1. Splash Screen

2. Lesson Menu Five-Minute Check (over Lesson 2–8) CCSS Then/Now New Vocabulary Example 1:Real-World Example: Mixture Problem Example 2:Real-World Example: Percent Mixture Problem Example 3:Real-World Example: Speed of One Vehicle Example 4:Real-World Example: Speed of Two Vehicles

3. Over Lesson 2–8 5-Minute Check 1 Solve 6r + t = r – 1 for r. A. B. C. D.

4. Over Lesson 2–8 5-Minute Check 2 Solve 4c – d = 4a – 2c + 1 for c. A. B. C. D.

5. Over Lesson 2–8 5-Minute Check 3 A. B. C. D. for h.

6. Over Lesson 2–8 5-Minute Check 4 A.4.6 cm B.5.8 cm C.6.2 cm D.6.4 cm

7. Over Lesson 2–8 5-Minute Check 5 A.236.25 miles B.472.5 miles C.945 miles D.1285.75 miles

8. CCSS Pg. 132 – 138 Obj: Learn how to solve mixture and uniform motion problems. Content Standards: A.REI.1 and A.REI.3

9. Why? –Baseball players’ performance is measure in large part by statistics. Slugging average (SLG) is a weighted average that measures the power of a hitter. The slugging average is calculated by using the following formula.

10. How is the slugging average weighted? How would you calculate a slugging average that is not weighted?

11. Then/Now You translated sentences into equations. Solve mixture problems. Solve uniform motion problems.

12. Vocabulary Weighted Average – found by multiplying each data value by its weight and then finding the mean of the new data set Mixture problems – problems in which two or more parts are combined into a whole Uniform Motion Problems or rate problems – are problems in which an object moves at a certain speed or rate – d=rt

13. Example 1 Mixture Problem PETS Mandisha feeds her cat gourmet cat food that costs $1.75 per pound. She combines it with cheaper food that costs $0.50 per pound. How many pounds of cheaper food should Mandisha buy to go with 5 pounds of gourmet food, if she wants the average price to be $1.00 per pound? Let w = the number of pounds of cheaper cat food. Make a table.

14. Example 1 Mixture Problem Write and solve an equation using the information in the table. 8.75 + 0.5w =1.00(5 + w)Original equation 8.75 + 0.5w =5 + 1wDistributive Property 8.75 + 0.5w – 0.5w=5 + 1w – 0.5wSubtract 0.5w from each side. 8.75=5 + 0.5wSimplify. Price of gourmet cat foodplus price of cheaper cat foodequals price of mixed cat food. 8.75 + 0.5w = 1.00(5 + w)

15. Example 1 Mixture Problem 8.75 – 5 =5 + 0.5w – 5Subtract 5 from each side. 3.75 =0.5w Simplify. Divide each side by 0.5. 7.5 = w Simplify. Answer: Mandisha should buy 7.5 pounds of cheaper cat food to be mixed with the 5 pounds of gourmet cat food so that the average price is $1.00 per pound of cat food.

16. Example 1 A.3.1 ounces B.5 ounces C.4.6 ounces D.2 ounces Cheryl bought 3 ounces of glass beads that cost $1.79 an ounce. The seed beads cost $0.99 an ounce. How many ounces of seed beads can she buy if she only wants the beads to be $1.29 an ounce for her craft project?

17. Example 2 Percent Mixture Problem AUTO MAINTENANCE A car’s radiator should contain a solution of 50% antifreeze. Bae has 2 gallons of a 35% antifreeze. How many gallons of 100% antifreeze should Bae add to his solution to produce a solution of 50% antifreeze? Let g = the number of gallons of 100% antifreeze to be added. Make a table.

