真二張四第日七初月四年尾巴歷室 WAH KIU YAT RO
日橋
(1)-2),
(p-q)a=y-
僑
5章橋
(8 p+q+h+o (A p q r m 1⁄21**
(p+q+r) ( p + q2+ ri=pq-gr−rp) =
+82 + r2 = 3pg r=0
四期星 日二廿月五年九六九一履公年八十五國民蔬中查教保藥
1969 EUR
(1)-(3),
將此两式相余
(Z - A) d
生物科
(4-3)(1-4)
***
$112,
prp - gr + pq -gr- rp
=3 4 gr
(二十九) ·益仲强·
(7) %% a+b+c=o, sta a+b 其逆定理成立否?試証明之. (32) A
a+b+c=
at
第廿八次預習題解答
白菜
a+2ab+
種項
a+b-c
zab
(1)某人步行,若每時速度較豫定速度多行一里,刻果 到李小時,若速度較豫定速度少行支里,則遲到 -chat. The 11 31 7 14 2 4
(解)設豫定速度每時X里,時間掣小時,则两地距
MI x4 = 1*11 £1}} * = xy = (x+1)( y − ) 1⁄2) -----(1)
= (x =) ( 4 + 1)--
(4)×2+(3)
5-24-1-
J
11
移項化簡
ab-abc-20a=404
(179) % abtrct
ca)
(a^+2ab+b")-(2c^a+&b2°c)+c2 4ab (a+b)=2023⁄4(a+b2)+(c^) — 4ab-o
(a+b2 (2)2 (zab)2 =
Q.E.
・油(株式
x+y==
由(2)式
x-ty-
(a+b-c2+208) (a+b=c-20f) ==
[(a+b)=c] [(a−b) c2]=0
#x(4),
(a+b+c)(a+b=c) (a−b+c) (a-b-c)=0
答:两地距離: 45 (一5×9)里豫定速度5$/時
向而行4 两地相距40公里甲乙二 甲行一公里需時間較乙行一公里特需時間少8分鐘 求二人速度...
We a+b+cmo atk-c
ot a-b+c=O
a-bu 口的能適合上式 可以逆定理不能成立 (8) (a) this y
G.E.D.
(解)詨甲每時行x公里,則行一公里需時2分鐘, 又詖乙每時行导公里,行一公里则需日
依題意 5x+5y=40
469x21 (276110−32+3
從横軸惧以一吋表一罩住) (台)利用此图像,以求方程式我 (解)先製成下表:y=
鐘
B
15y=axy (2)
2018 4-8-x,
15x−15 (8-x)=
)====2x (8-X)
分解
15x-120+15x=16x-2x2
X47X-60=o
(x-5)(x+2)=0
0-90278.
(廿九)
夢百鑒·
BIOLOGY (29)
(Answer:to tas questions of last week
What is
- Prastoolies arberan, luerrate the
external feature of a Named plant parasite. How is this parasite adapted to its mode of life? -
A parasite is an organism walon lives in or on other living organisms on which it is dependent and to which it usually does no harm. For its aurvival and life change partially or completely have to rely on the environment found within the body of that living organism,
In serious forms of parasitism it may damage tak body of the living organisa which is called the host while in milder forms of parasitism the damage may be merely that it occupies room on the host's body, but more usually it deprivas its host of food or actually feeds on the body tissue of the host, in addition to giving off poisonous metabolic by-products,
(b) Dodder is a plant parasite found on a variety of green plants such as Clover, Nettis, and Gorse. Its thread-like pink stems twine round the stem of its host. At frequent intervale haustoria penetrate the host a tissues and enable the parasite to obtain supplies of food and water. Although the pink flowers are so Cemall, but each one has sepals, petals, five
stamens and an ovary formed from the banes of two parpela. The early life of thể doddar ís normal, "germination resulting in the product 103 of a
root and a shoot, but, as soon as the shoot contanta the stem of the host plant and penstrates it with haustoria, the root dieš
Scale leat Haustoria
x=5 & X=-12 (742)
代入)将y=3.
答:甲每時5公里 乙哥時3公里
'3) 某人秉草車旅行24哩之路程,若每時速率多行,
2哩,则可早到一小時,求其原来速度。
(解)設原来速度每時x哩,则」
1-3 0
湯
4.3.2
10 4
3
hafi
24(x+2)-24x==
(X−6)(x+8)=0
x=6x=-8 (7239)
冬:原来速度毎時6月
) * 12 500 På ha 7 6 3 9 7 3 -WT&<%#
(解)至500間能被夕好整除的为6.14.18.
