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Excel in your academics by accessing McGraw Hill Math Grade 7 Answer Key PDF Lesson 5.4 Multiplying and Dividing with Negative Numbers existing for free of cost.

McGraw-Hill Math Grade 7 Answer Key Lesson 5.4 Multiplying and Dividing with Negative Numbers

Exercises Multiply or Divide

Question 1.
5 × (-3) = _________
Answer:
5 × (-3) = -15
Explanation:
The above two given numbers have different signs, one is positive and one is negative. So, there product is negative. Multiply 5 with -3 the product is equal to -15.

Question 2.
13 × (-10) = _________
Answer:
13 × (-10) = -130
Explanation:
The above two given numbers have different signs, one is positive and one is negative. So, there product is negative. Multiply 13 with -10 the product is equal to -130.

Question 3.
100 × (-1) = _________
Answer:
100 × (-1) = -100
Explanation:
The above two given numbers have different signs, one is positive and one is negative. So, there product is negative. Multiply 100 with -1 the product is equal to -100.

Question 4.
23 × (-3) = _________
Answer:
23 × (-3) = -69
Explanation:
The above two given numbers have different signs, one is positive and one is negative. So, there product is negative. Multiply 23 with -3 the product is equal to -69.

Question 5.
15 × (-15) = _________
Answer:
15 × (-15) = -225
Explanation:
The above two given numbers have different signs, one is positive and one is negative. So, there product is negative. Multiply 15 with -15 the product is equal to -225.

Question 6.
(-12) × 20 = _________
Answer:
(-12) × 20 = -240
Explanation:
The above two given numbers have different signs, one is negative and one is positive. So, there product is negative. Multiply -12 with 20 the product is equal to -240.

Question 7.
(-2) × (-4) = _________
Answer:
(-2) × (-4) = 8
Explanation:
The above two given numbers have same signs. So, there product is positive. Multiply -2 with -4 the product is equal to 8.

Question 8.
(-1) × 1 = _________
Answer:
(-1) × 1 = -1
Explanation:
The above two given numbers have different signs, one is negative and one is positive. So, there product is negative. Multiply -1 with 1 the product is equal to -1.

Question 9.
(-125) × (-1) = _________
Answer:
(-125) × (-1) = 125
Explanation:
The above two given numbers have same signs. So, there product is positive. Multiply -125 with -1 the product is equal to 125.

Question 10.
(-3) × 103 = _________
Answer:
(-3) × 103 = -309
Explanation:
The above two given numbers have different signs, one is negative and one is positive. So, there product is negative. Multiply -3 with 103 the product is equal to -309.

Question 11.
(-20) ÷ 20 = _________
Answer:
(-20) ÷ 20 = -1
Explanation:
The above two given numbers have different signs, one is negative and one is positive. So, there quotient is negative. Divide -20 by 20 the quotient is equal to -1

Question 12.
(-22) × 0 = _____
Answer:
(-22) × 0 = 0
Explanation:
Multiply -22 with 0 the product is equal to 0. Zero doesn’t have negative sign.

Question 13.
(-3) × (-1) = _____
Answer:
(-3) × (-1) = 3
Explanation:
The above two given numbers have same signs. So, there product is positive. Multiply -3 with -1 the product is equal to 3.

Question 14.
(-12) ÷ 4 = _____
Answer:
(-12) ÷ 4 = -3
Explanation:
The above two given numbers have different signs, one is negative and one is positive. So, there quotient is negative. Divide -12 by 4 the quotient is equal to -3.

Question 15.
15 ÷ (-3) = ____
Answer:
15 ÷ (-3) = -5
Explanation:
The above two given numbers have different signs, one is positive and one is negative. So, there quotient is negative. Divide 15 by -3 the quotient is equal to -5.

Question 16.
(-27) ÷ (-9 ÷ \(\frac{1}{9}\)) = ____
Answer:
(-27) ÷ (-9 ÷ \(\frac{1}{9}\)) = \(\frac{1}{3}\)
Explanation:
The above two given numbers have same signs. So, there quotient is positive. Divide (-27) by (-9 ÷ \(\frac{1}{9}\)) the quotient is equal to \(\frac{1}{3}\).

Excel in your academics by accessing McGraw Hill Math Grade 7 Answer Key PDF Lesson 5.3 Adding and Subtracting with Negative Numbers existing for free of cost.

McGraw-Hill Math Grade 7 Answer Key Lesson 5.3 Adding and Subtracting with Negative Numbers

Exercises Add

Question 1.
17 + (-3) = ____
Answer:
17 + (-3) = 14
Explanation:
Perform addition operation on above given numbers. Add 17 to -3 the sum is equal to 14.

