answer:First, let's evaluate each term separately. 1. Compute log_{3}9: Since 3^2 = 9, by the definition of logarithms, we have log_{3}9 = 2. 2. Compute 4^{-frac{1}{2}}: Recall that a^{-n} = frac{1}{a^n}. So, 4^{-frac{1}{2}} = frac{1}{4^frac{1}{2}}. Now, notice that sqrt{4} = 2, so 4^frac{1}{2} = 2. Therefore, 4^{-frac{1}{2}} = frac{1}{2}. Now, we can subtract these two values: 2 - frac{1}{2} = frac{4}{2} - frac{1}{2} = frac{4-1}{2} = frac{3}{2} Thus, the solution is boxed{frac{3}{2}}.

answer:1. **Calculate the initial lap time**: Initially, Elisa completed 15 laps in 30 minutes. So, the time for one lap would be computed as follows: [ text{Initial lap time} = frac{30 text{ minutes}}{15 text{ laps}} = 2 text{ minutes per lap} ] 2. **Calculate the current lap time**: Now, Elisa can complete 18 laps in 33 minutes. Thus, the time for one lap is: [ text{Current lap time} = frac{33 text{ minutes}}{18 text{ laps}} = 1.8333 text{ minutes per lap} ] which simplifies to: [ text{Current lap time} = frac{11}{6} text{ minutes per lap} ] 3. **Determine the improvement in lap time**: The improvement in lap time per lap can be found by subtracting the current lap time from the initial lap time: [ text{Improvement} = 2 text{ minutes per lap} - frac{11}{6} text{ minutes per lap} = frac{12}{6} - frac{11}{6} = frac{1}{6} text{ minutes per lap} ] 4. **Conclusion**: The improvement in her lap time is frac{1}{6} minutes per lap. [ frac{1{6} text{ minutes per lap}} ] The final answer is boxed{textbf{(A)} frac{1}{6}}

answer:Since a_5 = 2S_4 + 3 and a_6 = 2S_5 + 3, we have 2S_4 = a_5 - 3 and 2S_5 = a_6 - 3. Therefore, 2S_5 - 2S_4 = (a_6 - 3) - (a_5 - 3) = a_6 - a_5 = 2a_5. This implies 3a_5 = a_6. Thus, 3a_5 = a_5q. Solving for q, we get q = 3. Hence, the correct option is boxed{text{B}}.

answer:1. **Angle Bisectors Intersection**: Point M, where AD and CE intersect, is the incenter of triangle ABC. These bisectors divide the triangle into smaller triangles with common vertex M. 2. **Properties of Midpoint**: Since P is the midpoint of BC, triangle ABP and triangle ACP are equal in area. 3. **Analysis of triangle MPE**: By symmetry and properties of angle bisectors, M is equidistant from sides AB and AC, and PE is parallel to AC (as P is the midpoint). Triangle MPE will thus be similar to triangle MAB, but with area reduced by a scaling factor proportional to the square of frac{PM}{MB}. 4. **Calculation of Area Scaling**: - Since P is the midpoint, MB = PC, thus by the properties of bisectors and medians, MP is half of MB (as AD would bisect angle A roughly in half, assuming equal side lengths for simplicity). The area ratio of triangle MPE to triangle MAB is then left(frac{1}{2}right)^2 = frac{1}{4}. - Given that triangle MAB is about frac{1}{3} of triangle ABC (as angle bisectors partition the triangle in such a way), the area of triangle MPE is frac{1}{3} times frac{1}{4} = frac{1}{12} of triangle ABC. Conclusion: The ratio k = frac{1}{12}, so the area of triangle MPE is frac{1}{12} the area of triangle ABC. The final answer is boxed: frac{1{12}}. The final answer is The final answer, based on the problem and calculations, is boxed{textbf{(D)} frac{1}{12}}.