Chabot Mathematics
§1.3 Lines,
Linear Fcns
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Chabot College Mathematics
1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
Review §
1.2
 Any QUESTIONS About
• §1.2 → Functions Graphs
 Any QUESTIONS About
HomeWork
• §1.2 → HW-02
Chabot College Mathematics
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
§1.3 Learning Goals
 Review properties of lines: slope,
horizontal & vertical lines, and forms for
the equation of a line
 Solve applied problems involving linear
functions
 Recognize parallel (‖) and
perpendicular (┴) lines
 Explore a Least-Squares linear
approximation of Line-Like data
Chabot College Mathematics
3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
3 Flavors of Line Equations
 The SAME Straight Line Can be
Described by 3 Different, but Equivalent
Equations
• Slope-Intercept
(Most Common)
y  mx  b
– m & b are the slope and y-intercept Constants
• Point-Slope:
y  y1  mx  x1 
– m is slope constant
– (x1,y1) is a KNOWN-Point; e.g., (7,11)
Chabot College Mathematics
4
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
3 Flavors of Line Equations
3. General Form:
–
Ax  By  C  0
A, B, C are all Constants
 Equation Equivalence → With a little
bit of Algebra can show:
mA B
b  y1  mx1
b  C B
Chabot College Mathematics
5
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
Lines and Slope
 The slope, m , between two
y2  y1
m
points (x1,y1) and (x2,y2)
x2  x1
is defined to be:
 A line is a graph for which the slope is
constant given any two points on the line
 An equation that can be written as
y = mx + b for constants m (the slope)
and b (the y-intercept) has a line as its
graph.
Chabot College Mathematics
6
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
SLOPE Defined
 The SLOPE, m, of the line
containing points (x1, y1) and (x2, y2)
is given by
Change in y
m
Change in x
rise y2  y1


run x2  x1
Chabot College Mathematics
7
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
 Graph the line
containing the points
(−4, 5) and (4, −1)
& find the slope, m
 SOLUTION
Change in y
m
Change in x
rise y2  y1


run x2  x1
 1  5  6
m

4   4  8
Chabot College Mathematics
8
Change in y = −6
Example  Slope City
Change in x = 8
 Thus Slope
m = −3/4
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
Example  ZERO Slope
 Find the slope of the
line y = 3
 SOLUTION: Find
Two Pts on the Line
(3, 3)
(2, 3)
• Then the Slope, m
rise
33
m

run 2   3
0
m 0
5
 A Horizontal Line has ZERO Slope
Chabot College Mathematics
9
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
Example  UNdefined Slope
 Find the slope of
the line x = 2
 SOLUTION: Find
Two Pts on the Line
(2, 4)
• Then the Slope, m
rise 4   2 
m

run
22
6
m   ??
0
(2, 2)
 A Vertical Line has an UNDEFINED Slope
Chabot College Mathematics
10
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
Slope Symmetry
 We can Call
EITHER Point No.1
or No.2 and Get the
Same Slope
(−4,5) Pt1
 Example, LET
• (x1,y1) = (−4,5)
(4,−1)
rise y2  y1
m

run x2  x1
 1  5  6
3
m


4   4  8
4
Chabot College Mathematics
11
 Moving L→R
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
Slope Symmetry cont
 Now LET
(−4,5)
• (x1,y1) = (4,−1)
m
rise y2  y1

