2016 05-25- HPEDSB Making Math Contextual, Visual and Concrete

Making Math Contextual, Visual, and Concrete

@MathletePearcewww.tapintoteenminds.com

Making Math Contextual, Visual,

and Concrete



PROBLEM

ENGAGED YET?



ENGAGEMENTThere are several types/categories of engagement academic, cognitive, intellectual, institutional, emotional, behavioural, social, psychological, to name a few.

Parsons, J., & Taylor, L. (2011). Student Engagement:What do we know and what should we do? University Partners (pp 3). University of Alberta.

[email protected]


MathClass


4. Homework

3. Examples

2. Definitions, Formulae, Procedures/Algorithms

1. Take Up Homework

Math Class

Example 1(a) List all combinations of three out of the four colours: red, blue, white, and green.


My Result:

Successfulin Math

Unsuccessfulin Math


A Perception Develops

Good atMath

Poor atMath


Not

Confident in Math Class

Doubtful inMath Class

Students Feel


Not

EnjoyingMathematics

Disliking Mathematics

This May Lead To


Not

Willingnessto Engage InMath Class

Unwillingness to Engage In Math Class

This May Lead To


Not

Moving Ahead Falling Behind

This May Lead To


Two Groups of Students

Not

Good atMath

Poor atMath


Two Groups of StudentsStruggle with

unfamiliarproblems

Good at Math

Poor atMath


Interconnected

Visual and ConcreteContextual

VISION


Interconnected


CONFIDENCE


Interconnected


PERCEPTION


Interconnected


STUDENT SUCCESS

Change


Immune to

http://www.vancouversun.com/life/From+geek+chic/6296707/story.htmlPhotograph by: Maggie Wong


@MathletePearcewww.tapintoteenminds.com https://mathstrategies.wordpress.com/tag/numb3rs/

http://www.wykamath.com/

entertainment

engagement=



algorithm

whatsthis?

algorithm

s

[email protected]

algorithms

engagement=

understandingor


LEARNING


NEWOLD

PROGRESSIVETRADITIONALACTIVITYLECTURE

INQUIRY BASEDDIRECT INSTRUCTIONTECHNOLOGYPEN & PAPER

MATH WARS


BALANCENEW

OLD

PROGRESSIVETRADITIONAL

ACTIVITYLECTURE

INQUIRY BASEDDIRECT INSTRUCTION

TECHNOLOGYPEN & PAPER


What do you notice?What do you wonder?

-Part Math Lesson

How many packs of paper would it take to reach the ceiling?

4. Consolidate3. Connections

2. InquiryOUR FIRST QUESTION:QUESTIONSPREDICTIONS


PREDICTIONS


REQUEST INFORMATION


How many packs of paper would it take to reach the ceiling? 4. Consolidate3. Connections

273

cm24.7

5 cm

Answer Key: 55.15QUESTION


Mr. Pearce stacks reams of paper vertically against a wall. If the height of 5 reams is 24.75 cm and the height of the room is

273 cm, how many reams of paper would he need to reach the ceiling?

Answer Key: 55.15

HOW MANY SMALL VASES WILL IT TAKE TO FILL THE TALL VASE?

www.estimation180.com

http://www.estimation180.com

#MTBoS

The Best Teacher PD is FREE on Twitter

Follow the Math Twitter Blogosphere


3 Act Math Tasks

kylep.ca/curiousmath1p


iTunes U

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Visual and ConcreteContextualTASKS

CONSOLIDATE INQUIRYMove Towards

Procedural Fluency

Connect to Prior Knowledge

CONNECTIONS

Construct Conceptual Understanding

VISION

@MathletePearcekylep.ca/curiousmath1p


Four Stages of

Mastery

COMPETENCE

CO

NS

CIO

US

NE

SS

2 3

41


Four Stages of

Mastery

COMPETENCE

CO

NS

CIO

US

NE

SS

2 3

4

Consciously Competent

Consciously Incompetent

1Unconsciously IncompetentUnconsciously

Competent


Consciously Incompetent

The Four Stages of

Mastery

COMPETENCEC

ON

SC

IOU

SN

ES

S

1

2 3

4Unconsciously Competent

Consciously Competent

Unconsciously Incompetent1

2

3

4

1111

-1-1

-1-1-1-1

-1-1

11

1111

-1-1

-1-1

111

1111

4 - 8 + 6 - 4 + 7= - 4 + 6 - 4 + 7= 2 - 4 + 7= - 2 + 7= 5

What equation am I?

