher absolute value.
Finally, assign the sign of the number with the higher absolute value to the answer.
Combine the terms with same sign. Then add the terms in the parenthesis.
10−19+5=(10+5)−19
10−19+5=15−19
Transfer the subtraction into addition by changing the sign of the second number and determine the absolute values of both numbers.
10−19+5=15+(−19)
∣15∣=15 and ∣−19∣=19
Subtract the number with a lower absolute value from the number with the higher absolute value and assign the sign of the number with the higher absolute value to the answer.
Here,19>15.
19−15=4
The larger number 19 has negative sign and therefore,
15+(−19)=−4
10−19+5=−4
The value of the given expression is−4.
HMH Middle School Grade 7 Practice Fluency Workbook 1st Edition Chapter 1 Page 8 Problem 12 Answer
Given is the expression 10−19+5.
It is required to find the sum of the given expression.
To solve this question, first regroup the expression and keep the integers with same sign together.
Then add the absolute values of the numbers with same sign and keep the same sign as of the number.
Then to add two integers with different signs, first determine the absolute values of both numbers.
Then subtract the number with a lower absolute value from the number with the higher absolute value.
Finally, assign the sign of the number with the higher absolute value to the answer.
Regroup the same sign numbers together and determine the absolute values of each integer.
10−19+5=10+5−19
∣10∣=10
∣−19∣=19
∣5∣=5
Add the absolute values of the numbers having same sign and keep the same sign as of the numbers.
Here 10 and 5 have same sign. So,
10+5−19
15−19
Transfer the subtraction into addition by changing the sign of the second number.
15−19
15+(−19)
Subtract the number with a lower absolute value from the number with the higher absolute value and assign the sign of the number with the higher absolute value to the answer.
Here 19>15.
The larger absolute value is of 19 having negative sign so,
= 15+(−19)
= −(19−15)
−(19−15) = −4
So the sum of the given expression is −4.
The value of the given expression 10−19+5 is −4.
Exercise 1.4 Adding And Subtracting Integers Solutions For HMH Middle School Grade 7 Workbook Page 8 Problem 13 Answer
Given is the expression −80+10−6.
It is asked to regroup the integers.
To solve the problem first regroup the numbers having positive sign and then put parenthesis around the integers and put addition sign between them and positive sign outside the parenthesis.
Then regroup the numbers having negative sign and then put parenthesis around the integers and put addition sign between them and negative sign outside the parenthesis.
First regroup the addition signed number but as there is only one number that is 10 having positive sign. So,
−80+10−6
10−80−6
Now regroup the numbers having negative sign that is 80 and 6, and then put parenthesis around the integers and put addition sign between them and negative sign outside the parenthesis.
10−80−6
10−(80+6)
The regrouped expression is10−(80+6).
The regrouped expression of the given expression −80+10−6 is 10−(80+6).
HMH Middle School Grade 7 Practice Fluency Workbook 1st Edition Chapter 1 Page 8 Problem 14 Answer
It is given the expression−80+10−6.
It is required to add and subtract −80+10−6.
In order to add and subtract −80+10−6, using the final answer from part(a) of this problem, and then simplify the brackets by adding the expression inside the brackets and then subtract the resulting expression with the expression outside.
From part (a) of this problem, −80+10−6=10−(80+6)
Simplify the brackets by adding the expression inside the brackets and then subtract the resulting expression with the expression outside.
10−(80+6)=10−86=−76
10−(80+6)=−76
HMH Middle School Grade 7 Practice Fluency Workbook 1st Edition Chapter 1 Page 8 Problem 15 Answer
It is given the expression −80+10−6.
It is required to write the sum of −80+10−6.
In order to write the sum of −80+10−6, rearrange the integers and then take same sign common and then solve the equation.
It is given that −80+10−6.
Rearrange the given expression.
−80+10−6=10−80−6
Take negative common from last two digits.
−80+10−6=10−(80+6)
Solve the bracket.
−80+10−6=10−86
Simplify the resultant expression.
−80+10−6=−76
The sum of−80+10−6
is given by −80+10−6=−76
Student Edition Chapter 1 Exercise 1.4 Adding And Subtracting Integers HMH Grade 7 Workbook Guide Page 8 Problem 16 Answer
It is given the expression 7−21+13.
It is required to regroup the given expression.
In order to regroup the given expression, rearrange the integers and then take same sign common
It is given that 7−21+13.
Re-grouping the given expression by taking same sign integers together.
7−21+13=(7+13)−21
The equation 7−21+13 after re-grouping is given by
7−21+13=(7+13)−21
HMH Middle School Grade 7 Practice Fluency Workbook 1st Edition Chapter 1 Page 8 Problem 17 Answer
It is given the expression 7−21+13.
It is required to add and subtract 7−21+13.