Question 2.
(-15) + 6 = _____
Answer:
(-15) + 6 = -9
Explanation:
Perform addition operation on above two given numbers. Add -15 with 6 the sum is equal to -9.

Question 3.
1 + (-12) = ____
Answer:
1 + (-12) = -11
Explanation:
Perform addition operation on above two given numbers. Add 1 with -12 the sum is equal to -11.

Question 4.
75 + (-36) = ____
Answer:
75 + (-36) = 39
Explanation:
Perform addition operation on above two given numbers. Add 75 with -36 the sum is equal to 39.

Question 5.
110 + (-56) + 14 = ____
Answer:
110 + (-56) + 14 = 68
Explanation:
Perform addition operation on above three given numbers. Add 110 with -56 and 14 the sum is equal to 68.

Question 6.
95 + (-65) + (-1) = _______________
Answer:
95 + (-65) + (-1) = 29
Explanation:
Perform addition operation on above three given numbers. Add 95 with -65 and -1 the sum is equal to 29.

Question 7.
(-20) + (-20) + 7 = ___
Answer:
(-20) + (-20) + 7 = -33
Explanation:
Perform addition operation on above three given numbers. Add -20 with -20 and 7 the sum is equal to -33.

Question 8.
51 + (-33) + 20 = __________________
Answer:
51 + (-33) + 20 = 38
Explanation:
Perform addition operation on above three given numbers. Add 51 with -33 and 20 the sum is equal to 38.

Question 9.
30 + (-22) = __________
Answer:
30 + (-22) = 8
Explanation:
Perform addition operation on above two given numbers. Add 30 to -22 the sum is equal to 8.

Question 10.
18 + 18 + (-18) = ________________
Answer:
18 + 18 + (-18) = 18
Explanation:
Perform addition operation on above three given numbers. Add 18 with 18 and -18 the sum is equal to 18.

Question 11.
(-25) + 14 = ____
Answer:
(-25) + 14 = -11
Explanation:
Perform addition operation on above two given numbers. Add -25 to 14  the sum is equal to -11.

Question 12.
80 + 13 + 29 + (-100) = ___
Answer:
80 + 13 + 29 + (-100) = 22
Explanation:
Perform addition operation on above four given numbers. Add 80 to 13, 29 and -100 the sum is equal to 22.

Question 13.
57 + (-39) + 10 = __________
Answer:
57 + (-39) + 10 = 28
Explanation:
Perform addition operation on above three given numbers. Add 57 with -39 and 10 the sum is equal to 28.

Question 14.
(-23) + 63 + (-9) = ______
Answer:
(-23) + 63 + (-9) = 31
Explanation:
Perform addition operation on above three given numbers. Add -23 with 63 and -9 the sum is equal to 31.

Question 15.
4 + 2 + (-7) + (-1) = ____________
Answer:
4 + 2 + (-7) + (-1) = -2
Explanation:
Perform addition operation on above four given numbers. Add 4 with 2,-7 and -1 the sum is equal to -2.

Question 16.
1 + (-9) + 9 = ____
Answer:
1 + (-9) + 9 = 1
Explanation:
Perform addition operation on above three given numbers. Add 1 with -9 and 9 the sum is equal to 1.

Excel in your academics by accessing McGraw Hill Math Grade 7 Answer Key PDF Lesson 5.2 Absolute Value existing for free of cost.

McGraw-Hill Math Grade 7 Answer Key Lesson 5.2 Absolute Value

Exercises Solve

Question 1.
|-9| = ____
Answer:
|-9| = 9
The absolute value of |-9| is 9.
Explanation:
The absolute value of 9 doesn’t tell whether the number is positive or negative, it just means that the number is 9 away from zero.

Question 2.
|-567| = ____
Answer:
|-567| = 567
The absolute value of |-567| is 567.
Explanation:
The absolute value of 567 doesn’t tell whether the number is positive or negative, it just means that the number is 567 away from zero.

Question 3.
|48| = ___
Answer:
|48| = 48
The absolute value of |48| is 48.
Explanation:
The absolute value of 48 doesn’t tell whether the number is positive or negative, it just means that the number is 48 away from zero.

Question 4.
|-0.24| = ____
Answer:
|-0.24| = 0.24
The absolute value of |-0.24| is 0.24.
Explanation:
The absolute value of 0.24 doesn’t tell whether the number is positive or negative, it just means that the number is 0.24 away from zero.

Question 5.
|\(\frac{-7}{8}\)| = ____
Answer:
|\(\frac{-7}{8}\)| = \(\frac{7}{8}\)
The absolute value of |\(\frac{-7}{8}\)| is \(\frac{7}{8}\).
Explanation:
The absolute value of \(\frac{7}{8}\) doesn’t tell whether the number is positive or negative, it just means that the number is \(\frac{7}{8}\) away from zero.

