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Excel in your academics by accessing McGraw Hill Math Grade 7 Answer Key PDF Lesson 4.3 The Distributive and Identity Properties existing for free of cost.

McGraw-Hill Math Grade 7 Answer Key Lesson 4.3 The Distributive and Identity Properties

Exercises Solve

Question 1.
0 + 6 =
Answer:
0 + 6 = 6
Explanation:
In addition, the identity element is 0. Any addend + 0 will not change the total.

Question 2.
4(4 + 3) =
Answer:
4(4 + 3) = (4 x 4) + (4 x 3) = 16 + 12 = 28
Explanation:
The Distributive property of multiplication over addition states that when we multiply numbers, we have to multiply the numbers each separately and then add their products.

Question 3.
7 + 0 =
Answer:
7 + 0 = 7
Explanation:
In addition, the identity element is 0. Any addend + 0 will not change the total.

Question 4.
4(2 + 3 + 4) =
Answer:
4(2 + 3 + 4)
= (4 x 2) + (4 x 3) + (4 x 4)
= 8 + 12 + 16
= 36
4(2 + 3 + 4) = 36
Explanation:
The Distributive property of multiplication over addition states that when we multiply numbers, we have to multiply the numbers each separately and then add their products as we can observe in the above answer.

Question 5.
15(7 + 3) =
Answer:
15(7 + 3)
= (15 x 7) + (15 x 3)
= 105 + 45
= 150
15(7 + 3) = 150
Explanation:
The Distributive property of multiplication over addition states that when we multiply numbers, we have to multiply the numbers each separately and then add their products as we can observe in the above answer.

Question 6.
6(5 + 0) =
Answer:
6(5 + 0)
= (6 x 5) + (6 x 0)
= 30 + 0
= 30
6(5 + 0) = 30
Explanation:
The Distributive property of multiplication over addition states that when we multiply numbers, we have to multiply the numbers each separately and then add their products as we can observe in the above answer.

Question 7.
3 × 4 + 2 × 9 =
Answer:
3 × 4 + 2 × 9 = 12 + 18 = 30 
Explanation:
Multiply 3 with 4 and 2 with 9 the products are equal to 12 and 18. After adding the products the result is 30.

Question 8.
(4 + 8) 4 =
Answer:
(4 + 8) 4
= (4 x 4) + (8 x 4)
= 16 + 32
= 48
(4 + 8) 4 = 48
Explanation:
Multiply 4 with 4 and 8 with 4 the products are equal to 16 and 32. After adding the products the result is 48.

Question 9.
34 + 0 + 7 =
Answer:
34 + 0 + 7 = 41
Explanation:
Perform addition operation on above three given numbers. Add 34 with 0 and 7 the sum is equal to 41.

Question 10.
35(2 + 3 + 0) =
Answer:
35(2 + 3 + 0)
= (35 x 2) + (35 x 3) + (35 x 0)
= 70 + 105 + 0
= 175
35(2 + 3 + 0) = 175
Explanation:
The Distributive property of multiplication over addition states that when we multiply numbers, we have to multiply the numbers each separately and then add their products as we can observe in above answer.

Question 11.
0 + 7 + 7 + 0 =
Answer:
0 + 7 + 7 + 0 = 14
Explanation:
Perform addition operation on above four given numbers. Add 0 with 7, 7 and 0 the sum is equal to 14.

Question 12.
5(33 – 11) =
Answer:
5(33 – 11)
= (5 x 33) – (5 x 11)
= 165 – 55
= 110
5(33 – 11) = 110
Explanation:
The Distributive property of multiplication over subtraction states that when we multiply numbers, we have to multiply the numbers each separately and then subtract their products as we can observe in the above answer.

Question 13.
(64 – 28) 2 =
Answer:
(64 – 28) 2
= (64 x 2) – (28 x 2)
= 128 – 56
= 72
(64 – 28) 2 = 72
Explanation:
The Distributive property of multiplication over subtraction states that when we multiply numbers, we have to multiply the numbers each separately and then subtract their products as we can observe in the above answer.

Question 14.
7 – 0 + 0 – 7 =
Answer:
7 – 0 + 0 – 7 = 7 – 7 = 0
Explanation:
Subtract 0 from 7 and 7 from 0 the differences are equal to 0.

Question 15.
8(15 + 2 – 15) =
Answer:
8(15 + 2 – 15)
= (8 x 15) + (8 x 2) – (8 x 15)
= 120 + 16 – 120
= 16
8(15 + 2 – 15) = 16

Excel in your academics by accessing McGraw Hill Math Grade 7 Answer Key PDF Lesson 4.2 Commutative and Associative Properties existing for free of cost.

McGraw-Hill Math Grade 7 Answer Key Lesson 4.2 Commutative and Associative Properties

Exercises Identify The Property

Question 1.
7 × 4 × 3 = 4 × 3 × 7
Answer:
The expression 7 × 4 × 3 = 4 × 3 × 7 is commutative property of multiplication.
Explanation:
The commutative property of multiplication states that the factors are multiplied in any order without changing the product.
7 × 4 × 3 = 84
4 × 3 × 7 = 84

Question 2.
2 + 9 + 22 = 22 + 9 + 2
Answer:
The expression 2 + 9 + 22 = 22 + 9 + 2 is commutative property of addition.
Explanation:
The commutative property of addition states that the addends are added in any order without changing the sum.
2 + 9 + 22 = 33
22 + 9 + 2 = 33

Question 3.
3 × 4 × 4 × 2 = 3 × (4 × 4) × 2
Answer:
The expression 3 × 4 × 4 × 2 = 3 × (4 × 4) × 2 is Associative property of Multiplication.
Explanation:
The Associative property of multiplication states that the factors are grouped in any way without changing the product.
3 × 4 × 4 × 2 = 96
3 × (4 × 4) × 2 = 96