18. Example 2 Percent Mixture Problem Write and solve an equation using the information in the table. 0.35(2) + 1.0(g) = 0.50(2 + g)Original equation 0.70 + 1g = 1 + 0.50gDistributive Property 0.70 + 1g – 0.50g = 1 + 0.50g – 0.50gSubtract 0.50g from each side. Amount of antifreeze in 35% solutionplus amount of antifreeze in 100% solutionequals amount of antifreeze in 50% solution. 0.35(2) + 1.0(g) = 0.50(2 + g)

19. Example 2 Percent Mixture Problem 0.70 + 0.50g = 1Simplify. 0.70 + 0.50g – 0.70 = 1 – 0.70Subtract 0.70 from each side. 0.50g = 0.30Simplify. Answer: Bae should add 0.6 gallon of 100% antifreeze to produce a 50% solution. Divide each side by 0.50. g = 0.6Simplify.

20. Example 2 A recipe calls for mixed nuts with 50% peanuts. pound of 15% peanuts has already been used. How many pounds of 75% peanuts needs to be added to obtain the required 50% mix? A.B. C.D. lb

21. Example 3 Speed of One Vehicle AIR TRAVEL Nita took a non-stop flight to visit her grandmother. The 750-mile trip took three hours and 45 minutes. Because of bad weather, the return trip took four hours and 45 minutes. What was her average speed for the round trip? UnderstandWe know that Nita did not travel the same amount of time on each portion of her trip. So, we will need to find the weighted average of the plane’s speed. We are asked to find the average speed for both portions of the trip.

22. Example 3 Speed of One Vehicle PlanFirst, find the rate of the going portion, and then the return portion of the trip. Because the rate is in miles per hour, convert 3 hours and 45 minutes to 3.75 hours and 4 hours 45 minutes to 4.75 hours. Going Formula for rate

23. Return Example 3 Speed of One Vehicle Formula for rate Because we are looking for a weighted average, we cannot just average the speeds. We need to find the weighted average for the round trip.

24. Example 3 Speed of One Vehicle Solve Substitution Simplify.

25. Example 3 Speed of One Vehicle Answer: The average speed was about 176 miles per hour. Check The solution of 176 miles per hour is between the going portion rate 200 miles per hour, and the return rate, 157.9 miles per hour. So, the answer is reasonable.

26. Example 3 A.24 miles per hour B.30 miles per hour C.15 miles per hour D.45 miles per hour In the morning, when traffic is light, it takes 30 minutes to get to work. The trip is 15 miles through towns. In the afternoon, when traffic is a little heavier, it takes 45 minutes. What is the average speed for the round trip?

27. Example 4 Speeds of Two Vehicles RESCUE A railroad switching operator has discovered that two trains are heading toward each other on the same track. Currently, the trains are 53 miles apart. One train is traveling at 75 miles per hour and the other 40 miles per hour. The faster train will require 5 miles to stop safely, and the slower train will require 3 miles to stop safely. About how many minutes does the operator have to warn the train engineers to stop their trains? Step 1Draw a diagram. 53 miles apart Takes 5 miles to stop Takes 3 miles to stop 53 – (5 + 3) = 45 miles

28. Example 4 Speeds of Two Vehicles Step 2Let m = the number of hours that the operator has to warn the train engineers to stop their trains safely. Make a table. Step 3Write and solve an equation using the information in the table. Distance traveled by fast trainplus distance traveled by other trainequals45 miles. 75m + 40m = 45

29. Example 4 Speeds of Two Vehicles 75m + 40m = 45 Original equation 115m = 45Simplify. Answer: The operator has about 23 minutes to warn the engineers. Divide each side by 115. m ≈ 0.39Round to the nearest hundredth. 0.39 × 60 = 23.4Convert to minutes by multiplying by 60.

30. Example 4 A.17 minutes B.15 minutes C.14 minutes D.30 minutes Two students left the school on their bicycles at the same time, one heading north and the other heading south. The student heading north travels 15 miles per hour, and the one heading south travels at 17 miles per hour. After about how many minutes will they be 7.5 miles apart?

31. End of the Lesson