answer:Let's denote the speed at which Ms. Walker drives home from work as ( v ) miles per hour. The time it takes for Ms. Walker to drive to work at a speed of 60 miles per hour for a distance of 24 miles is given by the formula: [ text{Time} = frac{text{Distance}}{text{Speed}} ] So, the time to get to work is: [ text{Time to work} = frac{24 text{ miles}}{60 text{ mph}} ] [ text{Time to work} = frac{2}{5} text{ hours} ] The time it takes for her to drive home at a speed of ( v ) miles per hour for the same distance of 24 miles is: [ text{Time to home} = frac{24 text{ miles}}{v text{ mph}} ] Since the round trip takes 1 hour, we can write the equation: [ text{Time to work} + text{Time to home} = 1 text{ hour} ] [ frac{2}{5} text{ hours} + frac{24}{v} text{ hours} = 1 text{ hour} ] Now we solve for ( v ): [ frac{24}{v} = 1 - frac{2}{5} ] [ frac{24}{v} = frac{5}{5} - frac{2}{5} ] [ frac{24}{v} = frac{3}{5} ] Cross-multiply to solve for ( v ): [ 24 cdot 5 = 3v ] [ 120 = 3v ] [ v = frac{120}{3} ] [ v = 40 text{ mph} ] So, Ms. Walker's speed when driving home from work is boxed{40} miles per hour.

answer:```markdown 1. **Given:** We want to prove the following inequality for a, b, c in mathbb{R}_+: [ sum_{text{cyc}} frac{(2a + b + c)^2}{2a^2 + (b + c)^2} leq 8. ] 2. **Normalization**: Since the inequality involves symmetric terms a, b, and c, we can use the normalization a + b + c = 3. This simplifies our analysis. 3. **Reformulating the Inequality**: With the normalization a + b + c = 3, we reframe the inequality as: [ sum_{text{cyc}} frac{(a+3)^2}{2a^2 + (3-a)^2} leq 8. ] 4. **Defining the Function**: Consider the function: [ f(x) = frac{(x + 3)^2}{2x^2 + (3 - x)^2}, quad text{for } 0 < x < 3. ] We will analyze the behavior of this function to establish the inequality. 5. **Simplifying the Function**: Let's simplify the expression inside the function: [ f(x) = frac{x^2 + 6x + 9}{2x^2 + (3 - x)^2}. ] Expanding the denominator: [ 2x^2 + (3 - x)^2 = 2x^2 + 9 - 6x + x^2 = 3x^2 - 6x + 9. ] Thus, [ f(x) = frac{x^2 + 6x + 9}{3(x^2 - 2x + 3)}. ] 6. **Further Simplification**: Factor out common terms: [ f(x) = frac{1}{3} left[ 1 + frac{8x + 6}{(x - 1)^2 + 2} right]. ] 7. **Finding the Upper Bound**: Notice that the term frac{8x + 6}{(x - 1)^2 + 2} achieves its maximum when (x - 1)^2 is minimized. Using the fact the minimum value of (x-1)^2 is zero, we get: [ frac{8x + 6}{(x - 1)^2 + 2} leq frac{8x + 6}{2} = 4x + 3. ] Therefore, [ f(x) leq frac{1}{3} [1 + (4x + 3)] = frac{4x + 4}{3}. ] 8. **Summarizing the Cyclic Sum**: We now sum up separately for a, b, c: [ sum_{text{cyc}} f(a) leq sum_{text{cyc}} frac{4a + 4}{3}. ] Substituting a + b + c = 3: [ sum_{text{cyc}} frac{4a + 4}{3} = frac{4(a + b + c) + 12}{3} = frac{4 cdot 3 + 12}{3} = 8. ] 9. **Conclusion**: We have thus shown that: [ sum_{text{cyc}} frac{(2a + b + c)^2}{2a^2 + (b + c)^2} leq 8. ] Therefore, the inequality is proven, and it follows that: [ boxed{8}. ] ```

answer:Since the inequality (frac{a}{e^{a}}-b)^{2}geqslant m-(a-b+3)^{2} holds for any real numbers a and b, we have mleqslant (a-b+3)^{2}+(frac{a}{e^{a}}-b)^{2} always holds. We need to find the minimum value of (a-b+3)^{2}+(frac{a}{e^{a}}-b)^{2} to maximize m. The geometric meaning of (a-b+3)^{2}+(frac{a}{e^{a}}-b)^{2} is the square of the distance between the point (a,frac{a}{e^{a}}) and the point (b-3,b). The point (a,frac{a}{e^{a}}) is on the curve y=frac{x}{e^{x}}, and the point (b-3,b) is on the line y=x+3. Therefore, our problem is equivalent to finding the minimum value of the square of the distance between a point on the line y=x+3 and a point on the curve y=frac{x}{e^{x}}. Taking the derivative of y=frac{x}{e^{x}}, we get y'=frac{1-x}{e^{x}}. Setting y'=1, we have frac{1-x}{e^{x}}=1, which gives x=0. This means that the slope of the tangent line to the curve y=frac{x}{e^{x}} at x=0 is 1, and the coordinates of the tangent point are (0,0). The distance from this point to the line y=x+3 is the minimum value we are looking for, which is frac{3}{sqrt{2}}=frac{3sqrt{2}}{2}. Squaring this, we get frac{9}{2}. Therefore, mleqslant frac{9}{2}, so the maximum value of m is boxed{frac{9}{2}}. This problem involves the concept of inequalities that always hold, the geometric meaning of derivatives, and the formula for the distance from a point to a line. The key is to convert the problem. It is a moderate-level problem.

answer:Each page x is now paired with the page number 64 - x for 1 leq x leq 63. To satisfy the condition that old and new page numbers share the same units digit, the equation: [ x equiv 64 - x pmod{10} ] should be true. Simplifying, we get: [ 2x equiv 64 pmod{10} ] [ 2x equiv 4 pmod{10} ] [ x equiv 2 pmod{5} ] The solution to this equation is x = 2, 7, 12, 17, 22, 27, 32, 37, 42, 47, 52, 57, 62. Thus, there are 13 such pages and the answers are: [ boxed{13} ]