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Brennomartins81
@Brennomartins81
August 2019
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Determine o valor de x nos casos:
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juanbomfim22
A)x^2+h^2=49
(10-x)^2+h^2=64
100-20x+x^2+h^2=64
x^2+h^2+100-20x=64
49+100-20x=64
20x= 149-64
x=85/20 = 4,25
b) h^2+(7+x)^2 = 4.29
h^2+49+14x+x^2 = 116
h^2+x^2+14x=116-49
h^2+x^2+14x=67
pitagoras no 2 triangulo
h^2+x^2=5^2
h^2+x^2=25
25+14x=67
14x=42
x= 3
2 votes
Thanks 5
Brennomartins81
Muito obrigado!!!
Brennomartins81
Ta ok
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(10-x)^2+h^2=64
100-20x+x^2+h^2=64
x^2+h^2+100-20x=64
49+100-20x=64
20x= 149-64
x=85/20 = 4,25
b) h^2+(7+x)^2 = 4.29
h^2+49+14x+x^2 = 116
h^2+x^2+14x=116-49
h^2+x^2+14x=67
pitagoras no 2 triangulo
h^2+x^2=5^2
h^2+x^2=25
25+14x=67
14x=42
x= 3