answer:Given the set of values for ( m ): [ m = -5, -4, -3, -1, 0, 1, 3, 23, 124, 1000 ] We need to determine the common solution ( (x, y) ) for the following set of equations: [ (2m + 1)x + (2 - 3m)y + 1 - 5m = 0 ] Let us analyze the components: - The coefficient of ( x ) is ( 2m + 1 ) - The coefficient of ( y ) is ( 2 - 3m ) - The constant term is ( 1 - 5m ) This can be rewritten as: [ (2m + 1)x + (2 - 3m)y + 1 - 5m = 0 ] for each ( m ). We are searching for ( x ) and ( y ) that satisfy this equation for all given values of ( m ). To find the common solution, we need to eliminate ( m ) by solving the system of equations for arbitrary pairs ( (m_i, m_j) ) and ensuring the solution holds for all provided ( m ). First, let's choose two different values from ( m ) to eliminate ( m ) and solve for ( x ) and ( y ). Consider ( m = -5 ) and ( m = -4 ): 1. For ( m = -5 ): [ (2(-5) + 1)x + (2 - 3(-5))y + 1 - 5(-5) = 0 ] [ (-10 + 1)x + (2 + 15)y + 1 + 25 = 0 ] [ -9x + 17y + 26 = 0 ] 2. For ( m = -4 ): [ (2(-4) + 1)x + (2 - 3(-4))y + 1 - 5(-4) = 0 ] [ (-8 + 1)x + (2 + 12)y + 1 + 20 = 0 ] [ -7x + 14y + 21 = 0 ] Now, we have the system of linear equations: [ -9x + 17y + 26 = 0 ] [ -7x + 14y + 21 = 0 ] Solve this system by elimination or substitution: Step 1: Multiply the first equation by 7 and the second by 9 to align the coefficients of ( x ): [ -63x + 119y + 182 = 0 ] [ -63x + 126y + 189 = 0 ] Step 2: Subtract the first from the second to eliminate ( x ): [ (-63x + 126y + 189) - (-63x + 119y + 182) = 0 ] [ 7y + 7 = 0 ] [ 7y = -7 ] [ y = -1 ] Step 3: Substitute ( y = -1 ) into one of the original equations: [ -7x + 14(-1) + 21 = 0 ] [ -7x - 14 + 21 = 0 ] [ -7x + 7 = 0 ] [ -7x = -7 ] [ x = 1 ] Conclusively, the common solution to the system of given equations is: [ x = 1 ] [ y = -1 ] [ boxed{x=1, y=-1} ]

answer:Given the number of diagonals, D, in a polygon, the formula for the number of diagonals in terms of the number of sides, n, is given by: D = frac{n(n-3)}{2}. To find n, we rearrange the formula and solve for n using the quadratic equation: begin{align*} 2D &= n^2 - 3n, 0 &= n^2 - 3n - 2D. end{align*} Substituting D = 15 into the equation, we get: 0 = n^2 - 3n - 30. Using the quadratic formula, n = frac{-b pm sqrt{b^2 - 4ac}}{2a} with a = 1, b = -3, and c = -30, we calculate: begin{align*} n &= frac{-(-3) pm sqrt{(-3)^2 - 4 cdot 1 cdot (-30)}}{2 cdot 1} &= frac{3 pm sqrt{9 + 120}}{2} &= frac{3 pm sqrt{129}}{2}. end{align*} The solutions are: n = frac{3 + sqrt{129}}{2} text{ and } n = frac{3 - sqrt{129}}{2}. Since the number of sides, n, must be a positive integer, we use only the positive root: n = frac{3 + sqrt{129}}{2} approx 8.175. Since n must be an integer (as it represents the number of sides in a polygon), we round frac{3 + sqrt{129}}{2} to the nearest integer which is 8. Thus, a polygon with fifteen diagonals has boxed{8} sides.

answer:We start by expressing the base 1000 in terms of base 10: [ 1000 = 10^3. ] Then, apply the power rule: [ 10^y = 1000^4 = (10^3)^4 = 10^{3cdot 4} = 10^{12}, ] so by comparing the exponents, we find: [ y = boxed{12}. ]

answer:1. **Identify the midpoint ( F ) and its distance from ( O_1 )**: Given one of the circles' chords of length ( d ) with midpoint ( F ), we need to find the distance ( overline{O_1 F} ): [ overline{O_1 F} = sqrt{r^2 - left(frac{d}{2}right)^2} = sqrt{r^2 - frac{d^2}{4}} ] Let's denote this distance as ( varrho ): [ varrho = sqrt{r^2 - frac{d^2}{4}} ] 2. **Determine the locus of point ( F )**: The collection of all possible midpoints ( F ) forms a sphere centered at ( O_1 ) with radius ( varrho ). 3. **Consider the triangle ( ABC )**: The side ( a ) of the triangle ( ABC ) will be tangent to both spheres centered at ( O_1 ) and ( O_2 ) with radius ( varrho ). The tangency points for these spheres will be ( B ) and ( C ). 4. **Relation between ( d ) and ( r )**: Since ( d ) is the length of the chord within the sphere, which cannot be greater than the diameter: [ 0 < d < 2r ] So, [ 0 < d < 2r ] 5. **Determination of the geometric place of the ( a ) side**: Let's consider an ( S ) plane parallel to the line segment ( O_1O_2 ). This plane will intersect the spheres creating two circles of equal radius ( r ). 6. **Angle of tangency**: The angle ( alpha ) between the tangent lines to these circles and the central axis ( O_1O_2 ) is determined by the angle of the common internal tangents. These tangents will also align with the angle ( alpha ). 7. **Rotation around the central axis**: By rotating these tangents around the central axis ( O_1O_2 ), it maintains the tangency condition, which forms a one-sheet hyperboloid. If we rotate a hyperbola around its imaginary axis, we get this hyperboloid surface. 8. **Special Cases**: - If ( alpha = 0 ), the hyperboloid transforms into a common tangent cylinder between the two spheres. - If ( alpha ) reaches its maximum, the hyperboloid becomes a common tangent cone." 9. **Radius of the hyperboloid neck circle**: The radius of the neck circle of this hyperboloid will vary between ( varrho ) and 0. 10. **Non-parallel section planes**: If the cutting plane ( S ) is not parallel to ( O_1O_2 ), the intersections are circles of different radii. The common tangents, if they form an angle ( alpha ) with the central axis, would be part of the hyperboloid's generating lines. # Conclusion: [ boxed{0 < d < 2r} ] The geometric place for ( a ) side of the triangle ( ABC ) considering the conditions forms part of a hyperboloid of one sheet.