1.)A cake is divided into four pieces in the ratio 2 : 3 : 4 : 6 by weight. If the cake weights 900 g, find the weight of the heaviest piece.
2.) The three sides of △ABC are in the ratio AB : BC : CA = 5 : 7 : 4.
If AB = 20 cm, find the permeter of △ABC.
中二數, 快!
2006-11-03 4:56 am
回答 (6)
2006-11-03 5:25 am
✔ 最佳答案
(1)Because the heaviest one is 6/(2+3+4+6) of the whole. Therefore the weight of the heaviest one is
900x6/(2+3+4+6)
=900x6/15
=360(kg)
(2)
AB:BC:CA=5:7:4 where AB=20cm
Therefore 5y=20
y=4
Therefore BC=4x7cm=28cm
CA=4x4cm=16cm
Therefore the perimeter is (28+16+20)cm=64cm
2006-11-03 5:34 am
1.)the weight of the heaviest piece
=900×6 / (2+3+4+6)
=900×6 / 15
=900×2 / 5
=360g
2.)Let y be the permeter of △ABC
y×5 / (5+7+4) = 20
y×5 / 16 = 20
y×5 = 20×16
y×5 = 320
y = 320 / 5
y = 64
Therefore the permeter(周界) of △ABC is 64cm.
=900×6 / (2+3+4+6)
=900×6 / 15
=900×2 / 5
=360g
2.)Let y be the permeter of △ABC
y×5 / (5+7+4) = 20
y×5 / 16 = 20
y×5 = 20×16
y×5 = 320
y = 320 / 5
y = 64
Therefore the permeter(周界) of △ABC is 64cm.
參考: The form two maths book
2006-11-03 5:01 am
1.) 900g x 6 / 2+3+4+6
= 900g x 6/15
= 360g
2.) Let X be the permeter △ABC
X x 5/5+7+4 = 20cm
X = 20cm x 16/5
X = 64cm
Therefore, the permeter of △ABC is 64cm
= 900g x 6/15
= 360g
2.) Let X be the permeter △ABC
X x 5/5+7+4 = 20cm
X = 20cm x 16/5
X = 64cm
Therefore, the permeter of △ABC is 64cm
2006-11-03 5:00 am
900*[6/(2+3+4+6)]=900*[6/15]=360g
20/5=4(1 unit)
there are total 5+7+4 units=15units, therefore permeter is 15*4=60 cm
20/5=4(1 unit)
there are total 5+7+4 units=15units, therefore permeter is 15*4=60 cm
2006-11-03 4:59 am
1.)120:180:240:360
2006-11-03 4:59 am
1.
900 x 6 / (2+3+4+6)
= 900 x 6 / 15
= 360 g
900 x 6 / (2+3+4+6)
= 900 x 6 / 15
= 360 g
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