青教備頁三第張七第 日二十月正年巳丁歷
WAH KIU YAT PO
郭日僑華
日一十月三年七七九一公年六十六國民章中
辦主組育人成署
會討研育人成 行舉起六十月下 詞致烈鴻胡席主任建陶
熱點
次致岚任對 「研詞胡主有 肚共討。陽席由
測會烈,教
此
發
討
,及全忸同工屈時,
・主任,主任講師
司署成人教育組主
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教育研討會將於下
舉行,成教祖屬下 各中心校長:監督
磡育才迫理工學院 十七日假 九龍紅 八四】月十六日至
五期里
一之水拳6.. 典雅,全日會握至下午五時結束。 月,是日由上午九時開始,九時三十分舉行開幕 成人教育根奏下所有E工汨參加十六日蜜 一使各同工互相討論 交流蒩聽,以再成人教育 „ *KERGEINKBER - «MBEIEC-
·務衆許賢狠、康破事员公署众李青茅女出等作
上街,大會將邀請立法局謚員孟家舉聽受,肚實报
应狄祖員工及各中心校長、監督、主任等共同策
⇒函來者讀
HREK - (8) 加十七日拳行之項目,由上午九岸三十分至下午 成人教育康樂中心同工及業務就導及强
主任紙會管學生長髮
化學新教師含糊不清
柳城先生:
這真是令我們極之失望, 我們的苦衷,雖快去改善 - 使我們能安心上課。 : 诀,我希望藉追封信給校長和主任知道 至於實大的道德和老癿教學的問題佃不聞不問, 請者,讓生11月31十九日
「以便不張揚- 可是他們只去管鄀些無部的小事,
小息脚踏钴家常也遭主任責駡。
|師和學生都無得一二楚,但因我們煒敬他們所 :事實上,主任們的「黃」、「黑」背景,老
PREN-BEDS24**BERKEGU
| 罪 - 長髮會被罰停課四天至一週不等,但打架凡
「他事情,長髮比吸烟說汚餘部,甚至打架還重
只管長髮的主任,絕少理其 [答一句:「你們不留心上課」。 善,但每次都不得其門而入,只是國
一我們便三番四次向主任訴苦,還求改 一的努力是很難考獲美術的成親,所 |際,如得不到老師的帮助,只自己 。試想想,我們現在正是尘碩之 化学方程式而解釋得拖泥帶水的老部 ,晚來一個說英語含開不棄和只是高
吉下,而漸有跑步,但數月前,他難
·我們的化學科在上任老師的接蕩
-來信,以求有所改善。 任的不理不哚的態度,所以便不得不 一千金,但現在因閧題日益楼凤加上主 一考班,本不想來;因爲現在時間值
·我們是就於xx英文中學的會
校巿jj慈主席胡东太騋任立法局委員全體合攝。
· MESAKENSSEKROOKHREX
BY
猴策由
LEVN 18°26′
36*52'
1977中學會考試題預習專欄
明德社主編
LEVF - 2 x 18°26)
=
(Ans.)
新數學(十八)、發榮家、
MODERN. MATHEMATICS (18)
to exercise 7 (cont.)
(ii) In one figure, X,I are.
the mid points of AB and CD respectively.
The angle between planea VAL
and VCD
(iv) * (9,3) (Ans,)
eosZFAF
AE -
AE-AF (AKIAF)
41 + 33
AF - 71 433
(13/3)2
x2(x+y)f(x,y)
(x+y)(x2r(x,y)
+
+
y2(x+y)(x-y)
(x-y)]
is divisible by x + y.
Sinem P(k) is true implies P(k+2) is also i true and that
P(2) is true, therefore,
the
AE - 4 +
5.898
AF -√72
= 7.616
statement is true for all
even; natural numbers,
AL-A
= 28 + 13 = 41
C08 LEAF -
5.898 x 7,616
LXVY
R ZXYN
XN
ww
BC
=3em
In rtà XAN
LEAF
0.9127
B.
2407 (Ana,)
XN
15.
tan XVN =
VN
(Ext.
4
cyclic
·25 TQ+2KTH=
quad.)
180°
(Adj Ls on st. line)
LETO = 90°
FA = AQ
(roved)
¤A=AT=AQ (mid-st. of Hyn.)
ABAT (Subst.)
12. Given: AOP BOS and COT
are st. lines.
To Prove 2
AABC
AB
AFST
PS
As shown in the diagram ab-
let
the set of students fa
in mathematica.
LXVN
LXVI
(iv) Are
of rect.
ABCD
the set of students fail
in physics.
the set of students
8 x 6 cm
48cm
vX
.VN
B(A)
0,25 x n(1): 0.15
VX
122
xn(1)
× 0,1 x n(1)
the probability required
12.37 cm
Area of triangle VAB
4× ABX VX
(A (3)
n(8)
-0.ì x_n(1)
x 8 x 12.37
49.48 cm2
0.15 x N(1)
(1)
0.1
0.15
The probability required
{n(A_1}\B)
n(A)
0.1 x n(1)
0.25 x (1). (Ang.)
2
VE
-EN
VE-
12
12.65 cm
area of triangle VAD
4x VE 1 AD
1x 12.65 x 6 cm2 37.95 cn"
(i) If (a,1)
2
1. R
28 +
(Ans.)
