氮二第張六第
日十三月十年申屣
WAH KIU YAT PO 報日僑
報日橋華四期星
四期星日九十月二十年八六九一曆公年七十五國民華中
育教備
三九六九年中文中學會考試題預習
依正弦定律
*6232
文中學會考試題預習
42
79865
759759
生物科
(七)
夢
1.207.2
(E)
鰲仲强
-
BC
ain 9th Fugite
BIOLOGY (7)
2,0081
數學科
第六次預習題解答
CD- BC sin 75′′ 17′′
42 sin 75*47 sin
sim 9°6′
10之(三位有效數)
(1)一人在塔用50呎處測塔頂之仰角第42又測塔恒上
答:塔高102呎
(5)在一塔之正東A奌,測得塔頂
之旗竿項,其仰角為540.求旗竿之長
(解)設如右图:BC為塔口為塔上旗竿,
A為测点,則在△ABC中
aten ar
BC= 50 Tam U17 mx 50 X 0.9009
KGAADCY
45.02.
DC AC Ian 56.
50×13764
-68.82
54
-DBDC-BC= 68.82-4603—23.8.
答:旗等長 23.8呎,
(2) 两园半径各為7吋,8吋,园心距離 10吋,求两园之交角,
(即由交奌者作两国之切线好成之角】
(解)設如右图:CD為 A园之
切线,CE為B目切线,則
ODLAC CEL BC.
(w #1 w‡ 12
LDCE=LACB
(西角之相当边彼此来直
則此两角相等或相補) 但在AACB中,依飾
COS LACE
物
13 LAC8= "83" 20"
答:西田之支用蓋 8320
3) LAABC+, A£IKŁÿ 2:3:4. †ainA, sin♬
及ainCż值(用根式表示之)
(M) 1 in A: sinë: sim ÇEL:
tabc23:4
1* r* ***
Las Am
同理可得
塔(A-180°故神国正号)
(4) PELA ABC 16 TAX A To to BCL *C, X SE
(ar) (@X «C BALA (ko k >10])
THE CD. AD,
D= LACB (3 to A)
AADC 4, ZDAC=ML
《半国周角笃直角):
inD
(4) A . C $41 (*«* * {B})
4112 CD, AD, 9! «DAC=90°
(半國國周角為宜高)
R&D=LACE = 180°-LACE:
AC
D
角為31在増※ 47° 84μ]z 9] 19A $ 25. 15 to $200 of 求:AB 距離
(解)設如右图,CD蔫塔高
ACB 90°= 2)*=== 49°
10 ACP
AC=200 cot: 31—200 × 16 6 4 3
-332.86
XA B C D
BC 200 cot 25” — 2008 21445=428.9.
在水平面之△ABC中,依飾弦定律
ABI AC+ BC2 – ZACY BC cos 47°
=110800+183900-187,4001
=107.300
ABS (107.300-32F
答:距離328呎
NO.
2
200 0.3010
·332.9 2·6223 428.9 2.63 zu
Cos 49" 7.81696
(話)本題計算時鎖利用東方表
平方报表及对數表等,
1874x16 5.2726
B.其 (7)在475呎高之崖上C耎,測得地面上两下
俯角季別 103214054 第2ACB-56013.求 AB 距離!!
(4)DACD+
ACCD cac 10°32′
===475 cac 10°32′
2578
X $ 4 BCD 4.
BCHS CD CAL 14° 54′′
by cac 14° 54′.
1847
Te s ABC &
"AB2= AC2+BC-ZACK BC CON L'ACH =2598 + 18472
2× 25 9 8 × 1847 cos 56°13′
6749,000+3412.000-5338.000
4823000
AB
F4, 823.000
=2196 (≥61NBT2200) 答:AB距離2200呎
(8)設如右图:0因為单位园 一日為第二象限角戲正
(a) tan &=
(c) cost
(SE) LEPAK PM LAB, §]
sing- CP — MP (SEK)
OP
联BP,則-B
OM (2141)
LOG 2.676.7
"Cecƒ0°12′ 0.73806
3.41.47
1847
6767 0.53981 42665
0.30/0 2598 3.6147 18473,abbt
cm 56713. 79451, $33.9000 6.7273
(因周角等於同孤园心角之
¶ AP §}<MPA=48=£ (H£<BPMGQ HA) tân & MPAMA
MP.
