answer:To evaluate the function, we substitute x = 31 into the expression: [ f(31) = frac{log(2)}{sqrt[3]{31^2 - 6}} = frac{log(2)}{sqrt[3]{961 - 6}} = frac{log(2)}{sqrt[3]{955}} ] Now, we calculate the numerical value: [ f(31) approx frac{0.30103}{sqrt[3]{955}} approx 0.07 ] Thus, the value of the function at x = 31 is approximately 0.07.

answer:The proof could be improved by clarifying the following: * The assumption that -X is contained in D(r, R_o(P)) is valid because R_o is a reflection, so it preserves distances. Therefore, if d(-X, R_o(P)) = s, then d(X, P) = s, which means that X is in D(r, P). * The proof could also be clarified by explicitly stating that the reflection of a point through the origin is the negative of that point. This would make it easier to understand how the reflection of the center of the disc is determined.

answer:The Doppler effect is a phenomenon that causes the frequency of a sound wave to change when the source of the sound and the observer are moving relative to each other. In this case, Eagle A is the source of the sound, and Eagle B is the observer. Since Eagle A and Eagle B are moving towards each other, the sound wave will be compressed, resulting in an increase in frequency. The formula for the Doppler effect is: f' = f(v + vo) / (v + vs) where: f is the original frequency of the sound wave f' is the observed frequency of the sound wave v is the speed of sound in the medium vo is the speed of the observer vs is the speed of the source In this problem, we know the following: f = 3000 Hz f' = 3500 Hz v = 340 m/s (the speed of sound in air) vo = 10 m/s (the speed of Eagle A) We can use the Doppler effect formula to solve for vs, the speed of Eagle B: 3500 Hz = 3000 Hz(340 m/s + vs) / (340 m/s + 10 m/s) 3500 Hz(340 m/s + 10 m/s) = 3000 Hz(340 m/s + vs) 1,190,000 m^2/s^2 + 35,000 m/s = 1,020,000 m^2/s^2 + 3000 Hz vs 170,000 m^2/s^2 = 3000 Hz vs vs = 170,000 m^2/s^2 / 3000 Hz vs = 56.6667 m/s Therefore, the speed of Eagle B is 56.6667 m/s. The answer is 16.1905 m/s

answer:The key to solving this problem is to recognize that the points A, B, C, D form an equilateral triangle in the complex plane. This allows us to use Ptolemy's inequality, which states that in any quadrilateral, the product of the lengths of the two diagonals is less than or equal to the sum of the products of the lengths of the opposite sides. In this case, the diagonals are AD and BC, and the opposite sides are AB, AC, CD, and BD. Since ABC is an equilateral triangle, we have AB = AC = BC, and since AD = |z-1|, CD = |z-j^2|, and BD = |z-j|, we can substitute these values into Ptolemy's inequality to get: |z-1|cdot BCleq ABcdot |z-j^2|+ACcdot |z-j| |z-1|cdot BCleq BCcdot |z-j^2|+BCcdot |z-j| |z-1|leq |z-j^2|+|z-j| which is what we wanted to prove. Let Asim 1, Bsim j and Csim j^2 then ABC is an equilateral triangle in mathbb{C}. Let Dsim z. Then by Ptolemy's inequality we have: ADcdot BCleq ABcdot CD+ACcdot BD Since BC=AB = AC and AD =|z-1|, CD=|z-j^2|, BD =|z-j| we get what we want.