answer:Options A, B, and D are symbolic representations of axioms 1, 2, and 3, respectively, so they are all correct. For option C, there are two possibilities for l notsubseteq alpha: l is parallel to alpha, or l intersects alpha. If they intersect at point A, then A in l and A in alpha. Therefore, it is incorrect. Hence, the correct choice is boxed{text{C}}.

answer:1. Given the function ([x]) denotes the greatest integer less than or equal to (x). 2. We need to determine the elements of the set (left{ n leftlvert, n=left[ frac{k^{2}}{2005} right] right., 1 leqslant k leqslant 2004, k in mathbf{N} right}). 3. Start by analyzing the difference between consecutive terms: [ frac{(k+1)^{2}}{2005} - frac{k^{2}}{2005} = frac{2k+1}{2005} ] We want this difference to be less than or equal to 1: [ frac{2k+1}{2005} leqslant 1 ] Solving for (k): [ 2k + 1 leqslant 2005 implies 2k leqslant 2004 implies k leqslant 1002 ] 4. For (k = 1, 2, ldots, 1002), the values of (left[ frac{k^2}{2005} right]) can repeat or increase by 1. Specifically, if (left[ frac{k^2}{2005} right] = n), then: [ left[ frac{(k+1)^2}{2005} right] = n quad text{or} quad left[ frac{(k+1)^2}{2005} right] = n + 1 ] 5. Compute the extreme value of (n) when (k = 1002): [ left[ frac{1002^2}{2005} right] = left[frac{1004004}{2005} right] = 500 ] Also, [ left[ frac{1^2}{2005} right] = 0 ] So, for ( k = 1, 2, ldots, 1002 ), the values of ( left[ frac{k^2}{2005} right] ) range from (0) to (500). 6. For (k = 1003, 1004, ldots, 2004), the difference: [ frac{(k+1)^2}{2005} - frac{k^2}{2005} > 1 ] 7. This means: [ left[ frac{(k+1)^2}{2005} right] geqslant left[ frac{k^2}{2005} right] + 1 ] 8. Since each (left[ frac{k^2}{2005} right]) for ( k = 1003, 1004, ldots, 2004 ) are distinct and the values increment, there are (2004 - 1002 = 1002) distinct numbers. 9. The smallest value for ( k = 1003 ): [ left[ frac{1003^2}{2005} right] = 501 ] 10. Therefore, the total number of distinct elements in the set is: [ 501 text{ (values from 0 to 500) } + 1002 text{ (from 1003 to 2004) } = 1503 ] # Conclusion: [ boxed{1503} ]

answer:To solve the quadratic equation 3x^{2}-2x-1=0 using the formula method, we first identify the coefficients: - a = 3 - b = -2 - c = -1 The discriminant Delta of a quadratic equation ax^{2} + bx + c = 0 is given by Delta = b^{2} - 4ac. Substituting the values of a, b, and c into this formula, we get: [ begin{align*} Delta &= b^{2} - 4ac &= (-2)^{2} - 4 times 3 times (-1) &= 4 + 12 &= 16. end{align*} ] Therefore, the result of calculating b^{2} - 4ac for the given quadratic equation is boxed{16}, which corresponds to choice boxed{D}.

answer:To solve for the length of segment AD in quadrilateral ABCD, where AB = 5, BC = 8, and CD = 20 units, and both angles B and C are right angles, we proceed as follows: 1. Construct a segment AE parallel to CB. This makes AE = BC = 8 units because parallel lines preserve distance and angles, making ABCE a rectangle. 2. Since CD = 20 units and AB = 5 units, and AE is parallel and equal to BC, the length of segment DE can be found by subtracting the length of AB from CD: DE = CD - AB = 20 - 5 = 15 text{ units}. 3. Now, we have a right-angled triangle ADE with legs AE = 8 units and DE = 15 units. Applying the Pythagorean Theorem to find the length of the hypotenuse AD: begin{align*} AD^2 &= AE^2 + DE^2 &= 8^2 + 15^2 &= 64 + 225 &= 289. end{align*} 4. Taking the square root of both sides to find AD: begin{align*} AD &= sqrt{289} &= 17 text{ units}. end{align*} Therefore, the length of segment AD is boxed{17} units.