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Lesson 1-1 Expressions and Formulas Lesson 1-2 Properties of Real Numbers Lesson 1-3 Solving Equations Lesson 1-4 Solving Absolute Value Equations Lesson 1-5 Solving Inequalities Lesson 1-6 Solving Compound and Absolute Value Inequalities Chapter Menu

Main Ideas and Vocabulary Key Concept: Order of Operations Five-Minute Check Main Ideas and Vocabulary Key Concept: Order of Operations Example 1: Evaluate Algebraic Expressions Example 2: Use a Formula Lesson 1 Menu

Use the order of operations to evaluate expressions. Use formulas. variable power polynomial term like terms trinomial binomial formula algebraic expression order of operations monomial constant coefficient degree Lesson 1 MI/Vocab

Lesson 1 KC1

Evaluate Algebraic Expressions A. Evaluate (x – y)3 + 3 if x = 1 and y = 4. (x – y)3 + 3 = (1 – 4)3 + 3 Replace x with 1 and y with 4. = –27 + 3 Find (–3)3. = –24 Add –27 and 3. Answer: The value is –24. Lesson 1 Ex1

Evaluate Algebraic Expressions B. Evaluate s – t(s2 – t) if s = 2 and t = 3.4. s – t(s2 – t) = 2 – 3.4(22 – 3.4) Replace s with 2 and t with 3.4. = 2 – 3.4(4 – 3.4) Find 22. = 2 – 3.4(0.6) Subtract 3.4 from 4. = 2 – 2.04 Multiply 3.4 and 0.6. = –0.04 Subtract 2.04 from 2. Answer: The value is –0.04. Lesson 1 Ex1

Evaluate Algebraic Expressions x = 5, y = –2, and z = –1 Evaluate the numerator and the denominator separately. Multiply 40 by –2. Simplify the numerator and the denominator. Then divide. Answer: The value is –9. Lesson 1 Ex1

A. What is the value of (y – x)3 – 12 if x = –3 and y = –4? B. –11 C. –13 D. 13 A B C D Lesson 1 CYP1

B. What is the value of x – y2(x + 5) if x = 2 and y = 4? C. –54 D. –25 A B C D Lesson 1 CYP1

C. A. –23 B. –19 C. 19 D. 23 A B C D Lesson 1 CYP1

Answer: The area of the trapezoid is 152 square meters. Use a Formula Find the area of a trapezoid with base lengths of 13 meters and 25 meters and a height of 8 meters. Answer: The area of the trapezoid is 152 square meters. Lesson 1 Ex2

A. 450 cm3 B. 75 cm3 C. 50 cm3 D. 10 cm3 A B C D Lesson 1 CYP2

End of Lesson 1

Five-Minute Check (over Lesson 1-1) Main Ideas and Vocabulary Key Concept: Real Numbers Example 1: Classify Numbers Key Concept: Real Number Properties Example 2: Identify Properties of Real Numbers Example 3: Additive and Multiplicative Inverses Example 4: Real-World Example Example 5: Simplify an Expression Lesson 2 Menu

Use the properties of real numbers to evaluate expressions. Classify real numbers. Use the properties of real numbers to evaluate expressions. real numbers rational numbers irrational numbers Lesson 2 MI/Vocab

Lesson 2 KC1

Answer: irrationals (I) and reals (R) Classify Numbers Answer: irrationals (I) and reals (R) Lesson 2 Ex1

B. Name the sets of numbers to which 5 belongs. Classify Numbers B. Name the sets of numbers to which 5 belongs. Answer: naturals (N), wholes (W), integers (Z), rationals (Q), reals (R) Lesson 2 Ex1

Answer: rationals (Q) and reals (R) Classify Numbers C. Answer: rationals (Q) and reals (R) Lesson 2 Ex1

D. Name the sets of numbers to which –43 belongs. Classify Numbers D. Name the sets of numbers to which –43 belongs. Answer: integers (Z), rationals (Q), and reals (R) Lesson 2 Ex1

E. Name the sets of numbers to which –23.3 belongs. Classify Numbers E. Name the sets of numbers to which –23.3 belongs. Answer: rationals (Q) and reals (R) Lesson 2 Ex1

A. The number belongs to which sets? A. irrationals (I) and reals (R) B. rationals (Q) and reals (R) C. naturals (N), wholes (W), integers (Z), rationals (Q), and reals (R) D. none of the above A B C D Lesson 2 CYP1