x

1 11 1

=1 11 1

1 1 1 11 1 1 11 1 1 1

11111

x = 20+ 5

x

1 11 1

=1 11 1

1 1 1 11 1 1 11 1 1 1

11111

x

1 11 1=

1 11 1

1 1 1 11 1 1 11 1 1 1

11111

x = 20+ 5

-1-1-1-1

-1-1-1-1-1

-1

- 5 - 5

x

1 11 1=

1 11 1

1 1 1 11 1 1 11 1 1 1

11111

x = 20+ 5+ 5 + 5

11111

11111

x

1 11 1=

1 11 1

1 1 1 11 1 1 11 1 1 1

11111

x = 20+ 5+ x + x

xx

x

1 11 1

-x =1 11 1

1 1 1 11 1 1 11 1 1 1

11111

x = 20+ 5- x - x

-x

x =

x = 9- 3

-1-1-1

-1-1

-1

-

-1-1

-1-1-1

-1

x =111

x = 9- 3

-1-1-1

-1-1

-1

+ 3 + 3-

-1-1

-1-1-1

-1 111

x =

x = 9- 3

-1-1-1

-1-1

-1

- 3 - 3-

-1-1

-1-1-1

-1-1-1-1

-1-1-1

x =

x = 9- 3

-1-1-1

-1-1

-1

+ x-

-1-1

-1-1-1

-1

+ x

xx

x

-x =

x = 9- 3

-1-1-1

-1-1

-1

- x-

-1-1

-1-1-1

-1

- x

-x

What equation am I?

x 11

-1

=xx

1 11 1

1 1 11 1 1 1

111

-1

-1 -1

-1 -1 1

x 11

-1

=xx

1 11 1

1 1 11 1 1 1

111

-1

-1 -1

-1 -1 1

3(x - 2) = 17

3 x 23 groups of 2

1

11

3 x 23 groups of 2

1 1

1

11

3 x 23 groups of 2

1 1

1

11

3 x 23 groups of 2

1 1

1 1

1

11

3 x 23 groups of 2

1 11 1

1 1

1

11

3 x 23 groups of 2

1 11 11 1

1 1

1

11

3 x 23 groups of 2

1 11 11 1

= 6

1 1

1

11

5 x 3 = ?

1

11

5 groups of 3

1 1

1

11

5 x 35 groups of 3

1 11 11 1

= ?

1

11

111

1 11 1

11

1 1

1

11

5 x 35 groups of 3

1 1 1 1

= ?

1

11

1 11 1 1 11 11 1 1 11 11 1 1 11 11 1 1 11 1

1 1

1

11

5 x 35 groups of 3

1 1 1 1

= ?

1

11

1 11 1 1 11 11 1 1 11 1

1 1

1

11

5 x 35 groups of 3

1 1

= ?

1

11

1 11 11 1

5 x 14 = ?5 groups of 14

11111

1 1 1 1 1 1 1 1 1 1 1 1 1 1

5 x 14 = ?5 groups of 14

11111

1 1 1 1 1 1 1 1 1 1 1 1 1 110 4

5

5 groups of 10 plus 5 groups of 4or

?

5 x 14 = ?5 groups of 14

11111

1 1 1 1 1 1 1 1 1 1 1 1 1 110 4

5


? ?

5 x 14 = 705 groups of 14

11111

1 1 1 1 1 1 1 1 1 1 1 1 1 110 4

5


50 20

5 x 14 = 705 groups of 14

11111

1 1 1 1 1 1 1 1 1 1 1 1 1 110 4

5


20 50

5 x 14 = 245 groups of 14

11111

1 1 1 1 1 1 1 1 1 1 1 1 1 110 4

5


15 9

5 x 14 = 1405 groups of 14

11111

1 1 1 1 1 1 1 1 1 1 1 1 1 110 4

5


70 70

9 x 12 = ?9 groups of 12

111111111

1 1 1 1 1 1 1 1 1 1 1 1

9 x 12 = ?9 groups of 12 1

11111111

1 1 1 1 1 1 1 1 1 1 1 110 2

5

4

5 groups of 10 plus

4 groups of 2

or

9 x 12 = ?111111111

1 1 1 1 1 1 1 1 1 1 1 110 2

5

4

?

9 groups of 12

5 groups of 10 plus

4 groups of 2

or

9 x 12 = ?111111111

1 1 1 1 1 1 1 1 1 1 1 110 2

5

4

? ?

9 groups of 12

5 groups of 10 plus

4 groups of 2

or

9 x 12 = ?111111111

1 1 1 1 1 1 1 1 1 1 1 110 2

5

4

?

?

?

