answer:Let's break down the problem step by step. 1. James invests 2000 a week into his bank account. There are 52 weeks in a year, so over the course of a year, he would invest: 2000/week * 52 weeks/year = 104,000 2. At the end of the year, he gets a windfall that is worth 50% more than what he has in his bank account. Let's call the amount he has in his bank account before the windfall "X". The windfall would then be: 1.5 * X 3. After receiving the windfall, James has 885,000 in his account. This total amount is the sum of the amount he had before the windfall (X) and the windfall itself (1.5 * X): X + 1.5 * X = 885,000 4. Combining like terms gives us: 2.5 * X = 885,000 5. To find the value of X, we divide both sides of the equation by 2.5: X = 885,000 / 2.5 X = 354,000 6. The amount X includes the money he invested over the year, which is 104,000. To find out how much he had at the beginning of the year, we subtract the amount he invested from X: 354,000 - 104,000 = 250,000 James had boxed{250,000} in his account at the beginning of the year.

answer:1. **Understanding the Problem**: Two fixed points A and B are given, and we need to determine the locus of the centers of all circles with a radius b that pass through both A and B. Given AB = 2b. 2. **Visualizing and Analyzing**: Consider any circle with radius b that passes through both A and B. Let O be the center of such a circle. By the triangle inequality and the properties of a circle, O must be equidistant from both A and B. The only position where O can exist while satisfying OA = OB = b and AB = 2b is along the perpendicular bisector of AB, and exactly at the midpoint, since AB = 2b matches the diameter of such circles. 3. **Forming the Locus**: The locus of all such points O is a single point, the midpoint of AB. Unlike the individual case where the distance a forms a circle around a point, the condition AB = 2b and the requirement for O to be equidistant to A and B restrict O to just one point. 4. **Conclusion**: The locus of the centers of all circles of radius b passing through fixed points A and B (where AB = 2b) is a single point, specifically the midpoint of line segment AB. Thus, the answer is textbf{(A) text{a point}}. The final answer is boxed{textbf{(A) }text{a point}}

answer:Given the inverse proportion function y=frac{1-m}{x}, and knowing that its graph is located in the first and third quadrants, we can analyze the conditions for m based on the behavior of the function in these quadrants. 1. For the graph to be in the first and third quadrants, the value of y must be positive when x is positive (first quadrant) and negative when x is negative (third quadrant). This behavior is determined by the sign of the numerator 1-m since x changes sign from positive to negative as we move from the first to the third quadrant. 2. Therefore, for y to maintain the correct sign in each quadrant, 1-m must be positive. This leads to the inequality: [ 1-m > 0 ] 3. Solving this inequality for m gives us: [ 1 > m ] 4. Simplifying, we find the range of values for m: [ m < 1 ] Thus, the correct answer, based on the given conditions, is boxed{C}.

answer:To solve the given system of inequalities step by step, we start with each inequality individually: 1. For the first inequality x - a > 2, we add a to both sides to isolate x: [x > a + 2] 2. For the second inequality b - 2x > 0, we first add 2x to both sides and then divide by 2 to isolate x: [b > 2x] [x < frac{b}{2}] Given that the solution set of the system is -1 < x < 1, we can equate the derived inequalities to the boundaries of this solution set: 3. From x > a + 2, and considering the lower bound -1 for x, we have: [a + 2 = -1] Solving for a gives us: [a = -1 - 2] [a = -3] 4. From x < frac{b}{2}, and considering the upper bound 1 for x, we have: [frac{b}{2} = 1] Solving for b gives us: [b = 2 cdot 1] [b = 2] Finally, we find left(a+bright)^{2021} using the values of a and b we found: 5. Substituting a = -3 and b = 2 into left(a+bright)^{2021}: [left(a+bright)^{2021} = left(-3 + 2right)^{2021}] [left(a+bright)^{2021} = left(-1right)^{2021}] [left(a+bright)^{2021} = -1] Therefore, the final answer is: [boxed{-1}]