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question:Express .overline{35} as a common fraction.
answer:Let x = .overline{35}. Then multiplying x by 100 (since the repeating part has 2 digits), we have 100x = 35.overline{35}. To remove the repeating decimal, subtract x from 100x: 100x - x = 35.overline{35} - .overline{35} This simplifies to: 99x = 35 Therefore, solving for x: x = frac{35}{99} Thus, .overline{35} as a common fraction is boxed{frac{35}{99}}.
question:Let a,b,c be real numbers such that a+b+c= (frac{1}{a})+(frac{1}{b})+(frac{1}{c}) and abc=1 . Prove that at least one of a,b,c is 1
answer:Given the conditions: 1. (a + b + c = frac{1}{a} + frac{1}{b} + frac{1}{c}) 2. (abc = 1) We need to prove that at least one of (a, b, c) is 1. 1. Start with the given equation: [ a + b + c = frac{1}{a} + frac{1}{b} + frac{1}{c} ] 2. Using the second condition (abc = 1), we can rewrite the right-hand side: [ frac{1}{a} + frac{1}{b} + frac{1}{c} = frac{bc + ac + ab}{abc} = bc + ac + ab ] Therefore, the equation becomes: [ a + b + c = bc + ac + ab ] 3. Rearrange the equation to isolate terms involving (a, b, c): [ a + b + c - ab - ac - bc = 0 ] 4. Add and subtract 1 to the left-hand side: [ a + b + c - ab - ac - bc + 1 - 1 = 0 ] 5. Group the terms to factorize: [ (a - 1)(b - 1)(c - 1) = 0 ] 6. The factored equation ((a - 1)(b - 1)(c - 1) = 0) implies that at least one of the factors must be zero. Therefore, at least one of (a, b, c) must be 1. (blacksquare)
question:Melody planted sunflowers from two different seed packets, Packet A and Packet B, in her garden. She planted the sunflowers at different times and in different areas with varying sunlight exposure. The growth rates (in inches per day) for the two types of sunflowers are given by the following equations: Growth rate of Packet A sunflowers: R_A = 2x + y 0.1W Growth rate of Packet B sunflowers: R_B = 3x y + 0.2W Where x is the number of days since planting, y is the daily average sunlight exposure (measured in hours), and W represents the combined effect of competition with other plants for nutrients and space, as well as fluctuating weather conditions (measured on a 010 scale, with 10 being the highest competition and fluctuation). After observing for several days, Melody found that the sunflowers from Packet A were on average 20% taller than the sunflowers from Packet B. If on day 10, the sunflower from Packet A was 192 inches tall, calculate the average height of the sunflowers from Packet B on that day, assuming that the sunflowers received 6 hours of sunlight exposure on each day and the level of competition and weather fluctuation (W) was consistent at 5 throughout the 10 days.
answer:First, let's calculate the growth rate of the sunflowers from Packet A (R_A) using the given equation and the provided values for x, y, and W: R_A = 2x + y - 0.1W Given that x = 10 days, y = 6 hours of sunlight exposure, and W = 5, we have: R_A = 2(10) + 6 - 0.1(5) R_A = 20 + 6 - 0.5 R_A = 26 - 0.5 R_A = 25.5 inches per day Now, let's calculate the growth rate of the sunflowers from Packet B (R_B) using the given equation: R_B = 3x - y + 0.2W Again, using x = 10 days, y = 6 hours of sunlight exposure, and W = 5, we have: R_B = 3(10) - 6 + 0.2(5) R_B = 30 - 6 + 1 R_B = 24 + 1 R_B = 25 inches per day Now, we know that the sunflowers from Packet A are 20% taller than those from Packet B. Let's denote the height of the sunflowers from Packet B as H_B. According to the information given, the height of the sunflowers from Packet A (H_A) is: H_A = H_B + 0.2H_B H_A = 1.2H_B We are given that on day 10, the sunflower from Packet A was 192 inches tall, so: 192 = 1.2H_B To find H_B, we divide both sides by 1.2: H_B = 192 / 1.2 H_B = 160 inches Therefore, the average height of the sunflowers from Packet B on day 10 is boxed{160} inches.
question:John found that the average of 15 numbers is a certain value. If 14 is added to each number, then the mean of the numbers is 54. What was the initial average of the numbers?
answer:Let's denote the initial average of the 15 numbers as A. The sum of the 15 numbers is then 15 * A. If John adds 14 to each of the 15 numbers, the sum of the numbers will increase by 14 * 15. The new sum of the numbers will be 15 * A + 14 * 15. The new average after adding 14 to each number is given as 54. So, the new sum divided by the number of values (which is still 15) is 54: (15 * A + 14 * 15) / 15 = 54 Now, we can solve for A: 15 * A + 210 = 54 * 15 15 * A + 210 = 810 15 * A = 810 - 210 15 * A = 600 A = 600 / 15 A = 40 Therefore, the initial average of the numbers was boxed{40} .