9 groups of 12

5 groups of 10 plus

4 groups of 2

or

9 x 12 = ?111111111

1 1 1 1 1 1 1 1 1 1 1 110 2

5

4

?

?

?

?

9 groups of 12

5 groups of 10 plus

4 groups of 2

or

9 x 12 = ?111111111

1 1 1 1 1 1 1 1 1 1 1 110 2

5

4

50

40

10

8

9 groups of 12

5 groups of 10 plus

4 groups of 2

or

9 x 12 = ?111111111

1 1 1 1 1 1 1 1 1 1 1 110 2

5

4

15

14

7

6

9 groups of 12

5 groups of 10 plus

4 groups of 2

or

9 x 12 = ?111111111

1 1 1 1 1 1 1 1 1 1 1 110 2

5

4

40

50

20

8

9 groups of 12

5 groups of 10 plus

4 groups of 2

or

9 x 12 = ?111111111

1 1 1 1 1 1 1 1 1 1 1 110 2

5

4

10

8

50

40

9 groups of 12

5 groups of 10 plus

4 groups of 2

or

What equation am I?

x11

-1

=xx

1 11 1

1 1 1 1

111

-1

1

x

x

x

-1

-1 -1 -1

-1 -1 -1

1

x11

-1

=xx

1 11 1

1 1 1 1

111

-1

1

x

x

x

-1

-1 -1 -1

-1 -1 -1

1

3(2x - 3) = 15

Whats My Area Equation?


x x x x


x x x x111


x x x x111

Area = 3( )4x


x x x x111

x x x xx x x xx x x x

Area = 3( )4x= 12x units2


11

1 11 1

1 1 1 1

111

11

x x x x111

Area = 3( )4x= 45

111

111

11

1 11 1

1 1 1 1

111

11

111

111 1

11

units2


x

x x x x111

Area = 3( )4x= 15x units2

x x x x x x x x x x x x x x


x x x x111

Area = 3( )4x= 10x + 15 units2

x x x x x x x x x x

111

111

111

111

111

Find the Area of The Pool

x x x111

Area = 5( )3x

11

x x x


x x x

xx


x

x x x

x


x

x x x

x

Area = 2x groups

x

x x x

x

Area = 2x groups of 3x

x

x x x

x

Area = 2x ( )3x

x

x x x

xx2

Area = 2x ( )3x

x2x

x x x

xx2

Area = 2x ( )3x

x2x2x

x x x

xx2

Area = 2x ( )3x

x

x x x

xx2 x2 x2

x2 x2 x2

Area = 2x ( )3x

x

x x x

xx2 x2 x2

x2 x2 x2

6x2Area = 2x ( )3x =

xx x x

x

x2 x2 x2

x2 x2 x2

If the total area is 96 units2, solve for the value of x.

6x2Area = 2x ( )3x =

xx x x

x

x2 x2 x2

x2 x2 x2

96 = 6x2


x x x


x x x 1 1


x x x 1 11111


Area =

x x x 1 11111

Area =

x x x 1 11111

4 groups

Area =

x x x 1 11111

4 ofgroups 3x + 2

Area =

x x x 1 11111

4 )( 3x + 2

1 1111

x x x

1

1 1x x x1 1x x x1 1x x x1 1x x x

Area = 4 )( 3x + 2= 12x + 8

1 1111

x x x

1

1 1x x x1 1x x x1 1x x x1 1x x x

= 12x + 8If the total area is 80 units2, solve for the value of x.

80


x1 1x x x

x

x1 1x x x

x

Area = 2x groups

x1 1x x x

x

Area = 2x groups of 3x + 2

x1 1x x x

x

Area = 2x ( )3x + 2

x2x1 1x x x

x

Area = 2x ( )3x + 2

x2x2x1 1x x x

x

Area = 2x ( )3x + 2

x2 x xx2x2x1 1x x x

x

Area = 2x ( )3x + 2

x1 1x x x

x2 x2 x2 x x

x x2 x2 x2 x x

Area = 2x ( )3x + 2

x1 1x x x

x2 x2 x2 x x

x x2 x2 x2 x x

Area = 2x ( )3x + 2 = 6x2 + 4x


x1 1

1

1

x

1

x x

Area = (x + 3)(3x + 2)

x1 1

1

1

x

1

x x

x1 1

1

1

x

1

x x

x2 x2 x2

x x xx x xx x x

x x

1

11

1

11

Area = (x + 3)(3x + 2)

x1 1

1

1

x

1

x x

x2 x2 x2

x x xx x xx x x

x x

1

11

1

11

Area = (x + 3)(3x + 2) = 3x2 + 11x + 6

NOTICE? WONDER?

