answer:1. Start with the given equation: [ frac{5}{m} + frac{3}{n} = 1. ] 2. Multiply both sides by mn: [ 5n + 3m = mn. ] 3. Rearrange the terms: [ mn - 5n - 3m = 0. ] 4. Add 15 to both sides to aid with factoring: [ mn - 5n - 3m + 15 = 15. ] [ (m-3)(n-5) = 15. ] 5. Factorize 15 to find integer pairs that satisfy (m-3, n-5): [ 15 = 1 times 15, 15 = 3 times 5, 15 = 15 times 1, 15 = -1 times -15, 15 = -3 times -5, 15 = -15 times -1. ] Positive pairs are: [ (m-3, n-5) = (1, 15), (3, 5), (15, 1). ] And translating back to (m, n) gives: [ (m, n) = (4, 20), (6, 10), (18, 6). ] 6. Apply the condition that m must be odd: [ (m, n) = (4, 20), (6, 10), (18, 6) Rightarrow text{None of these have } m text{ as odd, so no solutions.} ] Conclusion: The condition that m is odd yields no valid pairs, thus there are 0 solutions. The final answer is boxed{A) 0}

answer:1. **Identify the given and required areas**: - Given: The radius of the smaller sphere ( r_1 = 4 ) cm. - Given: The radius of the larger sphere ( r_2 = 6 ) cm. - The area on the smaller sphere marked by the ball is ( 37 ) square cm. 2. **Understand the ratio of the spherical caps on the two spheres**: - The spherical caps left by the ball on both the smaller and larger spheres will have a proportional area based on the square of their respective radii. This is derived from the fact that for a given spherical cap created by an object maintaining its curvature, the area of the cap is proportional to the square of the radius of the sphere (( A propto r^2 )). 3. **Calculate the proportionality ratio**: - Since area ( A propto r^2 ), the ratio of the areas on the two spheres due to the same spherical cap is: [ frac{A_2}{A_1} = frac{r_2^2}{r_1^2} = frac{6^2}{4^2} = frac{36}{16} = frac{9}{4} ] 4. **Determine the area on the larger sphere**: - Let ( A_2 ) represent the area on the larger sphere. - By the ratio derived, ( A_2 = A_1 times frac{9}{4} ). - Substitute ( A_1 = 37 ) square cm: [ A_2 = 37 times frac{9}{4} = 37 times 2.25 = 83.25 ] **Conclusion**: [ boxed{83.25} ]

answer:To determine the value(s) of ( k ) such that the expression [ a x^{2}+b x+c+kleft(x^{2}+1right) ] is a perfect square, we need to manipulate the expression and set conditions for it to be a perfect square trinomial. 1. **Rewrite the expression**: [ a x^{2}+b x+c+k(x^{2}+1) ] Simplifies to: [ (a+k)x^{2} + b x + (c+k) ] 2. **Condition for a perfect square**: For the quadratic expression ((a+k)x^{2} + b x + (c+k)) to be a perfect square, it must be of the form: [ left( sqrt{a+k} x + sqrt{c+k} right)^2 ] When expanded, the squared form becomes: [ left( sqrt{a+k} x + sqrt{c+k} right)^2 = (sqrt{a+k})^2 x^2 + 2 sqrt{a+k} sqrt{c+k} x + (sqrt{c+k})^2 ] which simplifies to: [ (a+k) x^2 + 2 sqrt{(a+k)(c+k)} x + (c+k) ] By comparing coefficients, we see that: [ b = 2 sqrt{(a+k)(c+k)} ] 3. **Solving for ( k )**: Square both sides of the above equation: [ b^2 = 4(a+k)(c+k) ] 4. **Formulate and solve the quadratic equation in ( k )**: Rewrite the equation as: [ b^2 = 4ac + 4ak + 4ck + 4k^2 ] Move all terms to one side: [ 4k^2 + 4(a+c)k + 4ac - b^2 = 0 ] Divide the entire equation by 4: [ k^2 + (a+c)k + ac - frac{b^2}{4} = 0 ] 5. **Solve the quadratic equation using the quadratic formula**: The quadratic formula is: [ k = frac{-B pm sqrt{B^2 - 4AC}}{2A} ] Here, ( A = 1 ), ( B = a+c ), and ( C = ac - frac{b^2}{4} ): [ k = frac{-(a+c) pm sqrt{(a+c)^2 - 4 (ac - frac{b^2}{4})}}{2} ] 6. **Simplify the discriminant**: [ k = frac{-(a+c) pm sqrt{(a+c)^2 - 4ac + b^2 }}{2} ] Notice that: ((a+c)^2 - 4ac = a^2 + c^2 + 2ac - 4ac = a^2 + c^2 - 2ac = (a-c)^2) Hence: [ k = frac{-(a+c) pm sqrt{(a-c)^2 + b^2}}{2} ] 7. **Final values for ( k )**: [ k = frac{-(a+c) + sqrt{(a-c)^2 + b^2}}{2} ] and [ k = frac{-(a+c) - sqrt{(a-c)^2 + b^2}}{2} ] 8. **Conclusion**: Since (sqrt{(a-c)^2 + b^2}) is always a real number (non-negative), ( k ) has two real values. Therefore, ( k ) has indeed two real values. [boxed{k = -frac{a+c}{2} pm frac{1}{2} sqrt{(a-c)^2 + b^2}}]

answer:Let the first part of the sum be x rupees and the second part be (2795 - x) rupees. According to the problem, the interest on the first part for 8 years at 3% per annum is equal to the interest on the second part for 3 years at 5% per annum. Interest on the first part = Principal × Rate × Time / 100 Interest on the first part = x × 3% × 8 / 100 Interest on the first part = x × 3 × 8 / 100 Interest on the first part = 24x / 100 Interest on the second part = Principal × Rate × Time / 100 Interest on the second part = (2795 - x) × 5% × 3 / 100 Interest on the second part = (2795 - x) × 5 × 3 / 100 Interest on the second part = (2795 - x) × 15 / 100 Since the interests are equal, we can set them equal to each other: 24x / 100 = (2795 - x) × 15 / 100 Now, we can solve for x: 24x = (2795 - x) × 15 24x = 41925 - 15x 24x + 15x = 41925 39x = 41925 x = 41925 / 39 x = 1075 So, the first part of the sum is 1075 rupees. Now, we can find the second part of the sum: Second part = Total sum - First part Second part = 2795 - 1075 Second part = 1720 Therefore, the second sum is boxed{1720} rupees.