run x2  x1
5   1
6
3
m


 4  4   8
4
(4,−1)
Pt1
 Moving R→L
 Thus
Chg in y y2  y1 y1  y2
m


Chg in x x2  x1 x1  x2
Chabot College Mathematics
12

y2  y1
y1  y2
or
x1  x2
x2  x1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
Example  Application
 The cost c, in dollars, of shipping a
FedEx Priority Overnight package
weighing 1 lb or more a distance of
1001 to 1400 mi is given by
c = 2.8w + 21.05
• where w is the package’s weight in lbs
 Graph the equation and then use the
graph to estimate the cost of shipping a
10½ pound package
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
FedEx Soln: c = 2.8w + 21.05
 Select values for w and then calculate c.
 c = 2.8w + 21.05
• If w = 2, then c = 2.8(2) + 21.05 = 26.65
• If w = 4, then c = 2.8(4) + 21.05 = 32.25
• If w = 8, then c = 2.8(8) + 21.05 = 43.45
 Tabulating
the Results:
Chabot College Mathematics
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w
2
4
8
c
26.65
32.25
43.45
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
 The cost of shipping an
10½ pound package is
about $51.00
Chabot College Mathematics
15
Mail cost (in dollars)
 Plot the points.
 To estimate costs for a
10½ pound package, we
locate the point on the
line that is above 10½
lbs and then find the
value on the c-axis that
corresponds to that point
$51
FedEx Soln: Graph Eqn
10 ½ pounds
Weight (in pounds)
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
The Slope-Intercept Equation
 The equation y = mx + b is called
the slope-intercept equation.
 The equation represents a line of
slope m with y-intercept (0, b)
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
Example Find m & b
 Find the slope and the y-intercept of
each line whose equation is given by
3
a) y  x  2 b) 3x  y  7
c) 4 x  5 y  10
8
 Solution-a)
Slope is
3/8
Chabot College Mathematics
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3
y  x2
8
InterCept
is (0,−2)
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
Example  Find m & b cont.1
 Find the slope and the y-intercept of
each line whose equation is given by
3
a) y  x  2 b) 3x  y  7
c) 4 x  5 y  10
8
 Solution-b) We first solve for y to find an
equivalent form of y = mx + b.
y  3 x  7
 Slope m = −3
 Intercept b = 7
• Or (0,7)
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
Example  Find m & b cont.2
 Find the slope and the y-intercept of
each line whose equation is given by
3
a) y  x  2 b) 3x  y  7
c) 4 x  5 y  10
8
 Solution c) rewrite the equation in the
4
form y = mx + b.
y  x2
5
4 x  5 y  10
4 x  10  5 y
1
5 y  4 x  10
5
Chabot College Mathematics
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 Slope, m = 4/5
(80%)
 Intercept b = −2
• Or (0,−2) Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
Example  Find Line from m & b
 A line has slope −3/7 and y-intercept
(0, 8). Find an equation for the line.
 We use the slope-intercept equation,
substituting −3/7 for m and 8 for b:
3
y  mx  b   x  8
7
 Then in y = mx + b
Form
Chabot College Mathematics
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3
y   x 8
7
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
Example  Graph y = (4/3)x – 2
 SOLUTION: The
slope is 4/3 and the
y-intercept is (0, −2)
right 3
up 4 units
 We plot (0, −2) then
move up 4 units and
to the right 3 units.
Then Draw Line
 We could also move
down 4 units and to
the left 3 units. Then
draw the line.
Chabot College Mathematics
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(0, 2)
down 4
(3, 6)
left 3
4
y  x2
3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
(3, 2)
Parallel and Perpendicular Lines
 Two lines are parallel (||) if they lie in
the same plane and do not intersect
no matter how far they are extended.
 Two lines are perpendicular (┴) if
they intersect at a right angle
(i.e., 90°). E.g., if one line is vertical
and another is horizontal, then they
are perpendicular.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
Para & Perp Lines Described
 Let L1 and L2 be two distinct lines with
slopes m1 and m2, respectively. Then
• L1 is parallel to L2 if and only if
m1 = m2 and b1 ≠ b2
– If m1 = m2. and b1 = b2 then the Lines are CoIncident
• L1 is perpendicular L2 to if and only if
m1•m2 = −1.
• Any two Vertical or Horizontal lines are parallel
• ANY horizontal line is perpendicular to
ANY vertical line
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
Parallel Lines by Slope-Intercept
 Slope-intercept form allows us to quickly
determine the slope of a line by simply
inspecting, or looking at, its equation.
 This can be especially helpful when
attempting to decide whether two
lines are parallel
 These Lines All
Have the SAME
Slope
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
Example  Parallel Lines
 Determine whether the graphs of the
lines y = −2x − 3 and 8x + 4y = −6
are parallel.
8 x  4 y  6
 SOLUTION
• Solve General
Equation for y
• Thus the Eqns are
–
y = −2x − 3
–
y = −2x − 3/2
Chabot College Mathematics
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4 y  8 x  6
1
y   8 x  6 
4
3
y  2 x 
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
Example  Parallel Lines
 The Eqns y = −2x − 3 & y = −2x − 3/2
show that
• m1 = m2 = −2
• −3 = b1 ≠ b2 = −3/2
 Thus the Lines
ARE Parallel
• The Graph confirms
the Parallelism
Chabot College Mathematics
26
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
Example  ║& ┴ Lines
 Find equations in general form for the
lines that pass through the point (4, 5)
and are (a) parallel to & (b)
perpendicular to the line 2x − 3y + 4 = 0
 SOLUTION
2x  3y  4  0
• Find the Slope by
ReStating the
Line Eqn in
Slope-Intercept
Form
Chabot College Mathematics
27
3y  2x  4
2
4
y x
3
3
m  2 3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
Example  ║& ┴ Lines
 SOLUTION cont.
• Thus Any line parallel
to the given line must
have a slope of 2/3
• Now use the Given
Point, (4,5) in the
Pt-Slope Line Eqn
 Thus ║- Line Eqn
2 x  3 y  7
Chabot College Mathematics
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y  y1  m x  x1 
2
y  5  x  4 
3
3y  5   2 x  4 
3y  15  2x  8
3y  2x  7  0
2x  3y  7  0
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
Example  ║& ┴ Lines
 SOLUTION cont.
• Any line perpendicular
to the given line must
have a slope of −3/2
• Now use the Given
Point, (4,5) in the
Pt-Slope Line Eqn
 Thus ┴ Line Eqn
y  y1  m x  x1 
3
y  5   x  4 
2
2 y  5   3x  4 
2y  10  3x  12
3x  2y  22  0
3 x  2 y  22
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
Example  ║& ┴ Lines
 SOLUTION Graphically
Chabot College Mathematics
30
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
Scatter on plots on XY-Plane
 A scatter plot
usually shows how
an EXPLANATORY,
or independent,
variable affects a
RESPONSE, or
Dependent Variable
 Shown Below is a
Conceptual Scatter
plot that could
Relate the
RESPONSE to
some
EXCITITATION
 Sometimes the
SHAPE of the
scatter reveals a
relationship
Chabot College Mathematics
31
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
Linear Fit by Guessing
 The previous plot
looks sort of Linear
 We could use a
Ruler to draw a
y = mx+b
line thru the data
 But
• which Line is
BETTER?
• and WHY?
Chabot College Mathematics
32
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
Least Squares Curve Fitting
 Numerical Software
such as Scientific
Calculators,
MSExcel, and
MATLAB calc the
“best” m&b
 Almost All “Linear
Regression”
methods use the
“Least Squares”
Criterion
• How are these
Calculations Made?
Chabot College Mathematics
33
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
Least Squares
Best Guess-y
yL  mxk  b
h
y
 xk , y k 
data
x
yk  b
xL 
Best Guess-x
m
Chabot College Mathematics
34
 To make a Good Fit,
MINIMIZE the
|GUESS − data|
distance by one of
 yk  b

