answer:Since f(x)=2sin xcos x+2sqrt{3}cos^{2}x-sqrt{3} qquad= sin 2x + sqrt{3}cos 2x qquad= 2sin(2x + frac{pi}{3}), The smallest positive period of the function f(x) is T=frac{2pi}{2}=pi. When the graph of f(x) is shifted to the left by phi(frac{pi}{2}<phi<pi) units, the corresponding function becomes y=2sin[2(x+phi)+frac{pi}{3}]=2sin(2x+2phi+frac{pi}{3}). By requiring that y=2sin(2x+2phi+frac{pi}{3}) is an even function, we obtain: 2phi+frac{pi}{3}=kpi+frac{pi}{2}, where k in mathbb{Z}, Thus, phi=frac{kpi}{2}+frac{pi}{12}. Since frac{pi}{2}<phi<pi, We get phi=frac{7pi}{12}. Therefore, the answers are: boxed{pi} and boxed{frac{7pi}{12}}. This problem mainly tests the understanding of the graph transformation rules for the function y=Asin(omega x+phi), the even/odd properties of functions, the application of trigonometric identities, and the application of the trigonometric function period formula. It requires the ability to transform thinking and combine numerical and graphical approaches, making it a moderately difficult problem.

answer:Using the Factor Theorem, if x + 3 is a factor of Q(x), then Q(-3) = 0. Substitute -3 into the polynomial: Q(-3) = (-3)^3 - 4(-3)^2 + d(-3) - 8. Simplify the expression: Q(-3) = -27 - 4(9) - 3d - 8 = -27 - 36 - 3d - 8. Combine like terms: Q(-3) = -71 - 3d. Setting Q(-3) = 0 for it to be a factor: -71 - 3d = 0. Solve for d: 3d = -71, d = frac{-71}{3}. Therefore, the required value of d is boxed{frac{-71}{3}}.

answer:1. **Define the problem in terms of an arithmetic sequence:** Let the central angles of the sectors be represented by an arithmetic sequence with the first term a_1 and common difference d. The sequence will have 15 terms, and the sum of these terms must equal 360 degrees, the total degrees in a circle. 2. **Use the sum formula for an arithmetic sequence:** The sum S of the first n terms of an arithmetic sequence is given by: [ S = n cdot frac{a_1 + a_n}{2} ] Here, n = 15 and S = 360, so: [ 15 cdot frac{a_1 + a_{15}}{2} = 360 ] Simplifying, we get: [ a_1 + a_{15} = 48 ] 3. **Express a_{15} in terms of a_1 and d:** The last term a_{15} can be written as: [ a_{15} = a_1 + 14d ] Substituting into the equation a_1 + a_{15} = 48, we find: [ a_1 + (a_1 + 14d) = 48 ] Simplifying, we get: [ 2a_1 + 14d = 48 ] 4. **Solve for a_1 in terms of d**: Rearranging: [ a_1 = frac{48 - 14d}{2} ] 5. **Determine valid values for d:** a_1 must be an integer, thus 48 - 14d must be an even number. Also, a_1 must be positive. 6. **Check possible values for d:** Starting with smaller values of d: - If d = 2, then a_1 = frac{48 - 14 times 2}{2} = frac{48 - 28}{2} = 10 (an integer). 7. **Conclusion:** The smallest possible value for a_1 that satisfies all conditions is 10 degrees. Therefore, the degree measure of the smallest possible sector angle is 10. The final answer is boxed{10}

answer:To find the difference between their profit shares, we first need to determine the individual shares of X and Y based on the given ratio. The ratio of X's share to Y's share is 1/2 : 1/3. To make it easier to work with, we can convert these fractions to have a common denominator. The least common multiple of 2 and 3 is 6, so we can express the ratio as: X's share : Y's share = (1/2 * 6/6) : (1/3 * 6/6) X's share : Y's share = 3/6 : 2/6 Now, we can simplify the ratio to whole numbers by multiplying both parts by 6: X's share : Y's share = 3 : 2 The total parts in the ratio are 3 + 2 = 5 parts. Now, we can find the value of each part by dividing the total profit by the total number of parts: Value of each part = Total profit / Total parts Value of each part = Rs. 500 / 5 Value of each part = Rs. 100 Now we can find the individual shares of X and Y: X's share = 3 parts * Rs. 100/part = Rs. 300 Y's share = 2 parts * Rs. 100/part = Rs. 200 The difference between their profit shares is: Difference = X's share - Y's share Difference = Rs. 300 - Rs. 200 Difference = Rs. 100 Therefore, the difference between X's and Y's profit shares is Rs. boxed{100} .