Chabot Mathematics
§6.2 Rational Fcn
Add & Subtract
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Chabot College Mathematics
1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-29_Fa08_sec_6-1_Rational_Fcn_Mult-n-Div.ppt
Review § 6.1
MTH 55
 Any QUESTIONS About
• §6.1 → Rational Function
Simplification
 Any QUESTIONS About HomeWork
• §6.1 → HW-18
Chabot College Mathematics
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-29_Fa08_sec_6-1_Rational_Fcn_Mult-n-Div.ppt
Addition
 The Sum of Two Rational
Expressions
 To add when the denominators are the
same, add the numerators and keep the
common denominator:
K N KN


M M
M
Chabot College Mathematics
3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-29_Fa08_sec_6-1_Rational_Fcn_Mult-n-Div.ppt
Example  Rational Addition
Add. Simplify the result, if possible.
4 x 3x  2
5 6w

b)
a) 
w
c)
x7
w
3x 2  4 x  9 x 2  x  3

3x  1
3x  1
d)
x7
x 9
3
 2
2
x  36 x  36
 SOLUTION a)
5 6  w 11  w


w
w
w
Chabot College Mathematics
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The denominators are alike,
so we add the numerators
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-29_Fa08_sec_6-1_Rational_Fcn_Mult-n-Div.ppt
Example  Rational Addition
SOLUTION b)
4 x 3x  2 7 x  2


x7 x7
x7
The denominators are alike,
so we add the numerators
 SOLUTION c)
3x 2  4 x  9 x 2  x  3 (3x 2  4 x  9)  ( x 2  x  3)


3x  1
3x  1
3x  1
4 x 2  5 x  12

3x  1
Chabot College Mathematics
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Combining
like terms
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-29_Fa08_sec_6-1_Rational_Fcn_Mult-n-Div.ppt
Example  Rational Addition
SOLUTION d)
x 9
3
x6
Combining like terms
 2
 2
2
x  36 x  36 x  36
in the numerator
x6
Factoring

( x  6)( x  6)

1  ( x  6)
( x  6) ( x  6)
ReMove a
Multiplying “1”
1

x6
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-29_Fa08_sec_6-1_Rational_Fcn_Mult-n-Div.ppt
Subtraction
 The Difference of Two
Rational Expressions
 To subtract when the denominators are
the same, subtract the second
numerator from the first and keep the
common denominator:
K N KN


M M
M
Chabot College Mathematics
7
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-29_Fa08_sec_6-1_Rational_Fcn_Mult-n-Div.ppt
Example  Rational Subtraction
 Subtract. Simplify the result, if possible.
5x
x4
x2
x  30
b)


a)
x3
x3
x6
x6
 SOLUTION a)
5x
x  4 5 x  ( x  4)


x3 x3
x3
Chabot College Mathematics
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The parentheses are needed to make
sure that we subtract both terms.
5x  x  4

x3
Removing the parentheses and
changing the signs
(using the distributive law)
4x  4

x3
Combining like terms
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-29_Fa08_sec_6-1_Rational_Fcn_Mult-n-Div.ppt
Example  Rational Subtraction
b)
x2
x  30 x 2  ( x  30)


x6 x6
x6
x 2  x  30

x6
Removing the parentheses
(using the distributive law)
( x  6)( x  5)

x6

( x  6) ( x  5)
x6
Factoring, in hopes of simplifying
Removing a factor equal to 1
 x5
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-29_Fa08_sec_6-1_Rational_Fcn_Mult-n-Div.ppt
Least Common:
Multiples & Denominators
 To add or subtract rational expressions
that have different denominators, we
must first find EQUIVALENT rational
expressions that have a common
denominator.
 The least common denom must include
the factors of each number, so it must
include each prime factor the greatest
number of times that it appears in any of
the factorizations of any denom.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-29_Fa08_sec_6-1_Rational_Fcn_Mult-n-Div.ppt
Find the Least Common Denom
1.
2.
Write the prime factorization of each
denominator.
Select one of the factorizations and inspect
it to see if it contains the other.
a)
b)

If it does, it represents the
LCM of the denominators.
If it does not, multiply that factorization by any
factors of the other denominator that it lacks.
The final product is the LCM of the
denominators.
The LCD is the LCM of the denominators.
It should contain each factor the greatest
number of times that it occurs in any of the
individual factorizations.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-29_Fa08_sec_6-1_Rational_Fcn_Mult-n-Div.ppt
Example  LCD
7
3
and
.
 Find the LCD of:
2
3
6x
4x
SOLUTION
1. Begin by writing the prime factorizations:
Note that each factor appears
6x2 = 2  3  x  x
the greatest number of times
4x3 = 2  2  x  x  x that it occurs in either of these
factorizations.
2. LCM = 2  2  3  x  x  x
 The LCM of the denominators is thus
22  3  x3, or 12x3, so the LCD is 12x3.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-29_Fa08_sec_6-1_Rational_Fcn_Mult-n-Div.ppt
Example  LCD cont.
7
3
7
3
+ 3