Question 4.
4 × 2 × 3 × 4 = 2 × 3 × 4 × 4
Answer:
The expression 4 × 2 × 3 × 4 = 2 × 3 × 4 × 4 is commutative property of multiplication.
Explanation:
The commutative property of multiplication states that the factors are multiplied in any order without changing the product.
4 × 2 × 3 × 4 = 96
2 × 3 × 4 × 4 = 96

Question 5.
9 × 2 × 4 = 2 × 4 × 9
Answer:
The expression 9 × 2 × 4 = 2 × 4 × 9 is commutative property of multiplication.
Explanation:
The commutative property of multiplication states that the factors are multiplied in any order without changing the product.
9 × 2 × 4 = 72
2 × 4 × 9 = 72

Question 6.
9 × 7 × 9 × 9 = (9 × 7) × (9 × 9)
Answer:
The expression 9 × 7 × 9 × 9 = (9 × 7) × (9 × 9) is Associative property of Multiplication.
Explanation:
The Associative property of multiplication states that the factors are grouped in any way without changing the product.
9 × 7 × 9 × 9 = 5,103
(9 × 7) × (9 × 9) = 63 x 81 = 5,103

Question 7.
9 + 7 + 1 = 7 + 9 + 1
Answer:
The expression 9 + 7 + 1 = 7 + 9 + 1 is commutative property of addition.
Explanation:
The commutative property of addition states that the addends are added in any order without changing the sum.
9 + 7 + 1 = 17
7 + 9 + 1 = 17

Question 8.
4 × 11 × 9 × 4 = 4 × 4 × 9 × 11
Answer:
The expression 4 × 11 × 9 × 4 = 4 × 4 × 9 × 11 is commutative property of multiplication.
Explanation:
The commutative property of multiplication states that the factors are multiplied in any order without changing the product.
4 × 11 × 9 × 4 = 1,584
4 × 4 × 9 × 11 = 1,584 

Question 9.
2 + 8 + 6 + 4 = (2 + 8) + (6 + 4)
Answer:
The expression 2 + 8 + 6 + 4 = (2 + 8) + (6 + 4) is Associative property of addition.
Explanation:
The Associative property of addition states that the addends are grouped in any way without changing the sum.
2 + 8 + 6 + 4 = 20
(2 + 8) + (6 + 4) = 10 + 10 = 20

Question 10.
6 × 4 × 2 = 4 × 2 × 6
Answer:
The expression 6 × 4 × 2 = 4 × 2 × 6 is commutative property of multiplication.
Explanation:
The commutative property of multiplication states that the factors are multiplied in any order without changing the product.
6 × 4 × 2 = 48
4 × 2 × 6 = 48

Question 11.
22 + 21 + 24 + 22 = 22 + (21 + 24) + 22
Answer:
The expression 22 + 21 + 24 + 22 = 22 + (21 + 24) + 22 is Associative property of addition.
Explanation:
The Associative property of addition states that the addends are grouped in any way without changing the sum.
22 + 21 + 24 + 22 = 89
22 + (21 + 24) + 22 = 22 + 45 + 22 = 89

Question 12.
6 × 8 + 6 × 0 = 8 × 6 + 0 × 6
Answer:
The expression 6 × 8 + 6 × 0 = 8 × 6 + 0 × 6 is commutative property of multiplication.
Explanation:
The commutative property of multiplication states that the factors are multiplied in any order without changing the product.
6 × 8 + 6 × 0 = 48 + 0 = 48
8 × 6 + 0 × 6 = 48 + 0 = 48

Excel in your academics by accessing McGraw Hill Math Grade 7 Answer Key PDF Lesson 4.1 Order of Operations existing for free of cost.

McGraw-Hill Math Grade 7 Answer Key Lesson 4.1 Order of Operations

Exercises Calculate

Question 1.
(4 + 2) × (4 – 2) – (2 × 3) + 2^4-2 =
Answer:
We have to calculate the above given expression in order. The order of operation is PEMDAS. These letters stands for Parenthesis, Exponents, Multiplication, Division, Addition, Subtraction.
(4 + 2) × (4 – 2) – (2 × 3) + 2^4-2 = ?
First solve the parts that are inside parenthesis.
6 × 2 – 6 + 2^4-2 = ?
Second solve the parts that have exponents.
6 × 2 – 6 + 2² = ?
6 × 2 – 6 + 4 = ?
Third perform multiplication operation.
12 – 6 + 4 = ?
Fourth perform Addition and subtraction operation.
6 + 4 = 10
So, the expression (4 + 2) × (4 – 2) – (2 × 3) + 2^4-2  is equal to 10.

Question 2.
(5 – 4) × (10 – 6) – 2² + 4 =
Answer:
We have to calculate the above given expression in order. The order of operation is PEMDAS. These letters stands for Parenthesis, Exponents, Multiplication, Division, Addition, Subtraction.
(5 – 4) × (10 – 6) – 2² + 4 = ?
First solve the parts that are inside parenthesis.
1 × 4 – 2² + 4 = ?
Second solve the parts that have exponents.
1 × 4 – 4 + 4 = ?
Third perform multiplication operation.
4 – 4 + 4 = ?
Fourth perform Addition and subtraction operation.
0 + 4 = 4
So, the expression (5 – 4) × (10 – 6) – 2² + 4  is equal to 4.

Question 3.
(4 – 2)³ + (4 – 2)² + 1 – 2 + 2² =
Answer:
We have to calculate the above given expression in order. The order of operation is PEMDAS. These letters stands for Parenthesis, Exponents, Multiplication, Division, Addition, Subtraction.
(4 – 2)³ + (4 – 2)² + 1 – 2 + 2² = ?
First solve the parts that are inside parenthesis.
2³ + 2² + 1 – 2 + 2² = ?
Second solve the parts that have exponents.
8 + 4 + 1 – 2 + 4 = ?
Third perform Addition and subtraction operation.
12 + 1 – 2 + 4 = ?
13 – 2 + 4 = ?
11 + 4 = 15
So, the expression (4 – 2)³ + (4 – 2)² + 1 – 2 + 2² is equal to 15.

