answer:To solve this problem, we need to calculate the number of ways we can light up 3 LEDs out of 7 without having two adjacent LEDs lit, and then consider the color combinations for these LEDs. Firstly, let's determine the positions. There are 7 LEDs, and we need to choose 3 of them such that no two are adjacent. As there are not many LEDs, we can enumerate all allowed positions. Let the LEDs be represented as positions 1 to 7. If LED 1 is lit, the next possible positions for the second lit LED are 3, 4, 5, or 6, since 2 is adjacent to 1 and thus not allowed. For each choice of the second lit LED, there are fewer choices for the third one, ensuring no two are adjacent. Now, let's enumerate these combinations: - If we light LED 1 and LED 3, then the third LED can only be 5 or 6, giving us 2 combinations. - If we light LED 1 and LED 4, then the third LED can only be 6 or 7, giving us 2 combinations. - If we light LED 1 and LED 5, the third LED can only be 7, giving us 1 combination. Since the row is symmetric, lighting LED 1 will give us the same number of combinations as lighting LED 7. Therefore, we can double the counted combinations for LED 1 to account for LED 7. This gives 2 times (2 + 2 + 1) = 10 combinations when considering either LED 1 or LED 7 as the first lit LED. Next, consider when LED 2 is the first LED that is lit: - If we light LED 2 and LED 4, then the third LED can only be 6 or 7, giving us 2 combinations. - If we light LED 2 and LED 5, then the third LED can only be 7, giving us 1 combination. Again, by symmetry, lighting LED 2 will give us the same number as lighting LED 6. This adds another 2 times (2 + 1) = 6 combinations. Now, consider when LED 3 is the first LED that is lit: - If we light LED 3 and LED 5, the third LED can only be 7, giving us 1 combination. Adding this up, we have a total of 10 + 6 + 1 = 17 combinations. Next, we need to consider the color combinations. Each lit LED can be either red or green. Since we're lighting 3 LEDs, each of them can be in one of 2 colors, giving us 2^3 = 8 color combinations for each position combination. Multiplying the position combinations by the color combinations gives us the total number of pieces of information the row of LEDs can represent: 17 times 8 = 136. However, we must realize that the total of 136 includes cases where all LEDs are of the same color, which do not depend on the position. There are 2 cases of all red and 2 cases of all green, so we must subtract them: 136 - 4 = 132. But, we previously considered as different cases for the three LEDs to be lit with alternative colors, like red-green-red and green-red-green. Since the particular order of the colors doesn't matter but the pattern does, we must also divide these cases by 2. So, the corrected total number of different information pieces is frac{132}{2} = 66. However, this is not one of the available answers. At this point, we realize that the calculation for the total number of patterns must be revisited, as there is likely a mistake. In recounting the number of valid combinations, we verify our calculations: - Lighting LED 1 or 7 as the first LED (by symmetry): 2 times (2 + 2 + 1) = 10 combinations. - Lighting LED 2 or 6 as the first LED (by symmetry): 2 times (2 + 1) = 6 combinations. - Lighting LED 3 as the first LED (no symmetry needed here): 1 combination. Let's reevaluate our counting of LED positions, especially for LED 2 or 6 as the first lit LED: Upon reviewing our calculation, we notice an oversight. If LED 2 is lit, there are actually more possibilities for the other two lit LEDs: - LEDs 2, 4, and 6 can be lit. - LEDs 2, 4, and 7 can be lit. - LEDs 2, 5, and 7 can be lit. That's 3 combinations (not just 2 as previously counted), and the same goes when LED 6 is the first lit LED because of symmetry. So the corrected number of combinations considering symmetrical positions is 2 times (3 + 5) + 1 = 17 combinations (not 17 as we had before, but 17 now). Multiplying this by the 8 color combinations for each position combination, we get 17 times 8 = boxed{136}.

answer:1. **Alternate Arithmetic Progression (AAP) Definition**: Assume b = a - d and c = a + d. 2. **Substitute and Simplify**: [ ax + (a-d)y = a + d ] [ ax + ay - dy = a + d ] [ a(x+y) - dy = a + d ] Rearrange to focus on d: [ -dy = d - a(x + y) ] 3. **Solve for All Values of d**: Dividing by d (assuming d neq 0), [ -y = 1 - frac{a}{d}(x + y) ] To hold for all d, the term frac{a}{d}(x + y) must vanish. Therefore, a(x + y) = 0. Since a neq 0, x + y = 0. 4. **Find the Intersection**: From x + y = 0, substituting y = -x into the equation -dy = d - a(x + y): [ d(x) = d - a(0) ] [ x = 1 ] Substituting back, y = -1. **Verification**: Substituting x = 1 and y = -1 into x+y = 0 checks out. Conclusion: The coordinates of the common point are (1, -1). Answer boxed: (1, -1). The final answer is The final answer given the choices is boxed{textbf{(A) } (1, -1)}.

answer:Solution: (1) The curve C_{1}: x^{2}+y^{2}=1 undergoes a scaling transformation begin{cases} x'=2x y'=y end{cases} to obtain the curve C_{2}, therefore The equation of curve C_{2} is dfrac{x^{2}}{4}+y^{2}=1 therefore The parametric equations of curve C_{2} are begin{cases} x=2cos alpha y=sin alpha end{cases}, (alpha is the parameter). because The polar equation of curve C_{3} is rho=-2sin theta, i.e., rho^{2}=-2rhosin theta, therefore The Cartesian equation of curve C_{3} is x^{2}+y^{2}=-2y, i.e., x^{2}+(y+1)^{2}=1, therefore The parametric equations of curve C_{3} are begin{cases} x=cos beta y=-1+sin beta end{cases}, (beta is the parameter). (2) Let P(2cos alpha,sin alpha), then the distance from P to the center of curve C_{3} (0,-1) is: d= sqrt{4cos^{2}alpha+(sin alpha+1)^{2}}= sqrt{-3(sin alpha- dfrac{1}{3})^{2}+ dfrac{16}{3}}. because sin alphain[-1,1], therefore when sin alpha= dfrac{1}{3}, d_{max}= dfrac{4sqrt{3}}{3}. therefore |PQ|_{max}=d_{max}+r= dfrac{4sqrt{3}}{3}+1= boxed{dfrac{4sqrt{3}+3}{3}}.

answer:Let's set up an equation to represent the situation: 3x = (62 - x) + 26 Now, let's solve for x: 3x = 62 - x + 26 3x = 88 - x 3x + x = 88 4x = 88 x = 88 / 4 x = 22 So, the value of x is boxed{22} .