answer:- By the QM-AM inequality: [ sqrt{frac{x^2 + (1 - y)^2}{2}} ge frac{x + (1 - y)}{2} ] Thus, sqrt{x^2 + (1 - y)^2} ge frac{1}{sqrt{2}} (x + (1 - y)) and similarly for the other terms. Adding these gives: [ sqrt{x^2 + (1 - y)^2} + sqrt{y^2 + (1 - z)^2} + sqrt{z^2 + (1 - w)^2} + sqrt{w^2 + (1 - x)^2} ge 2sqrt{2}. ] - Using the triangle inequality: [ sqrt{x^2 + (1 - y)^2} le sqrt{x^2 + 2x(1 - y) + (1 - y)^2} = x + 1 - y, ] and similarly for the other terms. Adding these gives: [ sqrt{x^2 + (1 - y)^2} + sqrt{y^2 + (1 - z)^2} + sqrt{z^2 + (1 - w)^2} + sqrt{w^2 + (1 - x)^2} le 4. ] - Special case when x = y = z = w = t: [ sqrt{x^2 + (1 - y)^2} + sqrt{y^2 + (1 - z)^2} + sqrt{z^2 + (1 - w)^2} + sqrt{w^2 + (1 - x)^2} = 4 sqrt{t^2 + (1-t)^2}. ] Since t in [0,1], the expression varies from 2sqrt{2} to 4. Conclusion: The possible values of the expression are in the interval boxed{[2 sqrt{2}, 4]}.

answer:To find the rate of interest, we can use the formula for simple interest: Simple Interest (SI) = Principal (P) * Rate (R) * Time (T) / 100 We are given: Simple Interest (SI) = Rs. 90 Principal (P) = Rs. 642.8571428571428 Time (T) = 4 years We need to find the Rate (R). Plugging the given values into the formula: 90 = (642.8571428571428) * R * 4 / 100 To solve for R, we first multiply both sides by 100 to get rid of the denominator: 90 * 100 = (642.8571428571428) * R * 4 9000 = (642.8571428571428) * R * 4 Now, we divide both sides by (642.8571428571428) * 4 to isolate R: 9000 / (642.8571428571428 * 4) = R 9000 / 2571.428571428571 = R R ≈ 3.5 Therefore, the rate of interest is approximately boxed{3.5%} .

answer:First, substitute y = frac{4x-16}{3x-4} to simplify the equation: y^2 + y = 6. This rearranges to a quadratic equation: y^2 + y - 6 = 0. Factoring the quadratic, we find: (y - 2)(y + 3) = 0, which gives us y = 2 or y = -3. For y = 2, substituting back into the expression for x, we have: frac{4x-16}{3x-4} = 2. This simplifies to: 4x - 16 = 6x - 8, which results in: -16 + 8 = 6x - 4x, -8 = 2x, x = -4. For y = -3, substituting back into the expression for x, we have: frac{4x-16}{3x-4} = -3. This simplifies to: 4x - 16 = -9x + 12, which results in: 4x + 9x = 12 + 16, 13x = 28, x = frac{28}{13}. The greatest possible value of x is thus: boxed{frac{28}{13}}.

answer:1. **Claim**: The maximum number of points in ( M ) is 3. We will show that any set ( M ) with more than 3 points cannot satisfy the given conditions. 2. **Base Case**: Consider an equilateral triangle with vertices ( A, B, ) and ( C ). Clearly, ( |M| geq 3 ) since ( A, B, ) and ( C ) form an equilateral triangle, and for any two points among ( A, B, ) and ( C ), the third point completes the equilateral triangle. 3. **Assumption for Contradiction**: Assume there exists a finite set ( M ) with more than 3 elements satisfying the conditions. Choose two points ( A, B in M ) such that the distance ( |AB| ) is maximal. Without loss of generality, let ( |AB| = 1 ). 4. **Existence of Point ( C )**: By the problem's condition, there exists a point ( C in M ) such that ( triangle ABC ) is equilateral. Assume ( A, B, C ) are ordered counterclockwise around the perimeter of ( triangle ABC ). 5. **Reflections and Disks**: Reflect points ( A, B, ) and ( C ) with respect to the sides ( BC, CA, ) and ( AB ) to get points ( A', B', ) and ( C' ) respectively. Denote the disk with radius 1 centered at a point ( P ) as ( Gamma_P ). 6. **Additional Point ( D )**: Let ( D ) be a point in ( M ) different from ( A, B, ) and ( C ). All points of ( M ) must lie within the intersection of the disks ( Gamma_A cap Gamma_B cap Gamma_C ). 7. **Equilateral Triangle with ( D )**: There must be a point ( E in M ) such that ( triangle ADE ) is equilateral. Consider the two cases for the angle ( angle DAE ): - If ( angle DAE = 60^circ ), then ( D ) lies in ( Gamma_{C'} ) since ( E ) lies in ( Gamma_B ). - If ( angle EAD = 60^circ ), then ( D ) lies in ( Gamma_{B'} ) since ( E ) lies in ( Gamma_C ). 8. **Location of ( D )**: Thus, ( D ) must lie in either ( Gamma_{B'} ) or ( Gamma_{C'} ). Similarly, by considering other equilateral triangles involving ( D ), we find that ( D ) must also lie in ( Gamma_{C'} ) or ( Gamma_{A'} ) and ( Gamma_{A'} ) or ( Gamma_{B'} ). 9. **Intersection of Disks**: Therefore, ( D ) must lie in the intersection of two of the disks ( Gamma_{A'}, Gamma_{B'}, Gamma_{C'} ). Without loss of generality, assume ( D ) lies in ( Gamma_{B'} ) and ( Gamma_{C'} ). However, the intersection ( Gamma_{B'} cap Gamma_{C'} = {A} ), which contradicts ( D neq A ). 10. **Conclusion**: Hence, ( M ) cannot have more than 3 elements. Therefore, the maximum number of points in ( M ) is 3. (blacksquare) The final answer is ( boxed{ 3 } )