answer:(1) The expressions for the functions are f(x) = x^2 + 2x and g(x) = -x^2 - 2x. (2) The range of the real number lambda is boxed{(-infty, 0]}.

answer:1. **Define Variables:** Let ( M ) represent the total amount of money Elena has. Let ( B ) represent the total cost of all the books she wants to buy. 2. **Establish Relationships:** According to the new problem setup, Elena uses one-third of her money to buy half of the books. This is mathematically expressed as: [ frac{1}{3}M = frac{1}{2}B ] 3. **Solve for Total Cost of Books:** To determine the total cost of all the books, rearrange the equation to solve for ( B ): [ B = 2 left(frac{1}{3}Mright) = frac{2}{3}M ] This means the total cost of all the books is ( frac{2}{3}M ). 4. **Calculate Remaining Money:** After buying all the books, the amount of money Elena has left is: [ M - frac{2}{3}M = frac{3}{3}M - frac{2}{3}M = frac{1}{3}M ] 5. **Conclusion:** The fraction of her money that Elena will have left after buying all the books is ( frac{1}{3} ). Therefore, the answer is ( frac{1{3}} ). The final answer is boxed{**B)** ( frac{1}{3} )}

answer:(1) From the given conditions, we have c=frac{5}{2} and b=sqrt{6}. Using the formula 4a = 3sqrt{6}cos A = 3sqrt{6}left(frac{b^2+c^2-a^2}{2bc}right), we can substitute the values of b and c to obtain 12a^2 + 80a - 147 = 0. Solving this quadratic equation, we find a=frac{3}{2} or a=-frac{49}{6} (neglected since the length cannot be negative). So, a=boxed{frac{3}{2}}. (2) From part (1), we have cos A = frac{4}{3sqrt{6}}cdotfrac{3}{2} = frac{sqrt{6}}{3}. Therefore, sin A = sqrt{1-cos^2 A} = frac{sqrt{3}}{3}. Now, using the double angle formula, we obtain cos 2A = cos^2 A - sin^2 A = frac{1}{3}. Since a=frac{3}{2}, c=frac{5}{2}, and b=sqrt{6}, we have cos B = frac{a^2+c^2-b^2}{2ac} = frac{1}{3}. So, cos 2A = cos B. As c>b>a in triangle ABC, we can conclude that B=2A. Hence, lambda=boxed{2}.

answer:To find the largest quotient formed using two numbers from the set {-36, -6, -4, 3, 7, 9}, consider both maximizing the numerator and minimizing the denominator, while maintaining a positive quotient. 1. **Maximizing the Quotient**: - The largest quotient frac{a}{b} occurs where a is largest and b is the smallest positive value, or both a and b are negative with a large and b small in magnitude. 2. **Case 1: Both a and b are positive**: - Choose a = 9 (max positive number). - Choose b = 3 (min positive number but not the smallest absolute, which allows a higher ratio than using 7). - Compute frac{9}{3} = 3. 3. **Case 2: Both a and b are negative**: - Choose a = -36 (max in absolute among negatives). - Choose b = -4 (min in absolute among negatives but not the smallest overall, providing a larger ratio than using -6). - Compute frac{-36}{-4} = 9. 4. **Comparison and Conclusion**: - From Case 1, the quotient is 3. - From Case 2, the quotient is 9. - Therefore, the largest quotient is 9. Concluding, the largest quotient that can be formed using numbers from this revised set is 9. boxed{The correct answer is D) 9.}