Question 4.
(4 + 2) × (9 – 7) + 3² – (8 – 5)² =
Answer:
We have to calculate the above given expression in order. The order of operation is PEMDAS. These letters stands for Parenthesis, Exponents, Multiplication, Division, Addition, Subtraction.
(4 + 2) × (9 – 7) + 3² – (8 – 5)² = ?
First solve the parts that are inside parenthesis.
6 × 2 + 3² – 3² = ?
Second solve the parts that have exponents.
6 × 2 + 9 – 9 = ?
Third perform multiplication operation.
12 + 9 – 9 = ?
Fourth perform Addition and subtraction operation.
21 – 9 = 12
So, the expression(4 + 2) × (9 – 7) + 3² – (8 – 5)²  is equal to 12.

Question 5.
(6 – 4)³ – (6 – 4)³ + 2 – (2 – 1) =
Answer:
We have to calculate the above given expression in order. The order of operation is PEMDAS. These letters stands for Parenthesis, Exponents, Multiplication, Division, Addition, Subtraction.
(6 – 4)³ – (6 – 4)³ + 2 – (2 – 1) = ?
First solve the parts that are inside parenthesis.
2³ –  2³ + 2 – 1 = ?
Second solve the parts that have exponents.
8 – 8 + 2 – 1 = ?
Third perform Addition and subtraction operation.
0 + 2 – 1 = ?
2 – 1 = 1
So, the expression (6 – 4)³ – (6 – 4)³ + 2 – (2 – 1)  is equal to 1.

Question 6.
2² – (2³ – 2) + (2² + 4) =
Answer:
We have to calculate the above given expression in order. The order of operation is PEMDAS. These letters stands for Parenthesis, Exponents, Multiplication, Division, Addition, Subtraction.
2² – (2³ – 2) + (2² + 4) = ?
First solve the parts that are inside parenthesis.
2² – 6 + 8 = ?
Second solve the parts that have exponents.
4 – 6 + 8 = ?
Third perform Addition and subtraction operation.
– 2 + 8 = 6
So, the expression2² – (2³ – 2) + (2² + 4)  is equal to 6.

Question 7.
(5 – 2) + (6 – 4) – (3 – 1) =
Answer:
We have to calculate the above given expression in order. The order of operation is PEMDAS. These letters stands for Parenthesis, Exponents, Multiplication, Division, Addition, Subtraction.
(5 – 2) + (6 – 4) – (3 – 1) = ?
First solve the parts that are inside parenthesis.
3 + 2 – 2 = ?
Second perform Addition and subtraction operation.
5 – 2 = 3
So, the expression (5 – 2) + (6 – 4) – (3 – 1) is equal to 3.

Question 8.
(4 + 3) × (5 – 2) × (2 – 1)² =
Answer:
We have to calculate the above given expression in order. The order of operation is PEMDAS. These letters stands for Parenthesis, Exponents, Multiplication, Division, Addition, Subtraction.
(4 + 3) × (5 – 2) × (2 – 1)² = ?
First solve the parts that are inside parenthesis.
7 × 3 × 1² = ?
Second solve the parts that have exponents.
7 × 3 × 1 = ?
Third perform multiplication operation.
21 x 1 = 21
So, the expression (4 + 3) × (5 – 2) × (2 – 1)²  is equal to 21.

Question 9.
(4 – 2) + (5 × 2)² – 2 × 12 – 4² =
Answer:
We have to calculate the above given expression in order. The order of operation is PEMDAS. These letters stands for Parenthesis, Exponents, Multiplication, Division, Addition, Subtraction.
(4 – 2) + (5 × 2)² – 2 × 12 – 4² = ?
First solve the parts that are inside parenthesis.
2 + 10² – 2 × 12 – 4² = ?
Second solve the parts that have exponents.
2 + 100 – 2 × 12 – 16 = ?
Third perform multiplication operation.
2 + 100 – 24 – 16 = ?
Fourth perform addition and subtraction operation.
102 – 24 – 16 = ?
76 – 16 = 60 
So, the expression (4 – 2) + (5 × 2)² – 2 × 12 – 4²  is equal to 60.

Question 10.
(1 + 1 + 2) × 7 – 4² =
Answer:
We have to calculate the above given expression in order. The order of operation is PEMDAS. These letters stands for Parenthesis, Exponents, Multiplication, Division, Addition, Subtraction.
(1 + 1 + 2) × 7 – 4² = ?
First solve the parts that are inside parenthesis.
4 × 7 – 4² = ?
Second solve the parts that have exponents.
4 × 7 – 16 = ?
Third perform multiplication operation.
28 – 16 = ?
Fourth perform addition and subtraction operation.
28 – 16 = 12
So, the expression (1 + 1 + 2) × 7 – 4²  is equal to 12.

Question 11.
(13 – 2) – 3² + 10 – (2 × 4) =
Answer:
We have to calculate the above given expression in order. The order of operation is PEMDAS. These letters stands for Parenthesis, Exponents, Multiplication, Division, Addition, Subtraction.
(13 – 2) – 3² + 10 – (2 × 4) = ?
First solve the parts that are inside parenthesis.
11 – 3² + 10 – 8 = ?
Second solve the parts that have exponents.
11 – 9 + 10 – 8 = ?
Third perform addition and subtraction operation.
2 + 10 – 8 = ?
12 – 8 = 4
So, the expression (13 – 2) – 3² + 10 – (2 × 4) is equal to 4.

Question 12.
7² + (4 – 1) × 5 – 2 × (3 – 1)³ =
Answer:
We have to calculate the above given expression in order. The order of operation is PEMDAS. These letters stands for Parenthesis, Exponents, Multiplication, Division, Addition, Subtraction.
7² + (4 – 1) × 5 – 2 × (3 – 1)³ = ?
First solve the parts that are inside parenthesis.
7² + 3 × 5 – 2 × 2³ = ?
Second solve the parts that have exponents.
49 + 3 × 5 – 2 × 8 = ?
Third perform multiplication operation.
49 + 15 – 16 = ?
Fourth perform addition and subtraction operation.
64 – 16 = 48
So, the expression 7² + (4 – 1) × 5 – 2 × (3 – 1)³   is equal to 48.

