1. Mathematics Department Collaboration Outcome Brief Summary: Different math disciplines reviewed all District 4 benchmarks to see the growth. Although the growths were not very significant, yet the data revealed a steady improvement in every area. Benchmark 4 data given by Ms. Gumucio were analyzed and disaggregated by math faculty Algebra Essentials: 8 standards identified by the non-mastery item analysis were selected for re- teaching.Standards: 4,7,8,9,11,17,20,and 21. ALGEBRAI: Non-Master range of 22% to 32% related to power Standards #:5,7,12,13,15,20,21, and 23 Were identified for re-teaching. GEOMETRY:Standards # 4,7,16,17,18, 20, 21,and 22 were identified for re-teaching. ALGEBRA II: Standards# 2,7,10,11.2, 12,18,and 19 were identified for re-teaching. All teachers in different disciplines developed different exam questions based on the identified standards for re-teaching.

2. Mathematics Department Plan: All teachers shared and presented effective strategies for teaching identified standards.

3. NEXT STEPS TO SOLVE THE EXIGENCE Faculty based on the data unanimously decided to re- teach the designated standards starting Tuesday 4/20/2010 in order to be able to review all the concepts which were revealed as non-mastery areas. New CST style power point lessons created by Curriculum Specialist were given to all disciplines as another venue to teach.

4. TIMELINE Math department will implement the remediation and intervention as of Tuesday 4/20/10.in order to have time to review all the identified standards and concepts. Next year the starting date will be September.

5. WHO IS RESPONSIBLE ? Classroom Teacher Math Curriculum Specialist Discipline leader Math department chair AP Curriculum( Ms.Gumucio) AP Testing (Ms. Rubio)

6. To solve a quadratic eqn. by factoring, you must remember your factoring patterns!

7. 5.2 Solving Quadratic Equations by Factoring Goals: 1. Factoring quadratic expressions 2. Finding zeros of quadratic functions What must be true about a quadratic equation before you can solve it using the zero product property?

8. Zero Product Property Let A and B be real numbers or algebraic expressions. If AB=0, then A=0 or B=0. This means that If the product of 2 factors is zero, then at least one of the 2 factors had to be zero itself!

9. Example: Solve. x 2 +3x-18=0 x 2 +3x-18=0Factor the left side (x+6)(x-3)=0set each factor =0 x+6=0 OR x-3=0solve each eqn. -6 -6 +3 +3 x=-6 OR x=3check your solutions!

10. Example: Solve. 2t 2 -17t+45=3t-5 2t 2 -17t+45=3t-5Set eqn. =0 2t 2 -20t+50=0factor out GCF of 2 2(t 2 -10t+25)=0divide by 2 t 2 -10t+25=0factor left side (t-5) 2 =0set factors =0 t-5=0solve for t +5 t=5check your solution!

11. Example: Solve. 3x-6=x 2 -10 3x-6=x 2 -10Set = 0 0=x 2 -3x-4Factor the right side 0=(x-4)(x+1)Set each factor =0 x-4=0 OR x+1=0 Solve each eqn. +4 +4 -1 -1 x=4 OR x=-1 Check your solutions!

12. Finding the Zeros of an Equation The Zeros of an equation are the x- intercepts ! First, change y to a zero. Now, solve for x. The solutions will be the zeros of the equation.

13. Example: Find the Zeros of y=x 2 -x-6 y=x 2 -x-6Change y to 0 0=x 2 -x-6Factor the right side 0=(x-3)(x+2)Set factors =0 x-3=0 OR x+2=0Solve each equation +3 +3 -2 -2 x=3 OR x=-2Check your solutions! If you were to graph the eqn., the graph would cross the x-axis at (-2,0) and (3,0).

14. 5.2 Solving Quadratic Equations by Factoring Goals: 1. Factoring quadratic expressions 2. Finding zeros of quadratic functions What must be true about a quadratic equation before you can solve it using the zero product property?

15. To solve a quadratic eqn. by factoring, you must remember your factoring patterns!

16. Zero Product Property Let A and B be real numbers or algebraic expressions. If AB=0, then A=0 or B=0. This means that If the product of 2 factors is zero, then at least one of the 2 factors had to be zero itself!

17. Example: Solve. x 2 +3x-18=0 x 2 +3x-18=0Factor the left side (x+6)(x-3)=0set each factor =0 x+6=0 OR x-3=0solve each eqn. -6 -6 +3 +3 x=-6 OR x=3check your solutions!

18. Example: Solve. 2t 2 -17t+45=3t-5 2t 2 -17t+45=3t-5Set eqn. =0 2t 2 -20t+50=0factor out GCF of 2 2(t 2 -10t+25)=0divide by 2 t 2 -10t+25=0factor left side (t-5) 2 =0set factors =0 t-5=0solve for t +5 t=5check your solution!

