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question:the final game of the basketball season, four players scored points. Chandra scored twice as many points as did Akiko. Akiko scored 4 more points than did Michiko, and Michiko scored half as many points as did Bailey. If Bailey scored 14 points, how many points in total did the team score in the final game of the season?
answer:If Bailey scored 14 points, then Michiko scored half as many points, which is 14/2 = 7 points. Akiko scored 4 more points than Michiko, so Akiko scored 7+4 = 11 points. Chandra scored twice as many points as Akiko, so Chandra scored 11*2 = 22 points. The total points scored by the team in the final game of the season is 14+7+11+22 = 54 points. 54 The answer is: 54
question:What is the smallest positive integer x greater than 1 that is coprime with 120 (meaning the greatest common divisor of x and 120 is 1)?
answer:The prime factorization of 120 is 2^3 cdot 3 cdot 5. To find the smallest positive integer x that is coprime with 120, we need to find a number that does not have any factors of 2, 3, or 5. Since x must be greater than 1, the smallest such number is boxed{7}. The answer is: 7
question:Layla is feeding her pet fish. She has two Goldfish which each get one teaspoon of fish food. Her 3 Swordtails each get 2 teaspoons of food. Her 8 Guppies each eat half a teaspoon of food. How much food does she have to give to her fish in total?
answer:Layla has 2 Goldfish and each needs 1 teaspoon of food, so the Goldfish need 2 x 1 = 2 teaspoons of food. She also has 3 Swordtails and each needs 2 teaspoons of food, so the Swordtails need 3 x 2 = 6 teaspoons of food. And she has 8 Guppies and each needs half a teaspoon of food, so the Guppies need 8 x 0.5 = 4 teaspoons of food. In total, Layla needs to give her fish 2 + 6 + 4 = 12 teaspoons of food. 12 The answer is: 12
question:Given triangle triangle JKL as shown, with tan K = frac{3}{2}, what is the length of KL?
answer:We know that tan K = frac{3}{2} = frac{JK}{KL}. Therefore, JK = frac{3}{2} KL. Using the Pythagorean Theorem, we have JK^2 + KL^2 = JL^2. Substituting frac{3}{2} KL for JK, we have left(frac{3}{2} KLright)^2 + KL^2 = JL^2. Simplifying, we get frac{9}{4} KL^2 + KL^2 = JL^2. Combining like terms, we have frac{13}{4} KL^2 = JL^2. Taking the square root of both sides, we have frac{sqrt{13}}{2} KL = JL. Since JL = KL + JK, we have frac{sqrt{13}}{2} KL = KL + frac{3}{2} KL. Simplifying, we have frac{sqrt{13}}{2} KL = frac{5}{2} KL. Dividing both sides by KL, we have frac{sqrt{13}}{2} = frac{5}{2}. Therefore, KL = boxed{sqrt{13}}. The answer is: sqrt{13}