A. irrationals (I) and reals (R) B. rationals (Q) and reals (R) C. naturals (N), wholes (W), integers (Z), rationals (Q), and reals (R) D. none of the above A B C D Lesson 2 CYP1

C. The number belongs to which sets? A. irrationals (I) and reals (R) B. rationals (Q) and reals (R) C. naturals (N), wholes (W), integers (Z), rationals (Q), and reals (R) D. none of the above A B C D Lesson 2 CYP1

D. The number belongs to which sets? A. irrationals (I) and reals (R) B. rationals (Q) and reals (R) C. naturals (N), wholes (W), integers (Z), rationals (Q), and reals (R) D. none of the above A B C D Lesson 2 CYP1

E. The number 32.1 belongs to which sets? A. irrationals (I) and reals (R) B. rationals (Q) and reals (R) C. naturals (N), wholes (W), integers (Z), rationals (Q), and reals (R) D. none of the above A B C D Lesson 2 CYP1

Lesson 2 KC2

Identify Properties of Real Numbers Name the property illustrated by (–8 + 8) + 15 = 0 + 15. The Additive Inverse Property says that a number plus its opposite is 0. Answer: Additive Inverse Property Lesson 2 Ex2

What is the property illustrated by 3 + 0 = 3? A. Distributive Property B. Additive Inverse Property C. Identity Property of Addition D. Inverse Property of Multiplication A B C D Lesson 2 CYP2

Additive and Multiplicative Inverses Identify the additive inverse and multiplicative inverse for –7. Since –7 + 7 = 0, the additive inverse of –7 is 7. Answer: Lesson 2 Ex3

What is the additive inverse and multiplicative inverse for the number 5? A. B. C. D. A B C D Lesson 2 CYP3

There are two ways to find the total amount spent on stamps. Method 1 POSTAGE Audrey went to a post office and bought eight 39-cent stamps and eight 24-cent postcard stamps. What was the total amount of money Audrey spent on stamps? There are two ways to find the total amount spent on stamps. Method 1 Multiply the price of each type of stamp by 8 and then add. S = 8(0.39) + 8(0.24) = 3.12 + 1.92 = 5.04 Lesson 2 Ex4

Answer: Audrey spent a total of $5.04 on stamps. Method 2 Add the prices of both types of stamps and then multiply the total by 8. S = 8(0.39 + 0.24) = 8(0.63) = 5.04 Answer: Audrey spent a total of $5.04 on stamps. Notice that both methods result in the same answer. Lesson 2 Ex4

CHOCOLATE Joel went to the grocery store and bought 3 plain chocolate candy bars for $0.69 each and 3 chocolate-peanut butter candy bars for $0.79 each. How much did Joel spend altogether on candy bars? A. $2.86 B. $4.44 C. $4.48 D. $7.48 A B C D Lesson 2 CYP4

Simplify an Expression Simplify 4(3a – b) + 2(b + 3a). 4(3a – b) + 2(b + 3a) = 4(3a) – 4(b) + 2(b) + 2(3a) Distributive Property = 12a – 4b + 2b + 6a Multiply. = 12a + 6a – 4b + 2b Commutative Property (+) = (12 + 6)a + (–4 + 2)b Distributive Property = 18a – 2b Simplify. Answer: 18a – 2b Lesson 2 Ex5

Which expression is equivalent to 2(3x – y) + 4(2x + 3y)? A. 14x + 10y B. 14x + 2y C. 14x + y D. 11x + 2y A B C D Lesson 2 CYP5

End of Lesson 2

Five-Minute Check (over Lesson 1-2) Main Ideas and Vocabulary Example 1: Verbal to Algebraic Expression Example 2: Algebraic to Verbal Sentence Key Concept: Properties of Equality Example 3: Identify Properties of Equality Key Concept: Properties of Equality Example 4: Solve One-Step Equations Example 5: Solve a Multi-Step Equation Example 6: Solve for a Variable Example 7: Apply Properties of Equality Example 8: Write an Equation Lesson 3 Menu

Solve equations using the properties of equality. Translate verbal expressions into algebraic expressions and equations, and vice versa. Solve equations using the properties of equality. open sentence equation solution Lesson 3 MI/Vocab

Verbal to Algebraic Expression A. Write an algebraic expression to represent the verbal expression 7 less than a number. Answer: n – 7 Lesson 3 Ex1

