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『鄉同江龍律
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收掉感之成就表示冊及歌迎之忱、由各位 #KIFESIZERSIKERD) 大美食:爭民主自由而努力,同時並 糖妹團葵,以弘揚讚岸濟世活人意已任,烟護 CÆTERKI - BENDA - SELLEMEK 耕源-馬年好運,舊如業。首出議會理事長 始即相對行一肖躬義理並互蛸新年快樂理財 體理監事等六十餘人,會異常斯開,嵌式開 中野美研究院長林惠治、開完長調之義及至 副理事長林鐺、健保生、監事長鄒村、香港 潘文裕、程以正、區玉良、營法龍、葉百 、監事長蘇源、書员何家昌、批發、伍就光、 總監院定中、首席業長梁件仍、理事長吳英藝 大衆著返卷,出席者有越會執行轉監昪金- TIACZENIENCEBEKDEZEK (地図)香港中醫師公在所庚午年
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六期星日四廿月二(〇九九一年九十七國民華中(20)
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14
補數
與攝躪粹名當長(由蔡文浩出書代表接 揆〔右】張仲坑,监事長蔡伯勵發勝寒品、 碑:顺镞龍江同鄉會各節喵歅理事長黃淑 ** - *S***RELINE - Mestare 品,潘志泉主持,創技熱烈,此精选起,成積 點致敬詞衡,整行吼默賽會,席閶我没勝意的 四,共同發反務-爲鄉親謀幸郡。監事長蔡伯 部蛮作各項活動經費,更盼我會同人加强團結 各首長,會員鄉親踴躪股設,而所投得善狀全 宇,本會每年此日競投線盤及品,持希望 戚-生意興隆。注:今日業安服音開庫好
。〔本報輯]
Possible
combinations
of different
number
of bits
1. One bit
0
列職工
2 combinations
远程度負責一般 B6165
1990 中學會考預習專欄
2. Two bits
0 0
或以下需體阻耐 您每過工作五天
湯餸耐勞月薪
̇考羅業蚁理化三 可往理工學院進婷
經驗而定有慈誄 混街298 號校長
明德出版社
MILL & DALE PRESS
Computer (21)
0 1
4 combinations
1 @
3. Three hits
0 0 0
K. Choy
0.0 1
Meaning of decimal number
0 1 0
9 1 1
請
10
1990
+ 9 x 102 + 9 × 102 + a`
where 103, 102, 101 and 100 are the
placed values.
Meaning of binary number
1101
Conversion of number system
1. Convert 123.25 to binary system.
(a) Integral part
2 123
टाहर
230
215
27
8 combinations
10 C
1 0 1
1 1 0
11 1
熟中西裁剪
當値
更當値
歷電話近照寄
院助理行政主
ST
rt & Export
emale, F.5
e not
call 8936|11
cview.
+
Where 23, 22, 21 and 20 are the placed
values...
This is obvious that the same digit at different places has different values.
In binary number we have the binary digit. (Bit)
中
政科文憑教師
諜師三月廿
及雜條
天主教郭得勝
1990 中學會考預習連
遠出版社
MILL & DALE PRESS
Mathematics (21)
Exercise 21 Geometry (1)
Section, A
1. Find the values of a and b in the
following.
(a)
Binary addition
123.2510
Remark:
#1111011.012
(1) The integral part and decimal part can be done separately. (2) It can be easily understood by
the following:
101 +011
1000
Decimal equivalent
5
3
8
numbers
only.
2. Each bit is inverten.
1-0)
Example:
(0-1 and
Carry
1111011.012
Subtraction table for binary number
Subtraction borrow difference
lising 6-bit word to represent +1040 and -1010 in one's.complement method. Solution:
1x2+1x25x24x22x2+1x21+1x2°
0 0
0
0
0
+1910 is a positive number. complement method is not applicable.
One's
Integral part
1 0
1
0
1918. 001010 (six-bit)
Example:-
10:1002 111012
-1010 is a negative number.
One's
10:100
11101 1111
+0x2 +1x2-2
fractional part
64-32+16+8+0+2+1+0+0.25
= 123+0.25
= 125.25
2. Convert 1001.0112 to decimal number,
(a) Integral part
10012
1x23 + 0x22 + 0x21 1x20.
9:0
L-ADAS
Remainder }
LSD
(b) Fractional part
-0.0112
=
MSD
LSD: Least Significant Digit
MSD: Most Significant, Digit
12310 = 11110112
(b) Decimal
+
0x21 + 1x2 2+ 1x2-3
t
0.125
.1001.0112 = 9.37510
Addition table for binary number
Representation of numbers computer
within
1. It can be represented by electronic switches. That is the ON/OFF states of the switch.
2. The ON/OFF states can represent the binary integer system. It is converted to octal or hexadecimal as shorthand for easy reading.
3 bits of binary number car te changed to octal easily.
Complement is used.
-10110101. (in 1's complement)
2
Two's Complement representation
1. It is also used to represent
negative numbers only.
2. One's complement is used.
adged to the number.
3. It is most commonly used.
Example:
1 is
Using a 8-bit word to represent -109
by two's complement method.
109 - 011001112
(8-bit)
e.g.