Excel in your academics by accessing McGraw Hill Math Grade 7 Answer Key PDF Lesson 3.2 Estimating Quotients existing for free of cost.

McGraw-Hill Math Grade 7 Answer Key Lesson 3.2 Estimating Quotients

Exercises Estimate

Question 1.
4568 ÷ 8
Answer:
First look at the two highest digits in the dividend, 45. This cannot be evenly divided by 8. So, round to the closest compatible number, 48.
48 ÷ 8 = 6
Next, add placeholder zeros to the estimated quotient for the place values we ignored in the original dividend. Here we ignored two place values in original dividend. So, add two zeros to the estimated quotient.
4568 ÷ 8 is about 600

Question 2.
2112 ÷ 11
Answer:
First look at the two highest digits in the dividend, 21. This cannot be evenly divided by 11. So, round to the closest compatible number, 22.
22 ÷ 11 = 2
Next, add placeholder zeros to the estimated quotient for the place values we ignored in the original dividend. Here we ignored two place values in original dividend. So, add two zeros to the estimated quotient.
2112 ÷ 11 is about 200

Question 3.
674 ÷ 8
Answer:
First look at the two highest digits in the dividend, 67. This cannot be evenly divided by 8. So, round to the closest compatible number, 64.
64 ÷ 8 = 8
Next, add placeholder zeros to the estimated quotient for the place values we ignored in the original dividend. Here we ignored one place values in original dividend. So, add one zero to the estimated quotient.
674 ÷ 8 is about 80

Question 4.
4657 ÷ 15
Answer:
First look at the two highest digits in the dividend, 46. This cannot be evenly divided by 15. So, round to the closest compatible number, 45.
45 ÷ 15 = 3
Next, add placeholder zeros to the estimated quotient for the place values we ignored in the original dividend. Here we ignored two place values in original dividend. So, add two zeros to the estimated quotient.
4657 ÷ 15 is about 300

Question 5.
35734 ÷ 12
Answer:
First look at the two highest digits in the dividend, 35. This cannot be evenly divided by 12. So, round to the closest compatible number, 36.
36 ÷ 12 = 3
Next, add placeholder zeros to the estimated quotient for the place values we ignored in the original dividend. Here we ignored three place values in original dividend. So, add three zeros to the estimated quotient.
35734 ÷ 12 is about 3,000

Question 6.
4252 ÷ 9
Answer:
First look at the two highest digits in the dividend, 42. This cannot be evenly divided by 9. So, round to the closest compatible number, 45.
45 ÷ 9 = 5
Next, add placeholder zeros to the estimated quotient for the place values we ignored in the original dividend. Here we ignored two place values in original dividend. So, add two zeros to the estimated quotient.
4252 ÷ 9 is about 500

Question 7.
67891 ÷ 16
Answer:
First look at the two highest digits in the dividend, 67. This cannot be evenly divided by 16. So, round to the closest compatible number, 64.
64 ÷ 16 = 4
Next, add placeholder zeros to the estimated quotient for the place values we ignored in the original dividend. Here we ignored three place values in original dividend. So, add three zeros to the estimated quotient.
67891 ÷ 16 is about 4,000

Question 8.
321 ÷ 19
Answer:
First look at the two highest digits in the dividend, 32. This cannot be evenly divided by 19. So, round to the closest compatible number, 38.
38 ÷ 19 = 2
Next, add placeholder zeros to the estimated quotient for the place values we ignored in the original dividend. Here we ignored one place values in original dividend. So, add one zero to the estimated quotient.
321 ÷ 19 is about 20

Question 9.
682 ÷ 35
Answer:
First look at the two highest digits in the dividend, 68. This cannot be evenly divided by 35. So, round to the closest compatible number, 70.
70 ÷ 35 = 2
Next, add placeholder zeros to the estimated quotient for the place values we ignored in the original dividend. Here we ignored one place values in original dividend. So, add one zero to the estimated quotient.
682 ÷ 35 is about 20

Question 10.
92099 ÷ 34
Answer:
First look at the two highest digits in the dividend, 92. This cannot be evenly divided by 34. So, round to the closest compatible number, 102.
102 ÷ 34 = 3
Next, add placeholder zeros to the estimated quotient for the place values we ignored in the original dividend. Here we ignored three place values in original dividend. So, add three zeros to the estimated quotient.
92099 ÷ 34 is about 3,000

Question 11.
678 ÷ 22
Answer:
First look at the two highest digits in the dividend, 67. This cannot be evenly divided by 22. So, round to the closest compatible number, 66.
66 ÷ 22 = 3
Next, add placeholder zeros to the estimated quotient for the place values we ignored in the original dividend. Here we ignored one place values in original dividend. So, add one zero to the estimated quotient.
678 ÷ 22 is about 30

Question 12.
6578 ÷ 34
Answer:
First look at the two highest digits in the dividend, 65. This cannot be evenly divided by 34. So, round to the closest compatible number, 68.
68 ÷ 34 = 2
Next, add placeholder zeros to the estimated quotient for the place values we ignored in the original dividend. Here we ignored two place values in original dividend. So, add two zeros to the estimated quotient.
6578 ÷ 34 is about 200

Question 13.
789 ÷ 28
Answer:
First look at the two highest digits in the dividend, 78. This cannot be evenly divided by 28. So, round to the closest compatible number, 84.
84 ÷ 28 = 3
Next, add placeholder zeros to the estimated quotient for the place values we ignored in the original dividend. Here we ignored one place values in original dividend. So, add one zero to the estimated quotient.
789 ÷ 28 is about 30

