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Find each product.
1. 4 • 4 2. 7 • 7 3. 5 • 5 4. 9 • 9
Perform the indicated operations.
5. 3 + 12 – 7 6. 6 • 1 ÷ 2
7. 4 – 2 + 9 8. 10 – 5 – 4
9. 5 • 5 + 7 10. 30 ÷ 6 • 2
ALGEBRA 1 LESSON 1-2ALGEBRA 1 LESSON 1-2
Exponents and Order of OperationsExponents and Order of Operations
1-2
1. 4 • 4 = 16
2. 7 • 7 = 49
3. 5 • 5 = 25
4. 9 • 9 = 81
5. 3 + 12 – 7 = (3 + 12) – 7 = 15 – 7 = 8
6. 6 • 1 ÷ 2 = (6 • 1) ÷ 2 = 6 ÷ 2 = 3
7. 4 – 2 + 9 = (4 – 2) + 9 = 2 + 9 = 11
8. 10 – 5 – 4 = (10 – 5) – 4 = 5 – 4 = 1
9. 5 • 5 + 7 = (5 • 5) + 7 = 25 + 7 = 32
10. 30 ÷ 6 • 2 = (30 ÷ 6) • 2 = 5 • 2 = 10
ALGEBRA 1 LESSON 1-2ALGEBRA 1 LESSON 1-2
Exponents and Order of OperationsExponents and Order of Operations
Solutions
1-2
Simplify 32 + 62 – 14 • 3.
32 + 62 – 14 • 3 = 32 + 36 – 14 • 3 Simplify the power: 62 = 6 • 6 = 36.
= 32 + 36 – 42 Multiply 14 and 3.
= 68 – 42 Add and subtract in order from left to right.
= 26 Subtract.
ALGEBRA 1 LESSON 1-2ALGEBRA 1 LESSON 1-2
Exponents and Order of OperationsExponents and Order of Operations
1-2
Evaluate 5x = 32 ÷ p for x = 2 and p = 3.
5x + 32 ÷ p = 5 • 2 + 32 ÷ 3 Substitute 2 for x and 3 for p.
= 5 • 2 + 9 ÷ 3 Simplify the power.
= 10 + 3 Multiply and divide from left to right.
= 13 Add.
ALGEBRA 1 LESSON 1-2ALGEBRA 1 LESSON 1-2
Exponents and Order of OperationsExponents and Order of Operations
1-2
Find the total cost of a pair of jeans that cost $32 and have
an 8% sales tax.
total cost original price sales taxC = p + r • p
sales tax rate
C = p + r • p= 32 + 0.08 • 32 Substitute 32 for p. Change 8% to 0.08 and
substitute 0.08 for r.
= 32 + 2.56 Multiply first.
= 34.56 Then add.
The total cost of the jeans is $34.56.
ALGEBRA 1 LESSON 1-2ALGEBRA 1 LESSON 1-2
Exponents and Order of OperationsExponents and Order of Operations
1-2
Simplify 3(8 + 6) ÷ (42 – 10).
3(8 + 6) ÷ (42 – 10) = 3(8 + 6) ÷ (16 – 10) Simplify the power.
= 3(14) ÷ 6 Simplify within parentheses.
= 42 ÷ 6 Multiply and divide from left to right.
= 7 Divide.
ALGEBRA 1 LESSON 1-2ALGEBRA 1 LESSON 1-2
Exponents and Order of OperationsExponents and Order of Operations
1-2
Evaluate each expression for x = 11 and z = 16.
a. (xz)2
= (176)2 Simplify within parentheses. Multiply. = 11 • 256
= 2816= 30,976 Simplify.
(xz)2 = (11 • 16)2 Substitute 11 for x and 16 for z. xz2 = 11 • 162
ALGEBRA 1 LESSON 1-2ALGEBRA 1 LESSON 1-2
Exponents and Order of OperationsExponents and Order of Operations
1-2
b. xz2
Simplify 4[(2 • 9) + (15 ÷ 3)2].
4[(2 • 9) + (15 ÷ 3)2] = 4[18 + (5)2] First simplify (2 • 9) and (15 ÷ 3).
= 4[18 + 25] Simplify the power.
= 4[43] Add within brackets.
= 172 Multiply.
ALGEBRA 1 LESSON 1-2ALGEBRA 1 LESSON 1-2
Exponents and Order of OperationsExponents and Order of Operations
1-2
A carpenter wants to build three decks in the shape of
regular hexagons. The perimeter p of each deck will be 60 ft. The
perpendicular distance a from the center of each deck to one of the
sides will be 8.7 ft.