Verbal to Algebraic Expression B. Write an algebraic expression to represent the verbal expression the square of a number decreased by the product of 5 and the number. Answer: x2 – 5x Lesson 3 Ex1

A. Write an algebraic expression to represent the verbal expression 6 more than a number. A. 6x B. x + 6 C. x6 D. x – 6 A B C D Lesson 3 CYP1

B. Write an algebraic expression to represent the verbal expression 2 less than the cube of a number. A. x3 – 2 B. 2x3 C. x2 – 2 D. 2 + x3 A B C D Lesson 3 CYP1

Algebraic to Verbal Sentence A. Write a verbal sentence to represent 6 = –5 + x. Answer: Six is equal to –5 plus a number. Lesson 3 Ex2

Algebraic to Verbal Sentence B. Write a verbal sentence to represent 7y – 2 = 19. Answer: Seven times a number minus 2 is 19. Lesson 3 Ex2

A. What is a verbal sentence that represents the equation n – 3 = 7? A. The difference between a number and 3 is 7. B. The sum of a number and 3 is 7. C. The difference of 3 and a number is 7. D. The difference of a number and 7 is 3. A B C D Lesson 3 CYP2

B. What is a verbal sentence that represents the equation 5 = 2 + x? A. Five is equal to the difference of 2 and a number. B. Five is equal to twice a number. C. Five is equal to the quotient of 2 and a number. D. Five is equal to the sum of 2 and a number. A B C D Lesson 3 CYP2

Lesson 3 KC1

Identify Properties of Equality A. Name the property illustrated by the statement. a – 2.03 = a – 2.03 Answer: Reflexive Property of Equality Lesson 3 Ex3

Identify Properties of Equality B. Name the property illustrated by the statement. If 9 = x, then x = 9. Answer: Symmetric Property of Equality Lesson 3 Ex3

A. Reflexive Property of Equality B. Symmetric Property of Equality A. What property is illustrated by the statement below? If x + 4 = 3, then 3 = x + 4. A. Reflexive Property of Equality B. Symmetric Property of Equality C. Transitive Property of Equality D. Substitution Property of Equality A B C D Lesson 3 CYP3

A. Reflexive Property of Equality B. Symmetric Property of Equality B. What property is illustrated by the statement below? If 3 = x and x = y, then 3 = y. A. Reflexive Property of Equality B. Symmetric Property of Equality C. Transitive Property of Equality D. Substitution Property of Equality A B C D Lesson 3 CYP3

Lesson 3 KC2

Solve One-Step Equations A. Solve s – 5.48 = 0.02. Check your solution. s – 5.48 = 0.02 Original equation s – 5.48 + 5.48 = 0.02 + 5.48 Add 5.48 to each side. s = 5.5 Simplify. Check: s – 5.48 = 0.02 Original equation 5.5 – 5.48 = 0.02 Substitute 5.5 for s. ? 0.02 = 0.02 Simplify.  Answer: The solution is 5.5. Lesson 3 Ex4

Solve One-Step Equations Original equation Simplify. Lesson 3 Ex4

Solve One-Step Equations Check: Original equation ? Substitute 36 for t.  Simplify. Answer: The solution is 36. Lesson 3 Ex4

A. What is the solution to the equation x + 5 = 3? B. –2 C. 2 D. 8 A B C D Lesson 3 CYP4

B. What is the solution to the equation C. 15 D. 30 A B C D Lesson 3 CYP4

Solve a Multi-Step Equation Solve 53 = 3(y – 2) – 2(3y –1). 53 = 3(y – 2) – 2(3y – 1) Original equation 53 = 3y – 6 – 6y + 2 Apply the Distributive Property. 53 = –3y – 4 Simplify the right side. 57 = –3y Add 4 to each side to isolate the variable. –19 = y Divide each side by –3. Answer: The solution is –19. Lesson 3 Ex5

What is the solution to 25 = 3(2x + 2) – 5(2x + 1)? B. C. D. 6 A B C D Lesson 3 CYP5

Solve for a Variable GEOMETRY The formula for the area of a trapezoid is where A is the area, b1 is the length of one base, b2 is the length of the other base, and h is the height of the trapezoid. Solve the formula for h. Lesson 3 Ex6

Divide each side by (b1 + b2). Solve for a Variable Area of a trapezoid Multiply each side by 2. Simplify. Divide each side by (b1 + b2). Simplify. Answer: Lesson 3 Ex6

