Kathleen Williams – Secondary Math Instructor

Kathleen Williams – Secondary Math Instructor

Michael Koelzer – Construction Technology Instructor

Three MethodsExtended Algebra IIMoodle Algebra IIEmbedded Algebra II

Extended Algebra IIStudents Enroll in both CTE class and Algebra IIStudents receive 1 credit per semester from CTE

classStudents receive ½ credit from Algebra IIStudents attend Algebra II every day for 50

minutes at WDFCTC and Math Instructors and follow Wayne-Westland district pacing guide for Extended Algebra II

Advantage – Students can attend CTE center.Disadvantage – CTE Instructors must modify

instruction time and/or requirements for missed “skill” time in CTE class.

Moodle Algebra IITraverse Bay Area Intermediate School District

www.tbaisd.k12.mi.usHealth Occupation, Medical Assisting and Child Care Students

Enroll in both CTE class and Moodle Algebra II.Students receive 1 credit per semester from CTE classStudents receive ½ credit from Algebra IIStudents attend Algebra II every day for 50 minutes at WDFCTC

and Math Instructors and follow a modified curriculum for 2nd year Extended Algebra II (Family of Functions is taught 1st semester as a general overview and Trig is not taught 2nd semester.)

Advantage – Students can attend CTE center and be out of Algebra II by early April so that they can attend their clinicals full-time.

Disadvantage – Math Instructor has to teach complete lessons to Moodle students (who are attending mini-clinicals which last up to two weeks) after school up to 4 days a week until the math has been covered.

Embedded Algebra IIStudents stay in their CTE class full-time.Math Instructor teaches once a week in the

CTE classroom CTE Instructor repeats instruction as needed

over the weekAdvantage – Students have some project

based lessonsDisadvantage – Depth of all Algebra II is not

attainable with this format.

A1.2.9 Know common formulas (e.g., slope, distance between two points, quadratic formula, compound interest, distance = rate*time), and apply appropriately in contextual situations.

CONSTRUCTION:

ELECTRONICS:

WELDING:

E = I*R R = E/I I = E/R RT = R1+R2+R3

P = E*I P = I2*R P = E2/R RT = (R1*R2)/(R1+R2)I = P/E R = P/I2 R = E2/PE = P/I I = P/R E = P*R

A

c

90o

45o

B C

Hypotenuse = leg 2

c = 5 2

45o-45o-90o Triangle

a = b c c = b 2 or a 2

b

a

Find the length of rafter c(NOTE: Do not include any overhang)

c 7’

17’

L1.2.1 Use mathematical symbols (e.g., interval notation 0 ≤ x ≤ 5; set notation {1, 2, 3}; summation notation) to represent quantitative relationships and situations.

CONSTRUCTION:

Solder copper pipes into a square (6”) 1/8” tolerance.

5 7/8 ≤ x ≤ 6 1/8

ELECTRONICS:

Resistor color code: Red, Black, Orange, GoldYellow, Green, Brown, Silver 20,000 ± 5% 450 ± 10%

21,000 19,000 495 405

WELDING:

1” top view of box Center the hole on the plate 1 1/8” ± 1 1/16”

4” 1 1/16” to 1 3/16”

1 1/16” ≤ x ≤ 1 3/16”

5”

L3.2.1 Determine what degree of accuracy is reasonable for measurement in a given situation; express accuracy through use of significant digits, error tolerance, or percent of error; describe how errors in measurement are magnified by computation; recognize accumulated error in applied situations.

CONSTRUCTION, ELECTRONICS, WELDING:

Tolerance example: Gap (in mm)

|x+0.5| < 0.5

-0.5 < x+0.5 < 0.5 = -1.0 < x < 0

L1.3.1 Describe, explain, and apply various counting techniques (e.g., finding the number of different 4-letter passwords; permutations; and combinations); relate combinations to Pascal’s triangle; know when to use each technique.CONSTRUCTION:

How many different combinations of 2 bits, 1 hammer, and 1 drill can be chosen?

The shop has:40 Drill Bits12 Hammers 40C2 * 12C1 * 10C1 = 112,320 Combinations10 Drills

ELECTRONICS:

WELDING:

How many different combinations of exactly 2 pieces of metal, 1 brush, 1 hammer, and 1 hood can be chosen?

40 Pieces of Metal 40C2 * 18C1 * 12C1 * 10C1 18 Brushes 12 Chipping Hammers using a calculator:10 Welding Hoods 780 * 18 * 12 * 10 = 1,684,800 Possible Combinations

How many different combinations of 2 resistors, 1 capacitor, and 1 inductor?The shop has:40 Resistors12 Capacitors 40C2 * 12C1 * 10C1 = 112,320 Combinations10 Inductors

S1.1.1 Construct and interpret dot plots, histograms, relative frequency histograms, bar graphs, basic control charts, and box plots with appropriate labels and scales; determine which kinds of plots are appropriate for different types of data; compare data sets and interpret differences based on graphs and summary statistics.


HOURLY WAGE COMPARISON

$9

$15

$30

$20

$0

$5

$10

$15

$20

$25

$30

$35

Entry LevelNon-Union

Entry LevelUnion

JourneymanUnion

ExperiencedNon-Union

Hourl

y W

age

HOME PRICES IN WESTLAND

0

20

40

60

80

100

120

140

160

1000 1100 1200 1300 1400

Square Feet

Pri

ce in

Thousa

nds

AGE OF WORKERS IN SKILLED TRADES

9

15

2018

12

0

5

10

15

20

25

18-25 26-32 33-40 41-50 50+

Age of Workers

# o

f W

ork

ers

50K 90K 130K

40K 150K

Median

Home Values

40K 60K 90K 110K 150K

When remodeling a home, keep costs at the low end of the box and do not exceed the

upper value.