= 3(261) Simplify the fraction.
= 783 Multiply.
The total area of all three decks is 783 ft2.
A = 3 ( )pa2
= 3 ( ) 60 • 8.7
2Substitute 60 for p and 8.7 for a.
= 3 ( )5222
Simplify the numerator.
ALGEBRA 1 LESSON 1-2ALGEBRA 1 LESSON 1-2
Exponents and Order of OperationsExponents and Order of Operations
1-2
Use the formula A = 3 ( ) to find the total area of all three decks.pa2
ALGEBRA 1 LESSON 1-2ALGEBRA 1 LESSON 1-2
Simplify each expression.1. 50 – 4 • 3 + 6
2. 3(6 + 22) – 5
3. 2[(1 + 5)2 – (18 ÷ 3)]
Evaluate each expression.4. 4x + 3y for x = 2 and y = 4
5. 2 • p2 + 3s for p = 3 and s = 11
6. xy2 + z for x = 3, y = 6 and z = 4
44
25
60
20
51
112
Exponents and Order of OperationsExponents and Order of Operations
1-2
Write each decimal as a fraction and each fraction as a decimal.
1. 0.5 2. 0.05 3. 3.25 4. 0.325
5. 6. 7. 8. 3
(For help, go to skills handbook page 725.)
ALGEBRA 1 LESSON 1-3ALGEBRA 1 LESSON 1-3
Exploring Real NumbersExploring Real Numbers
1-3
25
38
23
59
ALGEBRA 1 LESSON 1-3ALGEBRA 1 LESSON 1-3
Exploring Real NumbersExploring Real Numbers
1-3
1. 0.5 = = =
2. 0.05 = = =
3. 3.25 = 3 = 3 = 3 or
4. 0.325 = = =
5. = 2 ÷ 5 = 0.4
6. = 3 ÷ 8 = 0.375
7. = 2 ÷ 3 = 0.6
8. 3 = 3 + (5 ÷ 9) = 3.5
5 10
12
5 • 15 • 2
5 100
5 • 1 5 • 20
25 100
14
134
25 • 125 • 4
325 1000
25 • 1325 • 40
1340
25
38
23
59
Solutions
1 20
Name the set(s) of numbers to which each number belongs.
a. –13 b. 3.28
integers
rational numbers
rational numbers
ALGEBRA 1 LESSON 1-3ALGEBRA 1 LESSON 1-3
Exploring Real NumbersExploring Real Numbers
1-3
Which set of numbers is most reasonable for displaying
outdoor temperatures?
integers
ALGEBRA 1 LESSON 1-3ALGEBRA 1 LESSON 1-3
Exploring Real NumbersExploring Real Numbers
1-3
Determine whether the statement is true or false. If it is false,
give a counterexample.
All negative numbers are integers.
The statement is false.
A negative number can be a fraction, such as – . This is not an integer.23
ALGEBRA 1 LESSON 1-3ALGEBRA 1 LESSON 1-3
Exploring Real NumbersExploring Real Numbers
1-3
Write – , – , and – , in order from least to greatest.
– = –0.75 Write each fraction as a decimal.
– = –0.583
– = –0.625
34
7 12
58
From least to greatest, the fractions are – , – , and – .34
7 12
58
–0.75 < –0.625 < –0.583 Order the decimals from least to greatest.
ALGEBRA 1 LESSON 1-3ALGEBRA 1 LESSON 1-3
Exploring Real NumbersExploring Real Numbers
1-3
34
7 12
58
Find each absolute value.
a. |–2.5| b. |7|
–2.5 is 2.5 units from 0 on a number line.
7 is 7 units from 0 on a number line.
|–2.5| = 2.5 |7| = 7
ALGEBRA 1 LESSON 1-3ALGEBRA 1 LESSON 1-3
Exploring Real NumbersExploring Real Numbers
1-3
ALGEBRA 1 LESSON 1-3ALGEBRA 1 LESSON 1-3
Name the set(s) of numbers to which each given number belongs.
1. –2.7 2. 11 3. 16
Use <, =, or > to compare.
4. 5.
6. Find |– |.
34
58
rational numbers irrational numbers natural numbers, whole numbersintegers, rational numbers
– –
7 12
> <
7 12
Exploring Real NumbersExploring Real Numbers
1-3
34
58
Evaluate – – 4z2 for x = 4, y = –2, and z = –4.
