BMayer@ChabotCollege.edu MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical.
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BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt1
Bruce Mayer, PE Chabot College Mathematics
Bruce Mayer, PELicensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Chabot Mathematics
§8.2 Quadratic§8.2 QuadraticEquation AppsEquation Apps
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt2
Bruce Mayer, PE Chabot College Mathematics
Review §Review §
Any QUESTIONS About• §8.2 → Complete-the-Square
Any QUESTIONS About HomeWork• §8.2 → HW-37
8.2 MTH 55
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt3
Bruce Mayer, PE Chabot College Mathematics
§8.2 Quadratic Formula§8.2 Quadratic Formula
The Quadratic Formula
Problem Solving with the Quadratic Formula
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt4
Bruce Mayer, PE Chabot College Mathematics
The Quadratic FormulaThe Quadratic Formula
The solutions of ax2 + bx + c = 0 are given by
a
acbbx
2
42
This is one of theMOST FAMOUSFormulas in allof Mathematics
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt5
Bruce Mayer, PE Chabot College Mathematics
Example Example Circular WalkWayCircular WalkWay
A Circular Pond 10 feet in diameter is to be surrounded by a “Paver” walkway that will be 2 feet wide.
Find the AREA of the WalkWay Familiarize: Recall the
Formula for the area, A ,of a Circle based on it’s radius, r
2rA
Also the diameter, d, is half of r. Thus A in terms of d 42
22dd
A
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt6
Bruce Mayer, PE Chabot College Mathematics
Example Example Circular WalkWayCircular WalkWay
Familiarize: Makea DIAGRAM
Translate: Use Diagram of Subtractive Geometry
4
2dAcirle
2 10
14 10
= −
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt7
Bruce Mayer, PE Chabot College Mathematics
Example Example Circular WalkWayCircular WalkWay
Translate: Diagram to Equation
4
10
4
14 22 walkA
14 10
= −
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt8
Bruce Mayer, PE Chabot College Mathematics
Example Example Circular WalkWayCircular WalkWay
CarryOUT: Solve Eqn for Awalk
24964
1001964
101444
10
4
14 2222
walk
walk
A
A
Using π ≈ 3.14 find
367514324 .. walkA
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt9
Bruce Mayer, PE Chabot College Mathematics
Example Example Circular WalkWayCircular WalkWay
Check: Use Acircle = πr2
367557886153
251434914357 22
...
..
chk
chk
A
A
State: The Area of the Paver Walkway is about 75.4 ft2
• Note that UNITS must be included in the Answer Statement
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt10
Bruce Mayer, PE Chabot College Mathematics
Example Example Partition Bldg Partition Bldg
A rectangular building whose depth (from the front of the building) is three times its frontage is divided into two parts by a partition that is 45 feet from, and parallel to, the front wall. If the rear portion of the building contains 2100 square feet, find the dimensions of the building.
Familiarize: REALLY needs a Diagram
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt11
Bruce Mayer, PE Chabot College Mathematics
Example Example Partition Bldg Partition Bldg
Familiarize: by Diagram
Now LET x ≡ frontage of building, in feet.
Translate: The other statements into Equations involving x
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt12
Bruce Mayer, PE Chabot College Mathematics
Example Example Partition Bldg Partition Bldg
The Bldg depth is three times its frontage, x→ 3x = depth of building, in feet
The Bldg Depth is divided into two parts by a partition that is 45 feet from, and parallel to, the front wall → 3x – 45 = depth of rear portion, in ft
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt13
Bruce Mayer, PE Chabot College Mathematics
Example Example Partition Bldg Partition Bldg
Now use the 2100 ft2 Area Constraint• Area of rear = 2100
• Area = x(3x−45), so
x 3x 45 2100
3x2 45x 2100 0
x2 15x 700 0
x 35 x 20 x 35 x 20
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt14
Bruce Mayer, PE Chabot College Mathematics
Example Example Partition Bldg Partition Bldg
So x is either 35ft or −20ft But again Distances can
NOT be negative Thus x = 35 ft Check: Use 2100 ft2 Area
21004510535
21004535335
2100453
?
