****
《三第四
¤一廿月三年申戊魔專
TAJALE KIU YAT PA
CITY HALA
J
日八十月四年八六九一間公七十五國高中
1998
式題額習書調
1968
0463 英文中學會考試題預習專欄
Vane
數學科
(廿五)
喬仲强·
生物科 (二十五)
REV
vachis.
第廿四次預習題解答
Ibiology (25);
(1)在100 300之間,各整數以3除之餘之者,赭數之 *** **†?
(解)在100峂300間*各蟄數中以了除之餘之者
最小的數=01 最大的數=299,而任何相鄰二數 23, a‡'s★* 3 - 4 14 14=101, 末項=299, 公差=3 依公式 l= a + (n-1) d if
解
299=101+(n−1) = 3
*=67 (## § 67 =])
67* (101+299)
5- ncate
2.
答:其和為13,400.
2 15 400
Answer to the exercise of last week
1. (n) Give a labelled diagram of a bird, showing its
chief external features. In what ways in a bird specially adapted for flight?
Quill
- Aftershaft
AMG.(4)
Nostril
Beak
2
Thorax
(2)寶石債値、奥其重量之津方盛正地今羃人有償 值5000元之寶石一顆不慎裂為艾米定其蔓蕾之比為
2:3, Þ★ $7?
(b) With the help of diagrams, describe briefly,
the structure of a wing feather.
Interior umbilicus
A typical wing feather
(UÆTTI sa NOW #1214 54 amb 3a;
其の値分別巻xy依題龕
5000
(50)
**
*
(Bay
7
47
*= 4×200=800.
200--X--2
Y=91 200 14 492.
50092-8002 – 1800 7 —— 2400%
答:损失2400元
(5)
x=a*+ 3*
(M) 93 x2-5={24 3$)-5
छत्र
5
2 11
=(21)+ 3× (2$7×(3†)+ 3x(2†)x(3†)+(3†)2 5 =2+3×21x3*+ 3×2t×3$43-5
3 × 25 × 3 £ x (2 * + 3 *}
— 3× (23×31)
22-5 23x 37=6 (7)
(4)化簡:
+
Хуа ay +
+ጌ
at
+
【原式
*
ха
80+2+
x4{a**+Q*)+
(xa* yaŸ3
x4
(xa*)2+2(xa*)(4a)+(yal)
(x+4)
4297 = x42*+ xy aa+ X*a*a*+y*a*a*
**** (2¥ Q** + **a*a*)+(x¥ a¥+g*e*a")
• xa*(ya*+xa¥)+4a3⁄4(xa!+ya*)
=(ya*+xa3)(xa2+ga")
(yakiza¥) (xe*+40}}
**** ** a log.
整數,故得
(1)
- - - - -(2)
B^=(a+1} ( ©−1) = ac+c-a-1=qc+c=(a+s
log from log (ac+1)
[xw #a]
2log &== log (1041)
(UMMA 2x2}**5x9*+ %** – 5*»«©
(213
(3*~ 2) (2***=1}·
x log 3mm logi
1 X**
1.2010
►
Abdomen
read
Eye
Ear Neck
Back
Wing
I
Rachis
Tail
Leg
A diagram showing the external feature of a pigeon
The ways in which a bird is adapted to
its flight my be considered under a number of headings:
(1) The body is stream-lined so that the air flow
moothly past as it flies,
(ii) The strong breast-bone or keel, to which
are attached powerful muscles which main ALS much of the bird's weight. iii)The body is covered with light feathers
instead of sokles; the bones are hollow, and these reduce the density of the body. 1v) The temperature of the blood is high
this presumably “bans a high rate of mergy raleisk, corresponding with the powerful fusolar action,
and
(v) The fore-limbs are modified into wings with festhere which are light and strong and constructed to faciliate flight. The small feathers are ÿery good insulators by virtus the layer of air which they entrap, and so they help to keep up body temperature. "(vi) The" joints of the skull are entirely lost anai
the face prolonged to form a beak, There is only one pag at the junation of the head so ating, so that the bird can turn its head almost completely round.
vil) The third eyelid plays an important part is protnotion of the arm whilst the bird is daa flight.
