國民華
育日一廿月十年巳己腦夏
報日僑萋
of
problem Analysis
Discussion:
六期 日八十月一十(九八九一)年八十七國民華中( 30 )
פ
201
二第
七類出
印承及行額
中華民國僑務委員會頒發登記證台教新字第一一八號
受∶外地融鲿
t
2AB+.
惹分巧胜
誆
話
JEENAWA
市北台
ARRANNY+
1990 中學會考預習專欄
月 出版社
MILL & DALE PRESS
Computer (7)
Advantages of flowcharting
K. Choy
(1) The step by step nature of flowchart gives a logical analysis of the problem. A difficult problems is broken up into simples steps.
(2) Flowchart can be used as a tool for debugging. It is important to test the. logic of the algorithm. Flowchart tracing is easier locate the logical error.
Limitations of Flowcharting
is
• (1) Flowchart useful when the Complicated algorithm is simple..
flowchart of complicated algorithm is certain difficult to trace.
(2) Complicated algorithm should be represented by flowcharts, Each one of the flowcharts perform a simple task of the algorithm.
(3) Flowchart tracing is easy in the
direction of flow.
It is
impossible to trace the flowchart in the reverse direction.
Another kind Technique
Top-down Analysis
A task is divided into smaller and simpler steps until each step, can be solved. Each step can be refined until it is simple and easy.
Example: Perform a top-down analysis of
the following task:
Solution to a
quadratic
equation
Solve a quadratic equation 8x2+bx+c=0 by the formula
X =
to
2
3
to
Input the
coefficients!
Calculation!
of the
a, b, c
the
roots
(3) Computer program is written
control the computer to perform the required job.. But the logic tracing on the program is rather difficult. Therefore the flowchart acts as an effective communication. for all the people involved.
(4) Flowchart helps others
understand the algorithm of the solution. Therefore it is then a vital
of part
program
documentation.
roots by formula
3.2
(1) To solve the quadratic equation is Our aim, It is put at the top of the analysis.
(2) The séquence of solving the problem will be from top to botton and then from left to right. It may be performed 65 the
following
sequence:
1. Input the coefficients a, b, c.
1.1 Make sure that a # 01
2. Calculate
formula
the roots by the
2.1 Calculate The determinant
D = b2-4ac
2.1.1 Make sure that > 0.
2.2 Calculate
root
START
Input the
coefficients
, b, c
15
> 0
Determinant
the
-210
=
2.3 Calculate
the other
reet
-b- √D
2
=
2:
to
1.1
Make
sure
2.1 Calculate
the
2.2
Calculate
the root
2.3
Calculate
the other
Print Print
that
determinant
root
The
18 £ 0
_-b+
2a
Dox refinement.
2.1.1
are the third
b-
2a
The method Usec is called stepwise refinement.
the
(5) Flowchart is known to be portable. This means that flowchart is all machine independent.. It 1:5 a symbolic representation of A logical algorithm. It acts as a part in the problem solving procedure. Therefore it is also independent of any high level programming language.
it is always a bad habit to sit before મૈં computer without
preparation.
2.1.1 Make
Sure
that
0 % 0
3. Figure 3 shows
y=x2-4x+3.
the
graph
of
flence, or otherwise, find the value of x if the area of triangle DEF is minimum.
10 INPUT "THE
$700
מי Print
redl
3. Print the roots
3.1 Print
3.2 Print
(3) The boxes 1, 2 and 3 comprises the
first level of refinement.
The boxes 1.1, (2.1, 2.2 2.3) and 3.2) are the (3.1,
second refinement.
(4) Complicated step is broken up to
simpler steps.
The "owchart can be coded into BASIC
progran.
COEFFICIENTS OF QUADRATIC EQUATION": A, 9, C
20 IT A <= 0 THEN GOTO 10
30 DEA 2 4 AC
40 IF DO THEN GOTO 90
50 X =(-B SQR(D))/(2*A)
60 X= (-8- SOR(D))/(2*A] 70 PRINT Xq> X2
80 GOTO 100
90 PRINT NO REAL ROOT"
100 END
THE
The procedure may be represented by the Flowchart below.
收附加費 *
費璐就會議」的十四名議員,昨齊集旺角 (E) RAZMERAFOES.
eles tauru的半,利度夏人曉娟與一也門社主葉雅其也好去,契快將時間的乘 而他們到達地鐵公司後,向三個部門的經們絕不承可收取整忙附加費是斂財,其實,
換及東墬通車後的繁忙時間乘客數字及實際 地鐵公司公關經理梁陳智明則解釋;他一 他們又要求地做公司提供實施附加費前
·會議」約十四名議員,昨日到地鐵公司發展
圖:代表「各級法及反對地鐵附加費聯席
|FE & CHER):
1990 中學會考預習專欄
明德出版社
MILL & DALE PRESS
Mathematics (7)
Exercise 7:
H. K. LO
Graphical solutions of
equations (II)
1. The shape of the graph of the
equation
y=Ax2+Bx+C
(where A, B and C are integers) is sketched as shown in figure 1.
((0, 3)
(5, -2)
(2,-5)
Figure 1
It passes through the points (0, 3), (2, -S) and (5, -2).
