answer:1. Let's start by determining the total weight of all 6 bags. We are given weights of 13 kg, 15 kg, 16 kg, 17 kg, 21 kg, and 24 kg. Adding these together, we get: [ 13 + 15 + 16 + 17 + 21 + 24 = 106 text{ kg} ] 2. One bag contains turnip, and the rest contain either onions or carrots. The total weight of the carrots is twice the total weight of the onions. Let ( O ) represent the total weight of the onions, and ( C ) represent the total weight of the carrots. Then: [ C = 2O ] Since one bag contains turnip, the total weight of the onions and carrots will be: [ 106 - T ] where ( T ) is the weight of the bag with the turnip. Therefore, we have: [ C + O = 106 - T ] Substituting ( C = 2O ) into the equation, we get: [ 2O + O = 106 - T implies 3O = 106 - T ] 3. Since ( 3O ) must be a whole number, ( 106 - T ) must be divisible by 3. To find possible values for ( T ), let's check the divisibility of ( 106 - T ) by 3 for each possible weight ( T ) of the bag containing turnip: - If ( T = 13 ): [ 106 - 13 = 93 quad text{and} quad 93 div 3 = 31 quad (text{is an integer}) ] - If ( T = 15 ): [ 106 - 15 = 91 quad text{and} quad 91 div 3 neq text{integer} ] - If ( T = 16 ): [ 106 - 16 = 90 quad text{and} quad 90 div 3 = 30 quad (text{is an integer}) ] - If ( T = 17 ): [ 106 - 17 = 89 quad text{and} quad 89 div 3 neq text{integer} ] - If ( T = 21 ): [ 106 - 21 = 85 quad text{and} quad 85 div 3 neq text{integer} ] - If ( T = 24 ): [ 106 - 24 = 82 quad text{and} quad 82 div 3 neq text{integer} ] 4. From the above, we observe that the two valid options are: - When the turnip weighs 13 kg: - The total weight of onions and carrots is 93 kg: [ 3O = 93 implies O = 31 quad text{and} quad C = 2O = 62 ] - Now, let's verify if we can split the bags accordingly. Assign the weights such that the sum of some weights is 31 kg (onions) and the others sum to 62 kg (carrots): - Onions: 15 kg and 16 kg. - Carrots: 17 kg, 21 kg, and 24 kg. [ 15 + 16 = 31 implies O = 31 quad text{and} quad 17 + 21 + 24 = 62 implies C = 62 ] - When the turnip weighs 16 kg: - The total weight of onions and carrots is 90 kg: [ 3O = 90 implies O = 30 quad text{and} quad C = 2O = 60 ] - Now, let's verify if we can split the bags accordingly. Assign the weights such that the sum of some weights is 30 kg (onions) and the others sum to 60 kg (carrots): - Onions: 13 kg and 17 kg. - Carrots: 15 kg, 21 kg, and 24 kg. [ 13 + 17 = 30 implies O = 30 quad text{and} quad 15 + 21 + 24 = 60 implies C = 60 ] # Conclusion: Therefore, the turnip can be in the bag weighing ( boxed{13 text{ or } 16 text{ kg}} ).

answer:Given: Three cyclists ride in the same direction on a circular track of 300 meters length, each with different constant speeds. We need to determine the smallest distance ( d ) such that the photographer can capture all the cyclists within a segment of the track. 1. **Assumption on Speeds and Directions:** - Without loss of generality, assume all cyclists move counterclockwise. - Let ( v_1, v_2, ) and ( v_3 ) be the speeds of the first, second, and third cyclists respectively, where ( v_1 > v_2 > v_3 ). 2. **Relative Motion Analysis:** - Consider the relative positions of the cyclists in a reference frame moving with the second cyclist. - In this reference frame, the second cyclist is stationary at point ( A ) on the track. - The first cyclist appears to move with speed ( v_1 - v_2 ) counterclockwise, and the third cyclist with speed ( v_2 - v_3 ) clockwise. 3. **Meeting Points:** - The cyclists periodically meet each other. Let ( B_1, B_2, B_3, ldots ) denote these meeting points from the start. - Moving in mutually exclusive directions and distinct speeds ensures they meet at distinct points and intervals. 4. **Segment Length Analysis:** - Denote by ( beta_n ) the smaller arc between successive meeting points ( B_n ) and ( B_{n+1} ). - The length of ( beta_n ) is at most half the track length, i.e., ( 150 ) meters. 5. **Cyclic Coverage and Containment:** - The arcs ( beta_n ) together span the entire circumference over multiple overlaps due to periodic meetings. - Thus, one of these arcs ( beta_m ) must contain point ( A ). - Either segment ( B_m ) to ( A ) or ( A ) to ( B_{m+1} ) is guaranteed to be at most ( 75 ) meters in length. 6. **Conclusion:** - Periodic meetings ensure that two cyclists cover the segment formed with the second cyclist within ( 75 ) meters. Therefore, the photographer will definitely be able to capture a successful photograph if ( d geq 75 ) meters. [ boxed{75} ]

answer:To find the ratio in which point E divides the side BC in the given triangle ABC, we need to follow a series of steps that leverage basic properties of triangles and line segments. 1. **Identify Points and their Relationships:** - Point F divides AC in the ratio 1:2. Therefore, if AF = k, then FC = 2k. - Point G is the midpoint of BF, so BG = GF. - Point E is the intersection of line AG with BC. 2. **Construct Parallel Lines:** - Draw FH parallel to AE to ease our calculations. - Given B G = G F, we then have B E = E H by similar triangles nature and properties (this follows because AG is a median in triangle ABF divided into equal segments). 3. **Use Given Ratios and Relations:** - Since B G = G F, AG is median, it must also mean that G bisects BF hence making GF=BG. - Then, knowing AF = k and FC = 2k, we get that BC = BE + EC. - triangle ABF and triangle AEC have proportional sides. By given 2 AF = FC, this implies 2 EH=HC from parallel lines property and as F, H, A are collinear. 4. **Establish the Ratio around AG:** - Combine this to show E is dividing side BC into 1:3 ratio because: [ BH + HC = BC ] When, [ 2HE = HC quad text{and} quad HE = BE ] Combining give, [ 2 BE +2 EH = BC quad text{implies} quad 3 BE = BC ] [ BE:EC = 1:3. ] Therefore, the ratio frac{BE}{EC} ultimately concludes with [ 1:3. ]. # Conclusion: (boxed{B})

answer:First, shifting all points on the graph of the function y=sin x to the right by dfrac{pi}{10} units, we get the equation of the resulting graph as y=sin (x- dfrac{pi}{10}). Then, stretching the x-coordinates of the obtained points by a factor of 2 (the y-coordinates remain unchanged), we get the equation of the resulting graph as y=sin (dfrac{1}{2}x- dfrac{pi}{10}). Therefore, the correct option is boxed{text{C}}. This problem mainly examines the translation transformations of trigonometric functions. The principle of translation is to add to the left and subtract from the right for horizontal shifts, and to add to the top and subtract from the bottom for vertical shifts.