?
xx
21006035
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt15
Bruce Mayer, PE Chabot College Mathematics
Example Example Partition Bldg Partition Bldg
State: The Bldg
Frontage is 35ft The Bldg Depth
is 3(35ft) = 105ft105’
35’
60’
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt16
Bruce Mayer, PE Chabot College Mathematics
Example Example Bike Tire BlowOut Bike Tire BlowOut
Devon set out on a 16 mile Bike Ride. Unfortunately after 10 miles of Biking BOTH tires Blew Out. Devon Had to complete the trip on Foot.
Devon biked four miles per hour (4 mph) faster than she walked
The Entire journey took 2hrs and 40min
Find Devon’s Walking Speed (or Rate)
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt17
Bruce Mayer, PE Chabot College Mathematics
Example Example Bike Tire BlowOut Bike Tire BlowOut
Familiarize: Make diagram
LET w ≡ Devon’s Walking Speed
Recall the RATE Equation for Speed
Distance = (Speed)·(Time)
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt18
Bruce Mayer, PE Chabot College Mathematics
Example Example Bike Tire BlowOut Bike Tire BlowOut
Translate: The BikingSpeed, b, is 4 mphfaster than theWalking Speed →
From the Diagram note Distances by Rate Equation:• Biking Distance = 10 miles = b·tbike
• Walking Distance = 6 miles = w·twalk
4wb
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt19
Bruce Mayer, PE Chabot College Mathematics
Example Example Bike Tire BlowOut Bike Tire BlowOut
Translate: Now the Total Distance of16mi is the sum of theBiking & Walking Distances →
walkbike
walkbike
wttw
wbandwtbt
416
416
From the Spd Eqn: Time = Dist/Spd
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt20
Bruce Mayer, PE Chabot College Mathematics
Example Example Bike Tire BlowOut Bike Tire BlowOut
Translate: Thus bySpeed Eqn:
Next, the sum of the Biking and Walking times is 2hrs & 40min = 2-2/3hr →
wwtt walkbike
6
4
10
3
22
wt
wbt
walk
bike
mi64
mi10mi10
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt21
Bruce Mayer, PE Chabot College Mathematics
Example Example Bike Tire BlowOut Bike Tire BlowOut
CarryOut: Clear Fractions in the last Eqn by multiplying by the LCD of 3·(w+4) ·w:
GCFby 928072168
721830328
43631048
436
4
10
3
8
6
4
10
3
86
4
10
3
22
22
2
wwww
wwww
wwww
wwww
wwww
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt22
Bruce Mayer, PE Chabot College Mathematics
Example Example Bike Tire BlowOut Bike Tire BlowOut
CarryOut: Divide the last Eqn by 8 to yield a Quadratic Equation
920
8
9280 22
wwww
This Quadratic eqn does NOT factor so use Quadratic Formula with a = 1, b = −2, and c = −9
12
91422
2
422
a
acbbw
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt23
Bruce Mayer, PE Chabot College Mathematics
Example Example Bike Tire BlowOut Bike Tire BlowOut
CarryOut: find w by Quadratic Formula
2
1042
2
1042
2
402
2
3642
w
101
2
1012
2
1022
w
So Devon’s Walking Speed is
162101or164101
1012
1012
2
1022
..
ww
w
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt24
Bruce Mayer, PE Chabot College Mathematics
Example Example Bike Tire BlowOut Bike Tire BlowOut
CarryOut: Since SPEED can NOT be Negative find:
Check: Test to see that the time adds up to 2.67 hrs
mph 164101 .w
668244312261164
6
4164
10672
6
4
10
3
22
.....