x  
 xk 
 m

y 
h
2
mxk  b   yk 
2
x  y
x   y 
2
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
Least Squares cont.
n
 Almost All Regression
2
J   yk 
Methods minimize theSum
k 1
of the Vertical Distances, J:
 §7.4 shows that for Minimum “J”
mbest

x  y   n xy


 x   n x 
2
2

x  y    x  xy


n x    x 
2
bbest
2
• What a Mess!!!
– For more info, please take ENGR/MTH-25
Chabot College Mathematics
35
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
2
DropOut Rates  Scatter Plot
 Given Column Chart
 Read Chart to
Construct T-table
Year x = Yr-1970
1970
0
1980
10
1990
20
1996
26
1997
27
2000
30
2001
31
Chabot College Mathematics
36
y=%
15%
14.1%
12.1%
11.1%
11.0%
10.9%
10.7%
 Use T-table to Make
Scatter Plot on the
next Slide
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
SCATTER PLOT: % of USA High School Students Dropping Out
16%
y (% USA HiSchool DropOuts)
14%
12%
10%
8%
6%
 Zoom-in to more
accurately calc the
Slope
4%
2%
0%
0
2
4
6
M55_§JBerland_Graphs_0806.xls
Chabot College Mathematics
37
8
10
12
14
16
18
20
22
24
26
x (years since 1970)
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
28
30
32
SCATTER PLOT: % of USA High School Students Dropping Out
16%
Intercept  15.2%
y (% USA HiSchool DropOuts)
15%
(x1,y1) = (8yr, 14%)
“Best” Line
(EyeBalled)
14%
13%
Rise  3%
12%
11%
Run  20 yrs
10%
0
4
M55_§JBerland_Graphs_0806.xls
Chabot College Mathematics
38
8
12
16
20
24
x (years since 1970)
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
28
32
DropOut Rates  Scatter Plot
 Calc Slope from
Scatter Plot
Measurements
rise  3%
m

