WAH KIU YAT PO
郭日僑華
OMEGA
頁二第張六第日三十月正年巳丁歷
|
[]|晚上七時半及九時半,
ROKFES-TEC KRZE-ESELI FREE-
HE PIPS }該團將於利享受。如欲欣賞團之 DY KNIGHT AND THER-BABER 於11月八日一星期二.」「遊戲感冒耿之人出。 | 帶來姬熱合唱團 GLA 荜演出,親樂都可以案。 爲香港愛好音樂的朋友都供水
·泉及娛樂有限公司 音樂界健十八年,內裝
華星娛樂有限公司
辦婦女節的新貢獻
BKG-ZIKU
FR - K+RAC+R | L-200-BESC
開價則分爲:廿元、四
正「四種,姬絲合唱團名盛。
As shown in the above figure,
1977中學會考試題預習專欄
明德社主鳊
40, 30 are the radii of the
circle
= 60
新數學(十七)·魯榮家
MODERN MATHEMATICS (17)
Answers to Exercise 7
Section A
1. Let the roots of the equa-
tion bed and ß.
since, angle
area of sector LOB
60 xx (70)2 cm2
360
2.
2566.b6cm"
Area of quad, OBTA
- 2 x ares of rt.A ATO
- 2 x x OAXAT
**
三期星
亞陸
米國
陸國鎮榮獲 歐!
特榮
獎獲
日二月三年七七九一公年六十六國民華中 育教聞港
事局主席黃合攝於酒會上市 FLEX- *BE/PRESKEPAGES 八蛋由左起:信昌集團執行董事樂財
KERK 專爲觀察森那美附屬機構伯昌集團屬下各公司之 招待本港工商界及財經界人士。陳氏此行來黹乃 主席嫌售假於日前假座香港希爾頓酒店裝置酒幸 (EKEREEKAKZOTKAFE.
會見港財經界政要
森那美董事局主席
本港新聞
*鍋三百七十九分記錄,榮獲米茄表
表一只。 安天時举行在該場館設立之韓別獎-獲頒重米茄
·示:米茄代孟梁士,且利眞頒予
a = 1 (Ana,)
substitute into (3) and (1)
Band. C - 3 (Ans.)
Section B
9. Since m
the roots
the equation
にx
1.e. 25x2 + 30x
2. Solution:
Since f(x)
"
·1.20
2x3-kx2+6x
3k is exactly divisible by x+2
£ (-2)=2(~2)3 -k(-2)2+ 6(-2)-3x
=0.
28
-7k
--4.
3. Solution:
tanix+cotx=2.
tanx +
tank
2
min
tan2x 2tanx 1
282
+
(161) 2
cm
32.5 cm
M
36CM Ń K.
DA. x, AT
Since one of the ro ceeds the other by
PX-
A OT
AT
DA tan30
'the qua
2k-
(2k
k
2k
(d~ß)?
(2).
(α-p)2 - rdp
4(2k.
Aera of quad. OBTA
DA = HA – tan 30 (cm)
70 x 70 x 0,5774
2899.26 cm
area of the shaded region.
2 (2829.26 2566.66) cm
k
or k
(Ans
262.6 em
22
5.
y-
(Ans.)
- 2mm
(-2)2 - 2 (?)
7 .-3
(ii). (m.
-(-2)2.
-10=4 101
n) (m - n) 10 i)
(Ans:)
(tanx
tanx
1
x = 45° or 225°.
Solution: Let AB-BC-CD- 2x units.
Area of semicircle AD = —m(3x)2 sq. units
9
2
TX2 sq. uni
Area of semicircle
AC = 1TM(2x)2 sɑ. units
2
=2x sh, units Area of semicircle
(y-3)2.
let the equation
2m+n and 2nim be
with roots:
AB
Sun ar
3(min) 3(-2)
Above, fig.
the radius of the circle
be rem. Since triangle ABC, is
equilateral, therefore HO is the angular Disector of LABC
and AD is the perpendicular bisector of pl
Hence B =
multiply both sides by
·2( y − 3)2 ≤ 4(y − 3) (y = 3)2 - 2(y• 3) <o
3)( y − 5)≤0
3< y < 5
solution set = {y : 36y65}
[
Note it is wrong
multi
ply both sides by y for we don't know whether
(y-3) is positive or not.
2y) b. ( x-3y +5) + k{x
(1+ k)x+(2k 3)x the slope is
fom
but
BC
2re os 30°
COB 300
300
= 2 × 3.464 (Ans.)
therefore, the radius of the
circle is 1,464en.
3. Let the numbers be
A and ar
their sum is 39
*A+AT" "39.
their product is 729.