Question 14.
4591 ÷ 17
Answer:
First look at the two highest digits in the dividend, 45. This cannot be evenly divided by 17. So, round to the closest compatible number, 51.
51 ÷ 17 = 3
Next, add placeholder zeros to the estimated quotient for the place values we ignored in the original dividend. Here we ignored two place values in original dividend. So, add two zeros to the estimated quotient.
4591 ÷ 17 is about 300

Question 15.
7777 ÷ 44
Answer:
First look at the two highest digits in the dividend, 77. This cannot be evenly divided by 44. So, round to the closest compatible number, 88.
88 ÷ 44 = 2
Next, add placeholder zeros to the estimated quotient for the place values we ignored in the original dividend. Here we ignored two place values in original dividend. So, add two zeros to the estimated quotient.
7777 ÷ 44 is about 200

Question 16.
456 ÷ 7
Answer:
First look at the two highest digits in the dividend, 45. This cannot be evenly divided by 7. So, round to the closest compatible number, 49.
49 ÷ 7 = 7
Next, add placeholder zeros to the estimated quotient for the place values we ignored in the original dividend. Here we ignored one place values in original dividend. So, add one zero to the estimated quotient.
456 ÷ 7 is about 70

Question 17.
96291 ÷ 31
Answer:
First look at the two highest digits in the dividend, 96. This cannot be evenly divided by 31. So, round to the closest compatible number, 93.
93 ÷ 31 = 3
Next, add placeholder zeros to the estimated quotient for the place values we ignored in the original dividend. Here we ignored three place values in original dividend. So, add three zeros to the estimated quotient.
96291 ÷ 31 is about 3,000

Question 18.
91111 ÷ 47
Answer:
First look at the two highest digits in the dividend, 91. This cannot be evenly divided by 47. So, round to the closest compatible number, 94.
94 ÷ 47 = 2
Next, add placeholder zeros to the estimated quotient for the place values we ignored in the original dividend. Here we ignored three place values in original dividend. So, add three zeros to the estimated quotient.
91111 ÷ 47 is about 2,000

Question 19.
69103 ÷ 41
Answer:
First look at the two highest digits in the dividend, 69. This cannot be evenly divided by 41. So, round to the closest compatible number, 82.
82 ÷ 41 = 2
Next, add placeholder zeros to the estimated quotient for the place values we ignored in the original dividend. Here we ignored three place values in original dividend. So, add three zeros to the estimated quotient.
69103 ÷ 41 is about 2,000

Question 20.
13401 ÷ 11
Answer:
First look at the two highest digits in the dividend, 13. This cannot be evenly divided by 11. So, round to the closest compatible number, 11.
11 ÷ 11 = 1
Next, add placeholder zeros to the estimated quotient for the place values we ignored in the original dividend. Here we ignored three place values in original dividend. So, add three zeros to the estimated quotient.
13401 ÷ 11 is about 1,000

Practice the questions of McGraw Hill Math Grade 8 Answer Key PDF Lesson 23.4 Circle Graphs to secure good marks & knowledge in the exams.

McGraw-Hill Math Grade 8 Answer Key Lesson 23.4 Circle Graphs

Exercises
INTERPRET
Question 1.
What are the three major areas of consumer expenditures?

Answer:
The three major areas of consumer expenditures are Housing, other and Transportation because these percentages value of consumption is comparatively more than all.

Explanation:
Consumption percentage of Personal insurance and pensions = 9%.
Consumption percentage of Transportation = 18%.
Consumption percentage of Housing = 30%.
Consumption percentage of food = 16%.
Consumption percentage of other = 27%.

Question 2.
Name the two regions that represent more than 50% of the company sales.

Answer:
Percentage of sales by Region 2 and Region 1 represent more than 50% of the company sales.

Explanation:
Percentage of sales by Region 1 = 17.89%.
Percentage of sales by Region 2 = 36.38%.
Percentage of sales by Region 3 = 15.08%.
Percentage of sales by Region 4 = 11.06%.
Percentage of sales by Region 5 = 10.55%.
Percentage of sales by Region 6 = 9.05%.

Excel in your academics by accessing McGraw Hill Math Grade 7 Answer Key PDF Lesson 26.6 Volume of Solid Figures existing for free of cost.

McGraw-Hill Math Grade 7 Answer Key Lesson 26.6 Volume of Solid Figures

Exercises

CALCULATE VOLUME

Question 1.

Answer:
Volume = 175π cu in.

Explanation:
Volume (V) = πr²h
V = π x 5 x 5 x 7
V = 175π cu in.

Question 2.

Answer:
Volume = 1125π cu in.

Explanation:
Volume (V) = πr²h
V = π x 15 x 15 x 5
V = 1125π cu in.

Question 3.

Answer:
Volume = 100 cu in.

Explanation:
Volume of a cuboid = (length × width × height) cubic units.
Volume (V) = (l × w × h) cubic units.
V = 10 x 5 x 2
V = 100 cu in.

Question 4.

Answer:
Volume = 154 cu in.

Explanation:
Volume of a cuboid = (length × width × height) cubic units.
Volume (V) = (l × w × h) cubic units.
V = 4 x 3.5 x 11
V = 154 cu in.

Question 5.

Answer:
Volume = 108 cu in.
Explanation:
Volume of a Triangular Prism =(1/2) (length × width × height) cubic units.
B = length x base
Volume of a Triangular Prism =(1/2) (length × width × height) cubic units.
Volume (V) =(1/2) (B × H) cubic units.
V = (1/2) x 3 x 6 x 12
V = 108 cu in.

Question 6.

Answer:
Volume = 17 cu in.
Explanation:
B = length x base
Volume of a Triangular Prism =(1/2) (length × width × height) cubic units.
Volume (V) =(1/2) (B × H) cubic units.
V = (1/2) x 1 x 2 x 17
V = 17 cu in.