(ii)
(0,6)
2
•. Total surface aren
48 + 2 x ( 49.48 4 37.95)cm2
1 +
n5.042
(ii) u(AUB)
\\n(A); ++n(B) (AB)
- 222.86cm2
(Ans.).
2
(v) Volume
2(0) + 2
(Ans.)
(iii) From the graph of
5 (ADB.)
Volume of the box
(1)
ires of rec&ABC D x VN
* 12 x 48 cm3
- 192 cm3 (Ans.)
(iv)
0.25xn(1) + 0.15 x n
0.1 x
= 0.3 x n(F)
the probability required
D{AUS)
0.3 x n(x)
n1
0.3 (Ans.)
12. (1) Since A,B, C and D are the vertice of a parallelogram
+8
4
- 2 + 10 (Ans.)
✦ 21 +6
is the required lineær graph From the graph,
5 x
440 21
the fraction of the space
wasted
=
Area of ABCD
absinx
480.QA cose
AC.BD.sine
sine 2ab sinx
ba
a2
b2
•*. 2absinx-4BQ.OAsine
2absing af¬b'
coso i.e. tane =
1(x),
the mininum value of f(x) is
3r x 2 x 9r
54
-30
The figure shows the cross-
eetion of the box
1 -√(2) 22
Br
Let the maximum number
spheres be n
2r + (n−1}} <
(n 1)1< 7r
(n=1){35 & 7x
(n
Proof: 4 SET=LSOT.
(Ls in the same.
segment)
ZBAC =4BOC.
4SOT = 4BOC(vert.opm./s) 4SPT = LBAC (Subst.) Similarly LST LACE
LTS LABC (3rd Ls of As)
ABC
are similar.
ST (Equiangylar As)
AABC AB
HALST
13. Solution:
Taxsec ̃x-1
++
20
m+2m+1-4m
4m
2m
secx+tanx=(1+1)+
(+)크
2m
2m
= mor.
secx-tanx=(2)
14. Solation:
21
or m.
m
BO2+002-230.nccos(180°
=B021002+2Bn.Occose
= BO
BO2+0A2-280.DACOSO Since ABCD is a //gran
a
Volume of the spheres
а
2-62
gta ABC
AB
48
10 cm
AN
AD
5cm
x = −1 or x 2. (Ans.)
2
(~)
3
b
5 (Ans.)
2x + 2 = -2x 5
therefore
+
D is (4,5)
- 2x + 5 is drawn
(ii
In rt.VAN
☐ 12cm
AN
tanZAYN - AN
M
101 +63
In the above figure,
(C
5
= 0.4167
UD
41 +
ZAYN - 22°37' (Ans.). (ii) In the figure, E,P áre
the mid-points of AD and C respectively.
the angle between plane VBG
and VID
•ZEVP'
53
since CE: ED 21
... OF 205 + OC
- 2 LEVN
EN-
LAB - kem
In rtABYN
EN
tan ZEVN
VN
4
- 0.3333
From the graph, the solution of x2
- 3 is the value of £3.
£3 1.73 or -1.73 (Ans.)
14. Les P(n), be the statement
n
1′′ – y1 is divisible by x+y
för all even nåtural numbers
When n 2
2
x
ww
2 У J
(x + y)(x − y)
..it is true for n = 2. Assume the statement is true for n k (where k is an even natural number)
3
k
2(47
+
+ 107 + 67
y
in divisible.
by y
3
61 +
谢
i.c. x
y
-
(x. +
k
ог
E is (6,
)
(Ans.)
For n
-
(iii) If F is the mid-point
then
of HC
OF × 4( OB + OC )
*(81 + 21 +
101
*
#
97 +43
x
k + 2
y) f(x,y) (x + y)t(x,y) + y'
k + 2
-
K + 2
**** _yk+2
x2( (x+y}f(x,y) + y ̧
2 k
k
jky -pk+2 x2(x+y)f(x,y) + x2** - jk+2 \2 (x+y){(x,y) + (x2-y2)
54. r
440 21
54 r
$47
567 (Anm.)
壆(十八)・交長波·
Mathematics 18.
Solution to exercise 8
Section B
11.
Given:
お
- A
BKTH is a cyclic quad. PA = AQ = AB
To Prove:
a) HTK-90°
Б) АВ = АТ
Proof: Since PA=AQ=AB (Given)]
.LAB=LABS; LAQB=LABQ
(Base is isos A)
LAS B+ LABS+ZAQB+ZABQ = 180°
(sum of a)
‚ˆ‚¿‹BA + LABQ = 90°
i.e. L'BQ = 90°
Since BKTH is a cyclic quad.
'.LKTH = LFBQ = 90°
・
15. Solution
a) the total weight of the
box
= · 10kg+50x(1−20%)kg+180X(20%
kg.
= (10 + 40 + 36)kg·
- 86 kg
b) Let xkg be the weight of.
content.
20%x 180
+
80%x 50
10x + 144x = 9000
x =
9000
154
5834
the total of the box
(10+5834)ka
16. Solution:
V
6834 kg
- 6
(6-12x)1
The ratio of the volume of spirit and water at first is 30:6 or 5:1 Volume of water often 121. are drawn out the first time
= 41. polume of spirit after 121 are drawn out and filled up with spirit
(36-4)1 321. volume of spirit after 12lare drawn out the 2nd.
time=
20
(32-12×32)1=2137. the final percentage of Spirit in the cask
21
X 100%
36
5977
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