19. Example: Solve. 3x-6=x 2 -10 3x-6=x 2 -10Set = 0 0=x 2 -3x-4Factor the right side 0=(x-4)(x+1)Set each factor =0 x-4=0 OR x+1=0 Solve each eqn. +4 +4 -1 -1 x=4 OR x=-1 Check your solutions!

20. Finding the Zeros of an Equation The Zeros of an equation are the x- intercepts ! First, change y to a zero. Now, solve for x. The solutions will be the zeros of the equation.

21. Example: Find the Zeros of y=x 2 -x-6 y=x 2 -x-6Change y to 0 0=x 2 -x-6Factor the right side 0=(x-3)(x+2)Set factors =0 x-3=0 OR x+2=0Solve each equation +3 +3 -2 -2 x=3 OR x=-2Check your solutions! If you were to graph the eqn., the graph would cross the x-axis at (-2,0) and (3,0).

22. 5.2 Solving Quadratic Equations by Factoring Goals: 1. Factoring quadratic expressions 2. Finding zeros of quadratic functions What must be true about a quadratic equation before you can solve it using the zero product property?

23. To solve a quadratic eqn. by factoring, you must remember your factoring patterns!

24. Zero Product Property Let A and B be real numbers or algebraic expressions. If AB=0, then A=0 or B=0. This means that If the product of 2 factors is zero, then at least one of the 2 factors had to be zero itself!

25. Example: Solve. x 2 +3x-18=0 x 2 +3x-18=0Factor the left side (x+6)(x-3)=0set each factor =0 x+6=0 OR x-3=0solve each eqn. -6 -6 +3 +3 x=-6 OR x=3check your solutions!

26. Example: Solve. 2t 2 -17t+45=3t-5 2t 2 -17t+45=3t-5Set eqn. =0 2t 2 -20t+50=0factor out GCF of 2 2(t 2 -10t+25)=0divide by 2 t 2 -10t+25=0factor left side (t-5) 2 =0set factors =0 t-5=0solve for t +5 t=5check your solution!

27. Example: Solve. 3x-6=x 2 -10 3x-6=x 2 -10Set = 0 0=x 2 -3x-4Factor the right side 0=(x-4)(x+1)Set each factor =0 x-4=0 OR x+1=0 Solve each eqn. +4 +4 -1 -1 x=4 OR x=-1 Check your solutions!

28. Finding the Zeros of an Equation The Zeros of an equation are the x- intercepts ! First, change y to a zero. Now, solve for x. The solutions will be the zeros of the equation.

29. Example: Find the Zeros of y=x 2 -x-6 y=x 2 -x-6Change y to 0 0=x 2 -x-6Factor the right side 0=(x-3)(x+2)Set factors =0 x-3=0 OR x+2=0Solve each equation +3 +3 -2 -2 x=3 OR x=-2Check your solutions! If you were to graph the eqn., the graph would cross the x-axis at (-2,0) and (3,0).

30. #1) Which statement must be true about the triangle? P Q R A. P + R < Q B. P + Q > R C. Q + P < R D. P + R = Q

31. B. P + Q > R

32. #2) For the figure shown below, lines m // l -which numbererd angles are equal to each other? A. 1 and 2 B. 1 and 5 C. 3 and 7 D. 5 and 7

33. Corresponding angle D

34. #3) If two consecutive angles of a parallelogram measure (2x + 10) º and (x-10)º, then what must be the value of x? A. 20 B. 60 C. 90 D. 120

35. B. 60

36. #4) Point x, y, and z are points on the circle. Which of the following facts is enough to prove that < XZY is a right angle? A. XY is a chord of the circle B. XZ = YZ C. XY > XZ D. XY is a diameter of the circle.

37. D One side of the triangle is a diameter => opposite is a RIGHT angle

38. #5) In the circle below, AB CD, AB is a diameter, and CD is a radius. ┴ What is m < B? A. 45º B. 55º C. 65º D. 75º

39. Inscribed angle B is half of arc AC = 90º A. 45º

40. #6 Which of the following translations would move the point (5, -2) to (7, – 4)? A.(x, y)  (x + 2, y + 2) B.(x, y)  (x – 2, y + 2) C.(x, y)  (x – 2, y – 2) D.(x, y)  (x + 2, y – 2)

41. (5, -2) to (7, – 4)? D. (x, y)  (x + 2, y – 2)

42. Which drawing below shows a completed construction of the angle bisector of angle B? A. B. C.D. #7)

43. Put the steps in order to construct a line perpendicular to line l from point P. #8)

44. Geometry CST prep

45. #1) Which statement must be true about the triangle? P Q R A. P + R < Q B. P + Q > R C. Q + P < R D. P + R = Q

