answer:Since a=sin 14^circ + cos 14^circ = sqrt{2}sin(14^circ + 45^circ) = sqrt{2}sin 59^circ, b=2sqrt{2}sin 30.5^circ cos 30.5^circ = sqrt{2}sin 61^circ, c= frac{sqrt{6}}{2} = sqrt{2}sin 60^circ, and the function y=sqrt{2}sin x is increasing in the interval (0^circ, 90^circ), therefore, sqrt{2}sin 59^circ < sqrt{2}sin 60^circ < sqrt{2}sin 61^circ which means: a<c<b. Thus, the correct option is: boxed{D}. First, we convert a, b, and c into the form of sine functions of different angles, then use the monotonicity of trigonometric functions to compare their sizes. This question mainly tests the application of the monotonicity of trigonometric functions, trigonometric transformations, and comparison of inequalities, focusing on computational solving ability, reduction and transformation thinking. It is a basic question.

answer:**Analysis** Given that the math scores of 10,000 middle school students in a city are approximately normally distributed as N(70, 10^2), and there are 230 students scoring above 90, we can use the symmetry of the normal distribution to infer the number of students scoring between 50 and 90. Since there are 230 students scoring above 90, by symmetry, there are also approximately 230 students scoring below 50. Therefore, the number of students scoring between 50 and 90 is approximately 10,000 - 2 times 230 = 9540 students. The correct choice is boxed{C}.

answer:1. **Define Points and Segments:** - Let ( K ) and ( N ) be points on segment ( AB ). - Let ( L ) be a point on segment ( AC ). - Let ( M ) be a point on segment ( BC ). - Given: ( CL = AK ), ( CM = BN ), and ( ML = KN ). 2. **Introduce Auxiliary Points:** - Let ( P ) be a point on ( BC ) such that ( BP = CL = AK ). - Let ( Q ) be a point on ( AC ) such that ( AQ = CM = BN ). 3. **Congruent Triangles:** - Consider triangles ( triangle AQK ), ( triangle CML ), and ( triangle BNP ). - Each of these triangles has a ( 60^circ ) angle (since ( ABC ) is an equilateral triangle) and two equal side lengths: - ( AQ = CM ) and ( AK = CL ) for ( triangle AQK ). - ( CL = AK ) and ( CM = BN ) for ( triangle CML ). - ( BN = CM ) and ( BP = CL ) for ( triangle BNP ). 4. **Equal Segments:** - From the congruence of the triangles, we have: [ PN = KQ = LM = KN = QL = PM ] 5. **Extend ( NM ) to Intersect ( AC ):** - Extend ( NM ) to intersect ( AC ) at point ( R ). 6. **Angle Relationships:** - Consider the angles: [ angle ALK = frac{angle AQK}{2} = frac{angle BNP}{2} = frac{120^circ - angle BPN}{2} ] - Since ( angle BPN = 60^circ ) (as ( triangle BNP ) is equilateral), we have: [ angle ALK = frac{120^circ - 60^circ}{2} = 30^circ ] - Similarly, for ( angle ARN ): [ angle ARN = 60^circ - angle BMN ] - Since ( angle BMN = 30^circ ) (as ( triangle BMN ) is equilateral), we have: [ angle ARN = 60^circ - 30^circ = 30^circ ] 7. **Parallel Lines:** - Since ( angle ALK = angle ARN ), lines ( KL ) and ( MN ) are parallel by the Alternate Interior Angles Theorem. (blacksquare)

answer:If it takes Jack 4 minutes to put his own shoes on, and the total time to get ready is 18 minutes, then the time spent helping both toddlers with their shoes is 18 - 4 = 14 minutes. Since there are two toddlers, we divide the time spent helping them by 2 to find out how long it takes to help each toddler. So, 14 minutes ÷ 2 toddlers = 7 minutes per toddler. To find out how much longer it takes Jack to help each toddler with their shoes compared to the time he takes to put his own shoes on, we subtract the time it takes him to put on his shoes from the time it takes to help one toddler. So, 7 minutes (helping one toddler) - 4 minutes (putting on his own shoes) = boxed{3} minutes longer to help each toddler with their shoes.