dan-l
dan-l
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Comments
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8x7x6x5x4 = 6720 maneiras
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8x7x6x5x4 = 6720 maneiras
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8x7x6x5x4 = 6720 maneiras
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8x7x6x5x4 = 6720 maneiras
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8x7x6x5x4 = 6720 maneiras
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No. It does not factorize. To solve an equation of the form x^2 - Bx + C, you have to be able to factorize it as follows: (x - M)(x - N) where M + N = B, and M(N) = C. There is no solution for this. No pair of whole number factors of 2 add …
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No. It does not factorize. To solve an equation of the form x^2 - Bx + C, you have to be able to factorize it as follows: (x - M)(x - N) where M + N = B, and M(N) = C. There is no solution for this. No pair of whole number factors of 2 add …
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fica boa sim, mas tem umas do U2 como one, que ficam bem tb afinal de contas os caras são os mestres dessa politicagem musica/guerra
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fica boa sim, mas tem umas do U2 como one, que ficam bem tb afinal de contas os caras são os mestres dessa politicagem musica/guerra
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fica boa sim, mas tem umas do U2 como one, que ficam bem tb afinal de contas os caras são os mestres dessa politicagem musica/guerra
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No. It does not factorize. To solve an equation of the form x^2 - Bx + C, you have to be able to factorize it as follows: (x - M)(x - N) where M + N = B, and M(N) = C. There is no solution for this. No pair of whole number factors of 2 add …
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8x7x6x5x4 = 6720 maneiras
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8x7x6x5x4 = 6720 maneiras
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8x7x6x5x4 = 6720 maneiras