Chabot Mathematics
§6.7 Rational
Eqn Apps
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Chabot College Mathematics
1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
Review § 6.6
MTH 55
 Any QUESTIONS About
• §6.6 → Rational Equations
 Any QUESTIONS About HomeWork
• §6.6 → HW-27
Chabot College Mathematics
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
§6.7 Rational Equation Applications
 Problems
Involving Work
 Problems
Involving Motion
 Problems Involving
Proportions
 Problems involving Average Cost
Chabot College Mathematics
3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
Solve a Formula for a Variable
 Formulas occur frequently as
mathematical models. Many
formulas contain rational
expressions, and to solve such
formulas for a specified letter, we
proceed as when solving rational
equations.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
Solve Rational Eqn for a Variable
1. Determine the DESIRED letter (many times
formulas contain multiple variables)
2. Multiply on both sides to clear fractions or
decimals, if that is needed.
3. Multiply if necessary to remove parentheses.
4. Get all terms with the letter to be solved for on one
side of the equation and all other terms on the
other side, using the addition principle.
5. Factor out the unknown.
6. Solve for the letter in question, using the
multiplication principle.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
Example  Solve for Letter
 Solve this formula for y:
aT
R
T  ay
 SOLN:
aT
T  ay   R  T  ay  
T  ay
R T  ay   aT
RT  Ray  aT
aT  RT
y
Ra
Chabot College Mathematics
Multiplying both
sides by the LCD
Simplifying
Multiplying
Ray  aT  RT
6
aT
R
.
T  ay
Subtracting RT
Dividing both
sides by Ra
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
Example  Fluid Mechanics
 In a hydraulic system, a fluid is confined
to two connecting chambers. The
pressure in each chamber is the same
and is given by finding the
force exerted (F) divided by
the surface area (A).
Therefore, F
F2
1
 .
we know
A1
A2
 Solve this Eqn for A2
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
Example  Fluid Mechanics
 SOLUTION:
F1
F2
A1 A2   A1 A2 
A1
A2
Multiplying both sides
by the LCD
A2 F1  A1F2
A1F2
A2 
F1
Dividing both sides by F1
 This formula can be used to calculate
A2 whenever A1, F2, and F1 are known
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
Problems Involving Work
 Rondae and Marrisa work during the
summer painting houses.
• Rondae can paint an average size
house in 12 days
• Marrisa requires 8 days to do the
same painting job.
 How long would it take them,
working together, to paint an
average size house?
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
House Painting cont.
1. Familiarize. We familiarize ourselves
with the problem by exploring two
common, but incorrect, approaches.
a) One common, incorrect, approach is to
add the two times. → 12 + 8 = 20 
b) Another incorrect approach is to
assume that Rondae and Marrisa
each do half the painting.
– Rondae does ½ in 12 days = 6 days
– Marrisa does ½ in 8 days = 4 days
– 6 days + 4 days = 10 days. 
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
House Painting cont.
 A correct approach is to consider how
much of the painting job is finished in
ONE day; i.e., consider the work RATE
 It takes Rondae 12 days to finish painting a
house, so his rate is 1/12 of the job per day.
 It takes Marrisa 8 days to do the painting
alone, so her rate is 1/8 of the job per day.
 Working together, they can complete
1/8 + 1/12, or 5/24 of the job in one day.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
House Painting cont.
 Note That given a TIME-Rate
[Amount] = [Rate]•[TimeQuantity]
 Form a table to help organize the info:
Painter
Rate of
Work
Time
Amount
Completed
Rondae
Marrisa
1/12
1/8
t
t
t/12
t/8
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
House Painting cont.
2. Translate. The time that we want is
some number t for which
Portion of work done by
Rondae in t days
Or
1
1
t  t  1
12
8
 1 1
   t  1 or
 12 8 
Portion of work done by
Marrisa in t days
5
 t  1.
24
Portion of work done together in t days
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
House Painting cont.
3. Carry Out. We can choose any one of
the above equations to solve:
5
t  1
24
24 5
24
  t  1
5 24
5
24
4
t  , or 4 days
5
5
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
House Painting cont.
4. Check. Test t = 24/5 days
1 24 1 24 2 3 5
  
   1
12 5 8 5 5 5 5

5. State. Together, it
will take Rondae &
Marrisa 4 & 4/5
days to complete
painting a house.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
The WORK Principle
 Suppose that A requires a units of time
to complete a task and B requires b
units of time to complete the same task.
 Then A works at a rate of
1/a tasks per unit of time.
 B works at a rate of
1/b tasks per unit of time,
 Then A and B together work at a
rate of [1/a + 1/b] per unit of time.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
The WORK Principle
 If A and B,
working together,
require t units of
time to complete a
task, then their
combined rate is
1/t and the
following
equations hold:
Chabot College Mathematics
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1
1
t  t 1
a
b
1 1
t    1
a b
t t
 1
a b
1 1 1
 
a b t
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
Problems Involving Motion
 Because of a tail wind, a jet is able to fly
20 mph faster than another jet that is
flying into the wind. In the same time
that it takes the first jet to travel 90 miles
the second jet travels 80 miles. How
fast is each jet traveling?
r
 HEAD Wind
Chabot College Mathematics
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r+20
 TAIL Wind
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
HEADwind vs. TAILwind
1. Familiarize. We try a guess. If the
fast jet is traveling 300 mph because
of a tail wind the slow jet plane would
be traveling 300−20 or 280 mph.
•
•

