中二升中三數學題~
2007-08-18 11:04 pm
一輛私家車在正午十二時離開學校。這輛私家車以56km/h的速度向正東方行駛。與此同時，一輛貸車離開同一間學校。這貸車以42km/h速度向正北方行駛。問在下午三時這兩輛車相距多遠?
回答 (4)
2007-08-18 11:10 pm
✔ 最佳答案
(42km X 3)2次方 + (56km x3)2次方 = 開方(兩輛車相距)兩輛車相距 = 開方(15876+28224)
= 210km
2007-08-18 11:34 pm
這輛私家車下午三時與學校距離
56x3=168km
貸車下午三時與學校距離
42x3=126km
設兩車相距為y
y^2=168^2+126^2
y^2=28224+15876
y^2=44100
y=√44100
y=210km
56x3=168km
貸車下午三時與學校距離
42x3=126km
設兩車相距為y
y^2=168^2+126^2
y^2=28224+15876
y^2=44100
y=√44100
y=210km
2007-08-18 11:19 pm
let k be the distance between two car
k^2 = (56x3)^2 + (42x3)^2
k^2 = 28224 + 15876
k = 210km
k^2 = (56x3)^2 + (42x3)^2
k^2 = 28224 + 15876
k = 210km
2007-08-18 11:13 pm
Let A be the location of the school
Let AB be the distance of the car at 3 pm
Let AC be the distance of the lorry at 3 pm
Let BC be the distance between the car and the lorry at 3 pm
AB = 56 * 3 = 168 km
AC = 42 * 3 = 126 km
Since ABC is a right-angled triange with 90 degree at angle A,
the distance of the car and the lorry at 3 pm will be
BC = sqrt ( AB**2 + AC**2)
= sqrt (168**2 + 126**2)
= 210 km
Let AB be the distance of the car at 3 pm
Let AC be the distance of the lorry at 3 pm
Let BC be the distance between the car and the lorry at 3 pm
AB = 56 * 3 = 168 km
AC = 42 * 3 = 126 km
Since ABC is a right-angled triange with 90 degree at angle A,
the distance of the car and the lorry at 3 pm will be
BC = sqrt ( AB**2 + AC**2)
= sqrt (168**2 + 126**2)
= 210 km
參考: Myself
收錄日期: 2021-04-25 21:56:02
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https://hk.answers.yahoo.com/question/index?qid=20070818000051KK02596