Excel in your academics by accessing McGraw Hill Math Grade 7 Answer Key PDF Lesson 26.5 Surface Area of Solid Figures existing for free of cost.

McGraw-Hill Math Grade 7 Answer Key Lesson 26.5 Surface Area of Solid Figures

Exercises

CALCULATE SURFACE AREA

Question 1.

Answer:
SA = 54 sq in.
Explanation:
The surface area of a cuboid, add the areas of all 6 faces.
We can also label the length (l), width (w), and height (h) of the prism and use the formula,
SA = 2lw + 2lh + 2hw, to find the surface area.
SA = 2 l w + 2 l h + 2 h w
SA = 2 x 3 x 3 + 2 x 3 x 3 + 2 x 3 x 3
SA = 18 + 18 + 18
SA = 54 sq in.

Question 2.

Answer:
SA = 104 sq m.
Explanation:
The surface area of a cuboid, add the areas of all 6 faces.
We can also label the length (l), width (w), and height (h) of the prism and use the formula,
SA = 2lw + 2lh + 2hw, to find the surface area.
SA = 2 l w + 2 l h + 2 h w
SA = 2 x 6 x 5 + 2 x 5 x 2 + 2 x 2 x 6
SA = 60 + 20 + 24
SA = 104 sq in
Question 3.

Answer:
SA = 450.325 sq in.
Explanation:
The surface area of a cuboid, add the areas of all 6 faces.
We can also label the length (l), width (w), and height (h) of the prism and use the formula,
SA = 2lw + 2lh + 2hw, to find the surface area.
SA = 2 l w + 2 l h + 2 h w
SA = 2 x 7.25 x 10 + 2 x 10 x 8.85 + 2 x 8.85 x 7.25
SA = 145 + 177 + 128.325
SA = 450.325 sq in

Question 4.

Answer:
SA = 125.46 sq ft.
Explanation:
The surface area of a cuboid, add the areas of all 6 faces.
We can also label the length (l), width (w), and height (h) of the prism and use the formula,
SA = 2lw + 2lh + 2hw, to find the surface area.
SA = 2 l w + 2 l h + 2 h w
SA = 2 x 4.2 x 1.5 + 2 x 1.5 x 9.9 + 2 x 9.9 x 4.2
SA = 12.6 + 29.7 + 83.16
SA = 125.46 sq ft.

Question 5.

Answer:
SA = 856 sq cm.
Explanation:
The surface area of a cuboid, add the areas of all 6 faces.
We can also label the length (l), width (w), and height (h) of the prism and use the formula,
SA = 2lw + 2lh + 2hw, to find the surface area.
SA = 2 l w + 2 l h + 2 h w
SA = 2 x 10 x 12 + 2 x 12 x 14 + 2 x 14 x 10
SA = 240 + 336 + 280
SA = 856 sq cm.
Question 6.

Answer:
SA = 862 sq m.
Explanation:
The surface area of a cuboid, add the areas of all 6 faces.
We can also label the length (l), width (w), and height (h) of the prism and use the formula,
SA = 2lw + 2lh + 2hw, to find the surface area.
SA = 2 l w + 2 l h + 2 h w
SA = 2 x 11 x 12 + 2 x 12 x 13 + 2 x 13 x 11
SA = 264 + 312 + 286
SA = 862 sq m.

Question 7.

Answer:
SA = 478 sq m.
Explanation:
The surface area of a cuboid, add the areas of all 6 faces.
We can also label the length (l), width (w), and height (h) of the prism and use the formula,
SA = 2lw + 2lh + 2hw, to find the surface area.
SA = 2 l w + 2 l h + 2 h w
SA = 2 x 9 x 7 + 2 x 7 x 11 + 2 x 11 x 9
SA = 126 + 154 + 198
SA = 478 sq m

Question 8.

Answer:
SA = 114 sq in.

Explanation:
The surface area of a cuboid, add the areas of all 6 faces.
We can also label the length (l), width (w), and height (h) of the prism and use the formula,
SA = 2lw + 2lh + 2hw, to find the surface area.
SA = 2 l w + 2 l h + 2 h w
SA = 2 x 8 x 3 + 2 x 3 x 3 + 2 x 3 x 8
SA = 48 + 18 + 48
SA = 114 sq in

Question 9.

Answer:
SA = 370 sq ft.
Explanation:
The surface area of a cuboid, add the areas of all 6 faces.
We can also label the length (l), width (w), and height (h) of the prism and use the formula,
SA = 2lw + 2lh + 2hw, to find the surface area.
SA = 2 l w + 2 l h + 2 h w
SA = 2 x 5 x 25 + 2 x 25 x 2 + 2 x 2 x 5
SA = 250 + 100 + 20
SA = 370 sq ft

Keep your answer in a form of pi.

Question 10.

Answer:
SA = 60π sq cm.
Explanation:
surface area of cylinder = Area of curved surface + Area of two circular regions
= CSA of cylinder + πr² + πr²
= 2πrh + 2πr²
Therefore, Total surface area of cylinder = 2πr (h + r) sq. units
Where r is the radius of the base and h is the height of the cylinder.
SA of cylinder = 2πr (h + r) sq. units
SA = 2π x 3 (7 + 3)
SA = 60π sq. units

Question 11.

Answer:
SA = 170π sq in.
Explanation:
surface area of cylinder = Area of curved surface + Area of two circular regions
= CSA of cylinder + πr² + πr²
= 2πrh + 2πr²
Therefore, Total surface area of cylinder = 2πr (h + r) sq. units
Where r is the radius of the base and h is the height of the cylinder.
SA of cylinder = 2πr (h + r) sq. units
SA = 2π x 5( 12 + 5)
SA = 170π sq in.

Question 12.

Answer:
SA = 252π sq in.
Explanation:
surface area of cylinder = Area of curved surface + Area of two circular regions
= CSA of cylinder + πr² + πr²
= 2πrh + 2πr²
Therefore, Total surface area of cylinder = 2πr (h + r) sq. units
Where r is the radius of the base and h is the height of the cylinder.
SA of cylinder = 2πr (h + r) sq. units
SA = 2π x 9 ( 5 + 9)
SA = 252π sq in.

