🧠:<think>To determine the possible first term of the geometric sequence, we start by noting the given terms:- The second term is (3), so (ar = 3).- The fourth term is (12), so (ar^3 = 12).We can find the common ratio (r) by dividing the fourth term by the second term:[frac{ar^3}{ar} = frac{12}{3} = 4 implies r^2 = 4 implies r = 2 text{ or } r = -2]Case 1: (r = 2)[a cdot 2 = 3 implies a = frac{3}{2}]This value does not match any of the provided options.Case 2: (r = -2)[a cdot (-2) = 3 implies a = -frac{3}{2}]This value also does not match any of the provided options.Since neither of the calculated values for (a) matches the given options, we consider the possibility of a different common ratio or re-evaluate the problem statement for any potential errors or misinterpretations.</think>

🧠:<think>First, we determine the number of triangular and hexagonal faces. Let ( t ) be the number of triangular faces and ( h ) be the number of hexagonal faces. We know that:[t + h = 30]Each triangular face contributes 3 edges, and each hexagonal face contributes 6 edges. Since each edge is shared by two faces, the total number of edges ( E ) is:[E = frac{3t + 6h}{2}]Substituting ( t = 30 - h ) into the equation for ( E ):[E = frac{3(30 - h) + 6h}{2} = frac{90 - 3h + 6h}{2} = frac{90 + 3h}{2}]Using Euler's formula ( V - E + F = 2 ), where ( F = 30 ):[V - frac{90 + 3h}{2} + 30 = 2]Solving for ( V ):[V = frac{90 + 3h}{2} - 28 = frac{90 + 3h - 56}{2} = frac{34 + 3h}{2}]Since each vertex has 3 triangular faces and 2 hexagonal faces meeting, the number of faces around each vertex is ( T + H = 3 + 2 = 5 ). The total number of face-vertex incidences is ( 3V ) for triangles and ( 2V ) for hexagons. Therefore:[3V = 3t quad text{and} quad 2V = 2h]Solving these, we find:[t = V quad text{and} quad h = V]Substituting back into ( t + h = 30 ):[V + V = 30 implies V = 15]Finally, calculating ( 100H + 10T + V ):[100 times 2 + 10 times 3 + 15 = 200 + 30 + 15 = 245]</think>

🧠:<think>First, I need to understand the given diagram and the information provided. There is a sector ( AOB ) with center ( O ) and ( angle AOB = 120^circ ). The radius of the circle is 12 units, so both ( OA ) and ( OB ) are 12 units long. The line ( OY ) is drawn perpendicular to ( AB ) and intersects ( AB ) at ( X ).Since ( OY ) is perpendicular to ( AB ), it means that ( OX ) is the altitude from ( O ) to ( AB ). In triangle ( OAB ), which is an isosceles triangle with ( OA = OB = 12 ) and ( angle AOB = 120^circ ), I can split it into two congruent right triangles by drawing the altitude ( OX ).In each of these right triangles, the angle at ( O ) is ( 60^circ ) (since ( 120^circ ) divided by 2 is ( 60^circ )), and the angle at ( A ) is ( 30^circ ). This makes each right triangle a 30-60-90 triangle.In a 30-60-90 triangle, the sides are in the ratio ( 1 : sqrt{3} : 2 ). Here, the hypotenuse ( OA ) is 12 units. Therefore, the side opposite the ( 30^circ ) angle (which is ( OX )) is ( 6sqrt{3} ) units, and the side opposite the ( 60^circ ) angle (which is ( AX )) is 6 units.Finally, to find ( XY ), I subtract the length of ( OX ) from ( OY ):[XY = OY - OX = 12 - 6sqrt{3}]</think>

🧠:<think>To solve the problem, we need to determine the least possible number of faces on two dice combined, given the constraints on the probabilities of rolling specific sums.1. Define Variables: Let the number of faces on the first die be ( a ) and on the second die be ( b ), with ( a geq b geq 8 ).2. Analyze the Probability of Rolling a Sum of 9: The sum of 9 can be obtained by combinations ( (1,8), (2,7), (3,6), (4,5), (5,4), (6,3), (7,2), (8,1) ), giving 8 ways.3. Analyze the Probability of Rolling a Sum of 12: The probability of rolling a sum of 9 is half the probability of rolling a sum of 12. Therefore, there must be ( 2 times 8 = 16 ) ways to roll a sum of 12.4. Constraint on the Number of Faces for Sum 12: To have 16 ways to roll a sum of 12, both ( a ) and ( b ) must be at least 11 (to include combinations like ( (1,11), (2,10), (3,9), ldots, (11,1) )).5. Analyze the Probability of Rolling a Sum of 14: The probability of rolling a sum of 14 is ( frac{1}{6} ). Let ( n ) be the number of ways to roll a sum of 14. Then, ( frac{n}{ab} = frac{1}{6} ), giving ( n = frac{ab}{6} ).6. Substitute ( b = 11 ): Substituting ( b = 11 ) into the equation for ( n ), we have ( n = frac{11a}{6} ). To find ( a ), we need ( frac{11a}{6} ) to be an integer, implying ( a ) must be a multiple of 6.7. Choose the Smallest ( a ): The smallest ( a ) that is a multiple of 6 and ( geq 11 ) is 12.8. Calculate Total Number of Faces: The total number of faces on the two dice is ( a + b = 12 + 11 = 23 ).Thus, the least possible number of faces on the two dice combined is ( 23 ).The final answer is (boxed{C}).</think>