.?
ww
tt walkbike
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt25
Bruce Mayer, PE Chabot College Mathematics
Example Example Bike Tire BlowOut Bike Tire BlowOut
State: Devon’s Walking Speed was about 4.16 mph (quite a brisk pace)
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt26
Bruce Mayer, PE Chabot College Mathematics
Example Example Golden Rectangle Golden Rectangle
Let’s Revisit the Derivation of the GOLDEN RATIO
A rectangle of length p and width q, with p > q, is called a golden rectangle if you can divide the rectangle into a square with side of length q and a smaller rectangle that is geometrically-similar to the original one.
The GOLDEN RATION then = p/q
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt27
Bruce Mayer, PE Chabot College Mathematics
Example Example Golden Rectangle Golden Rectangle
Familiarize: Make a Diagram
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt28
Bruce Mayer, PE Chabot College Mathematics
Example Example Golden Rectangle Golden Rectangle Translate: Use Diagram
Longer side of A
Shorter side of A
Longer side of B
Shorter side of Bp
q
q
p qp p q qq
p2 pq q2
p
q
2
p
q1
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt29
Bruce Mayer, PE Chabot College Mathematics
Example Example Golden Rectangle Golden Rectangle
CarryOut: LETΦ ≡ GoldenRatio= p/q
p
q
2
p
q1
2 1 0
1 1 2 4 1 1
2 1
1 5
26181or6180 ..
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt30
Bruce Mayer, PE Chabot College Mathematics
Example Example Golden Rectangle Golden Rectangle
Carry Out: Since both p & q are distances they are then both POSITIVE
Thus Φ = p/q must be POSITIVE State: GOLDEN RATIO as defined by
the Golden Rectangle
1 5
21.618
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt31
Bruce Mayer, PE Chabot College Mathematics
Example Example Pythagorus Pythagorus
The hypotenuse of a right triangle is 52 yards long. One leg is 28 yards longer than the other. Find the lengths of the legs
1. Familarize. First make a drawing and label it. Let s = length, in yards, of one leg. Then s + 28 = the length, in yards, of the other leg. s + 28
s 52
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt32
Bruce Mayer, PE Chabot College Mathematics
Pythagorean TrianglePythagorean Triangle
2. Translate. We use the Pythagorean theorem:s2 + (s + 28)2 = 522
3. Carry Out. Identify the Quadratic Formula values a, b, & c
s + 28
s 52
s2 + (s + 28)2 = 522
s2 + s2 + 56s + 784 = 2704
2s2 + 56s − 1920 = 0
s2 + 28s − 960 = 0
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt33
Bruce Mayer, PE Chabot College Mathematics
Pythagorean TrianglePythagorean Triangle
3. Carry Out: With s2 + 28s − 960 = 0
Find: a = 1, b = 28, c = −960
Evaluate the Quadratic Formula2( ) 4( )(28 )
2
28 1 960
1( )s
28 784 3840
2s
28 4624
2s
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt34
Bruce Mayer, PE Chabot College Mathematics
Pythagorean TrianglePythagorean Triangle
3. Carry Out: Continue Quadratic Eval
28 4624
2s
28 68
2s
28 68 28 68 or
2 2s s
40 9620 or 48
2 2s s
s + 28
s 52
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt35
Bruce Mayer, PE Chabot College Mathematics
Pythagorean TrianglePythagorean Triangle
4. Check. Length cannot be negative, so −48 does not check.
5. State. One leg is 20 yards and the other leg is 48 yards.
s + 28yd = 48 yds
s = 20 yds
52 yds
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt36
Bruce Mayer, PE Chabot College Mathematics
Vertical BallisticsVertical Ballistics
In Physics 4A, you will learn the general formula for describing the height of an object after it has been thrown upwards:
20 0
1,
2h gt v t h
• Where– g ≡ the acceleration due to gravity
A CONSTANT = 32.2 ft/s2 = 9.81 m/s2
– t ≡ the time in flight, in s
– v0 ≡ the initial velocity in ft/s or m/s
– h0 ≡ the initial height in ft or m
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt37
Bruce Mayer, PE Chabot College Mathematics
Example Example X-Games Jump X-Games Jump
In an extreme games competition, a motorcyclist jumps with an initial velocity of 80 feet per second from a ramp height of 25 feet, landing on a ramp with a height of 15 feet.
