answer:# Problem: Given a triangle and a rhombus inscribed in it with a common angle, where the sides of the triangle enclosing this angle are in the ratio frac{m}{n}, find the ratio of the area of the rhombus to the area of the triangle. 1. **Understand the Ratio and Setup**: The sides of the triangle enclosing the common angle are in the ratio frac{m}{n}. 2. **Identify the Similar Triangles**: The sides of the rhombus divide the triangle into smaller, similar triangles with similarity ratios frac{n}{m+n} and frac{m}{m+n}. This similarity arises because the diagonals of a rhombus bisect each other and the angles of the triangle. 3. **Calculate the Areas of the Similar Triangles**: The areas of the triangles that the rhombus creates with the initial triangle can be denoted as: - The area of the first similar triangle is left(frac{n}{m+n}right)^2 cdot S, where S is the area of the original triangle. - The area of the second similar triangle is left(frac{m}{m+n}right)^2 cdot S. 4. **Express the Area of the Rhombus**: The area of the rhombus can be found by subtracting the areas of these smaller triangles from the area of the original triangle: [ text{Area of Rhombus} = S - left( frac{n^2}{(m+n)^2} cdot S right) - left( frac{m^2}{(m+n)^2} cdot S right) ] 5. **Simplify the Expression**: Bringing all the terms together, we have: [ text{Area of Rhombus} = S - frac{n^2}{(m+n)^2} S - frac{m^2}{(m+n)^2} S ] Factor out the area S: [ text{Area of Rhombus} = S left( 1 - frac{n^2}{(m+n)^2} - frac{m^2}{(m+n)^2} right) ] 6. **Combine the Fractions**: Notice that: [ 1 - frac{n^2}{(m+n)^2} - frac{m^2}{(m+n)^2} = 1 - frac{n^2 + m^2}{(m+n)^2} ] The identity 1 = frac{(m+n)^2}{(m+n)^2} can be used: [ 1 - frac{n^2 + m^2}{(m+n)^2} = frac{(m+n)^2 - (n^2 + m^2)}{(m+n)^2} ] Simplify the numerator: [ (m+n)^2 - (n^2 + m^2) = m^2 + 2mn + n^2 - n^2 - m^2 = 2mn ] Thus: [ 1 - frac{n^2 + m^2}{(m+n)^2} = frac{2mn}{(m+n)^2} ] 7. **Final Area Ratio**: The ratio of the area of the rhombus to the area of the triangle is: [ frac{text{Area of Rhombus}}{text{Area of Triangle}} = frac{S cdot frac{2mn}{(m+n)^2}}{S} = frac{2mn}{(m+n)^2} ] 8. **Conclude**: Therefore, the ratio of the area of the rhombus to the area of the triangle is: [ boxed{frac{2mn}{(m+n)^2}} ]

answer:1. **Calculate the gasoline usage for each car:** - Ray's car averages 50 miles per gallon, so the gasoline used by Ray's car is frac{150}{50} = 3 gallons. - Tom's car averages 20 miles per gallon, so the gasoline used by Tom's car is frac{300}{20} = 15 gallons. 2. **Calculate the total gasoline used:** [ text{Total gallons used} = 3 + 15 = 18 text{ gallons} ] 3. **Calculate the total distance driven by both cars:** [ text{Total distance driven} = 150 text{ miles} + 300 text{ miles} = 450 text{ miles} ] 4. **Calculate the combined miles per gallon:** The combined miles per gallon is the total distance driven divided by the total gallons used: [ text{Combined miles per gallon} = frac{450 text{ miles}}{18 text{ gallons}} = 25 ] 5. **Conclusion:** The cars' combined rate of miles per gallon of gasoline is 25. The final answer is boxed{textbf{(C)} 25}

answer:1. Let ( x, y, z ) be distinct integers. 2. We need to prove that ( (x-y)^5 + (y-z)^5 + (z-x)^5 ) is divisible by ( 5(y-z)(z-x)(x-y) ). 3. Begin by setting: [ u = x-y, quad v = y-z ] Therefore, [ w = z-x = -((x-y) + (y-z)) = -(u+v) ] 4. Then, the expression becomes: [ u^5 + v^5 + (-(u+v))^5 ] 5. By using the binomial theorem, expand ( (-(u+v))^5 ): [ (-(u+v))^5 = -big( (u+v)^5 big) ] Expand ( (u+v)^5 ) using binomial expansion: [ (u+v)^5 = sum_{k=0}^{5} binom{5}{k} u^{5-k} v^k = u^5 + 5u^4v + 10u^3v^2 + 10u^2v^3 + 5uv^4 + v^5 ] 6. Hence, [ (-(u+v))^5 = -(u^5 + 5u^4v + 10u^3v^2 + 10u^2v^3 + 5uv^4 + v^5) ] 7. Then, consider: [ u^5 + v^5 + (-(u+v))^5 ] Substituting the expanded form, we get: [ u^5 + v^5 - (u^5 + 5u^4v + 10u^3v^2 + 10u^2v^3 + 5uv^4 + v^5) ] 8. Simplify the terms: [ u^5 + v^5 - u^5 - 5u^4v - 10u^3v^2 - 10u^2v^3 - 5uv^4 - v^5 ] 9. Combine like terms: [ 0 - 5(u^4v + 2u^3v^2 + 2u^2v^3 + uv^4) ] 10. Factor out ( -5uv ): [ -5uv (u^3 + 2u^2v + 2uv^2 + v^3) ] 11. Notice that ( u = x-y ), ( v = y-z ), and ( w = -(u+v) ) implies: [ -5(x-y)(y-z)(x-y + y-z) ] 12. We see that the expression ( -5(x-y)(y-z)(z-x) ) clearly shows divisibility by ( 5(x-y)(y-z)(z-x) ). # Conclusion: [ boxed{(x-y)^{5} + (y-z)^{5} + (z-x)^{5} text{ is divisible by } 5(y-z)(z-x)(x-y)} ]

answer:1. Observe that we are adding the fraction (frac{1}{3}) a total of 7 times. frac{1}{3} + frac{1}{3} + frac{1}{3} + frac{1}{3} + frac{1}{3} + frac{1}{3} + frac{1}{3} 2. Realize that adding the same fraction multiple times is equivalent to multiplying the fraction by the number of times it is added. frac{1}{3} + frac{1}{3} + frac{1}{3} + frac{1}{3} + frac{1}{3} + frac{1}{3} + frac{1}{3} = 7 times frac{1}{3} 3. Perform the multiplication: 7 times frac{1}{3} = frac{7}{3} Conclusion: The expression (frac{1}{3} + frac{1}{3} + frac{1}{3} + frac{1}{3} + frac{1}{3} + frac{1}{3} + frac{1}{3}) simplifies to (7 times frac{1}{3}), which matches option (E). [ boxed{7 times frac{1}{3}} ]