32 X 6=192
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x^3y = 2, y = 4 Substitute 4 for y: (x^3)(4) = 2 x^3 = 8 x = 2
[2/(3/4)]/6 = (2/1 x 4/3)/6 = (8/3)/(6/1) = 8/3 x 1/6 =8/18 = 4/9 ≈ 0.44
(2 x 3 x 4 x 5 x 6 x 7 x 8 x 9) + 1 will have a remainder of 1 when divided by 2, 3, 4, 5, 6, 7, 8, or 9. (2 x 3 x 4 x 5 x 6 x 7 x 8 x 9) + 1 = 362,881
-1
(8 x 7 x 6 x 5 x 4 x 3)/(6 x 5 x 4 x 3 x 2) = 28 combinations