run 20 yrs
 m  0.15 % yr
 Read Intercept from
Measurement
b  y x  0   15.2%
Chabot College Mathematics
39
 Thus the Linear
Model for the Data in
SLOPE-INTER Form
 0.15% 
y  
x  15.2%

 yr 
 To Find Pt-Slp Form
use Known-Pt from
Scatter Plot
• (x1,y1) = (8yr, 14%)
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
DropOut Rates  Scatter Plot
 Thus the Linear
Model for the Data
in PT-SLOPE Form
y  y1  mx  x1  
 X for 2010 →
x = 2010 − 1970 = 40
 In Equation
y2010
 0.15% 
40 yr   15.2%
 

 yr 
 6%  15.2%
 0.15% 
x  8 yr 
y  14%   

y2010
 yr 
y2010  9.2%
 Now use Slp-Inter
Eqn to Extrapolate
to DropOut-% in
2010
Chabot College Mathematics
40
 The model Predicts
a DropOut Rate of
9.2% in 2010
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
SCATTER PLOT: % of USA High School Students Dropping Out
16%
y (% USA HiSchool DropOuts)
15%
14%
13%
12%
11%
10%
9%
(Actually 7.4%)
9.2%
8%
0
5
M55_§JBerland_Graphs_0806.xls
Chabot College Mathematics
41
10
15
20
25
30
x (years since 1970)
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
35
40
Replace EyeBall by Lin Regress
 Use MSExcel commands for LinReg
• WorkSheet → SLOPE & INTERCEPT
Comands
• Plot → Linear TRENDLINE
 By MSExcel
Slope →
Intercept →
R2 →
-0.0015
0.1518
-0.15% ← Slope in %
15.18% ← Intercept in %
0.9816
98.16% ←Goodness in %
M15_Drop_Out_Linear_Regression_1306.xlsx
Chabot College Mathematics
42
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
Official Stats on DropOuts
Status dropout rates of 16- through 24-year-olds in the civilian, noninstitutionalized
population, by race/ethnicity: Selected years, 1990-2010
Race/ethnicity
Native
Year
Total1
White
Black Hispanic
Asian Americans
1990
12.1
9.0
13.2
32.4
4.9!
16.4!
1995
12.0
8.6
12.1
30.0
3.9
13.4!
1998
11.8
7.7
13.8
29.5
4.1
11.8
1999
11.2
7.3
12.6
28.6
4.3
‡
2000
10.9
6.9
13.1
27.8
3.8
14.0
2001
10.7
7.3
10.9
27.0
3.6
13.1
2002
10.5
6.5
11.3
25.7
3.9
16.8
2003
9.9
6.3
10.9
23.5
3.9
15.0
2004
10.3
6.8
11.8
23.8
3.6
17.0
2005
9.4
6.0
10.4
22.4
2.9
14.0
2006
9.3
5.8
10.7
22.1
3.6
14.7
2007
8.7
5.3
8.4
21.4
6.1
19.3
2008
8.0
4.8
9.9
18.3
4.4
14.6
2009
8.1
5.2
9.3
17.6
3.4
13.2
2010
7.4
5.1
8.0
15.1
4.2
12.4
http://nces.ed.gov/fastfacts/display.asp?id=16
SOURCE: U.S. Department of Education, National Center for Education Statistics. (2012). The Condition of
Education 2012 (NCES 2012-045.
! Interpret data with caution. The coefficient of variation (CV) for this estimate is 30 percent or greater.
‡ Reporting standards not met (too few cases).
1 Total includes other race/ethnicity categories not separately shown.
Chabot College Mathematics
43
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
WhiteBoard Work
 Problem §1.3-56
Food Carb/oz (g) Prot/oz (g)
I
3
2
II
5
3
• For the “Foodies”
in the Class
 Mix x ounces of Food-I and y ounces of
Food-II to make a Lump of Food-Mix
that contains exactly:
• 73 grams of Carbohydrates
• 46 grams of Protein
Chabot College Mathematics
44
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
All Done for Today
USA
HiSchl
DropOuts
Chabot College Mathematics
45
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
–
Chabot College Mathematics
46
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
Chabot College Mathematics
47
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
Chabot College Mathematics
48
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
Chabot College Mathematics
49
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
Chabot College Mathematics
50
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
Chabot College Mathematics
51
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx
Chabot College Mathematics
52
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH15_Lec-03_sec_1-3_Lines_LinearFcns_.pptx