answer:Suppose Grace has d dimes. Since she has the same number of pennies, she also has d pennies. The total value of her dimes and pennies is then calculated by .10d + .01d = .11d. To find the maximum number of dimes, we need to solve for d in the equation: [ .11d = 4.80 ] Dividing both sides by .11 gives: [ d = frac{4.80}{0.11} = frac{480}{11} approx 43.636 ] Thus, the greatest integer number of dimes she could have is boxed{43}. This amount is obtainable; for instance, she could have 43 pennies making up the rest.

answer:Given x > 0, y > 0, and frac{2}{x} + frac{1}{y} = 1, we are tasked with finding the range of real number m for which 2x + y > m^2 + 8m always holds. Let's break down the solution into detailed step-by-step format: 1. We start with the given condition frac{2}{x} + frac{1}{y} = 1. Multiplying this equation by 2x + y, we get: [ (2x + y) left(frac{2}{x} + frac{1}{y}right) = 2x cdot frac{2}{x} + 2x cdot frac{1}{y} + y cdot frac{2}{x} + y cdot frac{1}{y} ] Simplifying the above, we have: [ = 4 + frac{2x}{y} + frac{2y}{x} + 1 ] This simplifies further to: [ = 5 + frac{2x}{y} + frac{2y}{x} ] 2. By applying the AM-GM inequality, where the arithmetic mean is always greater than or equal to the geometric mean, for the terms frac{2x}{y} and frac{2y}{x}, we have: [ frac{frac{2x}{y} + frac{2y}{x}}{2} geqslant sqrt{frac{2x}{y} cdot frac{2y}{x}} ] Simplifying, we get: [ frac{2x}{y} + frac{2y}{x} geqslant 2sqrt{2} ] Substituting this back into our equation, we find: [ 5 + frac{2x}{y} + frac{2y}{x} geqslant 5 + 2sqrt{2} = 9 ] 3. Given 2x + y > m^2 + 8m always holds, we compare it with our derived condition that 5 + frac{2x}{y} + frac{2y}{x} geqslant 9. This implies that: [ m^2 + 8m < 9 ] To find the range of m, we solve the inequality: [ m^2 + 8m - 9 < 0 ] Factoring the quadratic equation, we get: [ (m + 9)(m - 1) < 0 ] This inequality holds for m in (-9, 1). Therefore, the correct range for m where the inequality 2x + y > m^2 + 8m always holds is -9 < m < 1. Therefore, the answer is: [ boxed{B} ]

answer:1. **Introduction and Problem Restatement**: The problem asks us whether the fact that the segments of the medians of a tetrahedron falling inside the inscribed sphere are equal implies that the tetrahedron is regular. 2. **Counterexample Setup**: Consider a rectangular prism that is not a cube. In particular, let's denote such a rectangular prism by the vertices ABCD on the bottom face and EFGH on the top face, such that A, B, C, and D form a square, and similarly, E, F, G, and H form a square. The height between the two faces is not equal to the side of the square. 3. **Non-Regular Tetrahedron**: Choose tetrahedron ACFH formed by vertices A, C, F, and H. Note that A, C, F, and H do not form a regular tetrahedron because at least one pair of non-adjacent edges (e.g., AC and AH) are not equal in length. 4. **Symmetry Argument**: The rectangular prism has symmetry properties that involve reflections. By reflecting the tetrahedron across the medians, we can map any vertex to any other vertex (e.g., A xrightarrow{t_3} C xrightarrow{t_2} F xrightarrow{t_3} H xrightarrow{t_2} A). Here, t_1, t_2, and t_3 denote the reflections across the medians joining midpoint of opposite edges of the rectangular prism. 5. **Median Mapping**: The symmetries ensure that the medians of the tetrahedron ACFH map to themselves or other medians under these reflections. Since these medians intersect the inscribed sphere at segments which transform into each other, the lengths of these segments are preserved and thus equal. 6. **Conclusion**: Despite these equal segment lengths within the inscribed sphere for the medians, the tetrahedron ACFH is not regular. This disproves the idea that equal segments imply a regular tetrahedron. Therefore, the answer to the problem is: **No**. Hence, [ boxed{text{No}} ]

answer:This problem tests our understanding of the properties of geometric sequences. The key to solving this problem is to use the properties of geometric sequences to convert the given equation into left(a_2a_4 + 2a_3a_5 + a_4a_6 = a_3^2 + 2a_3a_5 + a_5^2 = (a_3 + a_5)^2 = 25right). Step 1: We know that in a geometric sequence, the ratio between any two adjacent terms is constant. Let's denote this ratio as r. Then we can express a_2, a_3, a_4, a_5, a_6 in terms of a_1 and r as follows: - a_2 = a_1r - a_3 = a_1r^2 - a_4 = a_1r^3 - a_5 = a_1r^4 - a_6 = a_1r^5 Step 2: Substitute these expressions into the given equation: a_2a_4 + 2a_3a_5 + a_4a_6 = 25. This gives us: (a_1r)(a_1r^3) + 2(a_1r^2)(a_1r^4) + (a_1r^3)(a_1r^5) = 25 Step 3: Simplify the equation: a_1^2r^4 + 2a_1^2r^6 + a_1^2r^8 = 25 Step 4: Factor out a_1^2r^4: a_1^2r^4(1 + 2r^2 + r^4) = 25 Step 5: Recall that a_3 + a_5 = a_1r^2 + a_1r^4 = a_1r^2(1 + r^2). We can rewrite the equation in Step 4 as: (a_3 + a_5)^2 = 25 Step 6: Solve for a_3 + a_5: a_3 + a_5 = 5 So, boxed{a_3 + a_5 = 5}.