1969-11-13 — Page 21

1418011959 -

CITY HALE

育教僑華頁三第張六第一日四初月十年己屦夏 WAH KIU YAT PO

報日僑華

ECO英文中學會考試題預習專欄

之學生 私及分

听政府!

子

语

在

生物科 (二)

橋

言

一多人,明年於新湖

一之實施,預料人數款來西

K

之國家、超過一百個。

一接受祥獎人在本以就藐

王幸利主張

【特別傘貼在內,達九百 |至七〇年把篇止,使用

一十一萬三千七百五十,新

獠與新瓶越非馬印國比:從東新酒駕然白熊

來 回集城司 南賓西尼 一國 案站南國尼

二期研究獎金計劃·

周期一年。其说网

·老四九年度·倫

一究生,就讀於另一家】克安識风

:史氏稱·在一九四

[1] 各種專家職務誌計表

: 百分比:

及獎學金資助十八名研究員

1HO

1出六岁

四天九〇

四九四九四三四比

E

五花

輪杜獎學基金現已研殺人魔 | 翻料及大學獎學金計槛,以能的研究

獎學金三千五百個 一出佈獎人往國

别

新聞界

慈康

四大九九

〇九三七四九

·枣强 全數百分比·正在研究中

百分此

:

〇八七中〇九九三

廢除升中試此慢火

「西南各本商正在研究與未來研究 正在研究中信分比,时间研究中百分比

| 升中試中藏萬八千

BIOLOGY (2)

Answer to the questions of last weeks

I. Write on the line at the right of each statement

the number preceding the word or expression that

best completes the statement.

Ans. 1.(2) 2 911 K

2.(1) 3.(1)

7.(3) 8.(1)

14.(5) 12.(5)

II. What are the characteristics of living organisms Try. to evaluate the similarities and differences:

between a simple plant (e.g. spirogyra) and a

(simple animal (e.g. Amoeba) in their method.of.

nutrition and reproduction.

AnE The characteristics of living organisma are:

(1) Nutritions

A living organism needs matters from external environment either for obtaining

energy to maintain life, or for building up ita

own living material in the body.

The method of nutrition for an animal is nolozoic and that for a green plant e

holophytic. Holozoic nutrition in to digest and absorb solid organic food in solution. Holophytic nutrition means to make food from carbon dioxide and water, under sunlight and 11 the presence of chlorophyll by the mechanism:of photosynthesis.

(2) Respirations

Living organisms require energy for all life activities. Respiration 18 the process of oxidizing or breaking down the adasimilated food materials or the body substances, with the help of oxygen, to liberate energy- Both animals and plants respire according to environmental conditions.

(3) Growthi

A living organism pas tae sbility of

growth, Growth begins as the intake of food 18 greater than the immediate demand in an animal or in a plant body. The body of a living organism may increase in Bize and in complexrty when the process of growth occura

(4): Excrationi

Excretion is the removal of waste materiais an the body of a living organism. The waste "substanges are got rid of as by-products of

metabolism)

In animals, usually the waste products are collected and exoreted by special. excretory organs

dio, while flowering plants give off carbon

and water vapour into surrounding air through stomata,

5) Reproductions

All living organisms have the ability to reproduce their new kinds or own kinds. Therefore reproduction is the way to give rise to naw generations ao as to prolong the offspringa of species and kinds in animals and planta Reproduction is as simple as the binary fission in an amoeba. But, mostly the higher animale and plants reproduce by secual means,

(6) Irritability:

All living organisms possess the ability to detect and react to changes in their surroundings. as well as within their bodies. Most external changes in the environment are humidity. temperature, pressure and Light intensity etc.

(7) Locomotions.

A living organism, both plant and animal, can move. However, a liying animal moves eithe: of its own will or in response to some external stimuli. Plants are not able to move more freel in environment than animals but may motivate according to the methods of nutrition and reproduction.

Both epicogyra and amoeba are different in the mode of nutrition but have the same. process of absorption and assimilation. Spriogy): and amoeba a both have asexual reproduction too. The amoeba, even up to now, has no report to carry out sexual reproduction.

An amoeba, by moving ite paaudopodia, encloses micro-organisms or decayed remains of larger organism by the formation of food vacuoles The food in the food vacuoles is slowly digested by secretions poured out by its surrounding oytoplasm. The food is then dissolved into simple diffusible material.