MOTOA=~ON+OA=
#A APM
EB MA
代入式得
ten &
kim
BAAPM
AM CHPAS
MA
·B01 KB BM
Answer to the questions of last week.
1. (a) What do you understand by aerobic and
anaerobic respiration? Compare as far as you can, the differences between respiration and photosynthesis.
(b) by what experiment would you show that heat 19
given off during the process of respiration?
(a) nerobic respiration is an oxidation process
which occurs in the presence of free oxygeti, During this process, the intake oxygen combines with the carbon of the carbohydrate, so that the products of the process are carbon dioxide and water. In other words, there is an entire breakdown of the carbohydrate and complete liberation of energy. The process can be summerized as the following equation:-
• 6C02 • 0820 • £œergy
In anaerobic respiration, it occurs only
in the absence of a free oxygen supply. So tha the carbohydrate is broken down incompletely with only partial Liberation of energy.
The by-products of the process are carbon 110xide and ethyl alcohol. The precess of anaerobic respiration is represented by an equation as:
→20250 2002 Energy
The following table is a comparison of the differences between respiration and photosynthesis.
Hespiration
1. It takes placa in all
living cells of plants. and animals.
2. Material necessary for the process are oxygep "and" organic compounds. 3. The chemical energy
stored in the organic compounds are converted into kinetic energy.
4. The process occurs in
both day and night.
5. It takes in oxygen and
gives out carbon dioxide.
6. The products are carbon
fioxide and water,
The process results in
decrease of matters
the body.
play
لعقم
Boiled pea Sonds
Photosynthesis
It can only take place in the chlorophyll containing cells of plants.
Material necessary, for the process are water ans carbon dioxide.
The light energy is stored
in the organic compounds as in the form of the chemical energy.
The process occurs in daytime only.
It takes in carbon dioxide and gives out. oxygen.
The products are oxygen land carbohydrates.
The process results in an increase of na in the body."
Respiration generates heat
p
Rainc
串)
(5)在42呎高之屋頂上,将一塔夏之仰角萬 14° 13.
若在此厘之下面測止,则仰角為25*19求塔高
(解)詖如右图: AB馬屋 CD笃塔
# CE=XR BD xxy of
-==cot 144*1.3°
*9.9471%
XEA CODE,
مورد
2.9202% + 97, 4484
it #3 (0),(2) 18 8.947/x=2:3202x499,6684 #24.4-1
1-62692 97. 4484
$99+42=10!,? (=12****&102)
(MIM=) BAABC), LABCH
狸理 BM≈ 80¬MO — 80+OM=1+0018 代入得
RÍO ASP¶, & LAPB-st.
AP
BA
cos &
(1).(3)相乐
第七次預習題
*888131})
·(4)
()利用餘數定理,求2x344元 5x+ 6 14 % x−3 <HE (2) 3§ f(x)==x+AX
UXX+1 PKZ A 10, ka b± 11 (3) 31 ax2+bx2+ | Shi (X-1)2 =
74, #abz 11.
(4) 利用餘數定理,分解下列為因式:一
(b) X-10x+157-6
(5)解下到高次方程式:
(a) (X-1) (x-3) (X−4)(x−6)+8=0.
(8) X"+3x3-2x+3x+1=0.
(6)解聯立方程式
ras {
PIBUS BOISE QUELOS-W001 in a tinrMoc flask and cover it with some germinating pea Seeds. A thermometer is introduced. Inte the flask and is fixed in place with a wad of sotton-wool in the neck. Set up a control with pea seeds killed by boiling water or Foaking for a few hours in a solution of 0.15 mercuric calorida.
After 2-3 days the temperature inside the erimental fläsk bad increased, whereas the temperature inside the control flask remained incanged.
HOW WOLLUS YOu aistinguish between pollination and fertilisation in a flower? Describe ià a table form, the changes occur in a flower after the process of fertilization.
(b) What are the characteristics or an insect
pollinated flower? Mama ONE Tlover which is normally pollinated by inasets and carefully
Moribed the mohandas of its pollination.
Follination is the process' or 'transreRENAEL CRIN V IZIET pollen grains from the anthara of ónei flöwan to the stigma of another flovanj ofiska i dubine species. If the two flowers are borne on tele same individual plant, self-pollinationidai mas ko bave occurred. But, in the crode
laxmxy
【未完脚入第六第三眞)