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dynamic transition from the supercritical flow in the steep channel to the subcritical flow in the pool.
This situation differs from that shown in Figure 2 because the flow approaching the pool in Figure 4 is
supercritical and the total head in the approa
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x3 + x2 6x =0 xx2 + x 6 =0x (x 2)(x + 3) = 0
So, the solutions are x =0 , x = 3 and x =2. It follows that the xintercepts are (3,0), (0,0) and (2,0).
3. Since the zeros all have multiplicity 1, it follows that the graph of the polynomial will cross the x
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F (x)=Cx(x + 3)(x 1)(x 5)
where C is a constant.
Example 29 Find all the polynomials of degree 3 having 1, 1, and2 as zeros. Since the polynomial is
of degree 3, we are given all its zeros. So, we know all its linear factors. The polynomials we want are
o
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4
Figure 4: Fourth degree polynomial
Figure 5: Fourth degree polynomial
5
2.2 End Behavior of a Polynomial
Here, we study the behavior of the polynomial as x gets very large (we write as x ) and asx gets
very small (we write as x ). Geometrically, we are
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consecutive valuesa and b for which P (a) and P (b) have opposite signs, then P has a zero between a
and b.
Example 37 Find a zero forP (x)=x2 2, correct to 2 decimal places. It is easy to see that P (1) = 1 and
P (2) = 2. Thus,P has a zero between 1 and
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CONTROL SYSTEM
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Example 17 Consider the polynomial P (x)=( x 1)(x 3)(x + 3). This polynomial has three
zeros of odd multiplicity. They are x = 3, x =1and x =3. This means that the graph of P will
cross the xaxis at (3,0), (1,0) and (3,0). This can be veried on the graph
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CONTROL SYSTEM
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Remark 24 If (x c)2 is a factor of P (x), we say that c is a double zero. If (x c)3 is a factor of P (x), we
say that c is a triple zero. And so on.
This theorem is very useful. Consider the examples below.
17
Example 25 Factor f (x)=2x2 +5x 3 By the theo
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Polynomials
Dr. philippe B. laval Kennesaw State University
April 3, 2005
Abstract
Handout on polynomials. The following topics are covered:
Polynomial Functions
End behavior
Extrema
Polynomial Division
Zeros of a polynomial
Sign of a polynomial
Fa
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CONTROL SYSTEM
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Example 4 f (x)=x3 +7x +6is a third degree polynomial. The leading term is x3, the leading
coecient is 1.
Example 5 f (x)=5is a constant polynomial, its degree is 0.The degree of any constant polynomial is 0.
2 Shape of the Graph
2.1 Practice
In this sect
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9
Example 11 Select the graph from the four below that best represents the graph of f (x)=2x4 9x3 +
11x2 4
The degree off (x) is even (4) and the leading coecient is positive, this reduces the choices to graphs
2 and 3. Since the degree of f (x) is 4, it
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3 Real Zeros of a Polynomial
If f (x) is a polynomial, then we said earlier that c was a root or a zero of f if f (c)=0. Geometrically,
when we are nding the real zeros of a polynomial, we are nding its xintercepts. Finding the zeros
of a polynomial is a
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CONTROL SYSTEM
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y y 1 gy c + =gy y y 1 gy c 2/1 > + =
The larger the amplitude, the faster the wave travel. The larger the amplitude, the faster the wave
travel.
21
Continuous Continuoussinusoidal shape sinusoidal shapeWave Speed Wave Speed 1/2 1/2
A more general descrip
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and thus the flow is said to be under downstream control. If an obstruction, such as a culvert, causes
ponding, the water surface above the obstruction will be a smooth curve asymptotic to the normal
water surface upstream and to the pool level downstream
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CONTROL SYSTEM
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w
2V
K
2
w =
K is a constant dependent upon the roughness of the pipe K is a constant dependent upon the
roughness of the pipe
(Chapter 8) (Chapter 8)
53
The The Chezy Chezy& Manning Equation & Manning Equation 4/6 4/6
(16) (16)(17) (17)
C is termed the
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Example 10.10 Example 10.10 Solution Solution1/3 1/3
For the rectangular weir with For the rectangular weir with P Pw w=1. =1.
2/3
2/3
32
w
2/3
32
wr
H)H075.0 611.0(91.5 Q
bHg2
PH
075.0 611.0 bHg2 C Q
+=
+ = = (30)+(31) (30)+(31)
For the triangular weir
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Classification of Open Channel Flow Channel Flow
For open For open channel flow, the existence of a free surface allows channel flow, the existence
of a free surface allows additional types of flow. additional types of flow. The extra freedom that
all
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Water flows under the sluice gate in a constant width rectangula Water flows under the sluice gate in