What do you


VISION



VISION INQUIRY


Interconnected


CONNECTIONS

INQUIRYVISION


Interconnected


CONNECTIONS

INQUIRYVISIONCONSOLIDATE


@MathletePearce@MrOrr_Geek

Clap as fast as you can


Notice? Wonder?

INQUIRY


Will he beat the record?

INQUIRY


Act 3


CONSOLIDATE

Which One Doesnt Belong?

wodb.ca

http://wodb.ca

wodb.ca

@MaryBourassa

http://wodb.ca

wodb.ca

http://wodb.ca

Stage 1

Stage 2

?Stage 3

Stage 1

Stage 2

Stage 3

Stage 1

Stage 2

Stage 3

?Stage 4

Stage 1

Stage 2

Stage 3

?Stage 4

Stage 1

Stage 2

Stage 3

Stage 4

Stage 1 Stage 2

Stage 3 Stage 4

Stage 1 Stage 2

Stage 3 Stage 4

Initial Value:

Stage 1 Stage 2

Stage 3 Stage 4Initial Value:

Stage 0

No Squares

Stage 1 Stage 2


Stage 0

No Squares

0

Stage 1 Stage 2


Stage 0

No Squares

0

Rate of Change:

Stage 1 Stage 2


Stage 0

No Squares

0

Rate of Change:4

Stage 1 Stage 2


Stage 0

No Squares

0

Rate of Change:4

Equation:

Stage 1 Stage 2


Stage 0

No Squares

0

Rate of Change:4

Equation:y =

Stage 1 Stage 2


Stage 0

No Squares

0

Rate of Change:4

Equation:y = 0

Stage 1 Stage 2


Stage 0

No Squares

0

Rate of Change:4

Equation:y = 0 + 4x

Stage 1 Stage 2


Stage 0

No Squares

0

Rate of Change:

Equation:y = 0

Stage 1 Stage 2


Stage 0

No Squares

0

Rate of Change:

Equation:y = 0

4

+ 4x

Stage 1 Stage 2


Stage 0

No Squares

0

Rate of Change:

Equation:y =

4

4x

Stage 1 Stage 2

Stage 5 Stage 6

Stage 3 Stage 4

? ?

Stage 1 Stage 2

Stage 5 Stage 6

Stage 3 Stage 4

?

Stage 1 Stage 2

Stage 5 Stage 6

Stage 3 Stage 4


130

.75

cm

130

.75

cm


130

.75

cm

130

.75

cm

Notice? Wonder?


4.95 cm

130

.75

cm

?

130

.75

cm

How tall is the table?


4.95 cm

130

.75

cm

?

packs5

130

.75

cm



4.95 cm

130

.75

cm

?

packs5 x = 24.75 cm

130

.75

cm 24.75 cm



4.95 cm

130

.75

cm

?

packs5 x = 24.75 cm

130.75 cm24.75 cm-



4.95 cm

130

.75

cm

?

packs5 x = 24.75 cm

130.75 cm24.75 cm-

106 cm106 cm



4.95 cm

130

.75

cm packs5 x = 24.75 cm

130.75 cm24.75 cm-

106 cm

106

cm



-Part Math Lesson

4. Consolidate

Minds On

Inquiry2. -Part Math Lesson

Making Connections



4.95 cm

130

.75

cm packs5 x = 24.75 cm

130.75 cm24.75 cm-

106 cm

106

cm

How tall is the table?4.95 cmpacks5 x = 24.75 cm


130

.75

cm

130

.75

4.95 cmx = 24.75 cm

130.75 cm 24.75 cm 106 cm= +

packs5 4.95 cmx = 24.75 cmpacks5

106

cm


x = 24.75 cm

130.75 cm 24.75 cm 106 cm= +

packs5

4.95 cm (

= 24.75 cm

packs5

130.75 cm 24.75 cm 106 cm= +

4.95 cm

)( )

130

.75

cm

130

.75

106

cm


x = 24.75 cm

130.75 cm 24.75 cm 106 cm= +

packs5 = 24.75 cm

130.75 cm

24.75 cm

106 cm= +

4.95 cm

4.95 cm ( packs5 )( )

130

.75

cm

130

.75

106

cm


x = 24.75 cm

130.75 cm 24.75 cm 106 cm= +

packs5 = 24.75 cm

130.75 cm

24.75 cm

106 cm= +

height initial height of table= +number of packs

4.95 cm

height of a pack )

