answer:Line a is parallel to line b, for the following reasons: Since line a is parallel to plane alpha, and line a is parallel to plane beta, there must be a line in planes alpha and beta that is parallel to a, let's denote them as m and n respectively, which means a is parallel to m, and a is parallel to n. Therefore, m is parallel to n. Since alpha and beta intersect, m is within plane alpha, and n is within plane beta, which implies m is parallel to plane beta. Then m is parallel to l, and since a is parallel to m, it follows that a is parallel to l. Therefore, the correct choice is boxed{text{C}}.

answer:To demonstrate that widehat{BTM} = widehat{BTN}, we proceed with the following steps: 1. **Consider the homothety ( h ) centered at ( T )**: - Let ( Gamma ) be the larger circle and ( Gamma^{prime} ) the smaller circle with ( Gamma^{prime} ) inside ( Gamma ), and touching at the point ( T ). - Define ( h ) as the positive homothety centered at ( T ) that maps ( Gamma^{prime} ) to ( Gamma ). 2. **Map ( B ) to ( B^{prime} )**: - Let ( B ) be any point on ( Gamma^{prime} ) (other than ( T )). - Under the homothety ( h ), point ( B ) on ( Gamma^{prime} ) is mapped to a point ( B^{prime} ) on ( Gamma ). 3. **Map the tangent at ( B )**: - Consider the tangent to ( Gamma^{prime} ) at ( B ). - Under homothety ( h ), this tangent line at ( B ) is mapped to a line parallel to it, which is also tangent to ( Gamma ) at ( B^{prime} ). 4. **Points of intersection ( M ) and ( N )**: - Let ( M ) and ( N ) be the points where this tangent intersects ( Gamma ). These points are distinct and lie on the tangent to ( Gamma ). 5. **Introduce the images of ( M ) and ( N ) under ( h )**: - Under homothety ( h ), points ( M ) and ( N ) on ( Gamma ) map back to points ( B^{prime prime} ) and ( B^{prime prime prime} ) on ( Gamma^{prime} ). Here ( B^{prime prime} ) and ( B^{prime prime prime} ) coincide with ( B ). 6. **Angle chase (inscribed angles, tangent-secant angles, alternate interior angles)**: - Consider the angles formed by the tangents and secant lines at the point ( T ). Specifically, investigate the angles ( widehat{BTM} ) and ( widehat{BTN} ). - Since ( Gamma ) and ( Gamma^{prime} ) are tangent to each other at ( T ) and considering the properties of homothety (angles preserved), the triangles ( triangle BTB^{prime} ), ( triangle B^{prime}TM^{prime} ) and ( triangle B^{prime}TN^{prime} ) will share similar geometric properties. - By the properties of homothety and tangent-secant angles, it can be noted that the angles ( widehat{BTM} ) and ( widehat{BTN} ) must be equal. 7. **Conclusion**: - We can thus conclude that ( widehat{BTM} = widehat{BTN} ). Therefore, our final boxed answer is: [ boxed{widehat{BTM} = widehat{BTN}} ]

answer:From the given equation, [arcsin 3x = frac{pi}{4} - arcsin x.] Then [sin(arcsin 3x) = sinleft(frac{pi}{4} - arcsin xright).] Using the angle subtraction formula, [ 3x = sin frac{pi}{4} cos (arcsin x) - cos frac{pi}{4} sin (arcsin x) = frac{sqrt{2}}{2} sqrt{1 - x^2} - frac{sqrt{2}}{2} x. ] Thus, [ 3x = frac{sqrt{2}}{2} (sqrt{1 - x^2} - x). ] Solving for (x), we get: [ 6x = sqrt{2}(sqrt{1 - x^2} - x). ] Squaring both sides, [ 36x^2 = 2(1 - x^2 - 2xsqrt{1 - x^2}). ] This simplifies to: [ 38x^2 + 2xsqrt{1 - x^2} = 2. ] Isolating (xsqrt{1 - x^2}), [ xsqrt{1 - x^2} = 1 - 19x^2. ] Squaring again, [ x^2(1 - x^2) = 1 - 38x^2 + 361x^4, ] [ 361x^4 - 39x^2 + 1 = 0. ] Let ( y = x^2 ), [ 361y^2 - 39y + 1 = 0. ] Solving this quadratic, [ y = frac{39 pm sqrt{1521 - 1444}}{722} = frac{39 pm sqrt{77}}{722}. ] Thus, [ x = pm sqrt{frac{39 pm sqrt{77}}{722}}. ] Verification of each root within the domain of arcsine and the original equation is needed. Conclusion: The solutions are (x = boxed{pm sqrt{frac{39 pm sqrt{77}}{722}}}) subject to verification of being within the domain of arcsine and satisfying the original equation.

answer:To solve this, we first determine the chance of selecting each container. Each container is selected with equal chance, which is frac{1}{4} because there are four containers. - For container I: The probability of selecting a green ball, given that container I was selected is frac{4}{12} = frac{1}{3}. Since the probability to choose container I is frac{1}{4}, [ P(text{green from I}) = frac{1}{4} cdot frac{1}{3} = frac{1}{12} ] - For containers II and III: The probability of selecting a green ball from either container, given that either was selected, is frac{5}{7}. Since choosing either container II or III has a probability of frac{1}{4} each, [ P(text{green from II}) = frac{1}{4} cdot frac{5}{7} = frac{5}{28} ] [ P(text{green from III}) = frac{1}{4} cdot frac{5}{7} = frac{5}{28} ] - For container IV: The probability of selecting a green ball, given container IV was selected, is frac{4}{8} = frac{1}{2}. The probability to choose container IV is frac{1}{4}, [ P(text{green from IV}) = frac{1}{4} cdot frac{1}{2} = frac{1}{8} ] Adding these probabilities gives us the total probability of selecting a green ball: [ boxed{P(text{green}) = frac{1}{12} + frac{5}{28} + frac{5}{28} + frac{1}{8} = frac{7}{28} + frac{7}{28} + frac{14}{56} = boxed{frac{25}{56}}} ]