answer:To determine the possible values of xi, the random variable representing the number of defective products selected, we analyze the scenario step by step: 1. **Total Products**: There are 8 products in total. 2. **Defective Products**: Out of these, 2 are defective. 3. **Selection**: We are selecting 3 products randomly. Given these conditions, let's examine the possible outcomes for the number of defective products that can be selected: - **Minimum Defective Products**: The minimum number of defective products that can be selected is 0. This occurs if all selected products are non-defective. - **Maximum Defective Products**: The maximum number of defective products that can be selected is 2, since there are only 2 defective products available. Therefore, the possible values of xi are determined by the range of defective products that can be selected, which are: - If 0 defective products are selected. - If 1 defective product is selected. - If 2 defective products are selected. Given these considerations, the possible values of xi are 0, 1, and 2. Hence, the correct options are: boxed{text{A, B, and C}}.

answer:1. Start with the function g(x) = 8^x. We need to evaluate g(x+1) - g(x). 2. Substitute x+1 into the function: [ g(x+1) = 8^{x+1} ] 3. Calculate g(x+1) - g(x): [ g(x+1) - g(x) = 8^{x+1} - 8^x ] 4. Use the property of exponents, a^{m+n} = a^m cdot a^n, to rewite 8^{x+1}: [ 8^{x+1} = 8^x cdot 8 ] 5. Substitute this back into the expression for g(x+1) - g(x): [ g(x+1) - g(x) = 8^x cdot 8 - 8^x ] 6. Factor out 8^x from the expression: [ g(x+1) - g(x) = 8^x(8 - 1) = 8^x cdot 7 ] 7. Recognize that 8^x cdot 7 = 7g(x), since g(x) = 8^x. 8. Thus, the expression simplifies to: [ g(x+1) - g(x) = 7g(x) ] With this, we have 7g(x) as the answer. The final answer is (D) boxed{7g(x)}.

answer:First, simplify the quadratic equation: [ 3x^2 - kx + 2x + 24 = 0 ] [ 3x^2 + (2-k)x + 24 = 0 ] For a quadratic equation ax^2 + bx + c = 0, the condition for the roots to be real and equal is given by the discriminant Delta = b^2 - 4ac = 0. Here, a = 3, b = 2-k, and c = 24. 1. Set up the equation for the discriminant: [ (2-k)^2 - 4 cdot 3 cdot 24 = 0 ] [ (2-k)^2 - 288 = 0 ] 2. Solve for (2-k)^2: [ (2-k)^2 = 288 ] 3. Take the square root of both sides: [ 2-k = pm sqrt{288} ] [ 2-k = pm 12sqrt{2} ] 4. Solve for k in each case: - When 2-k = 12sqrt{2}: [ 2 - k = 12sqrt{2} ] [ -k = 12sqrt{2} - 2 ] [ k = 2 - 12sqrt{2} ] - When 2-k = -12sqrt{2}: [ 2 - k = -12sqrt{2} ] [ -k = -12sqrt{2} - 2 ] [ k = 2 + 12sqrt{2} ] Thus, the values of k for which the equation has real and equal roots are 2 - 12sqrt{2} and 2 + 12sqrt{2}. 2 - 12sqrt{2, 2 + 12sqrt{2}} The final answer is boxed{B}.

answer:1. Identify that point ( D ) is the ( A )-excenter of ( triangle ABC ). This means ( D ) is the center of the excircle opposite vertex ( A ). 2. Let ( I ) be the incenter of ( triangle ABC ), which is the center of the inscribed circle ( Omega ). 3. Note that the exterior angle bisectors of ( angle B ) and ( angle C ) intersect at ( D ), and ( D ) lies on the common side ( BC ). 4. Since ( D ) is the ( A )-excenter, the angles ( angle IBD ) and ( angle ICD ) are both ( 90^circ ). This is because the incenter and excenter are perpendicular to the sides of the triangle at the points of tangency. 5. Since ( angle IBD + angle ICD = 180^circ ), points ( I, B, C, D ) are concyclic (they lie on the same circle). 6. Given that ( DE ) is tangent to ( Omega ) at ( E ), we have ( angle IED = 90^circ ). 7. Since ( angle IED = 90^circ = angle IBD ), points ( I, E, B, D ) are concyclic. 8. Combining the two sets of concyclic points, we conclude that ( I, E, B, D, C ) are all concyclic. 9. To find ( angle BEC ), we use the fact that ( angle BEC = angle DEB + angle DEC ). 10. Since ( angle DEB = angle DBC ) and ( angle DEC = angle DCB ), we have: [ angle BEC = angle DBC + angle DCB ] 11. Using the properties of the excenter and incenter, we know: [ angle DBC = 90^circ - angle IBC quad text{and} quad angle DCB = 90^circ - angle ICB ] 12. Therefore: [ angle BEC = (90^circ - angle IBC) + (90^circ - angle ICB) ] 13. Since ( angle IBC = frac{angle ABC}{2} ) and ( angle ICB = frac{angle ACB}{2} ), we have: [ angle BEC = 180^circ - left( frac{angle ABC}{2} + frac{angle ACB}{2} right) ] 14. Using the angle sum property of a triangle, ( angle ABC + angle ACB = 180^circ - angle BAC ), we get: [ angle BEC = 180^circ - left( frac{180^circ - angle BAC}{2} right) ] 15. Substituting ( angle BAC = 40^circ ): [ angle BEC = 180^circ - left( frac{180^circ - 40^circ}{2} right) = 180^circ - left( frac{140^circ}{2} right) = 180^circ - 70^circ = 110^circ ] The final answer is (boxed{110^circ}).