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question:Betty has two containers. Initially, the first container is tfrac{3}{5} full of water and the second container is empty. She pours all the water from the first container into the second container, and finds that the second container is now tfrac{2}{3} full. What is the ratio of the volume of the first container to the volume of the second container? **A)** frac{5}{6} **B)** frac{7}{8} **C)** frac{9}{10} **D)** frac{10}{9} **E)** frac{11}{12}

answer:Let V_1 be the volume of the first container, and V_2 be the volume of the second container. From the problem, tfrac{3}{5}V_1 = tfrac{2}{3}V_2, because the amount of water moved from the first container fills the second container to this level. The requirement is to find the ratio frac{V_1}{V_2}. By manipulating the equation: [ frac{3}{5}V_1 = frac{2}{3}V_2 implies frac{V_1}{V_2} = frac{frac{2}{3}V_2}{frac{3}{5}V_1} = frac{2}{3} times frac{5}{3} = frac{10}{9}. ] Thus, the ratio of the volume of the first container to that of the second container is: [ frac{10{9}}. ] This means the first container is slightly larger than the second container. The final answer is **D)** boxed{frac{10}{9}}

question:Let f:(0,infty) rightarrow (0,infty) be a function such that [ 10cdot frac{x+y}{xy}=f(x)cdot f(y)f(xy)90 ] for every x,y in (0,infty) . What is f(frac 1{11}) ? textbf{(A)} 1 qquadtextbf{(B)} 11 qquadtextbf{(C)} 21 qquadtextbf{(D)} 31 qquadtextbf{(E)} text{There is more than one solution}

answer:1. Start by plugging in ( y = 1 ) into the given functional equation: [ 10 cdot frac{x + 1}{x cdot 1} = f(x) cdot f(1) - f(x) - 90 ] Simplifying the left-hand side: [ 10 cdot left( frac{x + 1}{x} right) = 10 left( 1 + frac{1}{x} right) = 10 + frac{10}{x} ] Thus, the equation becomes: [ 10 + frac{10}{x} = f(x) cdot f(1) - f(x) - 90 ] 2. Rearrange the equation to isolate ( f(x) ): [ 10 + frac{10}{x} + 90 = f(x) cdot f(1) - f(x) ] [ 100 + frac{10}{x} = f(x) cdot f(1) - f(x) ] [ 100 + frac{10}{x} = f(x) (f(1) - 1) ] 3. Now, plug in ( x = 1 ) and ( y = 1 ) into the original equation: [ 10 cdot frac{1 + 1}{1 cdot 1} = f(1) cdot f(1) - f(1) - 90 ] Simplifying the left-hand side: [ 10 cdot 2 = 20 ] Thus, the equation becomes: [ 20 = f(1)^2 - f(1) - 90 ] [ f(1)^2 - f(1) - 110 = 0 ] 4. Solve the quadratic equation for ( f(1) ): [ f(1) = frac{1 pm sqrt{1 + 4 cdot 110}}{2} = frac{1 pm sqrt{441}}{2} = frac{1 pm 21}{2} ] [ f(1) = 11 quad text{or} quad f(1) = -10 ] Since ( f ) maps ((0, infty)) to ((0, infty)), ( f(1) = -10 ) is not valid. Therefore, ( f(1) = 11 ). 5. Substitute ( f(1) = 11 ) back into the equation: [ 100 + frac{10}{x} = f(x) cdot 10 ] [ f(x) = frac{100 + frac{10}{x}}{10} = 10 + frac{1}{x} ] 6. Finally, find ( fleft( frac{1}{11} right) ): [ fleft( frac{1}{11} right) = 10 + 11 = 21 ] The final answer is ( boxed{21} ).

question:Expand (x+5)(3y+15).

answer:We use the distributive property (also known as the FOIL method for binomials) to expand the expression: [ (x+5)(3y+15) = x(3y+15) + 5(3y+15) ] Breaking it down further: [ = x cdot 3y + x cdot 15 + 5 cdot 3y + 5 cdot 15 ] [ = 3xy + 15x + 15y + 75 ] Thus, the expanded form of the expression is: [ boxed{3xy + 15x + 15y + 75} ]

question:gh school graduating class is made up of 466 students. There are 212 more girls than boys. How many boys are in the class?

answer:Let's denote the number of boys as B and the number of girls as G. According to the information given: G = B + 212 We also know that the total number of students in the class is 466, so: B + G = 466 Now we can substitute the first equation into the second one: B + (B + 212) = 466 Combine like terms: 2B + 212 = 466 Subtract 212 from both sides: 2B = 466 - 212 2B = 254 Now divide by 2 to solve for B: B = 254 / 2 B = 127 So there are boxed{127} boys in the class.