GEOMETRY The formula for the perimeter of a rectangle is where P is the perimeter, and w is the width of the rectangle. What is this formula solved for w? A. B. C. D. A B C D Lesson 3 CYP6

Apply Properties of Equality A B C D Read the Test Item You are asked to find the value of the expression 4g – 2. Your first thought might be to find the value of g and then evaluate the expression using this value. Notice that you are not required to find the value of g. Instead, you can use the Subtraction Property of Equality. Lesson 3 Ex7

Apply Properties of Equality Solve the Test Item Original equation Subtract 7 from each side. Answer: C Lesson 3 Ex7

If 2x + 6 = –3, what is the value of 2x –3? B. 6 C. –6 D. –12 A B C D Lesson 3 CYP7

the cost for installation Write an Equation HOME IMPROVEMENT Carl wants to replace the five windows in the bedrooms of his two-story house. His neighbor has agreed to help install them for $250. If Carl has budgeted $1000 for the total cost of replacing the windows, what is the maximum amount he can spend on each window? Explore Let c represent the cost of each window. Plan The number of windows times the cost per window plus the cost for installation equals the total cost. 5 ● c + 250 = 1000 Lesson 3 Ex8

Solve Original equation Write an Equation Solve Original equation Subtract 250 from each side. Simplify. Divide each side by 5. Simplify. Answer: Carl can afford to spend $150 on each window. Lesson 3 Ex8

Write an Equation Check The total cost to replace five windows at $150 each is 5(150) or $750. Add the $250 cost of the installation to that, and the total bill to replace the windows is 750 + 250 or $1000. Thus, the answer is correct. Lesson 3 Ex8

HOME IMPROVEMENT Kelly wants to repair the siding on her house HOME IMPROVEMENT Kelly wants to repair the siding on her house. Her contractor will charge her $300 plus $150 per square foot of siding. How much siding can she repair for $1500? A. 100 ft2 B. 10 ft2 C. 8 ft2 D. 4.5 ft2 A B C D Lesson 3 CYP8

End of Lesson 3

Five-Minute Check (over Lesson 1-3) Main Ideas and Vocabulary Key Concept: Absolute Value Example 1: Evaluate an Expression with Absolute Value Example 2: Solve an Absolute Value Equation Example 3: No Solution Example 4: One Solution Lesson 4 Menu

Evaluate expressions involving absolute values. Solve absolute value equations. absolute value empty set Lesson 4 MI/Vocab

Lesson 4 KC1

Evaluate an Expression with Absolute Value Replace x with 4. Simplify 2(4) first. Subtract 8 from 6. Add. Answer: The value is 4.7. Lesson 4 Ex1

A. 18.3 B. 1.7 C. –1.7 D. –13.7 A B C D Lesson 4 CYP1

Solve an Absolute Value Equation Case 1 a = b y + 3 = 8 y + 3 – 3 = 8 – 3 y = 5 or Case 2 a = –b y + 3 = –8 y + 3 – 3 = –8 – 3 y = –11 Check: |y + 3| = 8 |y + 3| = 8 ? |5 + 3| = 8 ? |–11 + 3| = 8 ? |8| = 8 ? |–8| = 8  8 = 8  8 = 8 Answer: The solutions are 5 or –11. Thus, the solution set is –11, 5. Lesson 4 Ex2

What is the solution to |2x + 5| = 15? B. –10, 5 C. –5, 10 D. –5 A B C D Lesson 4 CYP2

|6 – 4t| + 5 = 0 Original equation No Solution Solve |6 – 4t| + 5 = 0. |6 – 4t| + 5 = 0 Original equation |6 – 4t| = –5 Subtract 5 from each side. This sentence is never true. Answer: The solution set is . Lesson 4 Ex3

A. B. C. D. A B C D Lesson 4 CYP3

One Solution Case 1 a = b 8 + y = 2y – 3 8 = y – 3 11 = y Lesson 4 Ex4

One Solution Check:  Answer: Lesson 4 Ex4

A. B. C. D. A B C D Lesson 4 CYP4

End of Lesson 4

Five-Minute Check (over Lesson 1-4) Main Ideas and Vocabulary Key Concept: Properties of Inequality Example 1: Solve an Inequality Using Addition or Subtraction Example 2: Solve an Inequality Using Multiplication or Division Example 3: Solve a Multi-Step Inequality Example 4: Real-World Example: Write an Inequality Lesson 5 Menu

Solve inequalities with one operation. Solve multi-step inequalities. set-builder notation Lesson 5 MI/Vocab