46. B. P + Q > R

47. #2) For the figure shown below, lines m // l -which numbererd angles are equal to each other? A. 1 and 2 B. 1 and 5 C. 3 and 7 D. 5 and 7

48. Corresponding angle D

49. #3) If two consecutive angles of a parallelogram measure (2x + 10) º and (x-10)º, then what must be the value of x? A. 20 B. 60 C. 90 D. 120

50. B. 60

51. #4) Point x, y, and z are points on the circle. Which of the following facts is enough to prove that < XZY is a right angle? A. XY is a chord of the circle B. XZ = YZ C. XY > XZ D. XY is a diameter of the circle.

52. D One side of the triangle is a diameter => opposite is a RIGHT angle

53. #5) In the circle below, AB CD, AB is a diameter, and CD is a radius. ┴ What is m < B? A. 45º B. 55º C. 65º D. 75º

54. Inscribed angle B is half of arc AC = 90º A. 45º

55. #6 Which of the following translations would move the point (5, -2) to (7, – 4)? A.(x, y)  (x + 2, y + 2) B.(x, y)  (x – 2, y + 2) C.(x, y)  (x – 2, y – 2) D.(x, y)  (x + 2, y – 2)

56. (5, -2) to (7, – 4)? D. (x, y)  (x + 2, y – 2)

57. Which drawing below shows a completed construction of the angle bisector of angle B? A. B. C.D. #7)

58. Put the steps in order to construct a line perpendicular to line l from point P. #8)

59. Geometry CST prep

60. Sec. 10 – 6 Circles and Arcs Objectives: 1) To find the measures of central angles and arcs. 2) To find circumferences and arc lengths.

61. T C R D Circle – Set of all points equidistant from a given point Center ** Name the circle by its center. Radius – Is a segment that has one endpt @ the center and the other endpt on the circle. Ex. CD Diameter – A segment that contains the center of a circle & has both endpts on the circle.Ex. TR Central Angle – Is an  whose vertex is the center of the circle.Ex.  TCD C ** 360°

62. Finding measures of Central  s 25% 8% 27% 40% A B C D E m  BAE = = 40% of 360 = (.40) 360 = 144  m  CAD = 8% of 360 (.08)(360) = 28.8  m  DAE = 27% of 360 = 97.2 

63. More Circle terms P S R T Arc – Part of a Circle. * Measured in degrees ° Minor Arc – Smaller than a semicircle. (< 180°) * Named by 2 letters * Arc Measure = measure of central  * Ex: RS Major Arc – Greater than a semicircle. (> 180°) * Name by 3 letters * Order matters * Ex: RTS * Measure = Central  Semicircle – Half of a Circle. * Name by 3 letters * Ex: TRS = 180 

64. Arcs Continued A B C Adjacent Arcs – Are arcs of the same circle that have exactly one point in common. Ex: AB and BC mBCA = mBC + mCA Arc Addition!!

65. Ex 1 : Finding the measures of Arcs O B C D A 58  32  mBC = mDB = mAD = mAB = m  BOC = mBC + mCD mADC – mCD mABC – mBC 32 = 32 + 58 = 90 = 180 – 58 = 122 = 180 – 32 = 148 32° 122° 148°

66. Circumference Circumference – of a circle is the distance around the circle. C =  d or C = 2  r Pi = 3.14 Diameter of circle Radius of Cirlce

67. Ex. 2: Find the circumference of the following circle. 9cm C = 2  r =2  (9cm) =18  cm = 56.5cm

68. Example 2: Circumference The diameter of a bicycle wheel is 22in. To the nearest whole number, how many revolutions does the wheel make when the bicycle travels 100ft? Step 1: Convert diameter to feet. 12in in a foot C =  d =(1.83ft)  = 5.8ft Step 2 finish the prob 100ft/5.8ft = = 17.2 turns 22/12 = 1.83ft

69. Back to Arcs!! The measure of an arc is in degrees. Arc Length – Is a fraction of a circle’s Circumference. –It is the piece of string that would form the part of the circle. A B C

70. Length of AB = mAB 360 2r2r Measure of the arc. It is in Degrees. The Circumference Ex: An arc of 40  represents 40/360 or 1/9 of the circle. * Which means 1/9 of the Circumfernece.

71. Find the length of ADB in M. 18cm 150  A B M D mADB = 210 C = 2  r = 2  (18cm) = 113cm Length of ADB = (210/360) (113cm) = 66cm Length of ADB = mADB/360 2  r

72. Mathematics Department Plan: All teachers in different disciplines developed different exam questions based on the identified standards for re-teaching. All teachers shared and presented effective strategies for teaching identified standards.