answer:We are given the equation involving the absolute value function: [ |x + 6| = -(x + 6) ] We need to determine for which values of ( x ) this equation holds true. 1. **Understanding the Absolute Value Property**: By definition, the absolute value of a number ( a ) is defined as: [ |a| = begin{cases} a & text{if } a geq 0 -a & text{if } a < 0 end{cases} ] Therefore, in our case ( a = x + 6 ). 2. **Considering the Cases for Absolute Value**: The equation ( |x + 6| = -(x + 6) ) will be valid when ( x + 6 ) is negative. This corresponds to the second part of the absolute value definition: [ |x + 6| = -(x + 6) quad text{if} quad x + 6 < 0 ] 3. **Solving the Inequality**: Let us solve the inequality to find the values of ( x ): [ x + 6 < 0 ] Subtracting 6 from both sides, we obtain: [ x < -6 ] 4. **Conclusion**: The equality ( |x + 6| = -(x + 6) ) holds true for: [ x le -6 ] Thus, the required values of ( x ) that satisfy the given condition are: [ boxed{x leq -6} ]

answer:We use the Euclidean Algorithm to find ( gcd(13n+3, 7n+1) ). Start with: [ gcd(13n+3, 7n+1) ] Apply the Euclidean Algorithm: 1. Subtract ( 7n+1 ) from ( 13n+3 ): [ gcd(13n+3, 7n+1) = gcd(7n+1, (13n+3) - (7n+1)) = gcd(7n+1, 6n+2) ] 2. Subtract ( 6n+2 ) from ( 7n+1 ): [ gcd(7n+1, 6n+2) = gcd(6n+2, (7n+1)-(6n+2)) = gcd(6n+2, n-1) ] 3. Use a factor of n-1: [ gcd(6n+2, n-1) = gcd(n-1, 6n+2 - 6(n-1)) = gcd(n-1, 8) ] Thus, if ( n-1 ) is a multiple of 8, then the greatest common divisor of ( 13n+3 ) and ( 7n+1 ) is 8. Otherwise, the greatest common divisor is 1. Therefore, the maximum possible value for the greatest common divisor of ( 13n+3 ) and ( 7n+1 ) is ( boxed{8} ).

answer:If ||MF_1|-|MF_2||=24, then the trajectory of point M is a hyperbola with F_1(-13,0) and F_2(13,0) as its foci. The equation of this hyperbola is frac {x^{2}}{144}- frac {y^{2}}{25}=1. Since the line 5x+12y=0 is its asymptote, the entire line lies outside of the hyperbola, which means ||MF_1|-|MF_2||<24. Therefore, the correct choice is boxed{text{C}}. By applying the definition of a hyperbola, we can obtain the equation of the hyperbola and the equation of its asymptote, leading to the conclusion. This question tests the definition, equation, and properties of a hyperbola, as well as computational skills, and is considered a medium-level question.

answer:(text{I}) Solution: According to the problem, we know that frac {1}{2}absin C= frac { sqrt {3}}{4}times2abcos C. Therefore, tan C= sqrt {3}. Since 0 < C < pi, we have C= frac {pi}{3}. Thus, the size of angle C is boxed{frac {pi}{3}}. (text{II}) Solution: Given sin A+sin B =sin A+sin (pi-C-A) =sin A+sin ( frac {2pi}{3}-A) =sin A+ frac { sqrt {3}}{2}cos A+ frac {1}{2}sin A= frac {3}{2}sin A+ frac { sqrt {3}}{2}cos A= sqrt {3}sin (A+ frac {pi}{6})leqslant sqrt {3}. Equality holds when triangle ABC is an equilateral triangle, thus, the maximum value of sin A+sin B is boxed{sqrt {3}}.