Lesson 5 KC1

Solve an Inequality Using Addition or Subtraction Solve 4y – 3 < 5y + 2. Graph the solution set on a number line. 4y – 3 < 5y + 2 Original inequality 4y – 3 – 4y < 5y + 2 – 4y Subtract 4y from each side. – 3 < y + 2 Simplify. –3 – 2 < y + 2 – 2 Subtract 2 from each side. –5 < y Simplify. y > –5 Rewrite with y first. Lesson 5 Ex1

Solve an Inequality Using Addition or Subtraction Answer: Any real number greater than –5 is a solution of this inequality. A circle means that this point is not included in the solution set. Lesson 5 Ex1

Which graph represents the solution to 6x – 2 < 5x + 7? B. C. D. A B C D Lesson 5 CYP1

Lesson 5 KC2

Solve an Inequality Using Multiplication or Division Solve 12  –0.3p. Graph the solution set on a number line. Original inequality Divide each side by –0.3, reversing the inequality symbol. Simplify. Rewrite with p first. Lesson 5 Ex2

Solve an Inequality Using Multiplication or Division Answer: The solution set is p | p  –40. A dot means that this point is included in the solution set. Lesson 5 Ex2

What is the solution to –3x  21? A. x | x  –7 B. x | x  –7 C. x | x  7 D. x | x  7 A B C D Lesson 5 CYP2

Solve a Multi-Step Inequality Original inequality Multiply each side by 2. Add –x to each side. Divide each side by –3, reversing the inequality symbol. Lesson 5 Ex3

Solve a Multi-Step Inequality Lesson 5 Ex3

A. B. C. D. A B C D Lesson 5 CYP3

Explore Let g = the number of gallons of gasoline that Alida buys. Write an Inequality CONSUMER COSTS Alida has at most $15.00 to spend today. She buys a bag of potato chips and a can of soda for $1.59. If gasoline at this store costs $2.89 per gallon, how many gallons of gasoline, to the nearest tenth of a gallon, can Alida buy for her car? Explore Let g = the number of gallons of gasoline that Alida buys. Plan The total cost of the gasoline is 2.89g. The cost of the chips and soda plus the total cost of the gasoline must be less than or equal to $15.00. Write an inequality. Lesson 5 Ex4

Subtract 1.59 from each side. Write an Inequality The cost of chips & soda plus the cost of gasoline is less than or equal to $15.00. 1.59 + 2.89g  15.00 Original inequality Subtract 1.59 from each side. Simplify. Divide each side by 2.89. Simplify. Lesson 5 Ex4

Answer: Alida can buy up to 4.6 gallons of gasoline for her car. Write an Inequality Answer: Alida can buy up to 4.6 gallons of gasoline for her car. Check Lesson 5 Ex4

RENTAL COSTS Jeb wants to rent a car for his vacation RENTAL COSTS Jeb wants to rent a car for his vacation. Value Cars rents cars for $25 per day plus $0.25 per mile. How far can he drive for one day if he wants to spend no more that $200 on car rental? A. up to 700 miles B. up to 800 miles C. more than 700 miles D. more than 800 miles A B C D Lesson 5 CYP4

End of Lesson 5

Five-Minute Check (over Lesson 1-5) Main Ideas and Vocabulary Key Concept: “And” Compound Inequalities Example 1: Solve an “and” Compound Inequality Key Concept: “Or” Compound Inequalities Example 2: Solve an "or" Compound Inequality Example 3: Solve an Absolute Value Inequality (<) Example 4: Solve an Absolute Value Inequality (>) Key Concept: Absolute Value Inequalities Example 5: Solve a Multi-Step Absolute Value Inequality Example 6: Real-World Example: Write an Absolute Value Inequality Lesson 6 Menu

Solve compound inequalities. Solve absolute value inequalities. compound inequality intersection union Lesson 6 MI/Vocab

Animation: Compound Inequalities Lesson 6 KC1

Solve an “and” Compound Inequality Solve 10  3y – 2 < 19. Graph the solution set on a number line. Method 1 Write the compound inequality using the word and. Then solve each inequality. 10  3y – 2 and 3y – 2 < 19 12  3y 3y < 21 4  y y < 7 4  y < 7 Lesson 6 Ex1

Solve an “and” Compound Inequality Method 2 Solve both parts at the same time by adding 2 to each part. Then divide each part by 3. 10  3y – 2 < 19 12  3y < 21 4  y < 7 Lesson 6 Ex1