 Now Can Add
2
6x 4x
2 3 x  x 2  2  x  x  x
 To obtain equivalent expressions with the LCD,
we must multiply each expression by 1, using
the missing factors of the LCD to write the 1.
7
3
7
2x
3
3
+ 3
 

2
6x 4x
2  3 x  x 2x 2  2  x  x  x 3
Chabot College Mathematics
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14 x
9

+
3
3
12 x 12 x
14 x  9

12 x 3
The LCD requires
another factor of 3.
The LCD requires
additional factors of
2 and x
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-29_Fa08_sec_6-1_Rational_Fcn_Mult-n-Div.ppt
Example  Least Common Mult

For each pair of polynomials, find the
Least Common Multiple (LCM).
a) 16a and 24b
b) 24x4y4 and 6x6y2
c) x2 – 4 and x2 – 2x – 8
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-29_Fa08_sec_6-1_Rational_Fcn_Mult-n-Div.ppt
Example  LCM
 SOLUTION a)
16a is
16a = 2  2  2  2  a
a factor
of the LCM
24b = 2  2  2  3  b
The LCM = 2  2  2  2  a  3  b
24b is a factor of the LCM
The LCM is 24  3  a  b, or 48ab
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-29_Fa08_sec_6-1_Rational_Fcn_Mult-n-Div.ppt
Example  LCM
 SOLUTION b) LCM for 24x4y4 and 6x6y2
24x4y4 = 2  2  2  3  x  x  x  x  y  y  y  y
6x6y2 = 2  3  x  x  x  x  x  x  y  y
LCM =
2223xxxxyyyyxx
 Note that we used the highest power of
each factor. The LCM is 24x6y4
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-29_Fa08_sec_6-1_Rational_Fcn_Mult-n-Div.ppt
Example  LCM

SOLUTION c) LCM for
x2 – 4 and x2 – 2x – 8
x2 – 4 is a factor of the LCM
x2 – 4 = (x – 2)(x + 2)
x2 – 2x – 8 = (x + 2)(x – 4)
x2 – 2x – 8 is a factor of the LCM
LCM = (x – 2)(x + 2)(x – 4)
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-29_Fa08_sec_6-1_Rational_Fcn_Mult-n-Div.ppt
Example
 Find equivalent expressions
x5
that have the LCD for x  3
; 2
2
This Expression Pair x  4 x  2 x  8
 SOLUTION
From the previous example the LCD:
(x  2)(x + 2)(x  4)
x3
x3
x4


2
x  4 ( x  2)( x  2) x  4
Chabot College Mathematics
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( x  3)( x  4)

( x  2)( x  2)( x  4)
Multiply by the missing
expression
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-29_Fa08_sec_6-1_Rational_Fcn_Mult-n-Div.ppt
Example  Equiv. Expressions
x5
x5
x2


2
x  2 x  8 ( x  2)( x  4) x  2
Multiply by the
missing expression
( x  5)( x  2)

( x  2)( x  4)( x  2)
 We leave the results in factored form.
 In a later slides we will carry out the
actual addition and subtraction of such
rational expressions.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-29_Fa08_sec_6-1_Rational_Fcn_Mult-n-Div.ppt
To Add or Subtract Rational
Expressions Having Different
Denominators
1. Find the LCD.
2. Multiply each rational expression by a
Special form of 1 made up of the
factors of the LCD missing from that
expression’s denominator.
3. Add or subtract the numerators, as
indicated. Write the sum or difference
over the LCD.
4. Simplify, if possible
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-29_Fa08_sec_6-1_Rational_Fcn_Mult-n-Div.ppt
Example  Add
4 x2 5x

9 12
 SOLUTION
1. First, find the LCD:
9=33
LCD = 2  2  3  3 = 36
12 = 2  2  3
2. Multiply each expression by the
appropriate “form of 1” to get the LCD.
4 x2 5x 4 x2
5x



9 12 3  3 2  2  3
4 x2 4
5 x 3 16 x 2 15 x

 
 

33 4 2  2 3 3
36
36
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-29_Fa08_sec_6-1_Rational_Fcn_Mult-n-Div.ppt
Example  Add
4 x2 5x

9 12
3. Next we add the numerators:
2
2
16 x 15 x 16 x  15 x



36
36
36
4. Since 16x2 + 15x and 36 have no
common factor, [16x2 + 15x]/36
canNOT be simplified any further

