13 16 14 9 13 4 6 Bell Ringer Instructions: find the number that two other numbers add up to. Example – this would be 16 since 9 + 7 = 16 Bell Ringer 16 14 13 9 13 4 6

(over Lesson 6-1) Which of the following options represents the solution set and graph of the inequality y – 3 > 5? A. {y | y > 8} B. {y | y > 6} C. {y | y > 2} D. {y | y > 10} A B C D 5Min 2-1

(over Lesson 6-1) Which of the following options represents the solution set and graph of the inequality t + 9 6? A. {t | t < –3} B. {t | t < –3} C. {t | t < –4} D. {t | t < –2} A B C D 5Min 2-2

(over Lesson 6-1) Which of the following options represents the solution set and graph of the inequality 4n > 3n + 9? A. {n | n 10} B. {n | n 9} C. {n | n > 9} D. {n | n > 8} A B C D 5Min 2-3

(over Lesson 6-1) Write an inequality and solve the problem. The sum of a number and 7 is at least –5. A. n + 7 < –5; {n | n < –12} B. n + 7 > –5; {n | n > –12} C. n + 7 –5; {n | n –12} D. n + 7 > –5; {n | n > –12} A B C D 5Min 2-4

(over Lesson 6-1) Write the inequality and solve the problem. The sum of twice a number and 8 is more than 20. A. 20 < 2n + 8; {n | n > 6} B. 20 < 2n + 8; {n | n < 6} C. 20 < 2(n + 8); {n | n > 2} D. 20 < 2n – 8; {n | n > 14} A B C D 5Min 2-5

Solve the inequality A. {a | a > 4} B. {a | a > 16} (over Lesson 6-2) Solve the inequality A. {a | a > 4} B. {a | a > 16} C. {a | a > 20} D. {a | a > 64} A B C D 5Min 3-1

Solve the inequality p > –20. (over Lesson 6-2) Solve the inequality p > –20. A. {p | p > 28} B. {p | p > –28} C. {p | p < 28} D. {p | p < –28} A B C D 5Min 3-2

Solve the inequality –9v > –108. (over Lesson 6-2) Solve the inequality –9v > –108. A. {v | v 12} B. {v | v –12} C. {v | v 12} D. {v | v –12} A B C D 5Min 3-3

Solve the inequality 3x – 15 < 45. (over Lesson 6-3) Solve the inequality 3x – 15 < 45. A. {x | x < 10} B. {x | x > –10} C. {x | x < 20} D. {x | x > –20} A B C D 5Min 4-1

Solve the inequality 2p – 22 4p + 14. (over Lesson 6-3) Solve the inequality 2p – 22 4p + 14. A. {p | p < –18} B. {p | p < 18} C. {p | p > –18} D. {p | p > 18} A B C D 5Min 4-2

Solve the inequality < A. {x | x > –40} B. {x | x > 40} (over Lesson 6-3) Solve the inequality < A. {x | x > –40} B. {x | x > 40} C. {x | x > –64} D. {x | x > 64} A B C D 5Min 4-3

(over Lesson 6-3) Write an inequality for the sentence. Two times the difference of a number n and five is greater than seven. A. 2n – 5 > 7 B. 2(n – 5) > 7 C. 2(5 – n) > 7 D. 5 – 2n > 7 A B C D 5Min 4-4

Graph absolute value functions. Solve absolute value equations. piecewise function Lesson 5 MI/Vocab

Lesson 5 KC1

Solve an Absolute Value Equation A. WEATHER The average January temperature in a northern Canadian city is 1 degree Fahrenheit. The actual January temperature for that city may be about 5 degrees Fahrenheit warmer or colder. Solve |t – 1| = 5 to find the range of temperatures. Method 1 Graphing |t – 1| = 5 means that the distance between t and 1 is 5 units. To find t on the number line, start at 1 and move 5 units in either direction. The distance from 1 to 6 is 5 units. The distance from 1 to –4 is 5 units. The solution set is {–4, 6}. Lesson 5 Ex1

Solve an Absolute Value Equation Method 2 Compound Sentence Write |t –1| = 5 as t – 1 = 5 or t – 1 = –5. Case 1 Case 2 t – 1 = 5 t – 1 = –5 t – 1 + 1 = 5 + 1 Add 1 to each side. t – 1 + 1 = –5 + 1 Add 1 to each side. t = 6 Simplify. t = –4 Simplify. Answer: The solution set is {–4, 6}. The range of temperatures is –4°F to 6°F. Lesson 5 Ex1

Solve an Absolute Value Equation B. Solve |x – 1| = –7. Answer: |x – 1| = –7 means that the distance between x and 1 is –7. Since distance cannot be negative, the solution set is the empty set Ø. REMEMBER: An absolute value can NOT equal a negative number. Lesson 5 Ex1

A. WEATHER The average temperature for Columbus on Tuesday was 45ºF A. WEATHER The average temperature for Columbus on Tuesday was 45ºF. The actual temperature for anytime during the day may have actually varied from the average temperature by 15ºF. Solve to find the range of temperatures. A. {–60, 60} B. {0, 60} C. {–45, 45} D. {30, 60} A B C D Lesson 5 CYP1

B. Solve |x – 3| = –5. A. {8, –2} B. {–8, 2} C. {8, 2} D. A B C D Lesson 5 CYP1

Write an Absolute Value Equation Write an open sentence involving absolute value for the graph. Find the point that is the same distance from –4 as the distance from 6. The midpoint between –4 and 6 is 1. Lesson 5 Ex2

Write an Absolute Value Equation The distance from 1 to –4 is 5 units. The distance from 1 to 6 is 5 units. So, an equation is |y – 1| = 5. Answer: |y – 1| = 5 Lesson 5 Ex2

Homework: Worksheet 6-5 End of Lesson 6