130

.75

cm

130

.75

106

cm ( )(

4.95 cm ( packs5 )( )


x = 24.75 cm

130.75 cm 24.75 cm 106 cm= +

packs5 = 24.75 cm

130.75 cm

24.75 cm

106 cm= +

height initial height of table= +

4.95 cm13

0.7

5 c

m13

0.7

5

106

cm number

of packsheight of

a pack )( )(

4.95 cm ( packs5 )( )


x = 24.75 cm

130.75 cm 24.75 cm 106 cm= +

5 = 24.75 cm

130.75 cm

24.75 cm

106 cm= +


4.95 cm13

0.7

5 c

m13

0.7

5

106

cm

packs

number of packs

height of a pack )( )(

4.95 cm ( packs5 )( )


x = 24.75 cm

130.75 cm 24.75 cm 106 cm= +

5 = 24.75 cm

130.75 cm

24.75 cm

106 cm= +


4.95 cm13

0.7

5 c

m13

0.7

5

106

cm

packs

y b= +m x

number of packs

height of a pack )( )(

4.95 cm ( packs5 )( )

Minds On


-Part Math Lesson13

0.7

5 c

m

packs of paper heightIndependent Dependent

5packs13

0.7

55

4.95 cm



How tall is the table?13

0.7

5 c

m


5packs13

0.7

55

4.95 cmSlope



0.7

5 c

m


5packs

130.755

4.95 cmSlope



0.7

5 c

m


5packs

130.755

4.95 cmSlope )x y(



0.7

5 c

mpacks of

paper height5 packs130.7554.95 cm

Slope

)x y

( ,In

itia

l Val

ue

y -i

nter

cep

t



0.7

5 c

mpacks of

paper height

130.7554.95 cmSlope

)x y

( ,

y b= +mxInitia

l Val

ue

y -i

nter

cep

t

5 packs5130.7554.95

130

.75



0.7

5 c

mpacks of

paper height

130.7554.95 cmSlope

)x y

( ,

y b= +mxInitia

l Val

ue

y -i

nter

cep

t

5 packs5

130.75

5

130.75

b+

4.95



0.7

5 c

mpacks of

paper height

130.7554.95 cmSlope

)x y

( ,

y b= +m xInitia

l Val

ue

y -i

nter

cep

t

5 packs5

130.75

5

130.75 = )()( b+

4.95



0.7

5 c

mpacks of

paper height

130.7554.95 cmSlope

)x y

( ,

y b= +m xInitia

l Val

ue

y -i

nter

cep

t

5 packs

5130.75 5130.75 =)(

)()( b+

55)(4.95



0.7

5 c

mpacks of

paper height

130.7554.95 cmSlope

)x y

( ,

y = mInitia

l Val

ue

y -i

nter

cep

t

5 packs

130.75130.75 =)(

b+130.75 = b+x55 )(

)(55)(4.95

b+ b+54.95

54.95



0.7

5 c

mpacks of

paper height

130.7554.95 cmSlope

)x y

( ,

y = mInitia

l Val

ue

y -i

nter

cep

t

5 packs

130.75130.75 =)(

130.75 = b+24.75b+ b+

x55 )(

)(55)(4.95

b+ b+



0.7

5 c

mpacks of

paper height

130.7554.95 cmSlope

)x y

( ,

y = mInitia

l Val

ue

y -i

nter

cep

t

5 packs

130.75130.75 =)(

130.75 = b+b+ b+

x55 )(

)(55)(4.95

b+ b+

54.9524.7524.75-



0.7

5 c

mpacks of

paper height

130.7554.95 cmSlope

)x y

( ,

y = mInitia

l Val

ue

y -i

nter

cep

t

5 packs

130.75130.75 =)(

130.75 = b+b+ b+

x55 )(

)(55)(4.95

b+ b+

54.9524.7524.75-24.75-

b+



0.7

5 c

mpacks of

paper height

130.7554.95 cm

Slope )x y

( ,

y = mIn

itia

l Val

ue

y -i

nter

cep

t

5 packs

130.75130.75 =)(

130.75 = b+b+ b+

x55 )(

)(55)(4.95

b+ b+

54.9524.7524.75-24.75-

106 = b0 +

In the past


ExampleConsolidation


Task #1:

Minds On




5packs

130.755( )x y4.95 cmSlope

2 1 51 513

0.7

5 c

m

Minds On



packs of paper heightIndependent Dependentpacks

( )x ySlope

2 1 51 51


What does the point (1, 5) represent in the context of the Stacking Paper task?


( )x ySlope

2 1 51 51


What does the slope of 2 represent in the context of the Stacking Paper task?


