answer:To find the discounted price Smith paid for the shirt, we need to calculate the discount amount and subtract it from the original selling price. The discount rate is 32%, so we can calculate the discount amount as follows: Discount Amount = (Discount Rate) x (Original Selling Price) Discount Amount = (32/100) x Rs. 955.88 Now, let's calculate the discount amount: Discount Amount = 0.32 x Rs. 955.88 Discount Amount = Rs. 305.8816 Now that we have the discount amount, we can find the discounted price by subtracting the discount amount from the original selling price: Discounted Price = Original Selling Price - Discount Amount Discounted Price = Rs. 955.88 - Rs. 305.8816 Let's calculate the discounted price: Discounted Price = Rs. 955.88 - Rs. 305.8816 Discounted Price = Rs. 649.9984 Since prices are usually rounded to the nearest currency unit, the discounted price Smith paid for the shirt would be approximately Rs. boxed{650.00} .

answer:# Solution: Part 1: Finding A Given asin(A-B)=(c-b)sin A, we apply the sine rule to transform the equation: 1. By the sine rule, we have frac{a}{sin A} = frac{b}{sin B} = frac{c}{sin C}. 2. Therefore, asin(A-B)=(c-b)sin A becomes sin Asin(A-B)=(sin C-sin B)sin A. 3. Since sin A > 0, we can simplify to sin(A-B)=sin C-sin B. 4. This further simplifies to sin(A-B)=sin(A+B)-sin B. 5. Expanding both sides gives sin Acos B-cos Asin B=sin Acos B+cos Asin B-sin B. 6. Simplifying, we get 2sin Bcos A=sin B. 7. Since 0 < B < pi, implying sin B > 0, we divide both sides by sin B to get cos A=frac{1}{2}. 8. Given 0 < A < pi, we conclude A=frac{pi}{3}. Therefore, the angle A is boxed{frac{pi}{3}}. Part 2: Finding the Perimeter of triangle ABC Given AD=3, ∠ADC=frac{pi}{3}, and the area of triangle ABC is 3sqrt{3}: 1. The area S can be expressed as S=frac{1}{2}ADcdot[CDsin∠ADC+BDsin(pi-∠ADC)]=frac{a}{2}cdot ADcdotfrac{sqrt{3}}{2}=frac{3sqrt{3}}{4}a. 2. Since S=3sqrt{3}, equating both expressions for S gives a=4. 3. Also, S=3sqrt{3}=frac{1}{2}bcdot csin A leads to bc=12. 4. Using the cosine rule a^{2}=b^{2}+c^{2}-2bccos A, we find a^{2}=(b+c)^{2}-3bc. 5. Solving for b+c gives b+c=sqrt{a^{2}+3bc}=2sqrt{13}. Therefore, the perimeter of triangle ABC is boxed{4+2sqrt{13}}.

answer:**Analysis** This question mainly tests the calculation of definite integrals and the application of derivative formulas in reverse, which is a basic type of problem. The key is to accurately find the antiderivative of the integrand. **Solution** The value of the integral is (x-cos x)|_{-1}^{1}=(1-cos 1)-(-1-cos (-1))=2-cos 1+cos 1=2. Therefore, the correct choice is boxed{B}.

answer:1. Let the sides of the triangle ABC be 3x, 4x, and 5x respectively. 2. Since the perimeter is the sum of all sides, we have: [ 3x + 4x + 5x = 60 text{ cm} ] 3. Simplify to find x: [ 12x = 60 text{ cm} implies x = frac{60}{12} = 5 text{ cm} ] 4. Now find the lengths of the sides: - Side AB = 3x = 3 times 5 = 15 cm - Side BC = 4x = 4 times 5 = 20 cm - Side AC = 5x = 5 times 5 = 25 cm Conclusion: The lengths of the sides are 15 cm, 20 cm, and 25 cm respectively, with 15text{ cm, 20text{ cm}, 25text{ cm}}. The final answer is boxed{A}