Question 6.
|\(\frac{1}{3}\)| = ____
Answer:
|\(\frac{1}{3}\)| = \(\frac{1}{3}\)
The absolute value of |\(\frac{1}{3}\)| is \(\frac{1}{3}\).
Explanation:
The absolute value of \(\frac{1}{3}\) doesn’t tell whether the number is positive or negative, it just means that the number is \(\frac{1}{3}\) away from zero.

Compare using <, >, or =

Question 7.
|—2| ____ |—8|
Answer:
|-2| < |-8|
Explanation:
The absolute value of |-2|  is 2.
The absolute value of |-8|  is 8.
So, |-2| is less than |-8|.

Question 8.
|56| ___ |-56|
Answer:
|56| = |-56|
Explanation:
The absolute value of |56|  is 56.
The absolute value of |-56|  is 56.
So, |56| is equal to |-56|.

Question 9.
|-12| ___ |-3|
Answer:
|-12| > |-3|
The absolute value of |-12|  is 12.
The absolute value of |-3|  is 3.
So, |-12| is greater than |-3|.

Question 10.
|-4| _______ |567|
Answer:
|-4| < |567|
Explanation:
The absolute value of |-4| is 4.
The absolute value of |567| is 567.
So, |-4| is less than |567|.

Question 11.
|35| _______ |—98|
Answer:
|35| < |-98|
Explanation:
The absolute value of |35| is 35.
The absolute value of |-98| is 98.
So, |35| is less than |-98|.

Question 12.
|6789| _______ |6799|
Answer:
|6789| < |6799|
Explanation:
The absolute value of |6789| is 6789.
The absolute value of |6799| is 6799.
So, |6789| is less than |6799|.

Solve

Question 13.
If Kim’s bank account balance is $ -12.48, how much would she need to deposit in order to bring her balance to $100.00? _________________________________________
Answer:
Kim’s bank account balance is $ -12.48.
To bring her balance to $100.00. She need to deposit $112.48.
$-12.48 + $100.00 + $12.48 = $100.00

Question 14.
If gas prices were $1.20 above average in March and $1.36 below average in April, which month was further from its average price?
Answer:
April month was further from its average price.

Question 15.
Bela’s bank account balance is $15.00. If she spends $12.86 on a shirt and $3.24 on a bracelet, what is her new balance? _______________________________________________
Answer:
Bela’s bank account balance is $15.00.
She want to spend$12.86 on a shirt and $3.24 on a bracelet.
$15.00 – $12.86 – $3.24 = -$1.10
Bela’s new balance is -$1.10.

Excel in your academics by accessing McGraw Hill Math Grade 7 Answer Key PDF Lesson 5.1 Negative Numbers existing for free of cost.

McGraw-Hill Math Grade 7 Answer Key Lesson 5.1 Negative Numbers

Exercises Add

Question 1.
(-4) + 4 =
Answer:
(-4) + 4 = 0
Explanation:
Perform addition operation on above two given numbers. When we add a negative number to its inverse the total is 0. Add -4 to 4 the sum is equal to 0.

Question 2.
(-110) + 210 =
Answer:
(-110) + 210 = 100
Explanation:
Perform addition operation on above two given numbers. Add -110 to 210 the sum is equal to 100.

Question 3.
(-9) + 17 + 9 + (-17) =
Answer:
(-9) + 17 + 9 + (-17) = 0
Explanation:
Perform addition operation on above given numbers. When we add a negative number to its inverse the total is 0. Add – 9 with 17, 9 and -17 the sum is equal to 0.

Question 4.
145 + (-125) =
Answer:
145 + (-125) = 20
Explanation:
Perform addition operation on above two given numbers. Add 145 to -125 the sum is equal to 20.

Question 5.
48 + (-38) =
Answer:
48 + (-38) = 10
Explanation:
Perform addition operation on above two given numbers. Add 48 to -38 the sum is equal to 10.

Question 6.
(-75) + 55 + 732 + (-712) =
Answer:
(-75) + 55 + 732 + (-712) = 0
Explanation:
Perform addition operation on above given numbers. Add -75 to 55, 732 and -712 the sum is equal to 0.

Question 7.
\(\frac{1}{4}\) + \(\left(-\frac{1}{4}\right)\) =
Answer:
\(\frac{1}{4}\) + \(\left(-\frac{1}{4}\right)\) = 0
Explanation:
Perform addition operation on above two given numbers. When we add a negative number to its inverse the total is 0. Add 1/4 to -1/4 the sum is equal to 0.

Question 8.
541 + 641 + (-741) + (-641) =
Answer:
541 + 641 + (-741) + (-641) = – 200
Explanation:
Perform addition operation on above given numbers. Add 541 with 641, -741 and -641 the sum is equal to -200.

