Sabendo que os números são diretamente proporcionais, resolva:A)x/3=10/6, B)x/4=91\52, C)x/7=18/21, D)x/105=2/30, E)x/8=36/32, F)x/3=128/48, G)x/3=35/15, H)x/112=3/42, i)x/2=18/12, J)x/6=49/42, M)x/90=8/5
Respostas
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A) X/3 = 10/6
PRODUTO DOS MEIOS = PRODUTO DOS EXTREMOS
6*X = 3*10
X = 30/6
X = 5
B)X/4 = 91/52
X*52 = 4*91
X = 364/52
X = 7
C) X/7 = 18/21
X*21 = 7*18
X = 126/21
X = 6
D)X/105 = 2/30
X*30 = 105*2
X = 210/30
X = 7
E)X/8 = 36/32
X*32 = 8*36
X = 288/32
X = 9
F)X/3 = 128/48
X*48 = 3*128
X = 384/48
X = 8
G) X/3 = 35/15
X*15 = 3*35
X = 105/15
X = 7
H) X/112 = 3/42
X*42 = 112*3
X = 336/42
X = 8
I)X/2 = 18/12
X*12 = 2*18
X = 36/12
X = 3
J) X/6 = 49/42
X*42 = 6*49
X = 294/42
X = 7
M)X/90 = 8/5
X*5 = 90*8
X = 720/5
X = 144
PRODUTO DOS MEIOS = PRODUTO DOS EXTREMOS
6*X = 3*10
X = 30/6
X = 5
B)X/4 = 91/52
X*52 = 4*91
X = 364/52
X = 7
C) X/7 = 18/21
X*21 = 7*18
X = 126/21
X = 6
D)X/105 = 2/30
X*30 = 105*2
X = 210/30
X = 7
E)X/8 = 36/32
X*32 = 8*36
X = 288/32
X = 9
F)X/3 = 128/48
X*48 = 3*128
X = 384/48
X = 8
G) X/3 = 35/15
X*15 = 3*35
X = 105/15
X = 7
H) X/112 = 3/42
X*42 = 112*3
X = 336/42
X = 8
I)X/2 = 18/12
X*12 = 2*18
X = 36/12
X = 3
J) X/6 = 49/42
X*42 = 6*49
X = 294/42
X = 7
M)X/90 = 8/5
X*5 = 90*8
X = 720/5
X = 144
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