answer:Let's denote Rohan's marks in Mathematics as M, in Physics as P, in Chemistry as C, and in Biology as B. According to the information given: 1. M + P + B = 90 2. C = P + 10 3. B = C - 5 We need to find the weighted average of M, C, and B. The weighted average (WA) can be calculated as follows: WA = (M * weightage of M + C * weightage of C + B * weightage of B) / (sum of weightages) Given the weightages are: - Mathematics: 40% - Chemistry: 30% - Biology: 30% We can express the weightages as decimals for the calculation: - Mathematics: 0.4 - Chemistry: 0.3 - Biology: 0.3 Now, we need to express B and C in terms of P to solve for the marks. From equation 2: C = P + 10 From equation 3: B = C - 5 B = (P + 10) - 5 B = P + 5 Now we can substitute B and C in terms of P into equation 1: M + P + (P + 5) = 90 M + 2P + 5 = 90 M = 90 - 2P - 5 M = 85 - 2P Now we can calculate the weighted average using the expressions for M, C, and B in terms of P: WA = (M * 0.4 + C * 0.3 + B * 0.3) WA = ((85 - 2P) * 0.4 + (P + 10) * 0.3 + (P + 5) * 0.3) Now we can distribute the weightages: WA = (34 - 0.8P + 0.3P + 3 + 0.3P + 1.5) Combine like terms: WA = (34 + 3 + 1.5) + (-0.8P + 0.3P + 0.3P) WA = 38.5 - 0.2P We don't have the exact value of P, but we can find it using the information that the total marks in Mathematics, Physics, and Biology is 90. From equation 1: M + P + B = 90 (85 - 2P) + P + (P + 5) = 90 85 - P + 5 = 90 90 - P = 90 P = 0 Now we can find the exact marks for M, C, and B: M = 85 - 2*0 = 85 C = 0 + 10 = 10 B = 0 + 5 = 5 Now we can calculate the weighted average with the exact marks: WA = (M * 0.4 + C * 0.3 + B * 0.3) WA = (85 * 0.4 + 10 * 0.3 + 5 * 0.3) WA = (34 + 3 + 1.5) WA = 38.5 Therefore, the weighted average marks scored in Mathematics, Chemistry, and Biology together is boxed{38.5} .

answer:1. **Define variables**: Let ( x = frac{sqrt{sqrt{10}+3} - sqrt{sqrt{10}-3}}{sqrt{sqrt{10}+2}} ) and ( y = sqrt{6-4sqrt{2}} ). 2. **Simplify ( x )**: Multiply the numerator and the denominator of ( x ) by the conjugate ( sqrt{sqrt{10}-2} ) to rationalize the denominator: [ x = frac{sqrt{sqrt{10}+3} - sqrt{sqrt{10}-3}}{sqrt{sqrt{10}+2}} cdot frac{sqrt{sqrt{10}-2}}{sqrt{sqrt{10}-2}} = frac{(sqrt{10}+3) - (sqrt{10} - 3)}{2} = 3. ] 3. **Rewrite ( sqrt{6-4sqrt{2}} )**, known to be the square root of a special value: [ sqrt{6-4sqrt{2}} = sqrt{(sqrt{2}-2)^2} = 2-sqrt{2} quad text{(simplification of binomial squares)} ] 4. **Calculate ( N )**: [ N = x - y = 3 - (2-sqrt{2}) = 1 + sqrt{2}. ] Thus, the value of (N) is ( 1 + sqrt{2} ). The final answer is boxed{( boxed{text{(D) } 1+sqrt{2}} )}

answer:To solve this, we identify the premise in the question and then negate it according to the definition of the method of contradiction. Since the negation of the proposition "among natural numbers a, b, c, exactly one is even" is "all of a, b, c are odd or at least two of them are even", the correct choice is: boxed{text{D}}. This question tests the ability to prove mathematical propositions using the method of contradiction, by negating the conclusion to be proved, which is the breakthrough point for solving the problem.

answer:1. **Convert the logarithmic equation:** Start with log_2{x} = log_y{25}. Using the change of base formula: [ log_2{x} = frac{log_2{25}}{log_2{y}} = frac{4.5}{log_2{y}} ] since log_2{25} = 4.5. 2. **Express x in terms of y with the new product assumption:** We know xy = 81, so x = frac{81}{y}. Substitute this into the logarithmic equation: [ log_2left(frac{81}{y}right) = frac{4.5}{log_2{y}} ] Simplifying the left side we obtain: [ log_2{81} - log_2{y} = 6.5 - log_2{y} ] Equating this to the previous expression: [ 6.5 - log_2{y} = frac{4.5}{log_2{y}} ] 3. **Solve for log_2{y}:** Clear the fraction by multiplying through by log_2{y}: [ (6.5 - log_2{y})log_2{y} = 4.5 ] Rearranging gives us the quadratic equation: [ (log_2{y})^2 - 6.5log_2{y} + 4.5 = 0 ] Solve this using the quadratic formula: [ log_2{y} = frac{6.5 pm sqrt{(6.5)^2 - 4 cdot 1 cdot 4.5}}{2} = frac{6.5 pm sqrt{18.25}}{2} = frac{6.5 pm 4.27}{2} ] Choose log_2{y} = 3.535, the value that makes y positive. 4. **Calculate log_2{x}:** With log_2{y} = 3.535: [ log_2{x} = frac{4.5}{3.535} ] 5. **Calculate (log_2{tfrac{x}{y}})^2:** [ log_2{tfrac{x}{y}} = log_2{x} - log_2{y} approx 1.275 - 3.535 = -2.260 ] Squaring this result: [ (log_2{tfrac{x}{y}})^2 = (-2.260)^2 approx 5.11 ] So, the solution gives 5.11. The final answer is boxed{**C)** 5.11}