Find the time the motorcyclist is in the air.
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt38
Bruce Mayer, PE Chabot College Mathematics
Example Example X-Games Jump X-Games Jump
Familiarize: We are given the initial velocity, initial height, and final height and we are to find the time the bike is in the air. Can use the Ballistics Formula
Translate: Use the formula with h = 15, v0 = 80, and h0 = 25.
Use h = 15, v0 = 80, and h0 = 25
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt39
Bruce Mayer, PE Chabot College Mathematics
Example Example X-Games Jump X-Games Jump
CarryOut:
20 0
1
2h gt v t h
2115 ( 32.2) 80 25
2t t
215 16.1 80 25t t 20 16.1 80 10t t
Use the Quadratic Formula to find t• a = −16.1, b = 80 and c = 10
1162
1011648080
2
422
.
.
a
acbbt
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt40
Bruce Mayer, PE Chabot College Mathematics
Example Example X-Games Jump X-Games Jump
CarryOut:
Since Times can NOT be Negative t ≈ 5.09 seconds
Check: Check by substuting 5.09 for t in the ballistics Eqn. • The Details are left for later
State: The MotorCycle Flight-Time is very nearly 5.09 seconds
1220or095232
704480..
.
ttt
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt41
Bruce Mayer, PE Chabot College Mathematics
Example Example Biking Speed Biking Speed
Kang Woo (KW to his friends) traveled by Bicycle 48 miles at a certain speed. If he had gone 4 mph faster, the trip would have taken 1 hr less.
Find KW’s average speed
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt42
Bruce Mayer, PE Chabot College Mathematics
Example Example Biking Speed Biking Speed
Familiarize: In this can tabulating the information can help. Let r represent the rate, in miles per hour, and t the time, in hours for Kang Woo’s trip
Distance Speed Time
48 r t
48 r + 4 t – 1
48/r t48
41
rt
• Uses the Rate/Spd Eqn → Rate = Qty/Time
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt43
Bruce Mayer, PE Chabot College Mathematics
Example Example Biking Speed Biking Speed
Translate: From the Table Obtain two Equations in r & t
48r
t and
484 .
1r
t
Carry out: A system of equations has
been formed. We substitute for r from the first equation into the second and solve the resulting equation
48 484
1t t
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt44
Bruce Mayer, PE Chabot College Mathematics
Example Example Driving Speed Driving Speed
Carry out: Next Clear Fractions48 48
( 1) 4 ( 1)1
t t t tt t
Multiplying by the LCD
48 48( 1) ( 1) 4 ( 1)
1t t t t t t
t t
48( 1) 4 ( 1) 48t t t t
248 48 4 4 48t t t t 24 4 48 0t t
2 12 0t t
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt45
Bruce Mayer, PE Chabot College Mathematics
Example Example Driving Speed Driving Speed
Carry out: The Last Eqn is Quadratic in t:
2 12 0t t
Solve by Quadratic Forumula with:• a = 1, b = −1, and c = −12
12
121414
2
422
a
acbbt
2
6or
2
8
2
71
2
491
2
4811
t
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt46
Bruce Mayer, PE Chabot College Mathematics
Example Example Driving Speed Driving Speed
Carry out: By Quadratic Formula
Since TIMES can NOT be NEGATIVE, then t = 4 hours
Return to one of the table Eqns to find r
3or4 t
48 48
4r
t 12 mph.
