answer:Let x be one number, and y be the other number. We have: 1. x + y = 15 2. xy = 36 To solve for x and y, we can use the quadratic formula based on the idea that these are roots of the quadratic equation: [ t^2 - (x+y)t + xy = 0 ] Substituting the given information: [ t^2 - 15t + 36 = 0 ] Using the quadratic formula: [ t = frac{-b pm sqrt{b^2 - 4ac}}{2a} = frac{-(-15) pm sqrt{(-15)^2 - 4 times 1 times 36}}{2 times 1} ] [ t = frac{15 pm sqrt{225 - 144}}{2} ] [ t = frac{15 pm sqrt{81}}{2} ] [ t = frac{15 pm 9}{2} ] This gives us: [ t_1 = frac{15 + 9}{2} = 12 ] [ t_2 = frac{15 - 9}{2} = 3 ] Since we are asked for the smaller number, boxed{3} is the smaller number.

answer:Solution: (1)f(x)=(log_2x)^2-log_2x^2+3, Let log_2x=t, therefore g(t)=t^2-2t+3, tin[0,2] The maximum value m=3, the minimum value n=2, therefore P(3,2), therefore sin alpha= frac {2}{ sqrt {13}}, cos alpha= frac {3}{ sqrt {13}}, therefore sin alpha+cos alpha= frac {5}{ sqrt {13}}. (2)g(x)=3cos (2x+ frac {pi}{3})-2, h(x)=g(x)-k=3cos (2x+ frac {pi}{3})-2-k Rightarrow 3cos (2x+ frac {pi}{3})=2+k, When xin[0, frac {pi}{2}], 2x+ frac {pi}{3}in[ frac {pi}{3}, frac {4pi}{3}], therefore h(x)=g(x)-k has two different zeros x_1, x_2 Leftrightarrow y=3cos x, xin[0, frac{4pi}{3}] intersects with y=2+k at two points, therefore k+2in(-3,- frac {3}{2}], therefore kin(-5,- frac{7}{2}]. Thus, the answers are boxed{frac {5}{ sqrt {13}}} for part (1) and boxed{(-5,- frac{7}{2}]} for part (2).

answer:1. **Identify the vertices and the line dividing the triangle**: Let's label the vertices of the triangle as A = (0,0), B = (1,1), and C = (6m,0). The line y = mx is supposed to divide the triangle into two triangles of equal area. 2. **Determine the intersection point**: The line y = mx will intersect the line segment BC. Let D be the point of intersection. The coordinates of D can be found by setting the y-coordinates of the line y = mx and the line segment BC equal. The line segment BC can be represented by the equation y = -frac{1}{6m-1}(x-1) (using point-slope form and the points B and C). Setting mx = -frac{1}{6m-1}(x-1) and solving for x, we find the x-coordinate of D. 3. **Condition for equal areas**: The triangles ABD and ACD will have equal areas if D is the midpoint of BC. The midpoint of BC is given by left(frac{6m+1}{2}, frac{1}{2}right). For D to be on the line y = mx, the coordinates must satisfy y = mx. Substituting the midpoint coordinates into the line equation, we get: [ frac{1}{2} = m cdot frac{6m+1}{2} ] Simplifying this equation: [ 1 = m(6m+1) implies 6m^2 + m - 1 = 0 ] 4. **Solve the quadratic equation**: The quadratic equation 6m^2 + m - 1 = 0 can be solved using the quadratic formula: [ m = frac{-b pm sqrt{b^2 - 4ac}}{2a} quad text{where } a = 6, b = 1, c = -1 ] [ m = frac{-1 pm sqrt{1 + 24}}{12} = frac{-1 pm 5}{12} ] [ m = frac{4}{12} = frac{1}{3}, quad m = frac{-6}{12} = -frac{1}{2} ] 5. **Sum of all possible values of m**: Using Vieta's formulas, the sum of the roots of the equation 6m^2 + m - 1 = 0 is given by -frac{b}{a} = -frac{1}{6}. Thus, the sum of all possible values of m is boxed{textbf{(B)} - !frac {1}{6}}.

answer:To find out how many seconds the light glows for, we need to calculate the total time between 1:57:58 am and 3:20:47 am and then divide that by the number of times the light glowed. First, let's find the total time in seconds between the two timestamps: From 1:57:58 to 2:00:00 is 2 minutes and 2 seconds. From 2:00:00 to 3:00:00 is 1 hour, which is 60 minutes or 3600 seconds. From 3:00:00 to 3:20:47 is 20 minutes and 47 seconds. Now, let's convert the minutes to seconds and add all the seconds together: 2 minutes = 2 * 60 = 120 seconds 120 seconds (from 1:57:58 to 2:00:00) + 2 seconds = 122 seconds 3600 seconds (from 2:00:00 to 3:00:00) 20 minutes = 20 * 60 = 1200 seconds 1200 seconds + 47 seconds (from 3:00:00 to 3:20:47) = 1247 seconds Now, let's add all the seconds together: 122 seconds + 3600 seconds + 1247 seconds = 4969 seconds This is the total time in seconds between 1:57:58 am and 3:20:47 am. Now, let's divide this total time by the number of times the light glowed: 4969 seconds / 292.29411764705884 times = 17 seconds (rounded to the nearest whole number) So, the light glows for approximately boxed{17} seconds each time.