( )x ySlope

1 51 512 1 51 51



( )x y1


1 51 512 1 51 51



( )x y

2 1 51 5

1

5

1

52

y = m x + b

2

( )1 pack



( )x y

2 1 51 51 55

2 2( )

1

1y = m + b( )

51 pack



( )x y

2 1 51 51

5

5

2 2( )

1

1y = m + b( )51 pack

1pack



( )x y

2 1 51 51

5

5

2 2()

1

1y = + b()5 2

1pack



( )x y

2 1 51 51

5

5

2()

1

1y = + b()5 2-2-2

2

1pack



( )x y

2 1 51 51

5

5

2()

1

1y = +b()5 2-2-22

3

3

1pack



( )x y

2 1 51 51

5

5

2()

1

1y = +b()5 2-2-22

3

3y = 2x + 3



unfamiliarproblems

Good at Math

Poor atMath



Poor at Math

Struggle withunfamiliarproblems

Good at Math


How Many Are In The Bag?



13 13



b6


How Many Are In The Bag?b6 =



b



b 3



b 3936



b

3

9

3

6



b6

6



b6


How Many Are In The Bag?b

6 6

6



b6


How Many Are In The Bag?b6 =



b3



b3 =



b3 b b bb b b 6



3 bb b b 6 =+ + + b b b+ + ++



3 bbbb 6 =+ b b b+ + ++3



3 bbbb 6 =+ bbb4+3


How Many Are In The Bag?3 bbbb 6 =+ bbb4+33-


How Many Are In The Bag?bbbb 6 =+ bbb4+3



3-



3b-


How Many Are In The Bag?b3 =+ bbb4+


How Many Are In The Bag?b3 = bbb4+

3b-


How Many Are In The Bag?b3 = bbb

SOLVE FOR X

= 172x + 5

DOUBLE CLOTHESLINE

17

2x + 5

DOUBLE CLOTHESLINE

17

2x + 5

0

0x

DOUBLE CLOTHESLINE

17

2x + 5

0

0x 2x

DOUBLE CLOTHESLINE

17

2x + 5

0

0x 2x

12

DOUBLE CLOTHESLINE

17

2x + 5

0

0x 2x

12

1x

?

DOUBLE CLOTHESLINE

17

2x + 5

0

0x 2x

12

1x

6


Struggle withunfamiliarproblemsDigging Deeper


142

.75

cm

100

.75

cm

?142

.75

100

.75

?


100

.75

cm


142

.75

cm

100

.75

cm

?142

.75

100

.75

?


142

.75

cm

100

.75

cm


?142

.75

100

.75

?

142

.75

cm

100

.75

cm

Equations Given 2 Points


142

.75

cm

100

.75

cm


?

142.75 100.75

?

-


142

.75

cm

100

.75

cm


?

142.75 100.75

?

- = 42 cm42 cm


142

.75

cm

100

.75

cm


?

142.75 100.75

?

- = 42 cm42

cm


142

.75

cm

100

.75

cm


?

142.75 100.75

?

- = 42 cm42

cm

12p

ack

s

5p

ack

s

12p

ack

s

5p

ack

s


142

.75

cm

100

.75

cm


?

142.75 100.75

?

- = 42 cm42

cm

12p

ack

s

5p

ack

s12 packs 5 packs-


142

.75

cm

100

.75

cm


?

142.75 100.75

?

- = 42 cm42

cm

12p

ack

s

5p

ack

s12 packs 5 packs- = 7 packs7 packs


142

.75

cm

100

.75

cm


?

142.75 100.75

?

- = 42 cm42

cm

12p

ack

s

5p

ack

s12 packs 5 packs- = 7 packs

7p

ack

s


142

.75

cm

100

.75

cm


?

142.75 100.75

?

-=

42 cm42

cm

12p

ack

s

5p

ack

s12 packs 5 packs-

=7 packs

7p

ack

s


142

.75

cm

100

.75

cm


?

142.75 100.75

?

-=

42 cm42

cm

12p

ack

s

5p

ack

s12 packs 5 packs-

=

7 packs7

pa

cks

6 cm

1 packs=

6 cm


142

.75

cm

100

.75

cm


?

142.75 100.75

?

-=

42 cm

42

cm

12p

ack

s

5p

ack

s12 packs 5 packs-

=

7 packs7

pa

cks

6 cm

1 packs

= 6 cmSlope m,


142

.75

cm

100

.75

cm

??

42

cm

12p

ack

s

5p

ack

s7

pa

cks

= 6 cmSlope m,packs of

paper height

)x y

( ,

100

.75

5

100.75


142

.75

cm

100

.75

cm

??

42

cm

12p

ack

s

5p

ack

s7

pa

cks


paper height

5 )x y

( ,5 100.75

100

.75

5

6

5

100.75


142

.75

cm

100

.75

cm

??