'' Spirogyray a green alga, which is capable to manufacture carbohydrates from simple inorganic materials, the carbon dioxide and water under aunlight. This process of manufacturing of carbohydrates. is known as photosynthesis.

Questions for this week?

Huke a fully labelled drawing to show the external and internal structure of an Amoeba Briefly describe the process of reproduction. this animal..

2. Make a clear, fully labelled sectional diagram to:

show the gross structure of the body of a Hydra.

Describe the methods of nutrition and reproduction,

under adverse conditions, of Hydra. By means of simple diagrams illustrate the methods by which Hydra moves.

三六九五七一

三〇〇五四七六

20.1

中文中學

考試題預習專欄

數學科(二)

第一次預習題解答

喬仲强

分解下列為因式: (1) 12x2-23xy+10y.

(解)因之3萬奇數,故不能分解為2×6.

∴原式=(3x-2y) (4x-5y)〔答]

(2) 24d+224–21.

(解)

·原式=(Za+3)(120-7)[答] [Za

3)202-90-200.

-23x4

-7 360-144-224

(解)此題 200,不能分解為兩個偶數之因子,亦不能分

解為含有5因子的两相因子之

* 200=8×25

∴原式=(a+8)(20-25) [答]

(4) 12x2-683x+91

(解)原式=(2x-7)(6x-13) [答]

5) 6x3-38x-144X

【解)原式=2x (3x~19x-72)

22

示戒

四期星 日三十月一十年九六九一届公年八十五國民華中

由謝組葯扶慫

|

-鄎返奧斯龄大

|洲两個扶輪钍党

學位。頁合成小姐由澳

「出謝午醬例會之貝合成一宮為「弟之

部治小及前日 在過一段者

姐力例及史, 担燈會貝

米入的是船式一案

弱 四松?跟的税苦百冷弟 微端照主局,

了的年六 那本 英的了分款者上網區民

一 香港扶輪社現爲百份

香港扶輪社前日慶祝

扶輪社獎學基金現已頒發獎學金 千五百個使用 金額驁達一千萬美元

扶輪獎學金週

一詞中,詹 當

【金遜·並由挪威籍之舒德升巾就建毆,將予考

研究現况 (精)

因中項19不為偶數亦不為32因数 坂记板能分解. * 8x90x 1×7:

·原式=2x(3x+8)(x-9)[答]

6) 1202-7ab-12b2

(解)此題文 12 不能分解為之活 ∴原式=(30-46) (40+38) 【答

(7) 162 −218x+27

(解)因218不含3或42周3故么不能. 分解為4×427不分解為3×90

∴原式=(8x-1)(2x-27) [答] (8) 6x2-11 xy + 3y2+19%-114 +10

解

22-34

sixy!

答:原式

(解)

3x

2x

4+2)(2x-34+5)

-7x+y+6

答)原式=(x-2)(2x+y-3).

{10} 9a-12ax-p2¬q* −2pq+4x2.

(解)依ax及力.g集項,可得两平方之宏 Tax=(922-12ax+4x2)−(p‍+2fg+q2)

= ( 3a−2x) — (p+q)2

=[(30-2x)+(++g)][(30-2x)-(p+8)] -(3a−2x+†+z) (3a-zx-p−q) [4]

{ (11) |+2x+2Yz+x2-Y2 Z.

(解)原式=(1+2x+X)-(Y-2y3+子*)

(12) x4–2x

=(1+x)-(y-3)

=[(1+x)+(y+3)][(1+x)-(三)) (1+x+y-3)(1+xy+3) 〔答)

(解) 因各係數之和為零,顯然有x−1之因子,其餘à

因子,可用综合除法試之 1+0-2+3-2 L

++-+2

7 +1 −1 +2|-

(A) [EX=(x-1) (x+2) (x=x+1) --2+2-2

(13)(x+1)(x+2)(x+3) (x+4)-120

(解)原式=[(x+1)(x+4)][(x+2)(x+3)]-120

(x+5+)(+5276) 120

(5) Hot 24-120

=(x2+5x) + 10(x2+5x)-96 [Tilky2+10y=96]

4576)(x+52 +16)

(+6) (x+5+16) LÀ

(14)(x+1)(x+2)(x+3) (x+b)-15 (解)原式=[(x+1)Cx+6)][(x+2)(x+3)]-150

(X+7x+6)(x2+5x+6)~15x

(x2+6+7x) (x2+6+5x)−15x". (x+6)+12x(x+b)+35x156

(x2+6)+lax(x+6)+20x* [+20=2×10]

=[(x+6)+2x][(x+b)+10x7

(0+2x+b)(X+10X+6)[答]

15) 6X*+mxy-3y7+3x+10y-3可析為的一次因 式時,求加之值,又其式為何? (解)依x降冪序排列

能功

不凍甜 她抽可工長

單英時分有:

無實忙也生我有是脆一

18*=6x+(mg+3)x−(3y-loy+3)

想笑

3x

•6x2+(my+3)x−(y-3) (3y-1).

