answer:Let's denote the amount of flavoring in the standard formulation as ( F ), the amount of corn syrup as ( CS ), and the amount of water as ( W ). According to the given ratio in the standard formulation, we have: [ F : CS : W = 1 : 12 : 30 ] In the sport formulation, the ratio of flavoring to corn syrup is three times as great, and the ratio of flavoring to water is half that of the standard formulation. This means: [ frac{F_{sport}}{CS_{sport}} = 3 times frac{F}{CS} ] [ frac{F_{sport}}{W_{sport}} = frac{1}{2} times frac{F}{W} ] Given that ( frac{F}{CS} = frac{1}{12} ) and ( frac{F}{W} = frac{1}{30} ), we can write: [ frac{F_{sport}}{CS_{sport}} = 3 times frac{1}{12} = frac{1}{4} ] [ frac{F_{sport}}{W_{sport}} = frac{1}{2} times frac{1}{30} = frac{1}{60} ] This means that for every 1 part of flavoring in the sport formulation, there are 4 parts of corn syrup and 60 parts of water. We are given that the sport formulation contains 6 ounces of corn syrup. Using the ratio ( frac{F_{sport}}{CS_{sport}} = frac{1}{4} ), we can find the amount of flavoring: [ F_{sport} = frac{1}{4} times CS_{sport} ] [ F_{sport} = frac{1}{4} times 6 ] [ F_{sport} = 1.5 text{ ounces} ] Now, using the ratio ( frac{F_{sport}}{W_{sport}} = frac{1}{60} ), we can find the amount of water: [ W_{sport} = 60 times F_{sport} ] [ W_{sport} = 60 times 1.5 ] [ W_{sport} = 90 text{ ounces} ] Therefore, the sport formulation contains boxed{90} ounces of water.

answer:Let's denote the total number of boys in the camp as T. According to the information given, 20% of the total boys are from school A. So, the number of boys from school A is 0.20 * T. Out of these boys from school A, 30% study science. Therefore, 70% of the boys from school A do not study science (since 100% - 30% = 70%). We are told that there are 35 boys from school A who do not study science. So, we can set up the following equation: 0.70 * (0.20 * T) = 35 Now, let's solve for T: 0.70 * 0.20 * T = 35 0.14 * T = 35 T = 35 / 0.14 T = 250 Therefore, the total number of boys in the camp is boxed{250} .

answer:In general, a convex polygon with ( n ) sides has ( frac{n(n-3)}{2} ) diagonals. However, in this case, 5 out of 35 vertices do not contribute to creating diagonals directly. Thus, diagonals can only be formed between the remaining 35 - 5 = 30 vertices. 1. Calculate diagonals using 30 effective vertices: [ text{Diagonals} = frac{30 times (30-3)}{2} = frac{30 times 27}{2} = frac{810}{2} = 405 ] 2. Therefore, the number of diagonals can be represented by [ boxed{405} ]

answer:1. Let the sides of the triangle be (a), (b), and (c). The altitudes corresponding to these sides are (h_a), (h_b), and (h_c) respectively. Given that (h_a + h_b + h_c < 20) and all altitudes are integers. 2. The area ( Delta ) of the triangle can be expressed using any of the altitudes: [ Delta = frac{1}{2} a h_a = frac{1}{2} b h_b = frac{1}{2} c h_c ] Therefore, the altitudes can be written as: [ h_a = frac{2Delta}{a}, quad h_b = frac{2Delta}{b}, quad h_c = frac{2Delta}{c} ] 3. Given that (h_a + h_b + h_c < 20), we substitute the expressions for the altitudes: [ frac{2Delta}{a} + frac{2Delta}{b} + frac{2Delta}{c} < 20 ] Simplifying, we get: [ 2Delta left( frac{1}{a} + frac{1}{b} + frac{1}{c} right) < 20 ] [ Delta left( frac{1}{a} + frac{1}{b} + frac{1}{c} right) < 10 ] 4. Let (s) be the semiperimeter of the triangle, (s = frac{a + b + c}{2}). The inradius (r) is given by: [ r = frac{Delta}{s} ] Given that (r) is an integer, we can write: [ Delta = ks ] where (k) is a positive integer. 5. Using the inequality derived earlier: [ Delta left( frac{1}{a} + frac{1}{b} + frac{1}{c} right) < 10 ] Substituting (Delta = ks): [ ks left( frac{1}{a} + frac{1}{b} + frac{1}{c} right) < 10 ] 6. By the Cauchy-Schwarz inequality: [ (a + b + c) left( frac{1}{a} + frac{1}{b} + frac{1}{c} right) geq 9 ] Therefore: [ s left( frac{1}{a} + frac{1}{b} + frac{1}{c} right) geq frac{9}{2} ] Substituting this into the inequality: [ ks left( frac{1}{a} + frac{1}{b} + frac{1}{c} right) < 10 ] [ k cdot frac{9}{2} < 10 ] [ k < frac{20}{9} ] Since (k) is a positive integer, (k) can be 1 or 2. 7. If (k = 1), then (Delta = s). If (k = 2), then (Delta = 2s). 8. We need to find the possible values of (Delta) such that the altitudes are integers and their sum is less than 20. We test both cases: - For (k = 1): [ Delta = s ] Since (h_a + h_b + h_c < 20), we need to find integer values of (a), (b), and (c) such that the altitudes are integers and their sum is less than 20. This is possible for small integer values of (a), (b), and (c). - For (k = 2): [ Delta = 2s ] Similarly, we need to find integer values of (a), (b), and (c) such that the altitudes are integers and their sum is less than 20. This is also possible for small integer values of (a), (b), and (c). 9. By testing small integer values, we find that the possible values of (Delta) are 6 and 12. The final answer is (boxed{6 text{ and } 12}).