Respostas
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1)
B=(b)y 3+3, tal que bij= (i+j)²
B = (b₁₁ b₁₂ b₁₃)
(b₂₁ b₂₂ b₂₃)
(b₃₁ b₃₂ b₃₃)
b₁₁ = (i+j)² => (1+1)² => 2² => 4
b₁₂ = (i+j)² => (1+2)² => 3² => 9
b₁₃ = (i+j)² => (1+3)² => 4² => 16
b₂₁ = (i+j)² => (2+1)² => 3² => 9
b₂₂ = (i+j)² => (2+2)² => 4² => 16
b₂₃ = (i+j)² => (2+3)² => 5² => 25
b₃₁ = (i+j)² => (3+1)² => 4² => 16
b₃₂ = (i+j)² => (3+2)² => 5² => 25
b₃₃ = (i+j)² => (3+3)² => 6² => 36
B = (4 9 16)
(9 16 25)
(16 25 36)
2) A = B
(x+y 2) = (12 9)
(9 y²+y) (2 53)
x+y = 12
x = 12-y
y²+y = 2
y(y+1)=2
y'=2
y"=-1
p/ y=2
x = 12-y
x = 12 - 2
x = 10
p/ y =-1
x = 12-y
x = 12-(-1)
x = 13
pilgrimscott:
obrigado!
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