*之0.5x=x6x故可以有如下结果108

+(3y-1) | |

X~(y-3).

3x(3y-1)-2x(y-3)

7xy+3x

~(3y-1)

>+(y-3)

X(y-3)-6x(34-1)

*-19x4+3x

(i)若2=7.則原式=6x+7xy-3y+3x+1oY-3

(2x+3y-1)(3x-4+3)

m) 7/5m=-17, 8] *=6x-17x4-3y2+3x+104-3

=(x-3y+1)(6x+y-3). 上次我曾在本欄說過,用代數的運笑,可以解決. 幾何或者三角的問題,現在我們看看的例。

(例)两园外切,試証其外公切线之長,為两国直径之 比例中項。

(已知) O,P 两园外切於C, BAB尊其外公切线又

AD, BE各為西园直径

(求証) AB2="ADx BE

分析此題的証法,多

數人都会联AC, BC及

遇C奌作两园的公切线CT交ABT奌

然後利用“切线等長定理以証明<TAC=<TCA, LTBC LTCB.再利用三角形内角和定理,可以證得

∠ACB=90°,又因半园园周角為90,故BCD及ACE均爲 直线,由是証得MA ABE WADAB.於是得

AB:DA=BE:A日,

但我認為這一個証法,不僅繁雜,而且考生更会怒 略到先要証明 BCD及ACE均需直线這一但要点硦 在試看看下面的証法。

証明) 1.联OP,則遇C奌B

(两国相切,联心线必遇,

熱奌)

∠A=BA

(切线上切奌半径)

3.過P作OA之垂线

典DA安於Q,則ABPO為矩形

(周∠A=∠B=∠PQA

L,又回边形内角和篇360°故R=96)

4.設口园半=R, P园半狂二,則

QQ=OATRA= OA-PB=R-L(因矩形对比等代换)

·OP=OC+CP-R+r

(全量等於諸分量和)

在AOQP中因<OQP=t.L,

?Q= OP2=0Q" — (R+^}~ (R-^) — 4Rr= 2RX2A

(畢氏定理)

6 AB'=TQ"=28×21=AD×BE (@ AB, PQ STEM{WL)

(例=)已典 △ABC. 求你一直线典 AB, AC 各交於D,E两桌 使 AD=AE B △ ADE=△ABC

(芝知) △ABC

(本作)一直线 DE 各奖 AB. AC交拎D,E两奌使 AD AE

RAADE F÷AABC

(分析)額如右图 DE经已作出

級 AD=AE=x 則因

AADE

ADXAE

AABC ABXACT

80

(两AS有一角相等或相精其面積比等于乘積之比) 因犯萬已知主故可由比例中项之作法求得之!

(作法)1.平分AC柊M奌

2. 以 AB为直径作半园

3. 在AB上 AK=AM,

4.邊K

KHIAB典半国

古以A爲心, AH高半作

孤告交 AB ACD及E.

6. 联DE即求求

(証明)讀者可從分析中自行証明

(討論) AC'>AB>支AC, D, E两点在AB.AC两边上,

(註):AB<=AC, 則以RC為直狸作半园

第二次預習題

(1)求作一直线,垂直於三角形之底边,而平分其面積

(2)求作一直线,平行於三角形之分角线,而平分其面積 (3)求作一直线,平行於三角形之底边,而使用裁之三

角形為原三角形的号

(4)於△ABC內,求你一点P,使APAB, APBC,APCA

之比為2:3:4.

(5)巴奐两△ABC, ADEF, 求作第三ᅀPQR使之典

△ABC相似而典 △DEF等積

(6)求作△ABC,使AB=5厘米,AC=4厘米,A角平分

线 AD=3.5厘米.(要依尺寸作图)