(a) Find the value of C.
(b) Find the values of A and B.
(c) What additional linear, graph would be necessary to solve graphically the equation
x2-2x-2=0
by using the graph of
y=Ax2+Bx+C.
國
2. Figure 2
y=px2+qx+r.
as
Figure 3
(a) By drawing suitable straight
fines on the same graph, find the roots of 2x2-4x+1=0. (8.
(b) Using the given graph, explain x2-2x+2=0 has no real
roots.
4. (a) A piece of wire 4 m long is bent inte the shape of a complete rectangle. If the length of the rectangle is x m. Express the area of the rectangle in terms
(b) Draw a graph for values of x from 0 to 5 to show the relationship between the area
of the rectangle and the length
X.
(c) From your graph, determine
(i) the area of the rectangle when
x=1.5 m.
(ii) the greatest value of the area
of rectangle.
5. In figure 4, ABCD is â square.
AB=10 cm, BE=CF=x cm.
(a) Express the area of triangle
DEF in terms of x.
shown the
graph
of
0
Solutions to Exercise 7
1. (a) (0, 3) is on the graph
.. 3 A(0)2 + B(0) + C
. C3
{Ans:)
-10
(b)
the graph is y = Ax2 + Bx + 3
From the graph
(2. -5) is on the graph
x = -0.5 or x = 3
(Ans.)
-5= A(2) + B(2) + 3
2A + B + 4 = 0....(1)
-15
2. (a)
(5, -2) is on the graph
-2 A(5)2+ 8(5) + 3
25A + 58 + 5 = 0
5A + B + 1 = 0.....(2) (2) - (1)
3A.- 3 = 0
A = 1
(Ans.) Substitute the value of A into (2),
B = -6
(Ans.)
= 0
x2 - 6x + 3 = -4x + 5
= -4x + 5 is the required linear graph.
(Ans.)
(i) 2x2 - 4x + != 0
x2 - 2x + = 0
(0) (-1, 0), (0, 5) and (5, 0) are
on the given graph
0 = p(-1)2 + q(−1) + r
.. p-q+ r = 0.....(1)
-5 = (0)2 + q (0) + r
T = -5 .........(2)
0 = p(5) + q(5) + r
x2 - 4x + 3 = -2x + 3
y = -2x+ should be drawn.
From the graph,
x = 0.3 or x = 1.7 (Ans.)
(ii) The square root of 3 is the
roots of the equation
0
(c)
(i) From the graph,
when x 1.5
y = 0.75
.. the area of the rectangle is
0.75 m2
[Ans.) (ii) the maximum value of y is 1.
the greatest value of the area of the rectangle is 1 m2. (Ans.)
5. (a) Area of 4 FCD = {(FC)(CD) - 1x1 (10)
- 5x
Area of AED = (AE)(AD)
= (10-x) (10)
5[10-x)
Area of BFE = (BE) { BF)
++x)(10-x)
y Area of A DEF
=
-102-5x+5(10-x)+ 3{x}{10-x)]
專
欄
膠囊
醫藥的 突破
25p+ 5q + r. 0.....(3)
x2 - 3 = 0
From (1), (2) and (3)
or
x2 - 4x + 3 = -4x + 6
p = 1, q = -4 and r = -5
y = -4x+6 should be drawn.
A
E
X B
(Ans.)
From the graph,
Figure 4
(b) Plot the graph of the area
against x on the graph provided.
when
(ii) From the graph, y is negative
-1 < x < 5
x = -1.7 or x = 1.7 (Ans.)
(b)
(b) x2 - 2x + 2 = 0
50
x2 - 4x + 3 = -2x + 1
50
(b) The lowest point of the curve
(2, -9), therefore
..
y= -2x + 1 should be drawn.
2(2) - 3(-9) = k
k = 31
(Ans.)
Since the linear graph and the given curve
have intersection,
x2 - 2x + 2 = 0 has roots.
40
no real
37.5
по
=
19-[50+5x- x*}
- x2 -5x450
=
(Ans.)
病的免疫
全體內毒素
Figure 2
(a)
(i) Find the values of p, q and r.
40
(ii) Find the values of x such that
y<0.
港幣39元
『播電台
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(b) If the straight line 2x-3y=k, passes through the lowest point of the curve, find k.-
(c) On the given graph of
y=px2+qx+r, add. a suitable
straight line graph to solve
the equation
圖文傳真
5-594238
2x2-5x-3-0.
(e) from (a)
y = x2 - 4x - 5 Consider
yi
2x2- 5x - 3 = 0
= 0
4. (a) Let the area be y
width of the rectangle
2
-
= 2 - x
3
30
x2 - 4x-5=-
-
y = x(2 = x)
(Ans.)
3 4 5 6
A 9 10 x
(b) y = x(2 - x)
Figure 5
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3* - 7 should be drawn,
(關報日僑華
y = -x2 + 2x
(刋明) (出天)
30
0 7 2 3 4 5 6 7 8 9 10-x
From the graph, the minimum
value of the area of A DEF is 37.5 cm2.
x = 5 cm
(Ans.)
]]]
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