Question 13.

Answer:
SA = 275π sq m.
Explanation:
surface area of cylinder = Area of curved surface + Area of two circular regions
= CSA of cylinder + πr² + πr²
= 2πrh + 2πr²
Therefore, Total surface area of cylinder = 2πr (h + r) sq. units
Where r is the radius of the base and h is the height of the cylinder.
SA of cylinder = 2πr (h + r) sq. units
SA = 2π x11(1.5 + 11)
SA = 275π sq m.

Question 14.

Answer:
SA = 96.3π sq ft.
Explanation:
surface area of cylinder = Area of curved surface + Area of two circular regions
= CSA of cylinder + πr² + πr²
= 2πrh + 2πr²
Therefore, Total surface area of cylinder = 2πr (h + r) sq. units
Where r is the radius of the base and h is the height of the cylinder.
SA of cylinder = 2πr (h + r) sq. units
SA = 2π x 4.5(6.2 + 4.5)
SA = 96.3π sq ft.

Question 15.

Answer:
SA = 352π sq in.
Explanation:
surface area of cylinder = Area of curved surface + Area of two circular regions
= CSA of cylinder + πr² + πr²
= 2πrh + 2πr²
Therefore, Total surface area of cylinder = 2πr (h + r) sq. units
Where r is the radius of the base and h is the height of the cylinder.
SA of cylinder = 2πr (h + r) sq. units
SA = 2π x 8(14+ 8)
SA = 352π sq in.

Question 16.

Answer:
SA = 84π sq cm.
Explanation:
surface area of cylinder = Area of curved surface + Area of two circular regions
= CSA of cylinder + πr² + πr²
= 2πrh + 2πr²
Therefore, Total surface area of cylinder = 2πr (h + r) sq. units
Where r is the radius of the base and h is the height of the cylinder.
SA of cylinder = 2πr (h + r) sq. units
SA = 2π x 6( 1 + 6)
SA = 84π sq cm.

Question 17.

Answer:
SA = 138.125π sq in.
Explanation:
surface area of cylinder = Area of curved surface + Area of two circular regions
= CSA of cylinder + πr² + πr²
= 2πrh + 2πr²
Therefore, Total surface area of cylinder = 2πr (h + r) sq. units
Where r is the radius of the base and h is the height of the cylinder.
SA of cylinder = 2πr (h + r) sq. units
SA = 2π x 4.25 (12 + 4.25)
SA = 138.125π sq in.

Question 18.

Answer:
SA = 600π sq in.
Explanation:
surface area of cylinder = Area of curved surface + Area of two circular regions
= CSA of cylinder + πr² + πr²
= 2πrh + 2πr²
Therefore, Total surface area of cylinder = 2πr (h + r) sq. units
Where r is the radius of the base and h is the height of the cylinder.
SA of cylinder = 2πr (h + r) sq. units
SA = 2π x10 (20 + 10)
SA = 600π sq in.

Excel in your academics by accessing McGraw Hill Math Grade 7 Answer Key PDF Lesson 26.4 Circles existing for free of cost.

McGraw-Hill Math Grade 7 Answer Key Lesson 26.4 Circles

Exercises

IDENTIFY

Question 1.
What is the radius of the circle?

Answer:
5,
Explanation:
A radius is a line segment that begins at a circle’s origin and extends to tis circumference.
in the circle , all radii are equal in the length.
So, radius r = 5 units.

Question 2.
What is the radius of the circle?

Answer:
5 cm,
Explanation:
A radius is a line segment that begins at a circle’s origin and extends to tis circumference.
In the circle , all radii are equal in the length.
diameter d = 10 cm
radius r = 5 cm.

Question 3.
Identify the 2 radii and the one chord below.

Answer:
\(\overline{AX}\) and \(\overline{XB}\) radii, \(\overline{AB}\) chord.
Explanation:
line \(\overline{AX}\) and \(\overline{XB}\) are radii lines which are passing though center point.
\(\overline{AB}\) is chord, which is not touching center of the circle.
A line segment that has both the end points on the circumference of the circle is called a chord.

Question 4.
Identify the radius and chord in the figure below.

Answer:
\(\overline{VY}\) radius,
\(\overline{WX}\) is chord.
Explanation:
A radius is a line segment that begins at a circle’s origin and extends to tis circumference.
In the circle , all radii are equal in the length.
\(\overline{VY}\) is a radius line which is passing though center,
and \(\overline{WX}\) is chord.
A line segment that has both the end points on the circumference of the circle is called a chord.

Question 5.
What are the 5 chords formed by inscribing the pentagon inside of the circle below?

Answer:
\(\overline{XY}\), \(\overline{YZ}\), \(\overline{ZV}\), \(\overline{VW}\) and \(\overline{WX}\).
Explanation:
A line segment that has both the end points on the circumference of the circle is called a chord.
pentagon has five chords.
\(\overline{XY}\), \(\overline{YZ}\), \(\overline{ZV}\), \(\overline{VW}\) and \(\overline{WX}\) are chords of the circle.

Question 6.
Using the letters provided, can the diameter in the figure below be named? Explain your answer.

Answer:
No, \(\overline{CD}\) does not go through center point O.
Explanation:
A line segment that has both the end points on the circumference of the circle is called a chord.
\(\overline{AB}\) and \(\overline{CD}\) are chords of the circle

Question 7.
How many ways can you describe the radius in this circle?

Answer:
\(\overline{EA}\), \(\overline{AE}\), \(\overline{EC}\), \(\overline{EB}\), \(\overline{CE}\), and \(\overline{BE}\)
Explanation:
\(\overline{EA}\), \(\overline{AE}\), \(\overline{EC}\), \(\overline{EB}\), \(\overline{CE}\), and \(\overline{BE}\) are radii of the circle,
which are originating at center of the circle and end at on the circumference of the circle.