(+) (a) (ar) -.729
from (2)
Я
-729
(2)
3729-9
substitute into (1)
+
3r3
(r
for r
9
9
yr
*.39
+
- 39x
-10r
3)(3r
3 or r
the numbers are
. 9, 9(3)
3 = 0
1)
ile. 3, 9, 27 (Aus.)
.1/3
the numbers are
903)
27, 9 and 3 (Ans.)
-(2+)
(2k
since it is parallel to
swww
product of roots
(2n
4 m) (2m + n) 2(m2+
+5mn According to the
result of (
2
also ma
7
(2m+n) (2n+m)
? (-3)
k) = (2k -3) - 4 (Ans.)
therefore
the required
equation
6х
To be cont
* 12x + 23 = 0
next week)
(Ans
As shown in the graph above, the optimal solution required is the point (7,10).
the maximum value required
3(7) + (10)
+ 50 281 (Ans.)
*
8. Since
5x^~1 = A(x2+x+2} + {8x+C; ) { x=1).
• Ax ̃+Ax+2A
5x
1
2
+
Bx +Cx -DI
C
2
(A
4
B. + C)x
+ 21
.. A + B = 5
數學(十七)·女長波·
Mathematics 17
Solution to exercise
Section A
1. Solution: Since and q
are the roots of
5x2 + 2x - 4 = 0.
D + Q. =
-
and Do The sum of the roots of
the required equation is. (2p+a)+(p+26) 3(D+q)
3(-)
= 3(-
=
The product of the roots of of the required equation is (2D+n)(D+2a)= 202+500+202
2
So. units
Area of shaded part
= (2x2 - mx2) sq. units
을까? sa. units
Since 2x2
Of
Area of shaded part.
is one third of the area of the whole figure.
Solution="
The total distance travel- led by the motorist
(2X32+ — x 42)Km
= 85 Km
the average speed of the motorist for the whole 85.
journey
=
1+ 3/2
Km Der hour
56 km per hour
Ans: The average speed of the motorist for the whole
journey is 56 km/hour.
6. Solution: The external
volume of the box =25 X 14 X 8cu.cm
2800 cu.cm
The internal volume of the box
= (25~1)X(14-1)x(8−0.5).
cu.cm
2340 cu.cm.
The volume of wood (2800-2340)cu.cm.
= 460 cu.cm.
Ans: The amount of wood
used to make the open box is 460 cu.cm.
7. Given: ACLBD AK=20cm,
KC=36cm, AB=25cm
To find the length of
KB,
KD and the radius of the circle
Solution: Since ACIED
KB =
$252-202
= 15cm
.AK. KC = BK.KD
20 X 36 = 15KD
cm
KD = 48 cm
From 0, the centre of the circley drav OMIBD and ON LAC
BM MD = X(15+48) cm
31cm.
KM -
=
(312-15) cm 162cm
ON KM is a rectangle
A
- B+ C
(2)
2A
C-
1
(3)
5.
(4)
x2 +
6.
0x - 글
O
2(D+C)2 + 21-룩)24 (-1)
+ Da
=
12
25
ON=KM = 161c.
Thus, the required equa- tion is
CN=NA
=
1X(36+20)cm
the 28cm
radius of the
circle
Toom
(1) (2)
(3)
+
(4)
LA
B
Given: AB is the common tangent of the circles. To Eind: the size of LAPB+LAQB
Solution: join rQ Since AB is the common tangent of the circles
LAFQ = LAQB
205B = 2CBA.
i.€. LAB = LQAB + QBA Since AQB+¿QAB+ LQBA = 180°
LAB LAQB 180°. Section B
9. Solution: Let x be the number of $1 coins,"
Then 2x is the number of 50g coins and 10x is the number of 10g coins.
210 < x+2x+10x < 230 21013x < 230
230
<x<
1.3
<x<17-
I.e. 15-2<
Since x must be an integer
X = 17
The total number of coins
13 X 17 = 221
10. Solation:
the least value of y = -1 the range of values of x such that y>0 is x <0.5 or x >1.5
c) Since the graph y=nx2+ax
+3 Dass through the points (3,0) and (1,0)
= 20+ 10+31
+ 24 + 12 = -(1).
+ 3:
1.E. 90+60+12=0.
(1)X3 3D+6c+36=0 8.
(2)-(3) 6-24 = 0 Subst.
D= 4
4 into (1) 4+20+12 Q
C = 8
d) Since the given graph is
..
4x + 1 = 0
8x +3 = 1.
The solution of
4x2
-8x + 3
and
2x2 or 4x2
2x2
·4x + 1 0 are 0.29
or 1.7
4x2 4x2
9x + 4 = 0
8x + 3 =
X
or
3
The solution of
4x2 - 9x + 4 = O are 0.62 or 1.64.
No comments yet.
Private notes are available after approval.