1.(x + 10)(3 - 4/x)= 12
2. The sum of two numbers is 15.
a.if one of the numbers is x, find the other number in terms of x.
b.Given that the product of the two numbers is 56,find the two numbers
3.if the speed of a car increased by 15km/h in a journey of 240km , 80 minutes would have been saved. Find the original speed of the car.
中四數學題3題
2007-10-10 1:11 am
回答 (3)
2007-10-10 1:50 am
✔ 最佳答案
1. 3X+26-40/X=123X+14-40/X=0
multiply both sides by X
3X(X) +14X-40=0
factorise
(3X+20)(X-2)=0
X= -20/3 or 2
2. (a) (15-X)
(b) X(15-X)=56
X(X) -15X+56=0
factorise
(X-7)(X-8)
X=7 or X=8
3. Let X be the original speed, and 80 minutes =4/3 hours
240/X - 240/(X+15) = 4/3
{240(X+15) - 240X} /(X+15)X = 4/3
multply both sides by (X+15)X
(XX + 15X)4/3 = 240*15
X(X) + 15X -2700 =0
factorise
(X+60) (X-45)=0
X= -60 or +45
as the speed of the car can only by positive number,
the original speed is 45 km/h
2007-10-10 1:57 am
1.(x + 10)(3 - 4/x)= 12
3x-4+30-40/x=12
3x-40/x= -14
3x^2+14x-40=0
x=2or-20/3 ,,
2a.The other number is
=15-x,,
2b.Let the other no. be y.
15x=56-----(i)
x+y=15-----(ii)
From(ii) x+y=15
x=15-y-----(iii)
Sub.(iii)into(i)15(15-y)=56
225-15y=56
169=15y
y=169/15,,
Sub.y=169/15into(ii)x+(169/15)=15
x=56/15,,
3.計唔到
3x-4+30-40/x=12
3x-40/x= -14
3x^2+14x-40=0
x=2or-20/3 ,,
2a.The other number is
=15-x,,
2b.Let the other no. be y.
15x=56-----(i)
x+y=15-----(ii)
From(ii) x+y=15
x=15-y-----(iii)
Sub.(iii)into(i)15(15-y)=56
225-15y=56
169=15y
y=169/15,,
Sub.y=169/15into(ii)x+(169/15)=15
x=56/15,,
3.計唔到
參考: 我都係f.4
2007-10-10 1:29 am
1.X=2 or -20/3
2a 唔知
b x+y=15----------(1)
xy=56---------------(2)
when x=7,then y=8
when x=8,then y=7
3. (240/3)-(240/x+15)=(80/60)
x=45 or -60(rej.)
應該岩,有錯唔好怪我
2a 唔知
b x+y=15----------(1)
xy=56---------------(2)
when x=7,then y=8
when x=8,then y=7
3. (240/3)-(240/x+15)=(80/60)
x=45 or -60(rej.)
應該岩,有錯唔好怪我
收錄日期: 2021-04-15 22:42:38
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