S1.2.1 Calculate and interpret measures of center including: mean, median, and mode; explain uses, advantages and disadvantages of each measure given a particular set of data and its context..


Looking at the home prices from S1.1.1 write a scenario that would best fit selecting the mean. Repeat for median and mode.Example: Choose the mean if determining the average cost of a 1400 sq. ft. house with similar configurations such as 2 bedroom 1 ½ bath with 2 car garage.Choose the mode to determine what size (sq. ft.) of homes that were selling the most in a subdivision or particular school zoned area. Choose the median to determine the average house price in a subdivision or section of a city.

A1.1.4 Add, subtract, multiply, and simplify polynomials and rational expressions (e.g., multiply (x-1) (1-x2+3); simplify (9x-x3)/(x+3)).

CONSTRUCTION:

ELECTRONICS:

WELDING:

[2x2 – 6x + 4](x-1) 2x3 – 6x2 + 4x – 2x2 + 6x – 4 2x3 -8x2 + 10x – 4

[(5-2x) (4-2x)]x [20-8x-10x+4x2]x[20-18x+4x2]x20x-18x2+4x3

RT = (R1*R2)/(R1+R2) Let R1 = 2x and R2 = 6xRT = (2x*6x)/2x + 6x)RT = 12x2 / 8xRT = 3/2 x

A1.2.5 Solve polynomial equations and equations involving rational expressions (e.g., solve -2x(x2+4x+3) = 0; solve x-((1/(x+6)) = 3, and justify steps in the solution.

CONSTRUCTION:

ELECTRONICS:

WELDING:

V = L*W*H36 = (2x-2) (x-2) (x-1)36 = [2x2 - 2x - 4x +4](x-1) Foil binomials36 = [2x2 – 6x + 4](x-1) Combine like terms36 = 2x3 – 6x2 + 4x – 2x2 + 6x – 4 Distribution property 36 = 2x3 -8x2 + 10x – 4 Combine like terms0 = 2x3 - 8x2 + 10x + 32 Subtraction property

V = L*W*H6 = [(5-2x) (4-2x)]x 6 = [20-8x-10x+4x2]x Foil Binomials6 = [20-18x+4x2]x Combine like terms6 = 20x-18x2+4x3 Distribution property0 = 4x3-18x2+20x-6 Subtraction property

RT = (R1*R2)/(R1+R2) Let RT = 72 ohms, R1 = 2x and R2 = 6xRT = (2x*6x)/2x + 6x) Substitution PropertyRT = 12x2 / 8x Multiplication Property and combine like terms72 = 3/2 x Division Property of Exponents and Reducing FractionsX = 16 ohms Division Property of Fractions

A3.1.2 Adapt the general symbolic form of a function to one that fits the specifications of a given situation by using the information to replace arbitrary constants with numbers.

CONSTRUCTION:

ELECTRONICS:

WELDING:

Find the number of studs in a 37’ wallx = L*.75+1

x = 37*.75+1X = 29

c2 = a2+b2

52 = 42+22

25 = 16+425 > 20

Angle is obtuse

5

4 ?

2

Resonant Frequency Formula f = 1/(2лLC) Determine the frequency in KHz given an 330 µH Inductor and a .22 µF Capacitor. f = 159.2/330*.22f = 18.68 KHz

A1.1.7* Transform trigonometric expressions into equivalent forms using basic identities.

ELECTRONICS:

ELECTRONICS:

TAN = Xc/R

SIN = Xc/ZCOS = R/Z

Tp = E*I*COS

W = 2 fT = 1/ff = 1/T

Xc = 1/(2 f C)XL = 2 f L

FR= 1/(2л LC)

ZXc

R

A2.2.4* If a function has an inverse, find the expression(s) for the inverse.

A2.9.1 Write the symbolic form and sketch the graph of simple rational functions.

CONSTRUCTION:

ELECTRONICS:

WELDING:

Build a wooden box of 1” material. Outside dimensions are L = 2*H and L = 2*W. Total volume = 36 square inches.

V = L*W*H36 = (2x-2) (x-2) (x-1)

2x

x

x

Using a graphing calculator: Volume = 6 and x = 1

V = L*W*H6 = [(5-2x) (4-2x)]x

6 = [20-8x-10x+4x2]x6 = [20-18x+4x2]x6 = 20x-18x2+4x3

0 = 4x3-18x2+20x-6

x xx x

4”x x x x

5”

Embedded Algebra - Construction

A2.9.1

Using Rational Functions in Construction

Embedded Algebra - ConstructionA2.9.1 Using Rational

Functions in Construction

Do not lose this sheet! You must build a box for a customer that will hold exactly 162 cubic

inches of sand. The box will be made from ½ inch material, with 4 sides and a bottom, but no top. The length must be twice the width,

and the height will be the same as the width. You will first use algebra to determine the

size of the box. Next you will build the box. Finally you will test the size of the box with sand. Do your math on this page and turn

this in with your box.

Embedded Algebra - ConstructionEvaluation:Safety: 5Math: 15

Sand Test: 10Measurements: 10

Square: 5Cuts: 5

Total: 50 points (Test Grade)











And they all lived happily ever

after...

ReferencesMoodle Siteswww.tbaisd.k12.mi.uswww.moodle2.resa.net/ww

WDFCTC Website

http://ford.wwcsd.net/

Presenter’s [email protected]@wwcsd.net

Documents

Kathleen Williams – Secondary Math Instructor