answer:1. **Altitude Construction**: Construct the altitude from Y to XZ and name its length h. Both triangle XYW and triangle WYZ share this altitude. 2. **Area Calculation**: - The area of triangle XYW is frac{1}{2} times XW times h = frac{1}{2} times 8 times h. - The area of triangle WYZ is frac{1}{2} times WZ times h = frac{1}{2} times 12 times h. 3. **Ratio Determination**: - The ratio of the area of triangle XYW to the area of triangle WYZ is frac{frac{1}{2}times8times h}{frac{1}{2}times12times h}=frac{8}{12}=frac{2}{3}. Hence, the final answer is boxed{2:3}.

answer:1. Let's denote the radius of the smaller circle as ( r_1 = 1 ) and the radius of the larger circle as ( r_2 = 2 ). 2. The area of a circle is given by the formula ( pi r^2 ). Therefore, the area of the larger circle is: [ A_{text{large}} = pi r_2^2 = pi cdot 2^2 = 4pi ] 3. The given problem states that the shaded area is three times less than the area of the larger circle. So, let's denote the area of the shaded region as ( A_{text{shaded}} ). Therefore: [ A_{text{shaded}} = frac{1}{3} A_{text{large}} = frac{1}{3} cdot 4pi = frac{4pi}{3} ] 4. The shaded area is also a sector of the annulus (the region between the two circles). To find the angle ( angle AOB ) that subtends the shaded region at the center ( O ), we can use the area formula for the sector: [ text{Area of sector} = frac{theta}{2pi} cdot (text{Area of annulus}) ] The area of the annulus is the difference between the areas of the larger circle and the smaller circle: [ A_{text{annulus}} = A_{text{large}} - A_{text{small}} = 4pi - pi = 3pi ] Here, ( theta ) is the angle subtended at the center in radians. 5. Substitute the values into the sector area formula: [ A_{text{shaded}} = frac{theta}{2pi} cdot 3pi ] Simplify this equation: [ frac{4pi}{3} = frac{theta}{2pi} cdot 3pi implies frac{4pi}{3} = frac{3pi theta}{2pi} implies frac{4pi}{3} = frac{3 theta}{2} ] 6. Solve for ( theta ): [ frac{4pi}{3} = frac{3theta}{2} implies frac{8pi}{3} = 3theta implies theta = frac{8pi}{9} ] Therefore, the angle ( angle AOB ) in radians is: [ theta = frac{8pi}{9} ] # Conclusion: The angle ( angle AOB ) is ( boxed{ frac{8pi}{9} } ) radians.

answer:Let's denote the number of people who purchased only book B as ( x ). According to the information given, the number of people who purchased both books A and B is 500, which is twice the number of people who purchased only book B. Therefore, we can write: ( 500 = 2x ) Solving for ( x ), we get: ( x = frac{500}{2} = 250 ) So, 250 people purchased only book B. Now, we know that 1000 people purchased only book A (denoted as C). The total number of people who purchased book A (denoted as A) would be the sum of the people who purchased only book A and the people who purchased both books A and B: ( A = C + text{(both A and B)} ) ( A = 1000 + 500 ) ( A = 1500 ) The total number of people who purchased book B (denoted as B) would be the sum of the people who purchased only book B and the people who purchased both books A and B: ( B = x + text{(both A and B)} ) ( B = 250 + 500 ) ( B = 750 ) Now, we can find the ratio of the number of people who purchased book A to the number of people who purchased book B: ( text{Ratio of A to B} = frac{A}{B} ) ( text{Ratio of A to B} = frac{1500}{750} ) ( text{Ratio of A to B} = 2 ) Therefore, the ratio of the number of people who purchased book A to the number of people who purchased book B is boxed{2:1} .

answer:From (1), we have b_n=2 left( frac{1}{2} right)^{n-1} = frac{1}{2^{n-2}}, Therefore, c_n=nb_n= frac{n}{2^{n-2}}. The sum of the first n terms of the sequence {c_n}, S_n, is given by S_n = 2 + frac{2}{2} + frac{3}{2^2} + frac{4}{2^3} + ldots + frac{n}{2^{n-2}}. Multiplying S_n by frac{1}{2}, we get frac{1}{2}S_n = 1 + 1 + frac{3}{2^2} + frac{4}{2^3} + ldots + frac{n-1}{2^{n-2}} + frac{n}{2^{n-1}}. Subtracting this from the original sum, we obtain S_n - frac{1}{2}S_n = 2 + 1 + frac{1}{2} + frac{1}{2^2} + ldots + frac{1}{2^{n-2}} - frac{n}{2^{n-1}}. This simplifies to frac{1}{2}S_n = 2left(1 - frac{1}{2^n}right) div left(1 - frac{1}{2}right) - frac{n}{2^{n-1}} = 4 - frac{2+n}{2^{n-1}}, Therefore, we find the sum S_n to be S_n = boxed{8 - frac{2+n}{2^{n-2}}}.