At 300 mph the fast jet would have a 90
mile travel-time of 90/300, or 3/10 hr.
At 280 mph, the other jet would have a
travel-time of 80/280 = 2/7 hr.
Now both planes spend the
same amount of time traveling,
So the guess is INcorrect.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
HEADwind vs. TAILwind
2. Translate. Fill in the blanks using
[TimeQuantity]=[Distance]/[Rate]
Air
Craft
Jet 1
Jet 2
Distance Speed or Rate
Time
(miles) (miles per hour) (hours)
80
r
90
r + 20
r
Chabot College Mathematics
20
r+20
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
HEADwind vs. TAILwind
 Set up a RATE Table
[Distance]/[Rate] = [TimeQuantity]
Air Distance
Speed
Time
Craft (miles) (miles per hour) (hours)
Jet 1
80
r
80/r
Jet 2
90
r + 20
90/(r + 20)
The Times MUST be the SAME
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
HEADwind vs. TAILwind
 Since the times must be
the same for both planes,
we have the equation
80
90
t 
r
r  20
3. Carry Out. To solve the equation, we
first Clear-Fractions multiplying both
sides by the LCD of r(r+20)
80
90
r (r  20)   r (r  20) 
r
r  20
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
HEADwind vs. TAILwind
 Complete the “Carry Out”
80(r  20)  90r
80r  1600  90r
1600  10r
160  r
Simplified by Clearing Fractions
Using the distributive law
Subtracting 80r from both sides
Dividing both sides by 10
 Now we have a possible solution. The
speed of one jet is 160 mph and the
speed of the other jet is 180 mph
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
HEADwind vs. TAILwind
4. Check. ReRead the problem to confirm
that we were able to find the speeds. At
160 mph the jet would cover 80 miles in
½ hour and at 180 mph the other jet
would cover 90 miles in ½ hour. Since
the times are the same, the speeds Chk
5. State. One jet is traveling at
160 mph and the second jet is
traveling at 180 mph
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
Formulas in Economics
 Linear Production Cost Function
C x   variable cost   fixed costs 
 ax  b
• Where
– b is the fixed cost in $
– a is the variable cost of producing each unit in
$/unit (also called the marginal cost)
 Average
Cost ($/unit)
Chabot College Mathematics
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C x 
C x  
x
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
Formulas in Economics
 Price-Demand Function: p x   mx  d
Suppose x units
can be sold (demanded) or
at a price of
x  p   np  k
p dollars per units.
• Where
– m & n are SLOPE Constants in $/unit & unit/$
– d & k are INTERCEPT Constants in $ & units
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
Formulas in Economics
 Revenue Function
Revenue = (Price per unit)·(No. units sold)
Rx  p  x   mx  d x
 Profit Function
Profit = (Total Revenue) – (Total Cost)
P x   R x   C x 
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
Example  Average Cost

Metro Entertainment Co. spent $100,000
in production costs for its off-Broadway
play Pride & Prejudice. Once running,
each performance costs $1000
a) Write the Cost Function for conducting z
performances
b) Write the Average Cost Function for the z
performances
c) How many performances, n, result in an
average cost of $1400 per show
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
Example  Average Cost

SOLUTION a) Total Cost is the sum of
the Fixed Cost and the Variable Cost
$1k
C z   $100k 
z
show
 SOLUTION b) The Average Cost Fcn
$1k
$100k 
z
C z 
show
C z  

z
z
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
Example  Average Cost
SOLUTION c) In this case for “n” Shows
$1k
$100k 
n
$100k  $1k  n
show
C n   $1.4k 

n
n

$1.4k  n  100k  $1k  n
0.4k  n  100k
0.4k  n  100k
n  250
0.4k
 Thus 250 shows are needed to realize
a per-show cost of $1400
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
Problems Involving Proportions
 Recall that a RATIO of two quantities is
their QUOTIENT.
• For example, 45% is the ratio of 45 to 100,
or 45/100.
 A proportion is an equation stating that
two ratios are EQUAL:
An equality of ratios, A/B = C/D, is
called a proportion. The numbers
within a proportion are said to be
proportionAL to each other
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
Example  Triangle Proportions
 Triangles ABC and XYZ are “similar”
Y
B
x=8
a=7
A
b
C
X
Z
y = 12
• Note that “Similar” Triangles are
“In Proportion” to Each other
 Now Solve for b if
x = 8, y = 12 and a = 7
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
Example  Similar Triangles
 Set Up The
Proportions
B
a=7
A
b 7

12 8
C
b
Y
x=8
7
b  12
8
84
b
or 10.5
8
Chabot College Mathematics
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X
Z
y = 12
[b is to 12]
as
[7 is to 8]
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
Example  Similar Triangles
 Alternative
Proportions
B
a=7
A
b 12

