candiceyan

Q1.Area A of Japan is at the temperature of –3 °C, area B is 3 times as cold as
area A. Finally, area C is 9 °C warmer than area A.
(a)What are the temperatures of areas B and C?
(b)What is the average temperature of all the areas?


Q2. B10. Find the equation of the perpendicular bisector of the line segment joining
A(–5, 6) and B(9, –12).

Q3 A rectangular container is made up by cutting out squares of side x cm from
each corner of a piece of rectangular cardboard measuring 16 cm by 12 cm.

(a) If the area of the base of the completed box is 96 cm^2.write down an
equation in terms of x and show that it reduces to x^2 – 14x + 24 = 0.
(b) Solve the equation in part (a).      附圖了~

Q4. Refer to the figure below; given that 角CDB = 角ABD = 角AC'C = 90°, AB and
CD are of heights 45 m and 30 m respectively. The angle of elevation of A
from D is 30°.

(a) Find the length BD.
(b) Find the angle of elevation of A from C. 附圖了~






[ 本帖最後由 candiceyan 於 2010-5-17 08:36 PM 編輯 ]

felixtangc

Q1.Area A of Japan is at the temperature of –3 °C, area B is 3 times as cold as
area A. Finally, area C is 9 °C warmer than area A.
(a)What are the temperatures of areas B and C?
--->B: -3 x 3 = -9°C   C: -3+9 = 6°C

(b)What is the average temperature of all the areas?
--->Avg temp = [(-3) + (-9) + 6] / 3 = -2°C

有o羊唔明?計數? @@ (希望我冇理解錯,,,我十世未做過初中題目 )

[ 本帖最後由 felixtangc 於 2010-5-17 08:37 PM 編輯 ]

candiceyan

其他有邊個可以解答到我~

夜櫻

Q2. Mid-pt of AB=(2,-3)
     slope of AB=9/-7
    Required equation is  (y+3)/(x-2)=7/9
                                 7x-9y-41=0

hamburger

I will use(') as degree
a) tan 30' =45/DB
          DB =45/(tan30')
          DB=77.94228634

b) AC' = 45-30=15
    CC'=DB=77.94228634
    Let @ be the angle of elevation,
    tan@'= AC'/CC'
    tan@'=15/77.94228634
        @'=tan-1(15/77.94228634)
        @'=10.89339465

Hope this can help you.

a07222

a) the temperatures of areas B and C are -9°C and 6°c
b) the average temperature of all the areas is -2°c

a) x=4.898979486

a) the length BD is 77.9(corr. to.3 sig. fig.)
    [by using tan@]
b) the angle of elevation of A from C is 10.9(corr. to.3 sig. fig.)
    [tan@=15/BD]

peterp2509

Q3 A rectangular container is made up by cutting out squares of side x cm from
each corner of a piece of rectangular cardboard measuring 16 cm by 12 cm.

(a) If the area of the base of the completed box is 96 cm^2.write down an
equation in terms of x and show that it reduces to x^2 – 14x + 24 = 0.
(b) Solve the equation in part (a).

a.) area of the base of the completed box = (16-2x)(12-2x)=96
(16-2x)(12-2x)=96
192-56x+4x^2=96
x^2-14x+24=0

b)
solving x^2-14x+24=0
(x-12)(x-2)=0
x=12 or x=2
since 16-2x>0 and 12-2x>0,
so x<6 and x=2 is the only solution of x