answer:Since the intercept of line l on the x-axis is 3, and on the y-axis is -2, the equation of l can be written as frac{x}{3} + frac{y}{-2} = 1, which simplifies to 2x-3y-6=0. Therefore, the correct choice is: boxed{C}. This problem can be solved using the intercept form, and it is a basic question that tests understanding of the intercept form.

answer:1. **Draw a Circle on the Sphere**: Choose an arbitrary point A on the surface of the sphere as the center and draw a circle on the sphere using a compass. This circle can be defined with its center at A. 2. **Select Points on the Circle**: Select any three points ( M, N, P ) on this circle. Measure the distances between these points using a compass. 3. **Transfer Distances to a Plane**: Transfer these measured distances ( MN ), ( NP ), and ( MP ) to a flat plane, such as a sheet of paper, to draw the triangle ( M^{prime}N^{prime}P^{prime} ) that is congruent to the spherical triangle ( MNP ). 4. **Construct Circumcircle on the Plane**: Now, draw the circumcircle of the plane triangle ( M^{prime}N^{prime}P^{prime} ). The radius of this circumcircle will be the planar equivalent of a critical distance on the sphere. 5. **Identify Critical Lengths**: Note the length ( M^{prime}O_1^{prime} ) on the plane. This length is equivalent to the perpendicular distance from point ( M ) to the line ( AB ) on the sphere, where ( AB ) is the diameter of the sphere. 6. **Recreate Right Triangle**: With the hypotenuse ( A^{prime}M^{prime} ) (equal to the initial circle radius ( AM )), and the leg ( M^{prime}O_1^{prime} ) (equivalent to ( MO_1 )), construct the right-angled triangle ( M^{prime}O_1^{prime}A^{prime} ) on the plane. 7. **Extend Line Perpendicular from ( M^{prime} )**: Through the point ( M^{prime} ), draw a line perpendicular to ( M^{prime}A^{prime} ). Extend this line until it intersects with the length ( A^{prime}O_1^{prime} ) at point ( B^{prime} ). 8. **Find the Diameter**: The segment ( A^{prime}B^{prime} ) will be equal to the diameter ( AB ) of the sphere. This is because the relationships in the geometrical figures hold true as per construction on both the plane and the spherical surface. # Conclusion: The segment ( A^{prime}B^{prime} ) on the plane is equal to the diameter of the sphere. Hence, the radius of the material spherical object is: [ R = frac{A^{prime}B^{prime}}{2} ] [ boxed{R = frac{A^{prime}B^{prime}}{2}} ]

answer:To solve this problem, we need to determine the number of possible colorings of the integers (1, 2, ldots, n) using three colors (red, blue, yellow) under the given rules. Let's denote the number of such colorings by (a_n). # Step 1: Understanding the Rules 1. **Parity Rule**: A number (x) and the smallest number larger than (x) that is colored the same as (x) must have different parities. This means if (x) is even, the next number of the same color must be odd, and vice versa. 2. **Color Usage Rule**: If all three colors are used, exactly one color must have its smallest number as even. # Step 2: Establishing the Recursive Formula We need to show that: [ a_n = 2a_{n-1} + 3 ] # Step 3: Base Case For (n = 1), we have three possible colorings (one for each color): [ a_1 = 3 ] # Step 4: Recursive Step Assume the formula holds for (n). We need to show it holds for (n+1). Case 1: Using Two Colors If we use only two colors, the parity rule must be satisfied. For (n) numbers, there are (a_{n-1}) valid colorings. When we add the ((n+1))-th number, it can be colored in two ways (either of the two colors used). Thus, we have: [ 2a_{n-1} ] Case 2: Using Three Colors If we use all three colors, the smallest number in one of the colors must be even. For (n) numbers, there are (a_{n-1}) valid colorings. When we add the ((n+1))-th number, it can be colored in three ways (one for each color). However, we need to ensure that the smallest number in one color is even. This adds an additional 3 valid colorings: [ 3 ] Combining both cases, we get: [ a_n = 2a_{n-1} + 3 ] # Step 5: Solving the Recurrence To find a closed form for (a_n), we solve the recurrence relation: [ a_n = 2a_{n-1} + 3 ] We can use the method of characteristic equations or iterate the recurrence to find a pattern. Iteration: [ a_1 = 3 ] [ a_2 = 2a_1 + 3 = 2 cdot 3 + 3 = 9 ] [ a_3 = 2a_2 + 3 = 2 cdot 9 + 3 = 21 ] [ a_4 = 2a_3 + 3 = 2 cdot 21 + 3 = 45 ] We observe that: [ a_n = 3 cdot 2^n - 3 ] # Step 6: Verification We verify by induction: - Base case: (a_1 = 3) - Inductive step: Assume (a_k = 3 cdot 2^k - 3). Then: [ a_{k+1} = 2a_k + 3 = 2(3 cdot 2^k - 3) + 3 = 6 cdot 2^k - 6 + 3 = 6 cdot 2^k - 3 = 3 cdot 2^{k+1} - 3 ] Thus, the formula holds by induction. The final answer is ( boxed{ 3 cdot 2^n - 3 } )

answer:Let's break down the requirements and calculate the number of possible lock codes step by step. 1. The first, third, and fifth digits must be even. There are 4 even numbers from 1 to 9 (2, 4, 6, 8). Since no digit can be repeated, there are 4 choices for the first digit, 3 choices for the third digit, and 2 choices for the fifth digit. 2. The second digit must be a prime number. The prime numbers from 1 to 9 are 2, 3, 5, and 7. However, since the first digit is even and cannot be 2 (as it's already used), there are 3 remaining choices for the second digit (3, 5, 7). 3. No adjacent digits can be consecutive numbers. This means that after choosing the first two digits, the third digit cannot be consecutive with the second digit, the fourth digit cannot be consecutive with the third digit, and so on. Let's calculate the possibilities: - For the first digit, we have 4 choices (2, 4, 6, 8). - For the second digit, we have 3 choices (3, 5, 7) because it cannot be consecutive with the first digit. - For the third digit, we have 3 choices left (since one even number has been used and it cannot be consecutive with the second digit). - For the fourth digit, we have 5 choices left (since one prime number has been used and it cannot be consecutive with the third digit). - For the fifth digit, we have 2 choices left (since two even numbers have been used and it cannot be consecutive with the fourth digit). - For the sixth digit, we have 4 choices left (since five numbers have been used and it cannot be consecutive with the fifth digit). Now, let's multiply the number of choices for each position to find the total number of possible lock codes: 4 (first digit) * 3 (second digit) * 3 (third digit) * 5 (fourth digit) * 2 (fifth digit) * 4 (sixth digit) = 4 * 3 * 3 * 5 * 2 * 4 = 1440 Therefore, there are boxed{1440} possible lock codes that meet the given criteria.