answer:Given that three real numbers (x), (y), and (z) satisfy the following equality: [ frac{x + 4}{2} = frac{y + 9}{z - 3} = frac{x + 5}{z - 5} ] Let's denote the common value of the fractions by (k): [ frac{x + 4}{2} = k, quad frac{y + 9}{z - 3} = k, quad frac{x + 5}{z - 5} = k ] From these equations, we can express (x), (y), and (z) in terms of (k): 1. (frac{x + 4}{2} = k) [ x + 4 = 2k implies x = 2k - 4 ] 2. (frac{y + 9}{z - 3} = k) [ y + 9 = k(z - 3) implies y = k(z - 3) - 9 ] 3. (frac{x + 5}{z - 5} = k) [ x + 5 = k(z - 5) implies z - 5 = frac{x + 5}{k} implies z = frac{x + 5}{k} + 5 ] Now, we substitute (x = 2k - 4) into the expression for (z): [ z = frac{(2k - 4) + 5}{k} + 5 implies z = frac{2k - 4 + 5}{k} + 5 implies z = frac{2k + 1}{k} + 5 implies z = 2 + frac{1}{k} + 5 implies z = 7 + frac{1}{k} ] Next, we substitute (x = 2k - 4) and (z = 7 + frac{1}{k}) into the expression for (y): [ y = k(z - 3) - 9 implies y = kleft(7 + frac{1}{k} - 3right) - 9 implies y = k(4 + frac{1}{k}) - 9 implies y = 4k + 1 - 9 implies y = 4k - 8 ] Finally, we need to find the value of (frac{x}{y}): [ x = 2k - 4, quad y = 4k - 8 ] [ frac{x}{y} = frac{2k - 4}{4k - 8} = frac{2(k - 2)}{4(k - 2)} = frac{2}{4} = frac{1}{2} ] # Conclusion: (boxed{frac{1}{2}})

answer:1. Let's denote the velocities of Sasha, Lesha, and Kolya as v_S, v_L, and v_K, respectively. Since they all started at the same time and from the same place, we'll analyze their positions relative to each other at various points in the race. 2. Given that when Sasha finished the 100-meter race, Lesha was 10 meters behind, we can express the distances they ran: [ text{Sasha's distance} = 100 text{ meters} ] [ text{Lesha's distance} = 100 - 10 = 90 text{ meters} ] 3. When Lesha finished the 100-meter race, Kolya was 10 meters behind him. Thus: [ text{Kolya's distance when Lesha finished} = 100 - 10 = 90 text{ meters} ] 4. From the problem's condition, we infer the following speed ratios: [ v_L = frac{90 text{ meters}}{100 text{ meters} / v_S} = 0.9 v_S ] [ v_K = frac{90 text{ meters}}{100 text{ meters} / v_L} ] 5. Substituting v_L = 0.9 v_S into the equation for v_K, we get: [ v_K = frac{90}{100} v_L = 0.9 v_L = 0.9 cdot 0.9 v_S = 0.81 v_S ] 6. Now, using the speeds to find Kolya's position when Sasha finished: [ text{Kolya's position when Sasha finished} = v_K cdot t_S = 0.81 v_S cdot frac{100}{v_S} = 0.81 cdot 100 = 81 text{ meters} ] 7. Therefore, the distance between Sasha and Kolya when Sasha crossed the finish line is: [ text{Distance} = 100 text{ meters} - 81 text{ meters} = 19 text{ meters} ] # Conclusion: [ boxed{19 text{ meters}} ]

answer:- Calculate the multiples of 3 and 4 in the range 1 to 30. - Multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30. So, there are 10 multiples of 3. - Multiples of 4: 4, 8, 12, 16, 20, 24, 28. There are 7 multiples of 4. - Identify double-counting: - 12 and 24 are the multiples common between those of 3 and 4. There are 2 such multiples. - Calculate the total number of favorable outcomes (unique): [ 10 + 7 - 2 = 15 ] - Determine the probability: [ text{Probability} = frac{15}{30} = boxed{frac{1}{2}} ]

answer:The polynomial given is 3(x^8 - x^5 + 2x^3 - 6) - 5(x^4 + 3x^2) + 2(x^6 - 5). - Simplify each group: - 3(x^8 - x^5 + 2x^3 - 6) = 3x^8 - 3x^5 + 6x^3 - 18 - -5(x^4 + 3x^2) = -5x^4 - 15x^2 - 2(x^6 - 5) = 2x^6 - 10 - Combine like terms: - 3x^8 + 2x^6 - 3x^5 - 5x^4 + 6x^3 - 15x^2 - 28 - Setting x=1 to find the sum of the coefficients: - 3(1)^8 + 2(1)^6 - 3(1)^5 - 5(1)^4 + 6(1)^3 - 15(1)^2 - 28 - 3 + 2 - 3 - 5 + 6 - 15 - 28 = -40 The sum of the coefficients of the polynomial is boxed{-40}.