answer:Let's first find out how much work A and B can do in one day when working together. A can do the work in 15 days, so A's work rate is ( frac{1}{15} ) of the work per day. B can do the work in 20 days, so B's work rate is ( frac{1}{20} ) of the work per day. When A and B work together, their combined work rate per day is: ( frac{1}{15} + frac{1}{20} ) To find a common denominator, we can use 60 (the least common multiple of 15 and 20): ( frac{4}{60} + frac{3}{60} = frac{7}{60} ) So, A and B together can do ( frac{7}{60} ) of the work in one day. Let's call the number of days they worked together "d". The amount of work they would have completed together in "d" days is: ( frac{7}{60} times d ) We are given that the fraction of the work that is left is 0.5333333333333333 (which is ( frac{8}{15} ) when converted to a fraction). This means that the fraction of the work completed is ( 1 - frac{8}{15} = frac{7}{15} ). So, we have the equation: ( frac{7}{60} times d = frac{7}{15} ) To find "d", we can solve for it by dividing both sides of the equation by ( frac{7}{60} ): ( d = frac{frac{7}{15}}{frac{7}{60}} ) Simplifying the right side of the equation: ( d = frac{7}{15} times frac{60}{7} ) The 7's cancel out: ( d = frac{1}{15} times 60 ) So: ( d = 4 ) Therefore, A and B worked together for boxed{4} days.

answer:1. **Identify the relationship between the slopes and the angle**: Given that two lines intersect at a 45^circ angle, we can use the tangent addition formula to relate the slopes of the lines. Let the slope of one line be m_1 and the other be m_2. The tangent of the angle between two lines with slopes m_1 and m_2 is given by: [ tan(theta) = left|frac{m_2 - m_1}{1 + m_1 m_2}right| ] Since theta = 45^circ, tan(45^circ) = 1. Therefore, we have: [ left|frac{m_2 - m_1}{1 + m_1 m_2}right| = 1 ] 2. **Set up the equation for the slopes**: Given that one line's slope is 6 times the other, let m_1 = a and m_2 = 6a. Substitute these into the equation derived from the tangent of the angle: [ left|frac{6a - a}{1 + a cdot 6a}right| = 1 implies left|frac{5a}{1 + 6a^2}right| = 1 ] Solving for a, we get: [ 5a = 1 + 6a^2 implies 6a^2 - 5a + 1 = 0 ] 3. **Solve the quadratic equation**: The quadratic equation 6a^2 - 5a + 1 = 0 can be solved using the quadratic formula: [ a = frac{-(-5) pm sqrt{(-5)^2 - 4 cdot 6 cdot 1}}{2 cdot 6} = frac{5 pm sqrt{25 - 24}}{12} = frac{5 pm 1}{12} ] This gives us a = frac{1}{2} or a = frac{1}{3}. 4. **Calculate the product of the slopes for each case**: - If a = frac{1}{2}, then m_2 = 6 cdot frac{1}{2} = 3. The product of the slopes is m_1 m_2 = frac{1}{2} cdot 3 = frac{3}{2}. - If a = frac{1}{3}, then m_2 = 6 cdot frac{1}{3} = 2. The product of the slopes is m_1 m_2 = frac{1}{3} cdot 2 = frac{2}{3}. 5. **Determine the greatest product**: Comparing frac{3}{2} and frac{2}{3}, the greatest value is frac{3}{2}. Thus, the greatest possible value of the product of the slopes of the two lines is boxed{textbf{(C)} frac{3}{2}}.

answer:Since point M is in the fourth quadrant, its abscissa and ordinate are positive and negative numbers, respectively. Additionally, the distance from point M to the x-axis is 12, and the distance to the y-axis is 4. Therefore, the coordinates of point M are (4, -12). Hence, the answer is: boxed{text{A}}. Given that point M is in the fourth quadrant, the abscissa is greater than 0, and the ordinate is less than 0. By using the distances to the coordinate axes, we can determine the specific coordinates. This question primarily examines the signs of a point's coordinates when it is in the fourth quadrant, as well as the absolute values of the coordinates representing the distances to the x-axis and y-axis, respectively.

answer:First, we identify the common factor in each term. Here, the common factor is (x-4). begin{align*} x(x-4) + 2(x-4) &= x cdot (x-4) + 2 cdot (x-4) &= (x + 2)(x-4). end{align*} Conclusion: Thus, the factored form of the expression x(x-4) + 2(x-4) is boxed{(x+2)(x-4)}.