Solve an “and” Compound Inequality Graph the solution set for each inequality and find their intersection. 4  y < 7 y < 7 y  4 Lesson 6 Ex1

What is the solution to 11  2x + 5 < 17? B. C. D. A B C D Lesson 6 CYP1

Lesson 6 KC2

Solve an “or” Compound Inequality Solve x + 3 < 2 or – x  –4. Graph the solution set on a number line. Solve each inequality separately. –x  –4 or x + 3 < 2 x < –1 x  4 x < –1 or x  4 x < –1 x  4 Answer: The solution set is x | x < –1 or x  4. Lesson 6 Ex2

What is the solution to x + 5 < 1 or –2x  –6 What is the solution to x + 5 < 1 or –2x  –6? Graph the solution set on a number line. A. B. C. D. A B C D Lesson 6 CYP2

Solve an Absolute Value Inequality (<) Solve 3 < |d|. Graph the solution set on a number line. You can interpret 3 < |d| to mean that the distance between d and 0 on a number line is greater than 3 units. To make 3 < |d| true, you must substitute values for d that are greater than 3 units from 0. Notice that the graph of 3 < |d| is the same as the graph of d < –3 or d > 3. All of the numbers not between –3 and 3 are greater than 3 units from 0. Answer: The solution set is d | d < –3 or d > 3. Lesson 6 Ex3

What is the solution to |x| < 5? A. {x|x > 5 or x < –5} B. {x|–5 < x < 5} C. {x|x < 5} D. {x|x > –5} A B C D Lesson 6 CYP3

Solve an Absolute Value Inequality (>) Solve 3 > |d|. Graph the solution set on a number line. You can interpret 3 > |d| to mean that the distance between d and 0 on a number line is less than 3 units. To make 3 > |d| true, you must substitute numbers for d that are fewer than 3 units from 0. Notice that the graph of 3 > |d| is the same as the graph of d > –3 and d < 3. All of the numbers between –3 and 3 are less than 3 units from 0. Answer: The solution set is d | –3 < d < 3. Lesson 6 Ex4

What is the solution to |x| > 5? B. C. D. A B C D Lesson 6 CYP4

Lesson 6 KC3

Solve a Multi-Step Absolute Value Inequality Solve |2x – 2|  4. Graph the solution set on a number line. |2x – 2|  4 is equivalent to 2x – 2  4 or 2x – 2  –4. Solve the inequality. 2x – 2  4 or 2x – 2  –4 2x  6 2x  –2 x  3 x  –1 Answer: The solution set is x | x  –1 or x  3. Lesson 6 Ex5

What is the solution to |3x – 3| > 9 What is the solution to |3x – 3| > 9? Graph the solution set on a number line. A. B. C. D. A B C D Lesson 6 CYP5

The rent for an apartment can differ from the average Write an Absolute Value Inequality A. HOUSING According to a recent survey, the average monthly rent for a one-bedroom apartment in one city is $750. However, the actual rent for any given one-bedroom apartment in the area might vary as much as $250 from the average. Write an absolute value inequality to describe this situation. Let r = the actual monthly rent. The rent for an apartment can differ from the average by as much as $250. |750 – r|  250 Answer: |750 – r|  250 Lesson 6 Ex6

Write an Absolute Value Inequality B. Solve the inequality |750 – r|  250 to find the range of monthly rent. Rewrite the absolute value inequality as a compound inequality. Then solve for r. –250  750 – r  250 –250 – 750  –r  250 – 750 –1000  –r  –500 1000  r  500 Answer: The solution set is r | 500  r  1000. The actual rent falls between $500 and $1000, inclusive. Lesson 6 Ex6

A. HEALTH The average birth weight of a newborn baby is 7 pounds A. HEALTH The average birth weight of a newborn baby is 7 pounds. However, this weight can vary by as much as 4.5 pounds. What is an absolute value inequality to describe this situation? A. |4.5 – w|  7 B. |w – 4.5|  7 C. |w – 7|  4.5 D. |7 – w|  4.5 A B C D Lesson 6 CYP6

B. HEALTH The average birth weight of a newborn baby is 7 pounds B. HEALTH The average birth weight of a newborn baby is 7 pounds. However, this weight can vary by as much as 4.5 pounds. What is the range of birth weights for newborn babies? A. w | w  11.5 B. w | w  2.5 C. w | 2.5  w  11.5 D. w | 4.5  w  7 A B C D Lesson 6 CYP6