(144) The lungs áru well developā, ́and connected
with them AN ©air-suos which extend along the body contex and into some of the bones,
the body sam obtain a good supply, a and a large capacity for the
estinal region in relatively short
the intestins of minals. This svent too: moja waste substance
il the body, thus radia ing the
of excretory promueza ORE DE hiefly as urie soid which can be/ ted with little loss of water, This Ips, to reduɑn the body weight. \gi)_The_bird has lost the urinary bladder and
the ri
y and oviduct, possibly an
VORS-BBING of minimizing its weights,
{x}
VONSKONS
(tx 4)
(7)解联方程:云盖14的高一
(青)原石程式町化善
(2)× & −(3),
4-4-28! Gemin
Barb
-Troush in upper barbule
Barbula with nooks
·Barb
A simplified diagram to show interlocking
barbules
(2. (a) Describe the structure of a bird's »gg and
give the function performed by each part the
Ans.
you mention.
(b) Write a short account of the life history or
a lamed bird.
(a) The egg deposited by a female bird, is
enclosed by a hard covering, the shell. The shell, unde of chalk, is porous, allowing gases to pass through it. The oeutects, alon in a thin mecbrane, consists of a stere of thin food-solution, the albumen, and a slightl; chicker coloured yolk which contains a high proportion of fat. Hasting on the upper surface of the yolk is the living part of the *8ɛ, a flat germinal diso. The yolk is surrounded by the albumen. Through the albume are two twisting cards, each running from the yulk to the shell membrane, thus the yolk is beld in position. During the development of the Šembryo, both the yolk and the albussen: sarvo 48
the parts for food supplying, while the carmina) dise develops into the embryo,
Air chamber
Cord to hold yolk
Shell'
Shell membrane.
· Albumon”. -Garminat dis
·Yolk
Longitudinal section of a bird's egg
LIBRARIES
tập of the quill, wher
hole, the inferior
the central shaft is the feather, the vario. The valu
fios barbe emerging on azb baa omaller bárþuleo "At
the lower bärbulèn carry erleck with barbulss of the
the flat van_of_the
1
「答:两地相期 150%
(9)在一平面上,贯九條直線,用數學歸納法証明此策
直線五多将此平面分為古(n*+n+2)個區域.
(E) 設 D.表时分成區域之最大數目
(*) *** n=k
往乙地,若每小時多行5哩,用预定時 5oF, #1=67] ZOÉMI
线都相刻
响的溫
・予定時間、生地
(*+0=★(k+3k+4) \+1+&+/+2)=1[(k+1}+(k+1)+2]
(b) The_life history of spanyol de described as th
follows t
During «sting season, the mle bird sing all the time. The female and mala birds. goastrket their nest in a tree or close to our hourdes.
The eggs are fertilised in the body of the female bird and are deposited in their nest. Incubation takes about 12 days. The young birda are reared by the male. About a week latter, the young my leave their nest and follow the male shouting for food, whils the mother bird selects a suitable place to build a second nest. Finally, the young lear their marant and live independently.
questions for this weak: '
1. On a large scale make a fully labelled drawing to'
show the external features of a typical bony fish.,
How is the fish adapted to its mode of life i
respect of (a) locomotion, (b) respiration? ̧
即食+1時亦能成盍
42 dbtân -) ** mm), a ft & 174 & 6tɛ=) ▼ ko`rma, D 時亦能成立,如此類推,故此為任何正整數均可成立、 即是双條直线至多可将一界面分為士(2)&域
Q.E.D.
第五次預習題
(1)-RIA-STS LEMLEN SZ PLK 32, TER 距離為5呎求比索士長
(2)有两同心园,在韩国脚上谋奌與內閣上最近之與距離為6. 老包知内园圈長500呎,開两园周之差為若于吋?又笨外园时 畧5000 西岡之義・導干?
(3)說如右图:AT鴦0国切线,
Ž¢â¥* 4*, AT=10$. TP=Ja
試KAOP的度數.
(4)***$*:
( 14 sichu tan^8)( 1+ tan 6 sin20)mm( 1+ seep tot0) (1+tuĉasit()
(5) #L4 6A), ***
tan tan ♣
́(6) $.4 ABC 4, $XTE
(a)
tam A
tam:B
the carb) (tan A+ tam 8) = (*{ tanA – XanB). (a) (f ̃ c2) cst A+ (c2 a2) oot B+ (a+b) cot C=
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