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# Mathematics Department Collaboration Outcome - PowerPoint PPT Presentation

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.

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• 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.

• All teachers shared and presented effective strategies for teaching identified standards .

• 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.

• 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.

• Classroom Teacher

• Math Curriculum Specialist

• Discipline leader

• Math department chair

• AP Curriculum( Ms.Gumucio)

• AP Testing (Ms. Rubio)

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

### 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?

• 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. =0

2t2-20t+50=0 factor out GCF of 2

2(t2-10t+25)=0 divide by 2

t2-10t+25=0 factor left side

(t-5)2=0 set factors =0

t-5=0 solve for t

+5 +5

t=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!

• 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?

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

• 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. =0

2t2-20t+50=0 factor out GCF of 2

2(t2-10t+25)=0 divide by 2

t2-10t+25=0 factor left side

(t-5)2=0 set factors =0

t-5=0 solve for t

+5 +5

t=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!

• 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?

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

• 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. =0

2t2-20t+50=0 factor out GCF of 2

2(t2-10t+25)=0 divide by 2

t2-10t+25=0 factor left side

(t-5)2=0 set factors =0

t-5=0 solve for t

+5 +5

t=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!

• 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).

Which statement must be true about the triangle?

P

Q

R

• P + R < Q

• P + Q > R

• Q + P < R

• P + R = Q

B. P + Q > R

For the figure shown below, lines m // l

-which numbererd angles are equal to each other?

• 1 and 2

• 1 and 5

• 3 and 7

• 5 and 7

If two consecutive angles of a parallelogram measure (2x + 10) º and (x-10)º, then what must be the value of x?

• 20

• 60

• 90

• 120

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?

• XY is a chord of the circle

• XZ = YZ

• XY > XZ

• XY is a diameter of the circle.

One side of the triangle is a diameter

=> opposite is a RIGHT angle

#5)

In the circle below, ABCD, AB is a diameter, and CD is a radius.

• What is m < B?

• 45º

• 55º

• 65º

• 75º

Inscribed angle Bis half of arc AC = 90º

A. 45º

Which of the following translations

would move the point (5, -2) to (7, – 4)?

• (x , y)  (x + 2, y + 2)

• (x , y)  (x – 2, y + 2)

• (x , y)  (x – 2, y – 2)

• (x , y)  (x + 2, y – 2)

(5, -2) to (7, – 4)?

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.

Put the steps in order to construct a line perpendicular to line l from point P.

### Geometry

CST prep

Which statement must be true about the triangle?

P

Q

R

• P + R < Q

• P + Q > R

• Q + P < R

• P + R = Q

B. P + Q > R

For the figure shown below, lines m // l

-which numbererd angles are equal to each other?

• 1 and 2

• 1 and 5

• 3 and 7

• 5 and 7

If two consecutive angles of a parallelogram measure (2x + 10) º and (x-10)º, then what must be the value of x?

• 20

• 60

• 90

• 120

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?

• XY is a chord of the circle

• XZ = YZ

• XY > XZ

• XY is a diameter of the circle.

One side of the triangle is a diameter

=> opposite is a RIGHT angle

#5)

In the circle below, ABCD, AB is a diameter, and CD is a radius.

• What is m < B?

• 45º

• 55º

• 65º

• 75º

Inscribed angle Bis half of arc AC = 90º

A. 45º

Which of the following translations

would move the point (5, -2) to (7, – 4)?

• (x , y)  (x + 2, y + 2)

• (x , y)  (x – 2, y + 2)

• (x , y)  (x – 2, y – 2)

• (x , y)  (x + 2, y – 2)

(5, -2) to (7, – 4)?

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.

Put the steps in order to construct a line perpendicular to line l from point P.

CST prep

### Sec. 10 – 6 Circles and Arcs

Objectives:

1) To find the measures of central angles and arcs.

2) To find circumferences and arc lengths.

Circle – Set of all points equidistant from a given point

Center

** 360°

C

** Name the circle by its center.

C

D

R

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

B

mBAE =

= 40% of 360

= (.40) • 360

= 144

mCAD =

8% of 360

(.08)(360) = 28.8

mDAE =

27% of 360 = 97.2

25%

40%

A

C

8%

27%

D

E

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 

R

S

P

T

Semicircle – Half of a Circle.

* Name by 3 letters

* Ex: TRS = 180

Adjacent Arcs – Are arcs of the same circle that have exactly one point in common.

Ex: AB and BC

B

C

A

Arc Addition!!

mBCA = mBC + mCA

C

32°

B

mBC =

mDB =

mAD =

mAB =

mBOC =

mBC + mCD

mADC – mCD

mABC – mBC

32

58

32

D

O

= 32 + 58 = 90

148°

= 180 – 58 = 122

122°

A

= 180 – 32 = 148

• Circumference – of a circle is the distance around the circle.

C = d

C = 2r

or

Pi = 3.14

Radius of Cirlce

Diameter of circle

C = 2r

=2(9cm)

=18 cm

= 56.5cm

9cm

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

• 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

C

B

2r

Length of AB =

360

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.

mADB = 210

B

Length of ADB = mADB/360 • 2r

18cm

150

M

C = 2r

= 2(18cm) = 113cm

A

D

Length of ADB = (210/360) • (113cm)

= 66cm

• 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 .