answer:To solve for the smallest possible value of x + y given the equation frac{1}{x} + frac{1}{y} = frac{1}{12} and x neq y, we follow these steps: 1. Combine the fractions on the left-hand side: [ frac{x + y}{xy} = frac{1}{12} ] 2. Cross-multiply to eliminate the fraction: [ 12(x + y) = xy ] 3. Rearrange the equation to set it to zero: [ xy - 12x - 12y = 0 ] 4. Apply Simon's Favorite Factoring Trick by adding 144 to both sides to complete the square: [ xy - 12x - 12y + 144 = 144 ] 5. Factor the left-hand side: [ (x - 12)(y - 12) = 144 ] 6. To minimize x + y, we look for factors of 144 where x - 12 and y - 12 are closest to each other. The pairs of factors of 144 that are closest to each other are (18, 8) and (16, 9). Adding 12 back to each component to get the original x and y values, we have two pairs: (x, y) = (30, 20) and (x, y) = (28, 21). 7. Calculate the sum x + y for each pair to find the minimum: - For (x, y) = (30, 20), x + y = 50. - For (x, y) = (28, 21), x + y = 49. Therefore, the smallest possible value for x + y is boxed{49}.

answer:Since f(-1)=e^{-1}-1 < 0 and f(0)=1+0 > 0, and the function is continuous, by the existence theorem of roots, there exists a root of the function f(x)=e^{x}+x in the interval (-1,0). Since the endpoints are consecutive integers, the interval (k-1,k) where kin mathbb{Z} that contains the root is (-1,0). Therefore, k=0. Hence, the answer is: boxed{0}. The function f(x)=e^{x}+x has a root in the interval (-1,0), and by the root determination theorem, we can find that f(0)f(1) < 0, solving this inequality will give the range of k. This question tests the existence theorem of roots and examines the students' computational skills, which is quite basic.

answer:First, we need to convert the outlet rates from cubic inches per minute to cubic feet per minute to match the units of the tank's volume and the inlet pipes' rates. 1 cubic foot = 12 inches * 12 inches * 12 inches = 1728 cubic inches Now, let's convert the outlet rates: Outlet pipe 1: 9 cubic inches/min ÷ 1728 cubic inches/cubic foot = 0.00520833 cubic feet/min Outlet pipe 2: 8 cubic inches/min ÷ 1728 cubic inches/cubic foot = 0.00462963 cubic feet/min Outlet pipe 3: 6 cubic inches/min ÷ 1728 cubic inches/cubic foot = 0.00347222 cubic feet/min Now, let's calculate the total outlet rate by adding the rates of the three outlet pipes: Total outlet rate = 0.00520833 + 0.00462963 + 0.00347222 = 0.01331018 cubic feet/min Next, let's calculate the total inlet rate by adding the rates of the two inlet pipes: Total inlet rate = 5 cubic feet/min + 2 cubic feet/min = 7 cubic feet/min To find the net rate at which the tank is being emptied, we subtract the total outlet rate from the total inlet rate: Net emptying rate = Total inlet rate - Total outlet rate Net emptying rate = 7 - 0.01331018 = 6.98668982 cubic feet/min Now, we can calculate the time it takes to empty the tank by dividing the tank's volume by the net emptying rate: Time to empty = Tank volume / Net emptying rate Time to empty = 30 cubic feet / 6.98668982 cubic feet/min Time to empty ≈ 4.294 minutes Therefore, it takes approximately boxed{4.294} minutes to empty the tank when all pipes are opened.

answer:First, rewrite each term in the numerator and denominator: [ text{Numerator} = dfrac{16}{1} cdot dfrac{17}{2} cdot dfrac{18}{3} dotsm dfrac{32}{17} ] [ text{Denominator} = dfrac{18}{1} cdot dfrac{19}{2} cdot dfrac{20}{3} dotsm dfrac{32}{15} ] Next, recognize that these sequences can be expressed using factorials: [ text{Numerator} = frac{32!}{15!}, quad text{Denominator} = frac{32!}{17!} ] Now, simplify the expression: [ frac{frac{32!}{15!}}{frac{32!}{17!}} = frac{32!}{15!} cdot frac{17!}{32!} = frac{17!}{15!} = 16 times 17 = 272 ] [ boxed{272} ]