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt47
Bruce Mayer, PE Chabot College Mathematics
Example Example Driving Speed Driving Speed
Check: To see if 12 mph checks, we increase the speed 4 mph to16 mph and see how long the trip would have taken at that speed:
The Answer checks a 3hrs is indeed 1hr less than 4 hrs that the trip actually took
State: KW rode his bike at an average speed of 12 mph
hr 3hrmi 16
mi 4816 t
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt48
Bruce Mayer, PE Chabot College Mathematics
ReCall The WORK PrincipleReCall 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.
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt49
Bruce Mayer, PE Chabot College Mathematics
The WORK PrincipleThe WORK Principle
If A and B, working together, require t units of time to complete a task, then their rate is 1/t and the following equations hold:
111
tb
ta
111
bat
1b
t
a
t
tba
111
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt50
Bruce Mayer, PE Chabot College Mathematics
Example Example Empty Tower Tank Empty Tower Tank
A water tower has two drainpipes attached to it. Working alone, the smaller pipe would take 20 minutes longer than the larger pipe to empty the tower. If both drainpipes work together, the tower can be drained in 40 minutes.
How long would it take the small pipe, working alone, to drain the tower?
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt51
Bruce Mayer, PE Chabot College Mathematics
Example Example Empty Tower Tank Empty Tower Tank
Familiarize: Creating a Table helps to clarify the given data
Pipe Time to Complete the Job Alone
Work Rate
CombinedWorkingTime
Portion of Job Completed
Smaller t + 20 min. 40
Larger t min. 40
1
20t
1
t
40
t
40
20t
And the Job-Portions must add up to ONE complete Job:
40 401
20t t
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt52
Bruce Mayer, PE Chabot College Mathematics
Example Example Empty Tower Tank Empty Tower Tank
CarryOut:
40 40( 20) ( 20) 1
20t t t t
t t
40 40( 20) ( 20) ( 20) 1
20t t t t t t
t t
40 40( 20) ( 20)t t t t 240 40 800 20t t t t
280 800 20t t t 20 60 800t t
The Last Eqn is Quadratic
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt53
Bruce Mayer, PE Chabot College Mathematics
Example Example Empty Tower Tank Empty Tower Tank
Use Quadratic Formula:20 60 800t t
2( 60) ( 60) 4(1)( 800)
2t
60 3600 3200
2t
60 6800
2t
60 680071.2
2t
60 680011.2
2t
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt54
Bruce Mayer, PE Chabot College Mathematics
Example Example Empty Tower Tank Empty Tower Tank
Omit the negative solution as times cannot be negative
The amount of time required by the small pipe is represented by t + 20, it would take approximately 20 + 71.2 or 91.2 minutes
Check: Use the Work Eqn from Table
00415618043860271
40
192
401
271
40
20271
40...
....
?
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt55
Bruce Mayer, PE Chabot College Mathematics
Example Example Empty Tower Tank Empty Tower Tank
State: Working alone the SMALL pipe would empty the Water Tower in about 91.2 minutes
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt56
Bruce Mayer, PE Chabot College Mathematics
WhiteBoard WorkWhiteBoard Work
Problems From §8.2 Exercise Set• 74
The Arrhenius Rate Equation
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt57
Bruce Mayer, PE Chabot College Mathematics
All Done for TodayAll Done for Today
MotorCyleFatality
Statistics
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt58
Bruce Mayer, PE Chabot College Mathematics
Bruce Mayer, PELicensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Chabot Mathematics
AppendiAppendixx
–
srsrsr 22
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt59
Bruce Mayer, PE Chabot College Mathematics
Graph Graph yy = | = |xx||
Make T-tablex y = |x |
-6 6-5 5-4 4-3 3-2 2-1 10 01 12 23 34 45 56 6
x
y
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
file =XY_Plot_0211.xls
BMayer@ChabotCollege.edu • MTH55_Lec-50_sec_8-2b_Quadratic_Eqn_Apps.ppt60
Bruce Mayer, PE Chabot College Mathematics
-3
-2
-1
0
1
2
3
4
5
-3 -2 -1 0 1 2 3 4 5
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M55_§JBerland_Graphs_0806.xls
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