answer:We need to find the number of ways Bob can color five points within the set of lattice points ({(x, y) mid 1 leq x, y leq 5}) such that the distance between any two blue points is not an integer. 1. **Identify Constraints:** - The distance between any two points ((x_1, y_1)) and ((x_2, y_2)) should not be an integer. - For the distance (d = sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}) to not be an integer, no two points should lie on the same row or column, implying (x_1 neq x_2) and (y_1 neq y_2). 2. **Permutations of Coordinates:** - Given that each (x)-coordinate and (y)-coordinate must be unique across the selected points, we are looking for permutations of ({1, 2, 3, 4, 5}). - There are (5!) such permutations in total where (5! = 120). 3. **Forbidden Pairs (3-4-5 Triangles):** - We must exclude configurations in which any pair of selected points has integer distances according to the Pythagorean triple configurations ((3-4-5)). 4. **Counting Violations:** - The pairs of points whose distance is specifically (5) units while not lying on the same row or column are eight specific pairs. These can be identified using combinations of the Pythagorean triple coordinates. - Each such pair in the chosen five points means only (3) other coordinates are left to choose from the remaining, giving (3!) permutations. Thus: [ 8 text{ (pairs)} times 6 = 48 text{ (permutations violating the condition)} ] 5. **Overcounting Adjustments:** - If a configuration includes multiple violations, e.g., ((1,1)), ((4,5)), ((5,4)), termed a double-count, their permutations need to be subtracted. - For any configuration that includes exactly two violations, there are (4 times 2 = 8) such overcounted configurations. 6. **Valid Configurations Calculation:** - Subtract the overcounted pairs from the total violations: [ 48 - 8 = 40 text{ (double counted violations subtracted)} ] - Subtract the total non-compliant configurations from the total: [ 120 - 40 = 80 ] Therefore, the number of ways Bob can color the five points such that no two blue points have an integer distance between them is: [ boxed{80} ]

answer:To find the length of one side of the equilateral triangle, we first need to find the sum of the lengths of the four sides of the rectangle, which is the perimeter of the rectangle. The perimeter (P) of a rectangle is given by the formula: P = 2 * (length + width) Given that the rectangle is 125 cm wide and 115 cm long, we can calculate its perimeter as follows: P = 2 * (125 cm + 115 cm) P = 2 * 240 cm P = 480 cm Now, we know that the sum of the lengths of the three sides of the equilateral triangle is equal to the perimeter of the rectangle, which is 480 cm. Since all sides of an equilateral triangle are equal, we can divide the perimeter of the rectangle by 3 to find the length of one side of the equilateral triangle. Length of one side of the equilateral triangle = 480 cm / 3 Length of one side of the equilateral triangle = 160 cm Therefore, the length of one side of the equilateral triangle is boxed{160} cm.

answer:Since x > 2, It follows that x-2 > 0. Thus, y= frac {x^{2}-4x+8}{x-2}= frac {(x-2)^{2}+4}{x-2}=(x-2)+ frac {4}{x-2}. Applying the Arithmetic Mean-Geometric Mean Inequality (AM-GM inequality), we have: frac{(x-2) + frac{4}{x-2}}{2} geq sqrt{(x-2) cdot frac{4}{x-2}}, Which simplifies to (x-2) + frac{4}{x-2} geq 2 cdot sqrt{4} = 4. The equality holds when (x-2) = frac{4}{x-2}, which is true when x=4. Therefore, the minimum value of the function y= frac {x^{2}-4x+8}{x-2} is boxed{4}. This problem tests our understanding of the application of the AM-GM inequality in finding the minimum value of a function, and can be considered a basic-level problem.

answer:Let's denote Anne's savings as A and Katherine's savings as K. From the first statement, we have: A - 150 = 1/3 * K ...(1) From the second statement, we have: m * K = 3 * A ...(2) where m is the multiple of Katherine's savings. We also know that together they have saved 750, so: A + K = 750 ...(3) From equation (1), we can express K in terms of A: K = 3 * (A - 150) Substitute K in equation (3) with the expression from above: A + 3 * (A - 150) = 750 A + 3A - 450 = 750 4A = 750 + 450 4A = 1200 A = 1200 / 4 A = 300 Now that we have Anne's savings, we can find Katherine's savings using equation (3): 300 + K = 750 K = 750 - 300 K = 450 Now we can find the multiple m using equation (2): m * K = 3 * A m * 450 = 3 * 300 m * 450 = 900 m = 900 / 450 m = 2 So the multiple of Katherine's savings that would make her have 3 times Anne's savings is boxed{2} .