2條中三數學 (20點!)

1.
(a)
the volume of water poured
=the volume of the right prism with the base 15cm x 12cm and the height 5cm x2=10cm
-the volume of the ball
=15cm x 12cm x 10cm - (4*pi/3)*[(5 cm)^3]
=1276.401224 cm^3 (corr. to 6 decimal places)
=1280 cm^3 (corr. to 3 sig. fig.) (b)
the volume of 10 balls added
=10*(4*pi/3)*{[(2.4 cm)/2]^3}
=23.04*pi cm^3 let the length of the rise of water level be h cm.(h cm)*[(15 cm)*(12 cm)]=23.04*pi cm^3
h*15*12=23.04*pi
h=0.402123859 (corr. to 9 decimal places)
h=0.402 (corr. to 3 sig. fig.) the water level rises 0.402 cm2.
let the original number of raffle tickets be a.
let the number of prizes be b.
So,b/a=1/30
b=a/30 if 240 more tickets are added without prizes,
then the number of raffle tickets is (a+240)
the number of prizes is still b.b/(a+240)=1/90
b=(a+240)/90
a/30=(a+240)/90
(a/30)*90=[(a+240)/90]*90
3a=a+240
2a=240
a=120the original number of raffle tickets is 120

第一題要cover個metal ball既長方體既volume係15x12x10, 減去個metal ball就可以得到所需既水既volume

1a)

(15x12x10) - (4x兀x5^3)/3

=1800 - 523.59877

=1280 cm^3 (cor. to 3 sig fig)

1.2係變左做radius既數

1b)

{[(4x兀x1.2^3)/3]x10} / (15x12)

=72.38229 / 180

=0.402 cm (cor. to 3 sig fig)

第二題我會用linar eauation (我唔知有無更好既方法)

2)

let x be the number of ticket that can get a prize
let y be the number of ticket that can not get a prize

x/y = 1/30 ------------------ 1
x/(y+240) = 1/90 --------- 2

1. x/y = 1/30
x = y/30

sub 1 into 2

(y/30) / (y+240) = 1/90
(y/30) x [1/(y+240)] = 1/90
y/(30y+7200) = 1 /90 <-- 呢度交叉相乘到下一步
90y = 30y+7200
y(90-30)=7200 <-- 抽common factor
60y=7200
y=120

之後d step你識做