– – 4z2 = – 4(–4)2 Substitute 4 for x, –2 for y, and –4 for z.xy
–4–2
= – 4(16) Simplify the power.–4–2
= 2 – 64 Divide and multiply.
= –62 Subtract.
ALGEBRA 1 LESSON 1-6ALGEBRA 1 LESSON 1-6
Multiplying and Dividing Real NumbersMultiplying and Dividing Real Numbers
1-6
xy
Evaluate for p = and r = – .
= –2 Simplify.
= p ÷ r Rewrite the equation.pr
= ÷ Substitute for p and – for r.32
34
(– ) 32
34
= Multiply by – , the reciprocal of – .32
43(– ) 4
334
ALGEBRA 1 LESSON 1-6ALGEBRA 1 LESSON 1-6
Multiplying and Dividing Real NumbersMultiplying and Dividing Real Numbers
1-6
pr
32
34
ALGEBRA 1 LESSON 1-6ALGEBRA 1 LESSON 1-6
Simplify.
1. –8(–7) 2. –6(–7 + 10) – 4
Evaluate each expression for m = –3, n = 4, and p = –1.
3. + p 4. (mp)3 5. mnp
6. Evaluate 2a ÷ 4b – c for a = –2, b = – , and c = – .
56 – 22
–7 27 12
312
Multiplying and Dividing Real NumbersMultiplying and Dividing Real Numbers
1-6
8mn
13
12
ALGEBRA 1 LESSON 1-7ALGEBRA 1 LESSON 1-7
(For help, go to Lessons 1-2 and 1-6.)
Use the order of operations to simplify each expression.
1. 3(4 + 7) 2. –2(5 + 6) 3. –1(–9 + 8)
4. –0.5(8 – 6) 5. t(10 – 4) 6. m(–3 – 1)
The Distributive PropertyThe Distributive Property
1-7
12
ALGEBRA 1 LESSON 1-7ALGEBRA 1 LESSON 1-7
The Distributive PropertyThe Distributive Property
1-7
( )
1. 3(4 + 7) = 3(11) = 33
2. –2(5 + 6) = –2(11) = –22
3. –1(–9 + 8) = –1(–1) = 1
4. –0.5(8 – 6) = –0.5(2) = –1
5. t(10 – 4) = t(6) = (6)t = • 6 t = 3t
6. m(–3 – 1) = m(–4) = –4m
12
12
12
12
Solutions
Use the Distributive Property to simplify 26(98).
ALGEBRA 1 LESSON 1-7ALGEBRA 1 LESSON 1-7
26(98) = 26(100 – 2) Rewrite 98 as 100 – 2.
= 26(100) – 26(2) Use the Distributive Property.
= 2600 – 52 Simplify.
= 2548
The Distributive PropertyThe Distributive Property
1-7
Find the total cost of 4 CDs that cost $12.99 each.
4(12.99) = 4(13 – 0.01) Rewrite 12.99 as 13 – 0.01.
= 4(13) – 4(0.01) Use the Distributive Property.
= 52 – 0.04 Simplify.
= 51.96
The total cost of 4 CDs is $51.96.
ALGEBRA 1 LESSON 1-7ALGEBRA 1 LESSON 1-7
The Distributive PropertyThe Distributive Property
1-7
Simplify 3(4m – 7).
3(4m – 7) = 3(4m) – 3(7) Use the Distributive Property.
= 12m – 21 Simplify.
ALGEBRA 1 LESSON 1-7ALGEBRA 1 LESSON 1-7
The Distributive PropertyThe Distributive Property
1-7
Simplify –(5q – 6).
–(5q – 6) = –1(5q – 6) Rewrite the expression using –1.
= –1(5q) – 1(–6) Use the Distributive Property.
= –5q + 6 Simplify.
ALGEBRA 1 LESSON 1-7ALGEBRA 1 LESSON 1-7
The Distributive PropertyThe Distributive Property
1-7
Simplify –2w2 + w2.
–2w2 + w2 = (–2 + 1)w2 Use the Distributive Property.
= –w2 Simplify.
ALGEBRA 1 LESSON 1-7ALGEBRA 1 LESSON 1-7
The Distributive PropertyThe Distributive Property
1-7
Relate: –6 times the quantity 7 minus m
Write: –6 • (7 – m)
Write an expression for the product of –6 and the quantity 7
minus m.
