answer:The ratio of time Gwen jogged to time she walked is the amount of time she spent jogging divided by the amount of time she spent walking. Gwen jogged for 15 minutes and walked for 9 minutes. So, the ratio of jogging time to walking time is 15:9. To simplify the ratio, we can divide both numbers by their greatest common divisor, which is 3. 15 ÷ 3 = 5 9 ÷ 3 = 3 Therefore, the simplified ratio of jogging time to walking time is boxed{5:3} .

answer:Given the polynomial, we can simplify f(x) + f(-x) as follows: [ f(x) + f(-x) = (ax^7 - bx^3 + cx - 5) + (a(-x)^7 - b(-x)^3 + c(-x) - 5). ] Since (-x)^7 = -x^7 for odd powers, and (-x)^3 = -x^3 we have: [ f(x) + f(-x) = (ax^7 - bx^3 + cx - 5) + (-ax^7 + bx^3 - cx - 5) = -10. ] Therefore, [ f(x) + f(-x) = -10. ] Given f(2) = 3, we find f(-2) by: [ f(2) + f(-2) = -10 implies 3 + f(-2) = -10 implies f(-2) = -10 - 3. ] Thus, [ f(-2) = boxed{-13}. ]

answer:: Part (a): 1. **Setting Up the Problem**: - Let us consider a plane with two convex shapes Phi_1 and Phi_2. - We need to prove that there exists a line l that simultaneously divides both Phi_1 and Phi_2 into two regions of equal area. 2. **Choosing a Coordinate System**: - Select a direction O X on the plane to serve as a reference for measuring angles. - Any direction O Y can be defined by an angle alpha, which is measured counterclockwise from O X (Fig. 160). 3. **Finding Division Line for Phi_1**: - For each angle alpha, there is a unique line l(alpha) parallel to the direction O Y that divides the area of Phi_1 into two equal parts. 4. **Area Considerations of Phi_2**: - Suppose this line l(alpha) divides Phi_2 into two parts with areas S_1(alpha) (right of l(alpha)) and S_2(alpha) (left of l(alpha)): [ S_1(alpha) - S_2(alpha) > 0 ] 5. **Changing the Angle**: - If we replace alpha by alpha + 180^circ, the line l(alpha + 180^circ) coincides with l(alpha) but has an opposite direction. [ S_1(alpha + 180^circ) = S_2(alpha), quad S_2(alpha + 180^circ) = S_1(alpha) ] [ S_1(alpha + 180^circ) - S_2(alpha + 180^circ) = S_2(alpha) - S_1(alpha) < 0 ] 6. **Application of Intermediate Value Theorem**: - Since S_1(alpha) and S_2(alpha) are continuous functions of alpha, the intermediate value theorem assures us that there exists some alpha_0 for which: [ S_1(alpha_0) - S_2(alpha_0) = 0 ] 7. **Conclusion for Part (a)**: - Therefore, the line l(alpha_0) divides both Phi_1 and Phi_2 into two regions of equal area. Part (b): 1. **Defining the Convex Shape**: - Let Phi be any convex figure on the plane. - We have to prove that there exist two mutually perpendicular lines l and l^{*} that divide Phi into four regions of equal area. 2. **Choosing a Coordinate System**: - Select a direction O X as the initial reference direction. An angle alpha is measured from O X to generate the parallel direction O Y. 3. **Finding the First Division Line**: - For each direction O Y (defined by angle alpha), there exists a unique line l(alpha) parallel to O Y that divides the area of Phi into two equal parts. 4. **Finding the Perpendicular Division Line**: - There is also a unique line l^{*} perpendicular to l(alpha) (that is, direction O X becomes O Y and vice versa), which also divides the area of Phi into two equal parts. 5. **Area Considerations**: - Let S_1(alpha), S_2(alpha), S_3(alpha), S_4(alpha) be the areas of the four parts created by the intersection of l(alpha) and l^{*}: [ S_1(alpha) + S_2(alpha) = S_3(alpha) + S_4(alpha) ] [ S_1(alpha) + S_4(alpha) = S_2(alpha) + S_3(alpha) ] Therefore: [ S_2(alpha) = S_4(alpha) quad text{and} quad S_1(alpha) = S_3(alpha) ] 6. **Finding Equal Division**: - To ensure the four regions are equal, it must hold that S_1(alpha) = S_2(alpha). Assume for now S_1(alpha) > S_2(alpha): [ S_1(alpha) - S_2(alpha) > 0 ] 7. **Rotating the Division Lines**: - Rotating l(alpha) and l^{*} by 90^circ results in new lines. [ S_1(alpha + 90^circ) = S_2(alpha), quad S_2(alpha + 90^circ) = S_1(alpha) ] [ S_1(alpha + 90^circ) - S_2(alpha + 90^circ) = S_2(alpha) - S_1(alpha) < 0 ] 8. **Application of Intermediate Value Theorem**: There exists some angle alpha_0 for which: [ S_1(alpha_0) - S_2(alpha_0) = 0 ] 9. **Conclusion for Part (b)**: - Therefore, the mutually perpendicular lines l(alpha_0) and l^{*}(alpha_0) divide Phi into four equal parts. blacksquare

answer:Let's denote the son's current age as S and the father's current age as F. We are given that S = 27 and F = 48. We are also told that the sum of their ages is 75, which we can confirm: S + F = 27 + 48 = 75 Now, let's analyze the statement "I am some times as old as you were when I was as old as you are." This means that at some point in the past, when the father was the age the son is now (27), the son was a certain age (let's call this age X). The father is now a multiple of that age X. So, we can write the following equation: F = M * X where M is the multiplier we are trying to find. We also know that the difference in their ages is constant. So, when the father was 27, the son was X years old. The difference in their ages then was: 27 - X = F - S 27 - X = 48 - 27 27 - X = 21 X = 27 - 21 X = 6 So, when the father was 27, the son was 6 years old. Now, we can find the multiplier M: F = M * X 48 = M * 6 M = 48 / 6 M = 8 The multiplier for the father's age compared to the son's age at that time is boxed{8} .