a constant width rectangular r channel as shown in Fig. E10.1a. Describe this flow in terms of channel
as shown in Fig. E10.1a. Describe this flow in ter
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34
Determine Determine E Emin min
To determine the value of To determine the value of E Emin min
3/1 2
c32
gq
y0
gy q
1
dy dE
0
dy dE
=
1 Fr gy
yq
V
2 y3 Ec c c c c min= = = =
(11) (11)
Sub. (11) into (10) Sub. (11) into (10)
1. 1.The critical conditions
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81
Examples of Gradually Varies Flows Examples of Gradually Varies Flows 2/5 2/5
Typical surface configurations for Typical surface configurations for nonuniform nonuniformdepth
flow with a depth flow with a critical slope. S critical slope. S0 0= = S S0c
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2
2
222
1
211
+=
+
=+
s/ ft75.5 V y V y V y 2 2 2 1 1 2 2 = =
0 513.0 y90.1 y 2 2 3 2= +
(10.2 (10.2 1) 1)
The continuity equation The continuity equation
(10.2 (10.2 2) 2)ft 466.0 y ft638.0 ft72.1 y 2 2 = =
40
Example 10.2 Example 10.2 Solution Solu
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=
(10.6 (10.6 1) 1)
(10.6 (10.6 2) 2)
68
Example 10.6 Example 10.6 Solution Solution2/2 2/2
For given value of Fr, we pick For given value of Fr, we pick various value of y, determine the various
value of y, determine the corresponding value of S corr
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Figure 4 shows a special case for a steep channel discharging into a pool. A hydraulic jump makes a
3.2 General Instructions for Use of Charts 128 Charts 1n28 provide a solution of the Manning
equation for flow in open channels of uniform slope, cross se
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The upstream condition corresponds to The upstream condition corresponds to subcritical
subcriticalflow; the flow; the downstream condition is either downstream condition is either
subcritical subcriticalor supercritical, or supercritical, corresponding t
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tanh 0
y
=
=
Deep layer Deep layer
Shallow layer Shallow layer
23
Froude Number Effects Froude Number Effects 1/3 1/3
Consider an elementary wave Consider an elementary wave travelling travellingon the surface of a
fluid. on the surface of a fluid. If th
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intercepts as well as the coordinates of the max and min. Graphing polynomials with a calculator
usually involves playing with the viewing window.
(a) f (x)=3x2 +9x 10x3 4 (b) f (x)=x3 +7x2 + 36x 240 (c) f (x)=2x4 + x3 + 13x2 +8x 1 (d) f (x)=x4
2x3 +1
4.
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It is not dicult to see why a polynomial behaves like its term of highest degree. We simply have to
factor the leading term. Let f (x) be a polynomial of degree
7
Figure 8: n odd, an > 0
Figure 9: n odd, an < 0
8
n. Then, we have:
f (x)=anxn + an1xn1 + .
Kenyatta University Regional Centre for Capacity Development
CONTROL SYSTEM
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Spring 2016
regression analysis is a statistical process for estimating the relationships among variables. It
includes many techniques for modeling and analyzing several variables, when the focus is on
the relationship between a dependent variable and one or more ind
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Institute of technology
agricultural machinery, agricultural technology, farm machinery, ergonomics
Table 2. Chronology of the structural units, teaching cars and tractors
Period
Department (chair, institute, work group)
Subordination
Main groups of subje
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stochastic signals
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Summer 2016
COMPUTER LAB 1(USING TORA/EXCEL)
2
QUESTION 1
We let our constraints be,
X1 = number of juice cases
X2 = number of paste cases
Constraints: Number of cans of juice and paste, number of juice and paste cases
The objective function is given by
Maximum Z = 1
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stochastic signals
ENG ABE331

Summer 2016
COMPUTER LAB 2 (USING TORA/EXCEL)
1. There are three gravel pits with production capacity 6, 1 and 10 ton per day. From these
pits, gravel to be supported to four construction sites. The demands at the construction
site are 7, 5, 3 and 2 tons. The transpo
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stochastic signals
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Summer 2016
HOMEWORK 3
Network Models
Question 3a. Minimal Spanning Tree Algorithm
1. Solve the class example starting at node 5 (instead of node 1), and show that the
algorithm produces the same solution.
Figure 1: 3aQuestion 1 and 2
Iterations
C1 =cfw_ 5 , C 1=cf
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stochastic signals
ENG ABE331

Summer 2016
Stateoftheart in eMaintenance
Abstract
Emaintenance is one the rising concepts that can be described as a maintenance management
idea of monitoring and managing assets over the internet. However, there have been a number
of definitions that have emerg
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stochastic signals
ENG ABE331

Summer 2016
HOMEWORK 4
CPM and PERT
1a. Total and free floats of a critical activity
Total Float
Total float of a critical path is the amount of time that a scheduled activity can be delayed
without causing delay on the project completion date or breaching schedule c