End of Lesson 6

Five-Minute Checks Image Bank Math Tools Algebra Tiles Compound Inequalities CR Menu

Lesson 1-2 (over Lesson 1-1) Lesson 1-3 (over Lesson 1-2) 5Min Menu

1. Exit this presentation. To use the images that are on the following three slides in your own presentation: 1. Exit this presentation. 2. Open a chapter presentation using a full installation of Microsoft® PowerPoint® in editing mode and scroll to the Image Bank slides. 3. Select an image, copy it, and paste it into your presentation. IB 1

IB 2

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IB 4

Animation 1

Animation 2

Evaluate (12 – 9)3. A. 9 B. 27 C. 63 D. 909 A B C D 5Min 1-1

Simplify 5(4 + n) + 6n. A. 7n + 9 B. 7n + 20 C. 11n + 20 D. 35n + 20 A 5Min 1-2

A. B. C. D. A B C D 5Min 1-3

What is the area of a square with sides 15 centimeters? A. 1350 cm2 B. 225 cm2 C. 60 cm2 D. 30 cm2 A B C D 5Min 1-4

Write an expression for the area of a rectangle with length 2x and width x. A. 2x2 B. 3x C. 4x2 D. 6x A B C D 5Min 1-5

Three consecutive integers have a sum of 114 Three consecutive integers have a sum of 114. What is the second integer? A. 40 B. 39 C. 38 D. 37 A B C D 5Min 1-6

Find the value of the expression (13 + 5) ● 3 – 42. (over Lesson 1-1) Find the value of the expression (13 + 5) ● 3 – 42. A. 12 B. 18 C. 37 D. 38 A B C D 5Min 2-1

(over Lesson 1-1) A. B. –5 C. 2 D. A B C D 5Min 2-2

Evaluate 2(a + b) – 5w if a = –3, b = 2, and w = –1. (over Lesson 1-1) Evaluate 2(a + b) – 5w if a = –3, b = 2, and w = –1. A. –8 B. –5 C. 3 D. 6 A B C D 5Min 2-3

(over Lesson 1-1) A. 24 B. C. D. –24 A B C D 5Min 2-4

(over Lesson 1-1) One way to write the formula for the surface area of a rectangular prism is A = 2h(l + w) + 2lw. Find the surface area of a rectangular prism with a length of 6 inches, a width of 4 inches, and a height of 2 inches. A. 48 in2 B. 60 in2 C. 68 in2 D. 88 in2 A B C D 5Min 2-5

(over Lesson 1-1) Suppose x = 8 ÷ (9 + 2) – 3 and y = 8 ÷ [9 + 2(–4)]. Which statement is true? A. x > y B. x < y C. x = y D. x = –4y A B C D 5Min 2-6

A. I B. Z, N C. N, W, Z, Q, R D. R, I, W, Q, Z, N (over Lesson 1-2) A 5Min 3-1

A. W, Z, Q, R, N B. N, W, Z, I C. W, Z, Q D. Q, R (over Lesson 1-2) A 5Min 3-2

(over Lesson 1-2) Name the property illustrated by the equation a + (7 + c) = (a + 7) + c. A. Associative (+) B. Commutative (+) C. Distributive D. Inverse (×) A B C D 5Min 3-3

(over Lesson 1-2) Name the property illustrated by the equation 3(4 + 0.2) = 3(4) + 3(0.2). A. Associative (+) B. Commutative (+) C. Distributive D. Inverse (×) A B C D 5Min 3-4

Simplify (2c)(3d) + c + 5cd + 3c2. (over Lesson 1-2) Simplify (2c)(3d) + c + 5cd + 3c2. A. 3c2 + 6cd + c + 6 B. 3c2 + 5cd + 3c + 3d C. 3c2 + 10cd + c D. 3c2 + 11cd + c A B C D 5Min 3-5

(over Lesson 1-2) Which statements are true? I. All whole numbers are rational numbers. II. All natural numbers are integers. III. All real numbers are irrational. A. I only B. I and II only C. II and III only D. I, II, and III A B C D 5Min 3-6

(over Lesson 1-3) Write an algebraic expression to represent the verbal expression, three times the sum of a number and its square. A. 3x + x2 B. 3(x + x2) C. 3x2 + x D. x2 + x + 3 A B C D 5Min 4-1