42

cm

12p

ack

s

5p

ack

s7

pa

cks


paper height

5 )x y

( ,5

y b= +mx

100.75

100

.75

5

6

b+

100.75


142

.75

cm

100

.75

cm

??

42

cm

12p

ack

s

5p

ack

s7

pa

cks


paper height

5 )x y

( ,

5

y b= +mx100.75 6100.75 = )( )( 5 b+

100.75


142

.75

cm

100

.75

cm

??

42

cm

12p

ack

s

5p

ack

s7

pa

cks


paper height

5 )x y

( ,

5

y b= +mx100.75 6100.75 = )( )( 5 b+

5100.75 6100.75 = b+30

100.75


142

.75

cm

100

.75

cm

??

42

cm

12p

ack

s

5p

ack

s7

pa

cks


paper height

5 )x y

( ,

5

y b= +mx100.75 6100.75 = )( )( 5 b+

5100.75 6100.75 = b+3030-

100.75


142

.75

cm

100

.75

cm

??

42

cm

12p

ack

s

5p

ack

s7

pa

cks


paper height

5 )x y

( ,

5

y b= +mx100.75 6100.75 = )( )( 5 b+

5100.75 6100.75 = b+3030-30-

100.75


142

.75

cm

100

.75

cm

??

42

cm

12p

ack

s

5p

ack

s7

pa

cks


paper height

5 )x y

( ,

5

y b= +mx100.75 6100.75 = )( )( 5 b+

5100.75 6100.75 = b+3030-30-

70.75 = b0 +


142

.75

cm

100

.75

cm

?142

.75

100

.75

?

packs of paper height

)x y

( ,packs of

paper height

)x y

( ,12

pa

cks

5p

ack

s

142

.75

100

.75

12 512 5


142

.75

cm

100

.75

cm

?142

.75

100.75

?


)x y

( ,packs of

paper height

)x y

( ,12

pa

cks 5

pa

cks

142

.75

100.7512

512 5


142

.75

cm

100

.75

cm

?142

.75

100.75

?


)x y

( ,packs of

paper height

)x1 y1

( ,12

pa

cks 5

pa

cks

142

.75

100.7512

512 5

POINT 1


142

.75

cm

100

.75

cm

?

142.75100.75

?


)x y

( ,packs of

paper height

)x1 y1

( ,12

pa

cks 5

pa

cks

142.75100.75 125 12

5

POINT 1


142

.75

cm

100

.75

cm

?

142.75100.75

?


)( ,packs of

paper height

)x1 y1

( ,12

pa

cks 5

pa

cks

142.75100.75 125 12

5

POINT 1 POINT 2

x2 y2

142

.75

100

.75

12 5

142.75100.755 12x1 y1 x2 y2


142

.75

cm

100

.75

cm

?

142.75100.75

?


)( ,packs of

paper height

)x1 y1

( ,12

pa

cks 5

pa

cks

142.75100.75 125 12

5

POINT 1 POINT 2

x2 y212 5

142.75100.755 12x1 y1 x2 y2

142.75 100.75- = 42 cm

pa

cks

pa

cks


142

.75

cm

100

.75

cm

?

142.75100.75

?


)( ,packs of

paper height

)x1 y1

( ,12

pa

cks 5

pa

cks

142.75100.75 12

512

5

POINT 1 POINT 2

x2 y2142.75100.75

x1 y1 x2 y2

142.75 100.75- = 42 cm12 packs 5 packs- = 7 packs


142

.75

cm

100

.75

cm

?

142.75100.75

?


)( ,packs of

paper height

)x1 y1

( ,12

pa

cks 5

pa

cks

142.75100.75 12

512

5

POINT 1 POINT 2

x2 y2142.75100.75

x1 y1 x2 y2

142.75 100.75-=

42 cm12 packs 5 packs-

=7 packs


142

.75

cm

100

.75

cm

?

142.75100.75

?


)( ,packs of

paper height

)x1 y1

( ,12

pa

cks 5

pa

cks

142.75100.75 12

512

5

POINT 1 POINT 2

x2 y2142.75100.75

x1

y1

x2

y2

142.75 100.75- 42 cm12 packs 5 packs- 7 packs

==


142

.75

cm

100

.75

cm

?

142.75100.75

?


)( ,packs of

paper height

)x1 y1

( ,12

pa

cks 5

pa

cks

142.75100.75 12

512

5

POINT 1 POINT 2

x2 y2142.75100.75

x1y1

x2y2

142.75 100.75- 42 cm12 packs 5 packs- 7 packs

--

==


142

.75

cm

100

.75

cm

?