–6(7 – m)
ALGEBRA 1 LESSON 1-7ALGEBRA 1 LESSON 1-7
The Distributive PropertyThe Distributive Property
1-7
ALGEBRA 1 LESSON 1-7ALGEBRA 1 LESSON 1-7
Simplify each expression.
1. 11(299) 2. 4(x + 8) 3. – 3(2y – 7)
4. –(6 + p) 5. 1.3a + 2b – 4c + 3.1b – 4a
6. Write an expression for the product of and the quantity b minus .
3289 4x + 32 – 6y + 21
– 6 – p –2.7a + 5.1b – 4c
47
35b –( )
The Distributive PropertyThe Distributive Property
1-7
47
35
ALGEBRA 1 LESSON 1-8ALGEBRA 1 LESSON 1-8
(For help, go to Lessons 1-4 and 1-6.)
Simplify each expression.
1. 8 + (9 + 2) 2. 3 • (–2 • 5) 3. 7 + 16 + 3
4. –4(7)(–5) 5. –6 + 9 + (–4) 6. 0.25 • 3 • 4
7. 3 + x – 2 8. 2t – 8 + 3t 9. –5m + 2m – 4m
Properties of Real NumbersProperties of Real Numbers
1-8
ALGEBRA 1 LESSON 1-8ALGEBRA 1 LESSON 1-8
1. 8 + (9 + 2) = 8 + (2 + 9) = (8 + 2) + 9 = 10 + 9 = 19
2. 3 • (–2 • 5) = 3 • (–10) = –30
3. 7 + 16 + 3 = 7 + 3 + 16 = 10 + 16 = 26
4. –4(7)(–5) = –4(–5)(7) = 20(7) = 140
5. –6 + 9 + (–4) = –6 + (–4) + 9 = –10 + 9 = –1
6. 0.25 • 3 • 4 = 0.25 • 4 • 3 = 1 • 3 = 3
7. 3 + x – 2 = 3 + (–2) + x = 1 + x
8. 2t – 8 + 3t = 2t + 3t – 8 = (2 + 3)t – 8 = 5t – 8
9. –5m + 2m – 4m = (–5 + 2 – 4)m = –7m
Properties of Real NumbersProperties of Real Numbers
Solutions
1-8
Name the property each equation illustrates.
a. 3 • a = a • 3
b. p • 0 = 0
c. 6 + (–6) = 0
ALGEBRA 1 LESSON 1-8ALGEBRA 1 LESSON 1-8
Properties of Real NumbersProperties of Real Numbers
1-8
Commutative Property of Multiplication, because the order of the factors changes
Multiplication Property of Zero, because a factor multiplied by zero is zero
Inverse Property of Addition, because the sum of a number and its inverse is zero
Suppose you buy a shirt for $14.85, a pair of pants for
$21.95, and a pair of shoes for $25.15. Find the total amount you
spent.
14.85 + 21.95 + 25.15 = 14.85 + 25.15 + 21.95 Commutative Property of Addition
= (14.85 + 25.15) + 21.95 Associative Property of Addition
= 40.00 + 21.95 Add within parentheses first.
= 61.95 Simplify.
The total amount spent was $61.95.
ALGEBRA 1 LESSON 1-8ALGEBRA 1 LESSON 1-8
Properties of Real NumbersProperties of Real Numbers
1-8
Simplify 3x – 4(x – 8). Justify each step.
3x – 4(x – 8) = 3x – 4x + 32 Distributive Property
= (3 – 4)x + 32 Distributive Property
= –1x + 32 Subtraction
= –x + 32 Identity Property of Multiplication
ALGEBRA 1 LESSON 1-8ALGEBRA 1 LESSON 1-8
Properties of Real NumbersProperties of Real Numbers
1-8
ALGEBRA 1 LESSON 1-8ALGEBRA 1 LESSON 1-8
Name the property that each equation illustrates.
1. 1m = m 2. (– 3 + 4) + 5 = – 3 + (4 + 5)
3. –14 • 0 = 0
4. Give a reason to justify each step.
Iden. Prop. Of Mult. Assoc. Prop. Of Add.
Mult. Prop. Of Zero
a. 3x – 2(x + 5) = 3x – 2x – 10 Distributive Property
b. = 3x + (– 2x) + (– 10) Definition of Subtraction
c. = [3 + (– 2)]x + (– 10) Distributive Property
d. = 1x + (– 10) Addition
e. = 1x – 10 Definition of Subtraction
f. = x – 10 Identity Property of Multiplication
Properties of Real NumbersProperties of Real Numbers
1-8