answer:First, let's convert the weight limit of the box from pounds to ounces, since the weight of the plates is given in ounces. There are 16 ounces in a pound. 20 pounds * 16 ounces/pound = 320 ounces Hasan initially packed the box with 38 plates, each weighing 10 ounces. 38 plates * 10 ounces/plate = 380 ounces The box was too heavy, so he started removing plates one by one. We need to find out how many plates he needs to remove to get the weight of the box at or below 320 ounces. Let's subtract the weight limit from the initial weight of the box to find out how much weight he needs to remove. 380 ounces - 320 ounces = 60 ounces Each plate weighs 10 ounces, so we divide the excess weight by the weight of each plate to find out how many plates he needs to remove. 60 ounces / 10 ounces/plate = 6 plates Hasan needed to remove boxed{6} plates from the shipping box to meet the weight limit.

answer:In right triangle ABC, with angle A = 90^circ, we can denote the lengths of the sides opposite and adjacent to angle C as AB and AC, respectively. Given tan C = 3, we know that tan C = frac{AB}{AC}, so AB = 3AC. Using the Pythagorean theorem, we find that: [ (3AC)^2 + (AC)^2 = BC^2 ] [ 9AC^2 + AC^2 = 150^2 ] [ 10AC^2 = 22500 ] [ AC^2 = 2250 ] [ AC = sqrt{2250} = 15sqrt{10} ] Hence, side AB, which is 3 times AC, is: [ AB = 3 times 15sqrt{10} = 45sqrt{10} ] Therefore, the length of side AB is boxed{45sqrt{10}}.

answer:To demonstrate that a convex ( n )-gon with vertices at lattice points of a grid must have ( n leq 4 ) if no other lattice points lie inside it or on its edges, we proceed as follows: 1. **Consider the Polygon and Its Properties:** Given a convex polygon with vertices at lattice points ( left( x_i, y_i right) ) where ( i = 1, 2, ldots, n ): - The vertices are at integer coordinates, i.e., lattice points. - There should be no other lattice points strictly inside or on the edges of the polygon. 2. **Use Pick's Theorem:** Recall Pick's Theorem for lattice polygons, which states: [ A = I + frac{B}{2} - 1 ] where ( A ) is the area of the polygon, ( I ) is the number of interior lattice points, and ( B ) is the number of boundary lattice points. 3. **Apply the Conditions of the Problem to Pick's Theorem:** According to the problem: - There are no other lattice points inside the polygon, so ( I = 0 ). - There are no other lattice points on the edges apart from the vertices, making ( B = n ). 4. **Calculate the Area with Pick's Theorem:** Substituting the given conditions into Pick's Theorem, we get: [ A = 0 + frac{n}{2} - 1 = frac{n}{2} - 1 ] Rearrange to express ( n ) in terms of ( A ): [ A = frac{n}{2} - 1 implies frac{n}{2} = A + 1 implies n = 2A + 2 ] 5. **Analyze Area Limitations:** The area ( A ) of a convex polygon with its vertices at lattice points is a positive integer. For a simple convex polygon: [ A geq 1 ] Then, substituting this minimum area, we get: [ n = 2 cdot 1 + 2 = 4 ] 6. **Conclusions:** From the deduction above, we find: - The minimum number of sides is given by substituting the minimum area. - ( n ) cannot be greater than 4 without introducing additional lattice points inside or on the sides of the polygon. Thus, we have shown that for a convex ( n )-gon with vertices at lattice points and no other lattice points inside or on its sides, ( n leq 4 ). [ boxed{n leq 4} ]

answer:# Problem: On the sides ( BC, CA, AB ) of triangle ( ABC ), points ( X, Y, Z ) are taken such that the lines ( AX, BY, CZ ) intersect at a single point ( O ). Prove that among the ratios ( frac{OA}{OX}, frac{OB}{OY}, frac{OC}{OZ} ), at least one is not greater than 2 and at least one is not less than 2. 1. **Assumption for Contradiction:** Suppose that all the given ratios are less than 2, i.e., [ frac{OA}{OX} < 2, quad frac{OB}{OY} < 2, quad frac{OC}{OZ} < 2. ] 2. **Area Considerations:** Denote ( S_{triangle ABC} ) as the area of triangle ( ABC ). Similarly, let: [ S_{triangle ABO}, S_{triangle AOC}, S_{triangle OBC}, ] represent the areas of the respective triangles. Since ( O ) is the intersection of ( AX, BY, ) and ( CZ ), using the given ratios, we can express the areas relative to each other. For instance: [ S_{triangle AOB} + S_{triangle AOC} < 2S_{triangle OBC}, quad S_{triangle AOB} + S_{triangle OBC} < 2S_{triangle AOC}, quad S_{triangle AOC} + S_{triangle OBC} < 2S_{triangle AOB}. ] 3. **Summation of Inequalities:** Adding these inequalities together: [ (S_{triangle AOB} + S_{triangle AOC}) + (S_{triangle AOB} + S_{triangle OBC}) + (S_{triangle AOC} + S_{triangle OBC}) < 2(S_{triangle OBC} + S_{triangle AOC} + S_{triangle AOB}). ] Simplifying the left hand side, we have: [ 2(S_{triangle AOB} + S_{triangle AOC} + S_{triangle OBC}) < 2(S_{triangle AOB} + S_{triangle AOC} + S_{triangle OBC}). ] 4. **Contradiction:** This simplification implies: [ 2S_{triangle ABC} < 2S_{triangle ABC}, ] which is a contradiction. 5. **Conclusion from Contradiction:** Since our assumption that all the given ratios are less than 2 led to a contradiction, it follows that among the ratios ( frac{OA}{OX}, frac{OB}{OY}, frac{OC}{OZ} ), at least one must be not less than 2. Thus proven, at least one ratio is not greater than 2 and at least one is not less than 2. [ boxed{} ]