answer:Three-digit integers divisible by 21, which end in 7: - Numbers must be odd (since 7 is not a multiple of 2) and multiples of 21 (which are multiples of both 3 and 7). - Three-digit multiples of 21 range from 21 times 5 = 105 to 21 times 47 = 987. - We calculate the number of valid three-digit numbers in this range. All such numbers take the form 21k + 7, where k is an integer, because the number needs to end in 7. - The smallest integer k such that 21k + 7 is at least 105 can be found by solving 21k + 7 geq 105 or k geq 98/21; so k geq 5. - The largest integer k for which 21k + 7 leq 987 is found by solving 21k + 7 leq 987; so k leq 980/21; k leq 46. - We stepped by 10 (increasing k by 10 each time, as adding 210 to 21k + 7 will again end the number in 7), starting from k = 5. The sequence of k values in this range is 5, 15, 25, 35, 45. Count these integers: There are 5 three-digit integers divisible by 21 ending in 7. Thus, we have boxed{5} positive, three-digit integers with 7 in the units place that are divisible by 21.

answer:Let's denote the entrance fee for persons under the age of 18 as ( F ). Joe's 6-year-old twin brothers are both under the age of 18, so for both of them, the entrance fee would be ( 2F ). Since Joe is older, her entrance fee would be ( F + 0.20F = 1.20F ). The total cost for the entrance fees for all three of them would be ( 2F + 1.20F = 3.20F ). Each ride costs 0.50, and since they each took 3 rides, the total cost for the rides would be ( 3 times 3 times 0.50 = 9 times 0.50 = 4.50 ). The total amount Joe spent is the sum of the entrance fees and the ride costs, which is ( 3.20F + 4.50 = 20.50 ). Now we can solve for ( F ): ( 3.20F + 4.50 = 20.50 ) Subtract 4.50 from both sides to isolate ( F ): ( 3.20F = 20.50 - 4.50 ) ( 3.20F = 16.00 ) Now divide both sides by 3.20 to solve for ( F ): ( F = frac{16.00}{3.20} ) ( F = 5.00 ) So, the entrance fee for persons under the age of 18 is boxed{5.00} .

answer:1. We begin with the given function: [ f(x) = cos 3x + 4 cos 2x + 8 cos x. ] 2. We first use trigonometric identities to rewrite the function. One of the identities we can use is: [ cos 3x = 4 cos^3 x - 3 cos x. ] Substituting it into the function, we have: [ f(x) = 4 cos^3 x - 3 cos x + 4 cos 2x + 8 cos x. ] 3. Next, we use the identity for cos 2x: [ cos 2x = 2 cos^2 x - 1. ] Substituting it, we obtain: [ 4 (2 cos^2 x - 1) = 8 cos^2 x - 4. ] Incorporating this into our function gives: [ f(x) = 4 cos^3 x - 3 cos x + 8 cos^2 x - 4 + 8 cos x. ] 4. We now combine like terms: [ f(x) = 4 cos^3 x + 8 cos^2 x + (8 - 3) cos x - 4. ] Simplifying further: [ f(x) = 4 cos^3 x + 8 cos^2 x + 5 cos x - 4. ] 5. To find the minimum value of f(x), we analyze the function. Note that cos x ranges from -1 to 1. Let's rewrite the equation for clarity: [ f(x) = 4 cos x (cos^2 x + 2 cos x + frac{5}{4}) - 4. ] [ f(x) = (cos x + 1)(2 cos x + 1)^2 - 5. ] 6. Observe that the term ((2 cos x + 1)^2) is always non-negative as it is a square term. [ (cos x + 1) geq -1 ] and [ (2 cos x + 1)^2 geq 0. ] 7. From the factorization, the minimum occurs when ((cos x + 1)(2 cos x + 1)^2) is at its minimum. Both terms inside it should be minimized. This occurs when [ cos x = -1 quad text{or} quad cos x = -frac{1}{2}. ] 8. We evaluate each case: - If cos x = -1: [ f(x) = 4(-1)^3 + 8(-1)^2 + 5(-1) - 4 = -4 + 8 - 5 - 4 = -5. ] - If cos x = -frac{1}{2}: [ f(x) = 4 left(-frac{1}{2}right)^3 + 8 left(-frac{1}{2}right)^2 + 5 left(-frac{1}{2}right) - 4 = 4 left(-frac{1}{8}right) + 8 left(frac{1}{4}right) - frac{5}{2} - 4. ] [ f(x) = -frac{1}{2} + 2 - frac{5}{2} - 4 = -5. ] Therefore, the minimum value of the function f(x) is: [ boxed{-5}. ]

answer:1. **Given Setup**: We are given a 15 times 15 board where some pairs of neighboring cell centers are connected with lines to form a closed, non-self-intersecting polyline that is symmetric with respect to one of the board's diagonals. We need to prove that the length of this polyline does not exceed 200. 2. **Observation**: The polyline intersects the diagonal. Let A be a vertex of the polyline lying on the diagonal. We traverse the polyline until we first hit another vertex B that also lies on the diagonal. Due to the symmetry of the polyline with respect to the diagonal, if we move along the polyline from A in the opposite direction, B will be the first vertex on the diagonal we encounter. Therefore, the polyline forms a loop, and it cannot pass through any of the remaining 13 centers of cells on the diagonal. 3. **Coloring Argument**: Let us color the board in a checkerboard pattern such that the diagonal is black. On the polyline, the black and white cells alternate, so there are equal numbers of black and white cells on the polyline. 4. **Counting Cells**: The 15 times 15 board has 15^2 = 225 cells in total. Given the alternating coloring, there are left(frac{15^2 + 1}{2}right) = 113 black cells (since the diagonal is black). Since the polyline does not pass through 13 of these black diagonal cells, it passes through at most 113 - 13 = 100 black cells. 5. **Calculating the Length**: Each cell center on the polyline corresponds to a black cell and a symmetric pair of cells. Therefore, the polyline intersects both black and white cells creating a pairwise relation. Hence, the total number of cells the polyline passes through is at most twice the black cells it passes through: [ text{Total Length} leq 2 times 100 = 200 ] # Conclusion: The length of the polyline is at most 200. Thus, the claim is proven. blacksquare