Subtraction is performed in a very
similar Fashion
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-29_Fa08_sec_6-1_Rational_Fcn_Mult-n-Div.ppt
7
5

Example  Subtract
9 x 12 x 2
 SOLUTION: We follow the four steps as
shown in the previous example. First,
we find the LCD
9x = 3  3  x
LCD = 2  2  3  3  x  x = 36x2
12x2 = 2  2  3  x  x
 The denominator 9x must be multiplied
by 4x to obtain the LCD.
 The denominator 12x2 must be
multiplied by 3 to obtain the LCD.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-29_Fa08_sec_6-1_Rational_Fcn_Mult-n-Div.ppt
7
5

Example  Subtract
9 x 12 x 2
 Next, Multiply to obtain the LCD and
then subtract and, if possible, simplify
7
5
7 4x
5 3


 

2
2
9 x 12 x
9 x 4 x 12 x 3
28 x
15


2
36 x 36 x 2
28 x  15

36 x 2
Chabot College Mathematics
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Caution! Do not simplify
these rational expressions
or you will lose the LCD.
This cannot be simplified,
so we are done.
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-29_Fa08_sec_6-1_Rational_Fcn_Mult-n-Div.ppt
Example  Add
3a
2
 2
2
a  4 a  2a
 SOLUTION: First, we find the LCD:
a2 – 4 = (a – 2)(a + 2)
LCD = a(a – 2)(a + 2)
a2 – 2a = a(a – 2)
 Multiply by a form of 1 to obtain the
LCD in each expression:
3a
2
3a
a
2
a2
 2

 

2
a  4 a  2a (a  2)(a  2) a a(a  2) a  2
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-29_Fa08_sec_6-1_Rational_Fcn_Mult-n-Div.ppt
Example  Add
3a
2
 2
2
a  4 a  2a
 Continue the Reduction
3a
2
3a
a
2
a2
 2

 

2
a  4 a  2a (a  2)(a  2) a a(a  2) a  2
3a 2
2a  4


a(a  2)(a  2) a(a  2)(a  2)
3a 2  2a  4

a(a  2)(a  2)
 3a2 + 2a + 4 does not factor
so we are done
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-29_Fa08_sec_6-1_Rational_Fcn_Mult-n-Div.ppt
Example  Subtract
x  2 x 1

x4 x6
 SOLUTION: First, we find the LCD. It is
just the product of the denominators:
LCD = (x + 4)(x + 6).
 We multiply by a form of 1 to get the
LCD in each expression. Then we
subtract and try to simplify
x  2 x 1 x  2 x  6 x 1 x  4





x4 x6 x4 x6 x6 x4
x 2  8x  12
x 2  3x  4


( x  4)( x  6) ( x  4)( x  6)
Chabot College Mathematics
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Multiplying out
numerators.
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-29_Fa08_sec_6-1_Rational_Fcn_Mult-n-Div.ppt
Example  Continue Reduction
x 2  8x  12
x 2  3x  4


( x  4)( x  6) ( x  4)( x  6)
x 2  8x  12  ( x 2  3x  4)

( x  4)( x  6)
x 2  8 x  12  x 2  3x  4

( x  4)( x  6)
5 x  16

( x  4)( x  6)
Chabot College Mathematics
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When subtracting a numerator
with more than one term,
parentheses are important.
Removing parentheses and
subtracting every term.
5x + 16 does not factor so we
are finished
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-29_Fa08_sec_6-1_Rational_Fcn_Mult-n-Div.ppt
Example  Add
x 1
4
 2
2
x  4 x  4 x  3x  10
 SOLUTION: 1st Factor Denoms
x 1
4
x 1
4
 2


2
x  4 x  4 x  3x  10 x  2x  2 x  2x  5
 Next Put AddEnds over the LCD
x 1
x5
4
x2




( x  2)( x  2) x  5 ( x  2)( x  5) x  2
 Now can Start the Reduction
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-29_Fa08_sec_6-1_Rational_Fcn_Mult-n-Div.ppt
Example  Add
x 1
4
 2
2
x  4 x  4 x  3x  10
 The Reduction
x 1
x5
4
x2
x 1
4



