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Use Normal View for the Interactive Elements

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  1. Setting the PowerPoint View • Use Normal View for the Interactive Elements • To use the interactive elements in this presentation, do not select the Slide Show view. Instead, select Normal view and follow these steps to set the view as large as possible: • On the View menu, select Normal. • Close the Slides tab on the left. • In the upper right corner next to the Help button, click the ^ to minimize the ribbon at the top of the screen.  • On the View menu, confirm that Ruler is deselected. • On the View tab, click Fit to Window. • Use Slide Show View to Administer Assessment Items • To administer the numbered assessment items in this presentation, use the Slide Show view. (See Slide 21 for an example.)

  2. Table of Contents Click on the topic to go to that section Number Problems Age Problems Geometry Problems Percent Problems Mixture Problems Uniform Motion Problems Work Problems Proportionality Problems

  3. Problem Solving Strategies Most problems can be solved by incorporating one or more strategies. Work backwards Make a table, chart or diagram Solve a simpler or similar problem Guess and check Look for a pattern Eliminate possibilities Draw a picture

  4. Plan for Solving a Word Problem Plan: Read the problem several times. What do you know? What do you need to find? Eliminate any unnecessary information. Set Up: Define a variable. Making a chart or drawing a picture may be helpful. Write an open sentence. Solve the open sentence.

  5. Plan for Solving a Word Problem Check: Reread the problem. Did you answer the question? Did you state your answer clearly with the appropriate units? Is your answer consistent with the information given in the problem?

  6. Number Problems Return to Table of Contents

  7. Vocabulary Consecutive integers are obtained when you count by ones from any given integer. Examples 1, 2, 3, 4, 5 -6, -5, -4, -3 x, x + 1, x + 2 x - 1, x, x + 1

  8. Vocabulary Even integers are integers that are multiples of two. Consecutive even integers are obtained when you count by twos from any given even integer. Examples 2, 4, 6 -6, -4, -2 x, x + 2, x + 4 x - 2, x, x + 2

  9. Vocabulary Odd integers are integers that are not even. Consecutive odd integers are obtained when you count by twos from any given odd integer. Examples 3, 5, 7 -9, -7, -5 x, x + 2, x + 4 x - 2, x, x + 2

  10. Example 1 One number is 10 greater than another. If the lesser number is subtracted from three times the greater number, the difference is 42. Find the numbers. Plan: 2 numbers - one is 10 greater than the other Set Up: n + 10 = greater number n = lesser number Write an open sentence: 3(n + 10) - n = 42

  11. Solve: 3(n + 10) - n = 42 3n + 30 - n = 42 2n + 30 = 42 -30-30 2n = 12 2 2 n = 6 So n + 10 = 6 + 10 = 16 The two numbers are 6 and 16. Check: Reread the problem. Does your answer make sense?

  12. Example 2 One number is 12 greater than another. If the sum of the two numbers is 88, find the numbers. Plan: 2 numbers - one is 12 greater than the other Set Up: n + 12 = one number n = second number Write an open sentence: (n + 12) + n = 88 X

  13. Solve: (n + 12) + n = 88 2n + 12 = 88 2n + 12 = 88 -12-12 2n = 76 2 2 n = 38 So n + 12 = 38 + 12 = 50 The two numbers are 38 and 50. Check: Reread the problem. Does your answer make sense? X

  14. Example 3 Find three consecutive odd integers whose sum is 183. Plan: Find three consecutive odd integers Set Up: n = 1st consecutive odd integer n + 2 = 2nd consecutive odd integer n + 4 = 3rd consecutive odd integer Write an open sentence: n + n + 2 + n + 4 = 183 X

  15. Solve: n + n + 2 + n + 4 = 183 3n + 6 = 183 - 6- 6 3n = 177 3 3 n = 59 So n + 2 = 61 and n + 4 = 63 The three consecutive odd integers are 59, 61 and 63. Check: Reread the problem. Does your answer make sense? X

  16. Practice 1 One number is 70 greater than a second number. If the lesser number is subtracted from twice the greater number, the difference is 174. Find the numbers. Plan: Set Up: Write an open sentence: 2(n + 70) - n = 174 Solve: The two numbers are 34 and 104. Check Your Solution:

  17. Practice 2 Find three consecutive integers whose sum is -315. Plan: Set Up: Write an open sentence: x + x + 1 + x + 2 = -315 Solve: The three numbers are -106, -105 and -104. Check Your Solution:

  18. Practice 3 Find three consecutive even integers such that the sum of the least integer and the greatest integer is -180. Plan: Set Up: Write an open sentence: Solve: The three consecutive odd integers are -92, -90 and -88. Check Your Solution:

  19. 1 Find the largest of four consecutive integers whose sum is 130.

  20. The lengths of the sides of a triangle are consecutive odd integers. The perimeter is 27 meters. Find the length of the smallest side. 2

  21. 3 Find two consecutive even integers whose sum is 148.

  22. Sam has 6 more than twice as many newspaper customers as when he started selling newspapers. If he currently has 98 customers, how many did he have when he started? 4

  23. There are fifty coins in a jar that contains only dimes and quarters. The number of dimes in the jar is 2 less than three times the number of quarters. How many dimes are in the jar? 5

  24. Age Problems Return to Table of Contents

  25. Problem Solving Strategy Sometimes using a chart to organize the information given in a word problem can be helpful. You can use this strategy to solve age problems.