7 8
12
b7
8
84
b
or 10.5
8
Chabot College Mathematics
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C
b
Y
x=8
X
Z
y = 12
[b is to 7]
as
[12 is to 8]
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
Example  Quantity Proportions
 A sample of 186 hard drives contained 4
defective drives. How many defective drives
would be expected in a group of 1302 HDDs?
 Form a proportion in which the ratio of
defective hard drives is expressed in 2 ways.
defective drives
total drives
4
x

186 1302
186x  5208
x  28
Chabot College Mathematics
35
defective drives
total drives
 Expect to find 28
defective HDDs
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
Whale Proportionality
 To determine the number of humpback
whales in a pod, a marine biologist,
using tail markings, identifies 35
members of the pod.
 Several weeks later, 50 whales from the
SAME pod are randomly sighted. Of
the 50 sighted, 18 are from the 35
originally identified. Estimate the
number of whales in the pod.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
Tagged Whale Proportions
1. Familarize. We need to reread the
problem to look for numbers that could
be used to approximate a percentage
of the of the pod sighted.
 Since 18 of the 35 whales that were
later sighted were among those
originally identified, the ratio 18/50
estimates the percentage of the
pod originally identified.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
HumpBack Whales
2. Translate: Stating the Proportion
Whales originally identified
Entire pod
3. Carry
Out
35 18

w 50
Original whales sighted later
Total Whales sighted later
35 18
50w    50w
w 50
50  35  18  w
50  35
 w or 97.22
18
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
More On Whales
4. Check. The check is left to the student.
5. State. There are about 97 whales in the
Pod
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
One More Whale
 Another way to summarize [35 is to w]
the RANDOM-Tagging and
as
RANDOM-Sighting Relation: [18 is to 50]
 Thus the
Proportionality:
 Solve for w
w 50

35 18
35  50 1750
 w 50 
35    w 

 97.2
18
18
 35 18 
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
Example  Vespa Scooters
 Juan’s new scooter goes 4 mph
faster than Josh does on his
scooter. In the time it takes Juan to
travel 54 miles, Josh travels 48
miles.
 Find the speed of each scooter.
Chabot College Mathematics
41
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
Example  Vespa Scooters
 Familiarize. Let’s guess that Juan is
going 20 mph. Josh would then be
traveling 20 – 4, or 16 mph.
 At 16 mph, he would travel 48 miles in 3
hr. Going 20 mph, Juan would cover 54
mi in 54/20 = 2.7 hr. Since 3  2.7, our
guess was wrong, but we can see that if
r = the rate, in miles per hour, of Juan’s
scooter, then the rate of Josh’s scooter
= r – 4.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
Example  Vespa Scooters
 LET:
• r ≡ Speed of Juan’s Scooter
• t ≡ The Travel Time for Both Scooters
 Tabulate the data for clarity
Distance
Speed
Time
Juan’s
Scooter
54
r
t
Josh’s
Scooter
48
r4
t
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
Example  Vespa Scooters
 Translate. By looking at how we
checked our guess, we see that in the
Time column of the table, the t’s can be
replaced, using the formula
Time = Distance/Speed
Distance
Speed
Time
Juan’s
Scooter
54
r
54/ r
Josh’s
Scooter
48
r4
48 /(r  4)
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
Example  Vespa Scooters
 Since the Times are the
54
48

.
SAME, then equate the two
r
r

4
Time entries in the table as:
 Carry
Out
54
48

r
r4
54
48
r ( r  4)
 r (r  4)
r
r4
54r  216  48r
Chabot College Mathematics
45
216  6r
36  r.
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
Example  Vespa Scooters
 Check: If our answer checks, Juan’s
scooter is going 36 mph and Josh’s
scooter is going 36 − 4 = 32 mph.
Traveling 54 miles at 36 mph, Juan is
riding for 54/36 or 1.5 hours. Traveling
48 miles at 32 mph, Josh is riding for
48/32 or 1.5 hours. The answer checks
since the two times are the same.
 State: Juan’s speed is 36 mph,
and Josh’s speed is 32 mph
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
WhiteBoard Work
 Problems From §6.7 Exercise Set
• 16 (ppt), 34, 44

Mass Flow
Rate for a
Diverging
Nozzle
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
P6.7-16
 Given Avg Cost
Function Graph:
 Find Production
Quatity for
Avg Cost
of $425/Chair
 SOLUTION: Cast
Right & Down
20k
 ANS → 20k Chairs/mon
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
All Done for Today
Human
Proportions:
HeadLength
BaseLine
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
Example  Similar Triangles
 SOLUTION  Examine the drawing,
write a proportion, and then solve.
Y
B
x=8
a=7
A
b
C
X
Z
y = 12
 Note that side a is always opposite
angle A, side x is always opposite angle
X, and so on.
Chabot College Mathematics
50
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.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
51
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.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
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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
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.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
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file =XY_Plot_0211.xls
-10
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-35_sec_6-7_Rational_Equation_Applications.ppt
5
6