answer:Part (a) We need to find a 3 times 3 grid filled with (0)'s and (1)'s which has exactly (1) point, meaning exactly one of the rows, columns, or diagonals has an odd sum. One possible solution is: [ begin{pmatrix} 1 & 0 & 0 0 & 0 & 0 0 & 0 & 0 end{pmatrix} ] Here, only the first row sum is odd ((1+0+0=1)), while everything else sums to 0 (even). Part (b) Now, we want to find all 3 times 3 grids where exactly (8) of the (3) rows, (3) columns, and (2) diagonals have an odd sum. Clearly, (8) points covered means all combinations (rows, columns, and diagonals) must be odd. # Case 1: Middle number is (0) - Middle number is (0), and let (A, B, C, D) be the values at other four center points. - Each opposite term will have different value. Then we have a scenario where if (A) is 0, (bar{A} = 1), and similarly for (B, C, D). [ begin{pmatrix} X & A & B bar{X} & 0 & bar{D} X & bar{A} & bar{B} end{pmatrix} ] - Rows: Sum of ((A, B, bar{A})) and ((bar{D}, 0, D)) etc. It’s impossible because either top or bottom row will be even. Hence, no (8)-point grid is possible with middle as (0). # Case 2: Middle number is (1) - Middle number (1), let (A, B, C, D) be the values at other 4 center points. - Each opposite term will have the same value. [ begin{pmatrix} A & B & C B & 1 & D C & D & A end{pmatrix} ] - (A+B+C) and (A+D+C) both odd: (B = D). Checking for combinations that work with an overall odd sum: 1. ((A, B, C)) 2. Must sum to 1 or 3. Resulting: Possible pairs are: - ( (1,0,0)) - ( (0,1,0)) - ( (0,0,1)) - ( (1,1,1)) Configurations: 1. [ begin{pmatrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 end{pmatrix} ] 2. [ begin{pmatrix} 0 & 1 & 0 1 & 1 & 0 0 & 0 & 0 end{pmatrix} ] 3. [ begin{pmatrix} 0 & 0 & 1 0 & 1 & 1 1 & 1 & 0 end{pmatrix} ] 4. [ begin{pmatrix} 1 & 1 & 1 1 & 1 & 1 1 & 1 & 1 end{pmatrix} ] If these cover all grids summing 8, restricting all to 8 points only. Part (c) Let's prove (E = O) for even number (E) and odd (O). Consider the set of all (3 times 3) grids. Let’s generate grid pairs ((G, G^*)). Switch grid with transformation G^*. Flip top-left entry. - Transformation: Grid actions (0 leftrightarrow 1). For every odd or even sums (G/G^*): - Adding or removing one will flip parity. - Operators alter rows/cols and diagonals maintaining values. Thus, [ boxed{E=O} ] This equivalency implies same, noting smaller cases. Ensure fairness redistributed keeping point parity changes.

answer:Condition B: frac{1}{a} < frac{1}{b}, can be simplified to frac{1}{a} - frac{1}{b} < 0 iff frac{b-a}{ab} < 0 If Condition A: a > b > 0 is established, then Condition B is definitely established; However, when Condition B is established, it does not necessarily mean that Condition A: a > b > 0 is established Therefore, A is a sufficient but not necessary condition for B to be established Hence, the answer is: boxed{text{A}} First, simplify the inequality of Condition B, then judge whether the establishment of Condition A can lead to the establishment of Condition B; whether the establishment of Condition B can lead to the establishment of Condition A, use the definition of necessary and sufficient conditions to judge what kind of condition A is for B to be established. When judging whether one condition is another condition, it is necessary to simplify the two conditions first, and then use the definition of necessary and sufficient conditions to make a judgment.

answer:First, we need to subtract the cost of the adult's ticket from the total amount of money she has. 35 (total amount) - 8 (adult ticket) = 27 remaining for children's tickets. Now, we divide the remaining amount by the cost of a child's ticket to find out how many children she can take with her. 27 (remaining amount) ÷ 3 (child's ticket) = 9 children. Therefore, she can take boxed{9} children with her to the movies.

answer:Given that the line l has a direction vector overrightarrow{m}=(1,0,-1) and passes through the point A(1,1,1), and we need to find the distance from point P(1,-1,-1) to the line l. First, we calculate the vector overrightarrow{AP} which is the vector from A to P: [ overrightarrow{AP} = P - A = (1,-1,-1) - (1,1,1) = (0,-2,-2). ] The direction vector of line l is given as overrightarrow{m}=(1,0,-1). To find the distance from P to l, we first need to find the projection of overrightarrow{AP} onto overrightarrow{m}. The formula for the projection's length is: [ |text{proj}_{overrightarrow{m}}overrightarrow{AP}| = frac{{overrightarrow{AP}cdot overrightarrow{m}}}{{|overrightarrow{m}|}}. ] Calculating the dot product overrightarrow{AP}cdot overrightarrow{m} and the magnitude |overrightarrow{m}|: [ overrightarrow{AP}cdot overrightarrow{m} = (0,-2,-2) cdot (1,0,-1) = 0cdot1 + (-2)cdot0 + (-2)cdot(-1) = 2, ] [ |overrightarrow{m}| = sqrt{1^2 + 0^2 + (-1)^2} = sqrt{2}. ] Therefore, the length of the projection is: [ |text{proj}_{overrightarrow{m}}overrightarrow{AP}| = frac{2}{sqrt{2}} = sqrt{2}. ] The distance d from P to l can be found using the Pythagorean theorem, where d is the leg of a right triangle whose hypotenuse is |overrightarrow{AP}| and one leg is the projection length |text{proj}_{overrightarrow{m}}overrightarrow{AP}|: [ d = sqrt{|overrightarrow{AP}|^2 - (sqrt{2})^2} = sqrt{(0^2 + (-2)^2 + (-2)^2) - 2} = sqrt{8 - 2} = sqrt{6}. ] Therefore, the distance from point P to line l is boxed{sqrt{6}}.