mathematics department collaboration outcome brief summary: different math disciplines reviewed all...
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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.
Mathematics Department Plan:
• All teachers shared and presented effective strategies for teaching identified standards .
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.
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.
WHO IS RESPONSIBLE ?
• Classroom Teacher
• Math Curriculum Specialist
• Discipline leader
• Math department chair
• AP Curriculum( Ms.Gumucio)
• AP Testing (Ms. Rubio)
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?
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!
Example: Solve.x2+3x-18=0
x2+3x-18=0 Factor the left side
(x+6)(x-3)=0 set each factor =0
x+6=0 OR x-3=0 solve each eqn.
-6 -6 +3 +3
x=-6 OR x=3 check your solutions!
Example: Solve.2t2-17t+45=3t-5
2t2-17t+45=3t-5 Set eqn. =02t2-20t+50=0 factor out GCF of 22(t2-10t+25)=0 divide by 2t2-10t+25=0 factor left side(t-5)2=0 set factors =0t-5=0 solve for t+5 +5t=5 check your solution!
Example: Solve.3x-6=x2-10
3x-6=x2-10 Set = 0
0=x2-3x-4 Factor 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!
Finding the Zeros of an EquationFinding 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.
Example: Find the Zeros ofy=x2-x-6
y=x2-x-6 Change y to 0
0=x2-x-6 Factor the right side
0=(x-3)(x+2) Set factors =0
x-3=0 OR x+2=0 Solve each equation
+3 +3 -2 -2
x=3 OR x=-2 Check your solutions!
If you were to graph the eqn., the graph would cross the x-axis at (-2,0) and (3,0).
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?
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!
Example: Solve.x2+3x-18=0
x2+3x-18=0 Factor the left side
(x+6)(x-3)=0 set each factor =0
x+6=0 OR x-3=0 solve each eqn.
-6 -6 +3 +3
x=-6 OR x=3 check your solutions!
Example: Solve.2t2-17t+45=3t-5
2t2-17t+45=3t-5 Set eqn. =02t2-20t+50=0 factor out GCF of 22(t2-10t+25)=0 divide by 2t2-10t+25=0 factor left side(t-5)2=0 set factors =0t-5=0 solve for t+5 +5t=5 check your solution!
Example: Solve.3x-6=x2-10
3x-6=x2-10 Set = 0
0=x2-3x-4 Factor 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!
Finding the Zeros of an EquationFinding 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.
Example: Find the Zeros ofy=x2-x-6
y=x2-x-6 Change y to 0
0=x2-x-6 Factor the right side
0=(x-3)(x+2) Set factors =0
x-3=0 OR x+2=0 Solve each equation
+3 +3 -2 -2
x=3 OR x=-2 Check your solutions!
If you were to graph the eqn., the graph would cross the x-axis at (-2,0) and (3,0).
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?
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!
Example: Solve.x2+3x-18=0
x2+3x-18=0 Factor the left side
(x+6)(x-3)=0 set each factor =0
x+6=0 OR x-3=0 solve each eqn.
-6 -6 +3 +3
x=-6 OR x=3 check your solutions!
Example: Solve.2t2-17t+45=3t-5
2t2-17t+45=3t-5 Set eqn. =02t2-20t+50=0 factor out GCF of 22(t2-10t+25)=0 divide by 2t2-10t+25=0 factor left side(t-5)2=0 set factors =0t-5=0 solve for t+5 +5t=5 check your solution!
Example: Solve.3x-6=x2-10
3x-6=x2-10 Set = 0
0=x2-3x-4 Factor 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!
Finding the Zeros of an EquationFinding 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.
Example: Find the Zeros ofy=x2-x-6
y=x2-x-6 Change y to 0
0=x2-x-6 Factor the right side
0=(x-3)(x+2) Set factors =0
x-3=0 OR x+2=0 Solve each equation
+3 +3 -2 -2
x=3 OR x=-2 Check your solutions!
If you were to graph the eqn., the graph would cross the x-axis at (-2,0) and (3,0).
#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
#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
#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
#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.
#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
#6Which 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)
Which drawing below shows a completed construction of the angle bisector of angle B?
A. B.
C. D.
#7)
#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
#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
#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
#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.
#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
#6Which 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)
Which drawing below shows a completed construction of the angle bisector of angle B?
A. B.
C. D.
#7)
Sec. 10 – 6 Circles and Arcs
Objectives:1) To find the measures of central angles and arcs.2) To find circumferences and arc lengths.
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°
Finding measures of Central s
25%
8%
27%
40%A
B
C
D
E
mBAE =
= 40% of 360
= (.40) • 360
= 144
mCAD =
8% of 360
(.08)(360) = 28.8
mDAE =
27% of 360 = 97.2
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
Arcs Continued
A
BC
Adjacent Arcs – Are arcs of the same circle that have exactly one point in common.
Ex: AB and BC
mBCA = mBC + mCA
Arc Addition!!
Ex 1 : Finding the measures of Arcs
O
B C
D
A
5832
mBC =
mDB =
mAD =
mAB =
mBOC =
mBC + mCD
mADC – mCD
mABC – mBC
32
= 32 + 58 = 90
= 180 – 58 = 122
= 180 – 32 = 148
32°
122°
148°
Circumference
• Circumference – of a circle is the distance around the circle.
C = d or C = 2r
Pi = 3.14 Diameter of circle
Radius of Cirlce
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
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
BC
Length of AB = mAB360 2r
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.
Find the length of ADB in M.
18cm150
A
B
M
D
mADB = 210
C = 2r
= 2(18cm) = 113cm
Length of ADB = (210/360) • (113cm)
= 66cm
Length of ADB = mADB/360 • 2r