**¥*u* amb amb
代入
= a+ (n-1) d
499=== 6+ (n-1)x 6,
n(a+l) _ 83(6+498).
RV 5=9+18+27+
2498.
然後指出备兵,依次連,得图像如图将示。
* 4 ay-yx=36-4x
4x2 - yx - (36-93)=0.
14.
-Clusters of flowers
Stem of Dodder
Stem of Host plant
External Feature OF
Vascular
·Dodder
tisene
of
Dodder
Stem of Dodder Haustorium.
Vascular
bundle of host stem
A portion section (T.S.) of stems of Dodder and
its host
The dodder is adapted to its special mode of life by possessing the following important characteristics:-
(1) There is a reduction or xylen în ite stam,
bat this is no diadvantage, since it has not the ordinary full functions to carry. out.
ii) The eten is thread-like and twines round:
the stem of the hout, so that it is easily supported by the host,
20916..
495
195
9+ (n-1)x7 55x(9+495)
€860)
1-1, 1-1. 典求解方程式 3x-9=0 ## £1 X y=3
触符合,放巧加绘之直线為= 【園中交奌之橫坐標即為將求之根
久同時能被6及9除尽的
≈ 18+36+54 +
486 = 18+ (n-1) x 18,
-486
求证
27 × (18+486) 6804.3
S+ S = S, 20916+13860-6804-2747-
£4.3.27972.
(5) $1 x+4=4+3 = 4 * $z z + + = 1.
(10).& x+7=
依等量傳遞律些
又+!!
==
去分母
- 1) === z − 1 − x 3 + X
除以x
3+==== 1
Y Q.E.D.
(6)某等差根数え節が第8項等え項分別馬
q r s p. Ex š p2+z1+V=34gr.
(出正詖此等差級數之首項為
a+ (p-1)d=j at (8-1) dar 2+(-1)4=
(1):
Cy
(E)
答:x-1.20或1.95(至小數两位)
第廿九次預習題
(1) ABCDES AB P2 12 AP= AC, XPD,
AMF AN LPD, 1821 N. PN=2ND.
(2)過國內一定奌到任意两垂直,試證其平方和為
一定量
(3) AB, CD马团O内两平行弦,M马CD中美,BM之延
* BATE; X AE OMIZE £**?.or PAXPE=PO×PM.
(4)柃 △ABC之两边ABAC,各
形外作正方形ABDE
RACFG. 11 84=CE £27*#1
(5) 3£ § 0 ¥12 0A ZB /R AB— OA. % M BOATS,
CLB +41 32CM-CE.
(6) ABCD 17ŒLT; UK AB, AD 3 fo #3 3217
** ABHK, ADPQ. @QK=AC.
(7) ABC為已知奌;求以A,B,C為國心分别作三 自使A,B两国外切,而C园则均與A,B国内切。 (8)A,B两园外切於PC图则典A园内切於Q,而
★BE AWKR, <BAC=2<PRQ. (9)任意四边形ABCD之对角线AC平分BAD, A LACD=LABC, # BC = 45 & E CD45 $ F. ØJ
A, E, C, F..
(10) 0.PL to TE A. to AOB 1.01 ₤17 LB ** P** th£$R HSE <A@B=45°-£«ABQ.
(iii) Germination of the seed takes place in the
lats spring when the host plant has already developed.
(iv) The ovules and embryos are very simple, (v) The intervals haustoria penetrate the
vascular tissues of the host and enable it to obtain supplies of food and water. (vi) Numerous pink flowers result in the
production of numerous seeds. These compens➡ ate for the great loss during germination. (vii) Roots, and green leaves reduce, Since it
obtains all its food supply from the host.. viii) The seeds are relatively large and their
food-content is thus appreciable. This is important to ensure the survival of the Beadling during the fros living stage.
.:.
[2.(a) What is a saprophyte? 12.
What are the similarities between the parasitia and saprophytio modes of life? (c) Why are saprophytes important in natura? ¡ and.
(a) Saprophyte is a vegetable organism living ung
· dead plants and animals on which it is dependent and to which enzymes of it are secreted to digest the nutriment outside the "plant and animal. The digested and dissolved. nutrient, is then absorbad through and over the whole surface..
未完轉入第四張第三)