(over Lesson 1-3) Write an algebraic expression to represent the verbal expression, five less than the product of the cube of a number and –4. A. –4n3 – 5 B. (–4n)3 – 5 C. (–43)n – 5 D. –4(n – 5)3 A B C D 5Min 4-2

Write an equation to represent the sum of 23 and twice a number is 65. (over Lesson 1-3) Write an equation to represent the sum of 23 and twice a number is 65. A. 2(n + 23) = 65 B. n + 2(23) = 65 C. 2n + 23 = 65 D. (n + 2) + 23 = 65 A B C D 5Min 4-3

Solve 12f – 4 = 7 + f. A. –1 B. C. D. 1 (over Lesson 1-3) A B C D 5Min 4-4

Solve 10y + 1 = 3(–2y – 5). A. –4 B. –1 C. 1 D. 4 (over Lesson 1-3) A 5Min 4-5

If 3x – 2 = 16, what is the value of 15x –10? (over Lesson 1-3) If 3x – 2 = 16, what is the value of 15x –10? A. 6 B. 48 C. 80 D. 120 A B C D 5Min 4-6

Evaluate |4w + 3| if w = –2. A. 4 B. 5 C. 11 D. 20 (over Lesson 1-4) A 5Min 5-1

Evaluate |2x + y| if x = 1.5 and y = 4. (over Lesson 1-4) Evaluate |2x + y| if x = 1.5 and y = 4. A. 7 B. 7.5 C. 11 D. 14 A B C D 5Min 5-2

Evaluate 5|xy – w| if w = –2, x = 1.5, and y = 4. (over Lesson 1-4) Evaluate 5|xy – w| if w = –2, x = 1.5, and y = 4. A. 27.5 B. 20 C. 37.5 D. 40 A B C D 5Min 5-3

Solve |b + 20| = 21. A. {–41, –1} B. {–41, 1} C. {41, –1} D. {41, 1} (over Lesson 1-4) Solve |b + 20| = 21. A. {–41, –1} B. {–41, 1} C. {41, –1} D. {41, 1} A B C D 5Min 5-4

Solve –4|a + 5| = –8. A. {–7, –3} B. {–7, 3} C. {7, –3} D. {7, 3} (over Lesson 1-4) Solve –4|a + 5| = –8. A. {–7, –3} B. {–7, 3} C. {7, –3} D. {7, 3} A B C D 5Min 5-5

A. No value of n satisfies the equation. (over Lesson 1-4) Consider the equation |n – 8| = |8 – n|. Which of the following is true about the value(s) of n? A. No value of n satisfies the equation. B. Only positive values of n satisfy the equation. C. Only negative values of n satisfy the equation. D. All values of n satisfy the equation. A B C D 5Min 5-6

Which choice shows the solution set and the graph of 3x + 7 > 22? (over Lesson 1-5) Which choice shows the solution set and the graph of 3x + 7 > 22? A. {x | x > 5} B. {x | x ≥ 5} C. {x | x ≤ 5} D. {x | x ≥ 6} A B C D 5Min 6-1

Which choice shows the solution set and the graph of 3(3w + 1) ≥ 4.8? (over Lesson 1-5) Which choice shows the solution set and the graph of 3(3w + 1) ≥ 4.8? A. {w | w < 0.2} B. {w | w > 0.2} C. {w | w ≥ 0.2} D. {w | w ≥ 0.4} A B C D 5Min 6-2

(over Lesson 1-5) Which choice shows the solution set and the graph of 7 + 3y > 4(y + 2)? A. {y | y > –1} B. {y | y ≤ –1} C. {y | y ≤ –2} D. {y | y < –1} A B C D 5Min 6-3

A. {w | w < –3} B. {w | w ≤ –3} C. {w | w ≥ –3} D. {w | w > –3} (over Lesson 1-5) A. {w | w < –3} B. {w | w ≤ –3} C. {w | w ≥ –3} D. {w | w > –3} A B C D 5Min 6-4

(over Lesson 1-5) Tom makes $4.50 an hour. He worked 12 hours in one week. If at least one-third of this pay is taken out for taxes and other deductions, what is the greatest amount of money he can take home? A. $18 B. $27 C. $36 D. $54 A B C D 5Min 6-5

Which inequality is graphed on the number line in the figure? (over Lesson 1-5) Which inequality is graphed on the number line in the figure? A. 4x < –8 B. x > –2 C. 6 > –3x D. –2x < 4 A B C D 5Min 6-6

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