-109 *
100110002
(1's complement)
0 1 1 0 1 11 ·0
3-bit
= 10011001
2
[2's complement)
ocial
Example:
01 10
1 1 0
8-bit
-10310
Convert 10001111, in 2's complement
inte decimal number?
100011:12 = -128+8+4+2+1
2's
The first bit in complement can be treated as a aegative number and then add the following bit in decimal equivalent
positive
numbers.
as
hexadecimal
16
3. The word
by
limtec Compute
One's complement representation
word length are the bit size of the
Remark:
01
៥.
0 1
1
1 0 (carry = 1}
0.25,0 = 0.012
Example
1. It is used to
represent, bedative
Section B
# = 60°
X + y = 11
the figure, ABCD is square, ABP. is an equilateral triangle, Find the size of 2DPC.
Also, 9 = b + C
Y + z = -13
(1) (2)
but.
C = b
Z +
8
(3)
A = 26
60°
= 30°
2
(Ans.)
(1) - (3)
y?
3. (4)
¿
60° + @= 180°
(2) + (4)
a = 60"
(Ars.
2y = 16
y = &
(b)
RX & cm
[Ans.)
7AAs shown in the figure, AB, BC and
A touch the circle at X, Y and Z respectively.
if AB = 11 cn,
BC= 13 ch and CA = 8 cm, find the length of 8X.
70 a d
Let AP BY = CQ DX BROWN
AY = BQ = CX = OP
=(a+b) + (b-a)2
=a+b2+2ab+b+a-2b)
2
a = 35°
(Ans.)
C == 20
課
(b)
急急聘中文 文課教師乙 七月十三 袁歷地址電 美國華僑日
師
英文、地 貓,入職
無津校相
*2960
.0
2. As shown in the figure, 81 and LC
are the bisects of B and C of
ABC. If A 43°, find the size of [BIC.
3. In the figure, ABCD is a square, if
B1 bisects (BDE, find ¿AID.
8. In the figure, 0 is the centre of the circle and il triangle OBP is. equilateral.
(a) Find i, ZOBP
ii. ¿BAP
(b) If 0, P and Bare corcyclic,
prove that ZMOA = .90"
3. find LOMB
ift. hence, prove that 0, A, M
and P are coreyclic.
but c = 180° - (60°. + 70°)
PUBLIC
2.
茘
汝小學
名中四程 十九日到
4. In the figure, ABCD is a square and
AP = BY CQ = DX.
PQ2 = 2(AY - BY).
Prove that
9. In the figure, is the centre of the circle of radius r. DA, 03 and tangents at A, B and R If [POQ = 90°,
מים [!
respectively.
OP 2 cm and 001 cm. Find
(a) length of PQ in terms of r
(b) the value of r.
I
公時間洽)
文師
中文文
兼任美 意即電:
5. In
the
ZADR 90°,
figure, AC 20 cm, BC= 25 cm and CB = 4 cm Find the length of AB.
Solutions to exercise 21
1. (a)
(Proved)
ADE
врде ́(AC+CD)2 + 80*
(a) i.
(OAP = ZOPB = LOBP
(AC+CD)2 (80-004)
ZOUP = 50°
(Ans.)
= (20+4)2 + (254-47)
= 576 + 609
11. ¿BAD - K¿ROP
= 1185
AB = 34.42 cm
(Ans.]
글
x წეს
(b)
= 30"
Since 0, 8.
concyclic.
.*. ZMGB
هنر
ZAPE = 180°
But APB 90° (4 in semi-circle)
.". MOB + 90° = 180"
(Ans.)
and i are
Section B
6.
As shown in the figure
2x + 2y + 48° 180"
2(x + y) = 180° - 48°
(x + y) = 66%
/BIC + x + y = 160°
ZBIC + 66 = 180°
AB 8P AP
abx 60°
¿MOB = 90°
[MOA - 30°
(Proved)
11. ZOMB = 90* - LOBM
90 - 60°
= 30°
(Ans.)
ZBTC = 114°
{Ans.)
A ADP and A BCP are isosceles
f
ཀྭ + h
== 90°-60° 30°
180°-30° esfeh
2 ZDPC 360° - (f + x + g)
360* (75°00 + 15")
iii. Since ZOAP = LONB = 30°
.. A, 0, P and M are concyclic (equal is on the same side)
(Proved)
2
75°
9. (a) AC8C is a square.
.. 40 = RO CA C8 = r
AP - AO - PC
150"
(Ans.)
08 RO 00
Since ABCO is a square
.. ĐỤC = 45
2BDF = 45 + 90° = 135°
日
As Di bisects ¿BDE
../ED: 167.5°.
ZED! ZAID [DAI
LAID + 45°
'. 67.5°
ZAID
67.5° - 45°
= 22.5°
Since AB, BC and CA are. tangents to the circle.
Let AX = AL = x
BY= 8X y
(Ans.).
CY CZ = Z
(5)
But AP = PR and RQ = QB.
PRr - 2
RQ = r 1
PQ » PR + 10
= (1-2) + (r-1)
= (2r-2) cm
PQ
(2-3)
24+ 12
(2-3)= 5
(Ans:)
PO2 + 0Q2
2.- 32.236
2r = 5.236
= 2.62 cm
(Ans.)
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