answer:Solution: (Ⅰ) From 2cos C(acos B + bcos A) = c, By the sine rule, we have 2cos C(sin A cdot cos B + sin B cdot cos A) = sin C, 2cos C cdot sin (A + B) = sin C, Since A + B + C = pi and A, B, C in (0, pi), Therefore, sin (A + B) = sin C > 0, Thus, 2cos C = 1, cos C = frac{1}{2}, Since C in (0, pi), Therefore, C = frac{pi}{3}, So, boxed{C = frac{pi}{3}}. (Ⅱ) By the cosine rule, we have c^2 = a^2 + b^2 - 2ab cdot cos C = a^2 + b^2 - ab, = (a + b)^2 - 3ab geqslant (a + b)^2 - 3 frac{(a + b)^2}{4} = frac{(a + b)^2}{4}, Thus, a + b leqslant 2sqrt{3}, Also, a + b > c = sqrt{3}, sqrt{3} < a + b leqslant 2sqrt{3}, 2sqrt{3} < a + b + c leqslant 3sqrt{3}, The range of the perimeter is boxed{(2sqrt{3}, 3sqrt{3}]}.

answer:For n to be a perfect square, the exponent of 10 must be an even number because a perfect square is the product of an integer multiplied by itself, and the square of 10^k is 10^(2k). Given that n = 16 × 10^(-p), we know that 16 is already a perfect square (since 16 = 4^2). Therefore, we only need to consider the 10^(-p) part to determine when n will be a perfect square. The exponent -p must be even for 10^(-p) to be a perfect square. Since p is an integer and -4 < p < 4, the possible integer values for p are -3, -2, -1, 0, 1, 2, 3. Out of these, the values of p that will make -p even are -2, 0, and 2 (because -(-2) = 2, -0 = 0, and -2 = -2 are all even). Therefore, there are boxed{3} different integer values of p that will make n a perfect square.

answer:As with the original problem, consider the cross section containing the axes of symmetry. The biggest sphere will be tangent to the sides of the cones. Let A be the common vertex where the bases of the cones intersect, and C the center of the sphere. Define B and D as the vertices of the cones, with D being the center of the base of one cone. Let the sphere tangent to overline{AB} at E. Given that the similar triangles triangle ABD and triangle CBE share angle properties: - (ABD) is a right triangle, as AB is a radius (5) and BD is the slant height sqrt{12^2 + 5^2} = 13. - triangle CBE is similar to triangle ABD. Thus, applying the same reasoning, the sphere's radius r, which equals CE, can be determined by the ratio: [ r = BD cdotfrac{CD}{AB} = 5 cdot frac{BD-4}{13} ] where CD = 12 - 4 because the axis intersects 4 units above the base. Continue calculations: [ BD = sqrt{12^2 + 5^2} = 13, text{ so } BD - CD = 13 - 8 = 5 ] [ r = BD cdot frac{CD}{AB} = 5 cdot frac{5}{13} = frac{25}{13} ] Thus, r^2 = left(frac{25}{13}right)^2 = frac{625}{169}. Conclusion: [ text{Sum of numerator and denominator} = 625 + 169 = boxed{794} ]

answer:If each person gets Rs 1,950 and there are 22 persons, then the total amount distributed can be calculated by multiplying the amount each person gets by the number of persons. Total amount distributed = Amount each person gets × Number of persons Total amount distributed = Rs 1,950 × 22 Total amount distributed = Rs 42,900 So, the total amount distributed is Rs boxed{42,900} .