  26. Example 1 Jake is 12 years older than his dog. Next year he will be four times as old as his dog will be. How old is Jake now? Plan: Find Jake's age now. Set Up: Write an open sentence: Next year, Jake will be four times as old as his dog. (x + 12) + 1 = 4(x + 1) X

  27. Solve: (x + 12) + 1 = 4(x + 1) x + 13 = 4x + 4 -x -x 13 = 3x + 4 - 4- 4 9 = 3x 3 3 3 = x Jake's dog is 3 years old now and 3 + 12 = 15. Therefore, Jake is 15 years old now. Check: Reread the problem. Does your answer make sense? X

  28. Example 2 Erica is now four years older than her sister Alicia. In ten years, Erica will be twice Alicia’s present age. Find the age of each girl now. Plan: Find the age of each girl. Set Up: Write an open sentence: x + 14 = 2x X

  29. Solve: x + 14 = 2x -x-x 14 = x Alicia is 14 years old now and Erica is 18 years old now. Check: Reread the problem. Does your answer make sense? X

  30. Practice 1 Anthony is 9 years older than his sister Marie. Next year, he will be four times as old as his sister. How old is Anthony now? Plan: Find Anthony's age now. Set Up: Write an open sentence: Solve: Anthony is 11 years old now. Check:

  31. Practice 2 Cara is six years older than her brother. In three years, she will be twice as old as her brother will be. How old is Cara now? Plan: Set Up: Write an open sentence: Solve: Cara is 9 years old now. Check:

  32. Bebe is twice as old as Marcus. The sum of their ages is 57. How old is Bebe? 6

  33. 7 Each sister is two years older than the next. The oldest sister is twice the age of the youngest sister, with two sisters in between. How old are the sisters?

  34. Deanna's age is eight years greater than half of Metri's age. If the sum of their ages is 17, how old is Deanna? 8

  35. 9 Zach's age is three less than twice Matt's age. In five years, the sum of their ages will be 19. How old is Zach now?

  36. The son is 28 years younger than his father. The sum of their ages is 84. How old is dad? 10

  37. Geometry Problems Return to Table of Contents

  38. Formulas to Remember Perimeter is the distance around a figure. Area of a rectangle A = l w Area of a triangle A = 1/2 b h

  39. Example 1 - Area Find the measure of the area of the shaded region in the figure below. x + 6 x + 4 4x 2x Plan: Given the length and width of large and small rectangles. Set Up: A = area of shaded region Open Sentence: A = 4x(x + 6) - 2x(x + 4) X

  40. x + 6 x + 4 4x 2x Solve: 4x(x + 6) - 2x(x + 4) = A 4x2 + 24x - 2x2 - 8x = A 2x2 + 16x = A The area of the shaded region is (2x2 + 16x) units2 Check: X

  41. Example 2 - Perimeter/Area The measure of the perimeter of a square is 12a + 16b. Find the measure of the area of the square. Plan: Find the area of the square. Set Up: A = s2 s = (12a + 16b) ÷ 4 s = 3a + 4b Open Sentence: A = (3a + 4b)2 X

  42. Solve: (3a + 4b) (3a + 4b) 9a2 + 12ab +12ab +16b2 9a2 + 24ab + 16b2 The area of the square is (9a2 +24ab + 16b2) units2. Check: X

  43. Example 3 - Construction The length of a rectangular lot is 5 yards less than three times the width. If the length was decreased by 2 yards and the width increased by 5 yards, the area would be increased by 17 square yards. Find the original dimensions of the lot. Plan: Find the area of the original lot. It may help to draw a picture. Set Up: w = original width w + 5 = new width 3w - 5 = original length 3w - 7 = new length Open Sentence: w(3w - 5) + 17 = (w + 5)(3w - 7) X

  44. Solve: w(3w - 5) + 17 = (w + 5)(3w - 7) 3w2 - 5w + 17 = 3w2 + 15w - 7w - 35 3w2 - 5w + 17 = 3w2 + 8w - 35 -3w2 -3w2 -5w + 17 = 8w - 35 -8w-8w -13w + 17 = -35 -17-17 -13w = -52 -13 -13 w = 4 so, 3w - 5 = 7 The original dimensions were 4 yards by 7 yards. Check: X

  45. Practice 1 Melissa has a rectangular garden that is 10 feet longer than it is wide. A brick path that is 3 feet wide surrounds the garden. The total area of the path is 396 square feet. What are the dimensions of the garden? Plan: Set Up: Write an open sentence: Solve: The width is 25 feet and the length is 35 feet. Check Your Solution:

  46. Practice 2 A new athletic field is being sodded at Lawrence High School using 2-yard by 2-yard squares of sod. If the width of the field is 70 yards less than its length and its area is 6000 square yards, how many squares of sod will be needed? Plan: Set Up: Write an open sentence: Solve: 1500 squares of sod will be needed. Check Your Solution:

  47. The length of a garden is one more foot than twice the width. The area of the garden is 55 feet. What is the width of the garden? 11

  48. A 3 x 4 picture sits in a picture frame. The total area of the picture and frame is 56 inches. How wide is the frame? 12

  49. -----------------6x-------------------------------- 13 x + 15 The garage door is a square that measures 2x feet on each side. How many square feet of house surrounds the garage?

  50. A square has its side doubled in length. How much does the area of the square increase? 14 x