 2

2
x  4 x  4 x  3 x  10 ( x  2)( x  2) x  5 ( x  2)( x  5) x  2
x2  6 x  5
4x  8


( x  2)( x  2)( x  5) ( x  2)( x  2)( x  5)
x2  6x  5  4 x  8

( x  2)( x  2)( x  5)
x 2  10 x  3

( x  2)( x  2)( x  5)
Chabot College Mathematics
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Adding numerators
x2 + 10x – 3 does not factor
so we are finished
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-29_Fa08_sec_6-1_Rational_Fcn_Mult-n-Div.ppt
When Factors are Opposites
 When one denominator is the
opposite of the other, we can
first multiply either expression
by 1 using –1/ –1.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-29_Fa08_sec_6-1_Rational_Fcn_Mult-n-Div.ppt
Example  Add
y 4

3 3
 SOLUTION by Reduction
y 4
y 4 1

  
3 3 3 3 1
y 4
 
3 3
y  ( 4)

3
y4

3
Chabot College Mathematics
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Multiplying by 1 using −1/−1
The denominators are
now the same.
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-29_Fa08_sec_6-1_Rational_Fcn_Mult-n-Div.ppt
Example  Add
5x
2

x 3 3 x
 SOLUTION by Reduction
5x
2
5x
2 1




x  3 3  x x  3 3  x 1
5x
2


x  3 3  x
5x
2


x3 x3
5x  2

x 3
Chabot College Mathematics
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−3 + x = x + (−3) = x − 3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-29_Fa08_sec_6-1_Rational_Fcn_Mult-n-Div.ppt
Example  Add
x
5

2
x  36 6  x
 SOLUTION by Reduction
x
5
x
5



2
x  36 6  x ( x  6)( x  6) 6  x
x
5 1



( x  6)( x  6) 6  x 1
x
5


( x  6)( x  6) x  6
x
5 x  6



( x  6)( x  6) x  6 x  6
Chabot College Mathematics
34
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-29_Fa08_sec_6-1_Rational_Fcn_Mult-n-Div.ppt
Example  Add
x
5

2
x  36 6  x
 Complete the Reduction
x
5 x  6



( x  6)( x  6) x  6 x  6
x
5 x  30


( x  6)( x  6) ( x  6)( x  6)
x  5 x  30

( x  6)( x  6)
4 x  30

( x  6)( x  6)
Chabot College Mathematics
35
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-29_Fa08_sec_6-1_Rational_Fcn_Mult-n-Div.ppt
WhiteBoard Work
 Problems From §6.2 Exercise Set
• 82 (ppt), 28, 64, 84

Add Rational Expressions
Chabot College Mathematics
36
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-29_Fa08_sec_6-1_Rational_Fcn_Mult-n-Div.ppt
P6.2-82  Risking a Ticket
 If Drive-Time is to
8hr, how much over
the 70mph & 65mph
Spd Limits is
required
 ID 8hrs on Graph
and find the
OverSpeed needed
22
to make this time
 ANS → need to go about 22 mph over the
Speed Limit for the entire 8 hrs
Chabot College Mathematics
37
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-29_Fa08_sec_6-1_Rational_Fcn_Mult-n-Div.ppt
P6.2-82  Risking a Ticket
 Is 8hrs too fast for this trip?
 Find the average speed = Dist/Time
• Total Distance = 470mi + 250mi = 720mi
• Avg Speed = [720mi]/[8hrs] = 90 miles/hr
 WOW! Running at 90 mph for 8 straight
hours is NOT a realistic travel Plan
• Better to try 10 hrs for an
avg speed of 72 mph
Chabot College Mathematics
38
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-29_Fa08_sec_6-1_Rational_Fcn_Mult-n-Div.ppt
All Done for Today
A Different
Kind of
LCM
Chabot College Mathematics
39
 Landing Craft, Mechanized
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-29_Fa08_sec_6-1_Rational_Fcn_Mult-n-Div.ppt
Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
–
Chabot College Mathematics
40
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-29_Fa08_sec_6-1_Rational_Fcn_Mult-n-Div.ppt
Graph y = |x|
6
 Make T-table
x
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
Chabot College Mathematics
41
5
y = |x |
6
5
4
3
2
1
0
1
2
3
4
5
6
y
4
3
2
1
x
0
-6
-5
-4
-3
-2
-1
0
1
2
3
-1
-2
-3
-4
-5
file =XY_Plot_0211.xls
-6
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-29_Fa08_sec_6-1_Rational_Fcn_Mult-n-Div.ppt
4
5
6
y
5
2
y
1
x
4
0
-6
-5
-4
-3
-2
-1
3
0
1
2
3
4
-1
-2
2
-3
1
-4
x
-5
5
-6
0
-3
-2
-1
0
1
2
3
-1
4
-7
-8
-2
-9
M55_§JBerland_Graphs_0806.xls
-3
Chabot College Mathematics
42
file =XY_Plot_0211.xls
-10
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-29_Fa08_sec_6-1_Rational_Fcn_Mult-n-Div.ppt
5
6