answer:1. **Initial Setup and Substitution**: Start with x_0 = 7 and convert the recursive sequence using the substitution x_n = y_n + 3. Therefore, y_0 = x_0 - 3 = 4. 2. **Transforming the Recurrence Relation**: Utilizing the substitution, the recursive formula becomes: [ y_{n+1} + 3 = frac{(y_n + 3)^2 + 6(y_n + 3) - 2}{y_n + 3 + 8}. ] Simplifying, we find: [ y_{n+1} + 3 = frac{y_n^2 + 12y_n + 25}{y_n + 11}, ] therefore: [ y_{n+1} = frac{y_n^2 + 12y_n + 25}{y_n + 11} - 3 = frac{y_n^2 + 9y_n + 4}{y_n + 11}. ] 3. **Analyzing Behavior and Estimating Decay**: Notice y_n is decreasing and positive: [ y_{n+1} = frac{y_n^2 + 9y_n + 4}{y_n + 11} < y_n. ] We approximate the reduction ratio to measure decay: [ frac{3}{4} leq frac{y_n + 4}{y_n + 11} leq frac{8}{11}. ] The geometric-like decay is described by left(frac{8}{11}right)^n. 4. **Finding m**: We need y_m leq frac{1}{2^{10}}: [ left(frac{8}{11}right)^m leq frac{1}{2^{10}}, ] taking logarithms: [ m logleft(frac{8}{11}right) leq -10 log(2), ] solving for m: [ m geq frac{-10 log(2)}{logleft(frac{8}{11}right)} approx 35.0. ] So, the least integer m approx 35. [ 35 ] falls in the provided interval choice classification which will be added next. The final answer is boxed{textbf{(C) } [35, 60]}

answer:To solve this problem, we follow these steps: 1. **Identify the Center of the Circle**: The given equation of the circle is x^{2}+y^{2}-4x+6y-3=0. Completing the square for both x and y terms, we rewrite this equation as (x-2)^{2} + (y+3)^{2} = 4^2. Thus, the center of the circle, C, is at (2, -3). 2. **Identify a Point on the Line**: The line equation is given by mx+y+m-1=0. Substituting x=-1 into this equation, we get -m+1+m-1=0, which is satisfied for any m, indicating that the point A(-1, 1) lies on the line for any value of m. 3. **Maximizing the Distance**: The distance from the center of the circle to the line is maximized when the line passing through C and A is perpendicular to the given line l. This is because the perpendicular distance from a point to a line is the maximum distance when compared to any other non-perpendicular line passing through the same point. 4. **Calculating the Slope of CA**: The slope of the line segment CA is calculated using the coordinates of C(2, -3) and A(-1, 1). The slope, k_{AC}, is given by frac{1 - (-3)}{-1 - 2} = frac{4}{-3} = -frac{4}{3}. 5. **Condition for Perpendicular Lines**: For two lines to be perpendicular, the product of their slopes must be -1. Let k_l be the slope of line l. Since the slope of l is -m (from the equation mx+y+m-1=0, rearranging to y=-mx-m+1), we have k_{AC} cdot k_{l} = -1. Substituting the values, we get -frac{4}{3} cdot (-m) = -1. 6. **Solving for m**: Simplifying the equation -frac{4}{3} cdot (-m) = -1, we find that m = -frac{3}{4}. Therefore, the value of m that maximizes the distance from the center of the circle to the line l is -frac{3}{4}. Thus, the correct answer is boxed{C}.

answer:We need to prove that if two skew lines are perpendicular, there exists a unique plane passing through one of them that is perpendicular to the other. 1. **Setup and Assumption**: - Let ( a ) and ( b ) be two skew lines with ( a perp b ). - We need to show that through one of these lines, say ( b ), we can construct a unique plane that is perpendicular to the other line ( a ). 2. **Construct Perpendicular from a Point**: - Choose any point ( B ) on the line ( b ). - From ( B ), drop a perpendicular ( BA ) to the line ( a ) (since ( a perp b ), we can find such a perpendicular). 3. **Formation of the Plane**: - Now, consider the plane formed by the intersecting lines ( b ) and ( BA ). Let's call this plane ( alpha ). 4. **Plane Perpendicularity Proof**: - The plane ( alpha ) is perpendicular to the line ( a ) because the line ( a ) is perpendicular to both lines ( b ) and ( BA ), which lie in ( alpha ). 5. **Uniqueness Argument**: - Suppose there exists another plane ( beta ) passing through the line ( b ) which is also perpendicular to the line ( a ). - Take the line of intersection of ( beta ) and plane ( alpha ) (if such a ( beta ) exists, it must intersect ( alpha ) along ( b ), forming a unique line along ( b )). - Let's denote the intersection line as ( m ); this line lies on ( b ), and hence on both ( alpha ) and ( beta ). 6. **Contradiction**: - Through point ( M ) on line ( b ), there would be two distinct planes ( alpha ) and ( beta ) that are both perpendicular to line ( a ). - However, this leads to a contradiction because if two planes intersect in a line and are perpendicular to the same line at the point of intersection, they must coincide. Therefore, there's only one unique plane ( alpha ). # Conclusion: Thus, we have shown that there is a unique plane passing through the skew line ( b ) (which intersects with ( a )) that is perpendicular to the line ( a ). [ blacksquare ] --- # Proof of Parallelism: We need to prove the statement that a line and a plane are parallel if they are both perpendicular to the same line. 1. **Setup Assumptions**: - Let line ( a ) and plane ( alpha ) be perpendicular to the same line ( b ), i.e., ( a perp b ) and ( alpha perp b ). 2. **Constructing the Perpendicular Plane**: - Since line ( b ) is perpendicular to both ( a ) and the plane ( alpha ), there exists a plane ( beta ) that contains the line ( a ) and is perpendicular to ( b ). 3. **Planes' Relationship**: - The planes ( alpha ) and ( beta ) are both perpendicular to the line ( b ). 4. **Implication of Perpendicular Planes**: - Since two distinct planes perpendicular to the same line ( b ) must be parallel to each other, ( alpha parallel beta ). 5. **Line Parallel to the Plane**: - Consequently, since line ( a ) lies in plane ( beta ) and plane ( beta ) is parallel to plane ( alpha ), line ( a ) must be parallel to plane ( alpha ). - This follows from the fact that a line lying in one plane parallel to another plane does not intersect the other plane. # Conclusion: Thus, we have proven that line ( a ) and plane ( alpha ) are parallel if they are both perpendicular to the same line ( b ). [ blacksquare ]

answer:A century has 100 years, so Gideon has 100 marbles. If he gives 3/4 of his marbles to his sister, he gives away 3/4 * 100 = 75 marbles. This leaves him with 100 - 75 = 25 marbles. If he then multiplies the number of remaining marbles by 2, he gets 25 * 2 = 50. Since Gideon is 45 years old now, and multiplying the remaining marbles by 2 gives him his age a certain number of years from now, we can set up the equation: 45 + x = 50 where x is the number of years from now when he gets his age. Solving for x: x = 50 - 45 x = 5 So, Gideon gets his age boxed{5} years from now.