In order to add and subtract 7−21+13, using the final answer from part (a) of this problem, and then simplify the brackets by adding the expression inside the brackets and then subtract the resulting expression with the expression outside.
From part (a) of this problem, (7+13)−21
Simplify the brackets by adding the expression inside the brackets and then subtract the resulting expression with the expression outside.
(7+13)−21=20−21=−1
(7+13)−21=−1
HMH Middle School Grade 7 Practice Fluency Workbook 1st Edition Chapter 1 Page 8 Problem 18 Answer
It is given the expression 7−21+13.
It is required to write the sum of 7−21+13.
In order to write the sum of 7−21+13, rearrange the integers and then take same sign common and then solve the equation.
It is given that 7−21+13.
Rearrange the given expression.
7−21+13=7+13−21
=(7+13)−21
Solve the bracket.
(7+13)−21=20−21
Simplify the resultant expression.
20−21=−1
The sum of 7−21+13 is given by 7−21+13=−1
HMH Middle School Grade 7 Practice Fluency Workbook 1st Edition Chapter 1 Page 8 Problem 19 Answer
It is given the expression −5+13−6+2.
It is required to regroup the given expression.
In order to regroup the given expression, rearrange the integers and then take same sign common.
It is given that −5+13−6+2.
Re-grouping the given expression by taking same sign integers together.
−5+13−6+2=(−5−6)+(13+2)
Taking negative sign common from first bracket and rearrange the integers.
−5+13−6+2=(13+2)−(5+6)
The equation −5+13−6+2
after re-grouping is given by
−5+13−6+2=(13+2)−(5+6)
HMH Middle School Grade 7 Practice Fluency Workbook 1st Edition Chapter 1 Page 8 Problem 20 Answer
It is given the expression−5+13−6+2.
It is required to add and subtract −5+13−6+2.
In order to add and subtract 7−21+13, using the final answer from part(a) of this problem, and then simplify the brackets by adding the expression inside the brackets and then subtract the resulting expression with the expression outside.
From part (a) of this problem,
−5+13−6+2=(13+2)−(5+6)
Simplify the brackets by adding the expression inside the first and second brackets and then subtract the resulting expression.
−5+13−6+2=(13+2)−(5+6)
−5+13−6+2 =15−11
−5+13−6+2 =4
−5+13−6+2=4
HMH Middle School Grade 7 Practice Fluency Workbook 1st Edition Chapter 1 Page 8 Problem 21 Answer
It is given the expression −5+13−6+2.
It is required to write the sum of −5+13−6+2.
In order to write the sum of −5+13−6+2, rearrange the integers and then take same sign common and then solve the equation.
It is given that−5+13−6+2.
Rearrange the given expression.
−5+13−6+2=(−5−6)+(13+2)
=−(5+6)+(13+2)
Solve the bracket.
−(5+6)+(13+2)=−11+15
−(5+6)+(13+2) =4
The sum of −5+13−6+2 is given by −5+13−6+2=4
HMH Middle School Grade 7 Practice Fluency Workbook 1st Edition Chapter 1 Page 8 Problem 22 Answer
It is given the expression 18−4+6−30.
It is required to regroup the given expression.
In order to regroup the given expression, rearrange the integers and then take same sign common.
It is given that 18−4+6−30.
Re-grouping the given expression by taking same sign integers together.
18−4+6−30=(18+6)+(−4−30)
Taking negative sign common from second bracket and rearrange the integers.
18−4+6−30=(18+6)−(4+30)
The equation18−4+6−30 after re-grouping is given by
18−4+6−30=(18+6)−(4+30)
HMH Middle School Grade 7 Practice Fluency Workbook 1st Edition Chapter 1 Page 8 Problem 23 Answer
It is given the expression18−4+6−30.
It is required to add and subtract 18−4+6−30.
In order to add and subtract 18−4+6−30, use the final answer from part (a) of this problem, and then simplify the brackets by adding the expression inside the brackets and then subtract the resulting expression with the expression outside.
From part (a) of this problem, 18−4+6−30=(18+6)−(4+30)
Simplify the brackets by adding the expression inside the brackets and then subtract the resulting expression.
18−4+6−30=(18+6)−(4+30)
18−4+6−30 =24−34
18−4+6−30 =−10
18−4+6−30=−10
HMH Middle School Grade 7 Practice Fluency Workbook 1st Edition Chapter 1 Page 8 Problem 24 Answer
In this question, it is given that the expression is 18−4+6−30.
It is required to write the sum of the given expression.
Solving the given expression
=18−4+6−30
18−4+6−30 =18+6−30−4
18−4+6−30 =24−34=−10
as the magnitude of −34 is greater than that of 24, hence the answer is negative.
The sum is 18−4+6−30=−10.