answer:Given: - The outer dimensions of the frame are 100 cm (height) by 140 cm (width). - The frame width is 15 cm on the longer sides and 20 cm on the shorter sides. 1. Subtract twice the frame width from each dimension to find the dimensions of the picture: - For the height: (100 text{ cm} - 2 times 20 text{ cm} = 60 text{ cm}) - For the width: (140 text{ cm} - 2 times 15 text{ cm} = 110 text{ cm}) 2. Calculate the area of the picture: [ text{Area} = text{Height} times text{Width} = 60 text{ cm} times 110 text{ cm} = 6600 text{ cm}^2 ] So, the area of the picture is (boxed{6600 text{ cm}^2}).

answer:To find the largest positive real number s that satisfies at least one equation of the form x^3 + a_2x^2 + a_1x + a_0 = 0 with |a_i| leq 3 for each i, we need to maximize x by minimizing the right-hand side in the rearranged equation: [ x^3 = -(a_2x^2 + a_1x + a_0) ] 1. **Choosing Coefficients**: To minimize a_2x^2 + a_1x + a_0, choose a_2 = a_1 = a_0 = -3 (the minimum possible values within the new bounds). This leads to: [ a_2x^2 + a_1x + a_0 = -3x^2 - 3x - 3 ] 2. **Substituting and Simplifying**: [ x^3 = -(-3x^2 - 3x - 3) = 3x^2 + 3x + 3 ] [ x^3 - 3x^2 - 3x - 3 = 0 ] 3. **Finding the Roots**: The nature of this cubic equation suggests using numerical or graphical methods for exact roots. However, by testing values, we can approximate the location of the largest root. - **At x = 3**: [ f(3) = 3^3 - 3 cdot 3^2 - 3 cdot 3 - 3 = 27 - 27 - 9 - 3 = -12 ] This value is negative. - **At x = 4**: [ f(4) = 4^3 - 3 cdot 4^2 - 3 cdot 4 - 3 = 64 - 48 - 12 - 3 = 1 ] This value is positive. 4. **Conclusion**: Since f(3) is negative and f(4) is positive, the largest root s must be between 3 and 4. We conclude: [ 3 < s < 4 ] The final answer is boxed{textbf{(B)} 3 < s < 4}

answer:1. **Identify the Formulas for Sum:** The sums U_n and V_n of arithmetic sequences can be expressed as: U_n = frac{n}{2} (2c + (n-1)f), quad V_n = frac{n}{2} (2g + (n-1)h) where c and f are the first term and common difference of the first sequence, and g and h are those of the second sequence. 2. **Utilize the Ratio Given:** frac{U_n}{V_n} = frac{5n^2+3n+2}{3n^2+2n+30} Simplifying the ratio of U_n to V_n leads to: frac{n(2c + (n-1)f)}{n(2g + (n-1)h)} = frac{5n^2+3n+2}{3n^2+2n+30} 3. **Equate and Simplify:** frac{2c + (n-1)f}{2g + (n-1)h} = frac{5n^2+3n+2}{3n^2+2n+30} 4. **Solving for Specific Values:** Set n=1: frac{2c}{2g} = frac{10}{33} implies frac{c}{g} = frac{5}{33} Set n=2: frac{2c + f}{2g + h} = frac{28}{42} = frac{2}{3} 5. **Find f and h in Terms of c and g:** frac{2(frac{5}{33}g) + f}{2g + h} = frac{2}{3} Simplify and solve these equations to express f and h in terms of g. 6. **Calculate the 15th Terms:** c_{15} = c + 14f, quad g_{15} = g + 14h Substitute and simplify using values of c, f, g, and h from the above steps. 7. **Final Calculation for Ratio:** frac{c_{15}}{g_{15}} = frac{c + 14f}{g + 14h} Conclusion: The resulting ratio simplifies to frac{125{99}}. The correct answer is B) boxed{frac{125}{99}}.

answer:Since w, x, y, and z are nonnegative numbers whose sum is 200, we have that [wx + xy + yz + wz.] By similar logic as in the original problem, we consider the symmetric breaking by setting w + z and x + y. We express the sum as: [wx + xy + yz + wz = (w+z)(x+y) - wz - xy.] By the AM-GM inequality, [(w+z)(x+y) le left(frac{(w+z) + (x+y)}{2}right)^2 = 10000.] To maximize wx + xy + yz + wz, we aim to minimize wz + xy. Using AM-GM on each pair, [wz + xy geq 2sqrt{wz cdot xy},] the minimum value of wz + xy is 0 which occurs when w = y = 100 and x = z = 0. Then, [wx + xy + yz + wz = 10000 - 0 = 10000.] Thus, the largest value is boxed{10000}.