142.75100.75

?


)( ,packs of

paper height

)x1 y1

( ,12

pa

cks 5

pa

cks

142.75100.75 12

512

5

POINT 1 POINT 2

x2 y2142.75100.75

x1y1

x2y2

142.75 100.75- 42 cm

12 packs 5 packs- 7 packs

--

Slope m, =

= ==6 cm

1 packs==


142

.75

cm

100

.75

cm

?

142.75100.75

?

)( ,)x1 y1

( ,12

pa

cks

5

pa

cks

142.75100.75 12

5

POINT 1 POINT 2

x2 y2142.75100.75

Slope m, = 6 cm

y b= +mx b+


142

.75

cm

100

.75

cm

?

142.75100.75

?

)( ,)x1 y1

( ,12

pa

cks

5

pa

cks

142.75100.75 12

5

POINT 1 POINT 2

x2 y2142.75100.75

Slope m, = 6 cm

y b= +mx b+NOW, SUBSTITUTE

SLOPE (m) and EITHER POINT TO FIND INITIAL VALUE/Y-INTERCEPT!

b+


142

.75

cm

100

.75

cm

142.75100.75 )( ,)x1 y1

( ,12

pa

cks

5

pa

cks

142.75100.75 12

5

POINT 1 POINT 2

x2 y2142.75100.75

Slope m, = 6 cm

y b= +mx b+

b

12 142.756


142

.75

cm

100

.75

cm

142.75100.75 )( ,)x1 y1

( ,12

pa

cks

5

pa

cks

142.75100.75 12

5

POINT 1 POINT 2

x2 y2142.75100.75

Slope m, = 6 cm

y b= +mx b+

b12142.75 = 6 )( )( b+


142

.75

cm

100

.75

cm

142.75100.75 )( ,)x1 y1

( ,12

pa

cks

5

pa

cks

142.75100.75 12

5

POINT 1 POINT 2

x2 y2142.75100.75

Slope m, = 6 cm

y b= +mx b+

b12142.75 = 6 )( )( b+

12142.75 = 6 b+72


142

.75

cm

100

.75

cm

142.75100.75 )( ,)x1 y1

( ,12

pa

cks

5

pa

cks

142.75100.75 12

5

POINT 1 POINT 2

x2 y2142.75100.75

Slope m, = 6 cm

y b= +mx b+

b12142.75 = 6 )( )( b+

12142.75 = 6 b+7272-

72

6


142

.75

cm

100

.75

cm

142.75100.75 )( ,)x1 y1

( ,12

pa

cks

5

pa

cks

142.75100.75 12

5

POINT 1 POINT 2

x2 y2142.75100.75

Slope m, = 6 cm

y b= +mx b+

b12142.75 = 6 )( )( b+

12142.75 = 6 b+7272-72-

70.75 = b0 +

6


142

.75

cm

100

.75

cm

142.75100.75 )( ,)x1 y1

( ,12

pa

cks

5

pa

cks

142.75100.75 12

5

POINT 1 POINT 2

x2 y2142.75100.75

Slope m, = 6 cm

y b= +mx b+b 70.75b6

Therefore, the equation of the line is:


2x - 4 = 8Solve for x:



+ 4 + 4



+ 4 + 4

2x = 12



+ 4 + 4

2x = 122 2


2x - 4 = 8Solve for x:+ 4 + 4

2x = 122 2x = 6


GAMIFICATION

Making Connections



0

0


0

0-

+


0

0-

+ -

+


0

0

0

0


0

0

0 = 0

xx


0

0

x

0 = 0x+

-aa-

a

-

+


0

0

x

-a

+a

= 00 x a a+ -

b

b

+

-

+


0

0

x

-a

+a-b

+b

=0 x a a+ - + 0 b b+ -

c

d

+

+


Making Connections

0

0-a

+a-b

+b

x

+c

+d

0 x a a = 0 b b c d+ - + + - + +


Making Connections-a

+a-b

+b

x

x a a = b b c d- + - + +

+c

+d

+0 +

+c

+d


Making Connections

-a+a-b

+b

x

x aa = b b c d-+ - + +

0

0

+c

+d


Making Connections

-a+a-b+b

x

x aa = bb c d-+ - + +

0

0

0

0

+c

+d


Making Connectionsx

xaa = bbc d- - ++0 0


Making Connections

x = bbc d+


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2016 05-25- HPEDSB Making Math Contextual, Visual and Concrete

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Transcript of 2016 05-25- HPEDSB Making Math Contextual, Visual and Concrete