answer:We are given that two heights of a triangle are 10 and 6, and we are asked to prove that the third height is less than 15. Let's denote the sides of the triangle by (a), (b), and (c), and the corresponding heights by (h_a), (h_b), and (h_c). 1. Given: [ h_a = 10 quad text{and} quad h_b = 6 ] 2. The area of the triangle can be expressed using the sides and their corresponding heights: [ text{Area} = frac{1}{2} a h_a = frac{1}{2} b h_b = frac{1}{2} c h_c ] Setting these equal, we get: [ a h_a = b h_b = c h_c ] 3. Since (h_a = 10) and (h_b = 6), we find the ratio for sides (a) and (b): [ a cdot 10 = b cdot 6 implies b = frac{10a}{6} = frac{5a}{3} ] 4. We need to establish an inequality involving (c). Using the triangle inequality, we have: [ b - a < c ] Substituting (b = frac{5a}{3}): [ frac{5a}{3} - a < c ] Simplifying: [ frac{5a}{3} - a = frac{5a}{3} - frac{3a}{3} = frac{2a}{3} implies frac{2a}{3} < c ] 5. We want to express (h_c) in terms of (a) and (c): [ h_c = frac{a h_a}{c} = frac{a cdot 10}{c} = frac{10a}{c} ] 6. Using (frac{2a}{3} < c), we write: [ frac{a}{c} < frac{3}{2} ] Thus: [ h_c = frac{10a}{c} = 10 cdot frac{a}{c} < 10 cdot frac{3}{2} = 15 ] # Conclusion: [ boxed{h_c < 15} ]

answer:Let's calculate the number of canoes built each month, then sum them up. 1. Number of canoes built in January: 2. 2. In February (3 times January's production): 2 times 3 = 6. 3. In March (3 times February's production): 6 times 3 = 18. 4. In April (3 times March's production): 18 times 3 = 54. Adding them all together gives: [ 2 + 6 + 18 + 54 = boxed{80} ]

answer:Here, the number m can be expressed as 1000n + 100(n+1) + 10(n+2) + (n+3) = 1111n + 123. n needs to be a digit such that m is a four-digit number and digits are consecutive integers from n to n+3. The first digit, n, must be from 1 to 6 (as n+3 must still be a single digit): - From n=1, m=1234 = 1111 cdot 1 + 123 - From n=2, m=2345 = 1111 cdot 2 + 123 - Up to n=6, m=6789 = 1111 cdot 6 + 123 Considering division by 43, decompose 1111 and 123 in terms of 43: - 1111 = 25 cdot 43 + 36 - m = 1111n + 123 = 43(25n+2) + 36n + 43 - The remainder is then 36n + 43 Each n from 1 to 6 gives remainders: - For n=1, 36 cdot 1 + 43 = 79 equiv 36 pmod{43} - For n=2, 36 cdot 2 + 43 = 115 equiv 29 pmod{43} - And so forth up to n=6, 36 cdot 6 + 43 = 259 equiv 36 pmod{43} Summing these possible remainders: - Remainders: 36, 29, 22, 15, 8, 1 - Sum: 36 + 29 + 22 + 15 + 8 + 1 = boxed{111}

answer:The amount of foil required to wrap a candy is given by the surface area of the sphere, which is ( 4pi r^2 ). - Sally's candies: ( 4pi cdot 5^2 = 100pi ) square millimeters. - Dean's candies: ( 4pi cdot 7^2 = 196pi ) square millimeters. - June's candies: ( 4pi cdot 9^2 = 324pi ) square millimeters. To find when all three workers finish wrapping a candy simultaneously, calculate the least common multiple (LCM) of the surface areas: - Sally: ( 100 = 2^2 cdot 5^2 ) - Dean: ( 196 = 2^2 cdot 7^2 ) - June: ( 324 = 2^2 cdot 3^4 ) LCM of ( 100pi, 196pi, ) and ( 324pi ): - LCM of constants: ( 2^2 cdot 5^2 cdot 7^2 cdot 3^4 = 44100 ) Therefore, the LCM is ( 44100pi ). The number of candies Sally wraps by that point is ( frac{44100pi}{100pi} = 441 ). Conclusion: Sally will have wrapped (boxed{441}) candies by the time they all finish their tasks together.