Question 8.
Which two lines in the circle below are chords?

Answer:
\(\overline{AB}\) and \(\overline{CD}\) are chords of the circle.
Explanation:
A line segment that has both the end points on the circumference of the circle is called a chord.
\(\overline{AB}\) and \(\overline{CD}\) are chords of the circle.

Question 9.
Identify the chord and the diameter below.

Answer:
\(\overline{RA}\) is diameter of the circle,
\(\overline{QA}\) is the chord of the circle.
Explanation:
A line segment that has both the end points on the circumference of the circle is called a chord.
\(\overline{RA}\) is diameter of the circle,
\(\overline{QA}\) is the chord of the circle.

Question 10.
Calculate the area of the circle below. Leave your answer in the form of pi.

Answer:
100π in
Explanation:
Area of the circle = πr²
radius of the circle is 10 in
A = π x 10 x 10
A = 100π in

Calculate the circumference and the area.

Question 11.

Answer:
Circumference = __________________
Area = _____________
Answer:
Circumference = 10π  units.
Area = 25π sq units.
Explanation:
Circumference of the circle C = 2 π r
C = 2 π 5
C = 10π units
Area of the circle = πr²
radius of the circle is 10 in
A = π x 5 x 5
A = 25π sq units.

Question 12.

Circumference = ____________
Area = _____________
Answer:
Circumference = 14π units.
Area = 49π sq units.
Explanation:
Circumference of the circle C = 2 π r
C = 2 π 7
C = 14π units
Area of the circle = πr²
radius of the circle is 7 in
A = π x 7 x 7
A = 49π sq units.

Excel in your academics by accessing McGraw Hill Math Grade 7 Answer Key PDF Lesson 26.3 Polygons existing for free of cost.

McGraw-Hill Math Grade 7 Answer Key Lesson 26.3 Polygons

Exercises

IDENTIFY

Label the polygons.

Question 1.

Answer:
Heptagon.
Explanation:
Hepta means seven and gon means sides.
A Heptagon is a polygon with seven sides and seven angles.
It has seven straight sides and seven corners or vertices.

Question 2.

Answer:
Octagon.
Explanation:
Octa means eight and gon means sides.
So, octagon consists of 8 sides.
It has eight angles and eight corners.

Question 3.

Answer:
Pentagon.
Explanation:
Penta denotes five and gon denotes angle.
A pentagon is a simple polygon, which has five sides and five angles.

Question 4.

Answer:
Hexagon.
Explanation:
Hexa means six and gona means angles.
A hexagon is a closed two-dimensional polygon with six sides.
Hexagon has 6 vertices and 6 angles.

Question 5.

Answer:
Hexagon.
Explanation:
Hexa means six and gona means angles.
A hexagon is a closed two-dimensional polygon with six sides.
Hexagon has 6 vertices and 6 angles.

Question 6.

Answer:
Octagon.
Explanation:
Octa means eight and gon means sides.
So, octagon consists of 8 sides.
It has eight angles and eight corners.

Question 7.

Answer:
Heptagon.
Explanation:
Hepta means seven and gon means sides.
A Heptagon is a polygon with seven sides and seven angles.
It has seven straight sides and seven corners or vertices.

Question 8.

Answer:
Pentagon.
Explanation:
Penta denotes five and gon denotes angle.
A pentagon is a simple polygon, which has five sides and five angles.

Question 9.
Are these two figures congruent? Why or why not?

Answer:
Yes, sides and angle measures are same.
Explanation:
Two geometric figures are said to be congruent,
if they have same size and shape.
The mirror image of one shape is same as the other.

Question 10.
Are these two figures congruent? Why or why not?

Answer:
No, bases are of different lengths.
Explanation:
Two geometric figures are said to be congruent,
if they have same size and shape.
The above given images are not congruent,
As, their bases are different.

Excel in your academics by accessing McGraw Hill Math Grade 7 Answer Key PDF Lesson 26.2 Quadrilaterals existing for free of cost.

McGraw-Hill Math Grade 7 Answer Key Lesson 26.2 Quadrilaterals

Exercises

IDENTIFY

Label the shape as: square, rectangle, rhombus, trapezoid, or kite.

Question 1.

Answer:
Square,
Explanation:
A square is closed, two-dimensional shape with 4 equal sides.

Question 2.

Answer:
Kite,
Explanation:
A flat shape with 4 straight sides that has two pairs of sides,
which are two adjacent sides and angles are equal in length.

Question 3.

Answer:
Rectangle,
Explanation:
A Rectangle is a four sided-polygon, having all the internal angles equal to 90 degrees.
The two sides at each corner or vertex meet at the right angles.
The opposite sides of the rectangle are equal in length which makes it different from a square.

Question 4.

Answer:
Rhombus,
Explanation:
Rhombus is a quadrilateral with all equal sides.
Since opposite sides of a parallelogram are equal.
So, Rhombus is a special type of a parallelogram whose all sides are equal.

Question 5.

Answer:
Kite,
Explanation:
A flat shape with 4 straight sides that has two pairs of sides,
which are two adjacent sides that are equal in length.

Question 6.

Answer:
Rhombus,
Explanation:
Rhombus is a quadrilateral with all equal sides.
Since opposite sides of a parallelogram are equal.
So, Rhombus is a special type of a parallelogram whose all sides are equal.

Question 7.

Answer:
Trapezoid,
Explanation:
A trapezoid is a flat closed shape having 4 straight sides,
with one pair of parallel sides.

Question 8.

Answer:
Rectangle,
Explanation:
A Rectangle is a four sided-polygon, having all the internal angles equal to 90 degrees.
The two sides at each corner or vertex meet at the right angles.
The opposite sides of the rectangle are equal in length which makes it different from a square.

Question 9.

Answer:
Square,
Explanation:
A square is closed, two-dimensional shape with 4 equal sides and angles.