Question 9.
44 + (-77) + 0 =
Answer:
44 + (-77) + 0 = -33
Explanation:
Perform addition operation on above given numbers. Add 44 to -77 and 0 the sum is equal to -33.

Question 10.
16 + \(\frac{1}{4}\) + \(\left(-\frac{1}{4}\right)\) + (-10) =
Answer:
16 + \(\frac{1}{4}\) + \(\left(-\frac{1}{4}\right)\) + (-10) = 6
Explanation:
Perform addition operation on above given numbers. Add 16 to 1/4, -1/4 and -10 the sum is equal to 6.

Excel in your academics by accessing McGraw Hill Math Grade 7 Answer Key PDF Lesson 4.4 Zero Property, Equality Properties existing for free of cost.

McGraw-Hill Math Grade 7 Answer Key Lesson 4.4 Zero Property, Equality Properties

Exercises Solve

Question 1.
5 × 0 =
Answer:
5 × 0 = 0 
Explanation:
Any number multiplied with zero is equal to zero. This property is called as zero property of multiplication.

Question 2.
0(1 + 4) =
Answer:
0(1 + 4)
= (0 x 1) + (0 x 4)
= 0 + 0
= 0
0(1 + 4) = 0
Explanation:
Zero property of addition and multiplication states that any addend + 0 will not change the total and any number multiplied with 0 is equal to 0.

Question 3.
0 × 1.111 =
Answer:
0 × 1.111 = 0 
Explanation:
Any number multiplied with zero is equal to zero. This property is called as zero property of multiplication.

Question 4.
7(0 + 5) =
Answer:
7(0 + 5)
= (7 x 0) + (7 x 5)
= 0 + 35
= 35
7(0 + 5) = 35
Explanation:
Zero property of addition and multiplication states that any addend + 0 will not change the total and any number multiplied with 0 is equal to 0.

Question 5.
0 × 0 × 3 × 9 =
Answer:
0 × 0 × 3 × 9 = 0
Explanation:
Any number multiplied with zero is equal to zero. This property is called as zero property of multiplication.

Question 6.
233.31 × 0 =
Answer:
233.31 × 0 = 0
Explanation:
Any number multiplied with zero is equal to zero. This property is called as zero property of multiplication.

Question 7.
0 × 200.893 =
Answer:
0 × 200.893 = 0
Explanation:
Any number multiplied with zero is equal to zero. This property is called as zero property of multiplication.

Question 8.
0 × 0 + 2 =
Answer:
0 × 0 + 2 = 0 + 2 = 2
Explanation:
Multiply 0 with 0 the product is equal to 0. Add 0 with 2 the sum is equal to 2.

Question 9.
0 (2 + 0) =
Answer:
0 (2 + 0) = (0 x 2) + (0 x 0) = 0
0 (2 + 0) = 0
Explanation:
Zero property of addition and multiplication states that any addend + 0 will not change the total and any number multiplied with 0 is equal to 0.

For questions 10-13, answer yes or no and briefly explain your answer.

Question 10.
If 8 + 1 = 6 + 3, then does 4(8 + 1) = 4(6 + 3)?
Answer:
Yes, If 8 + 1 = 6 + 3, then 4(8 + 1) = 4(6 + 3).
Explanation:
In equality property of multiplication the equation is equal if we multiply both sides by the same number. Here we are multiplying both sides by 4. So the answer is yes.

Question 11.
If 6 × 9 = 54, then does 4 + (6 × 9) = 54 + 4?
Answer:
Yes, If 6 × 9 = 54, then 4 + (6 × 9) = 54 + 4.
Explanation:
In equality property of addition the equation is equal if we add both sides by the same number. Here we are adding both sides by 4. So the answer is yes.

Question 12.
If \(\frac{1}{5}\) = \(\frac{3}{15}\), then does \(\frac{1}{5}\) – 5 = \(\frac{3}{15}\) – 5 ?
Answer:
Yes, If \(\frac{1}{5}\) = \(\frac{3}{15}\), then \(\frac{1}{5}\) – 5 = \(\frac{3}{15}\) – 5.
Explanation:
In equality property of subtraction the equation is equal if we subtract both sides by the same number. Here we are subtracting both sides by 5. So the answer is yes.

Question 13.
If 5 – 1 = 20 × .2, then does \(\frac{(5-1)}{22}\) = \(\frac{(20 \times .2)}{22}\)?
Answer:
Yes, If 5 – 1 = 20 × .2, then does \(\frac{(5-1)}{22}\) = \(\frac{(20 \times .2)}{22}\).
Explanation:
In equality property of division the equation is equal if we divide both sides by the same number. Here we are dividing both sides by 22. So the answer is yes.

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%.