answer:Let's first calculate the cost price (CP) of the kite for the shopkeeper. Given that the shopkeeper gains 30% when selling the kite for Rs. 30, we can write the following equation: Selling Price (SP) = Cost Price (CP) + Gain 30 = CP + (30% of CP) 30 = CP + (0.30 * CP) 30 = 1.30 * CP CP = 30 / 1.30 CP = Rs. 23.08 (approximately) Now, during the clearance sale, the shopkeeper still gains but now at 17%. So, we can calculate the new selling price (SP') using the gain percentage during the sale: SP' = CP + (17% of CP) SP' = 23.08 + (0.17 * 23.08) SP' = 23.08 + 3.924 SP' = Rs. 27.004 (approximately) The marked price (MP) before the sale was Rs. 30, and the selling price during the sale is approximately Rs. 27.004. Now we can calculate the discount percentage (D%) on the marked price during the clearance sale: Discount = MP - SP' Discount = 30 - 27.004 Discount = Rs. 2.996 (approximately) Discount Percentage (D%) = (Discount / MP) * 100 D% = (2.996 / 30) * 100 D% = 0.09987 * 100 D% = 9.987% (approximately) Therefore, the discount percentage on the marked price during the clearance sale is approximately boxed{9.99%} .

answer:1. We need to find all three-digit numbers overline{Pi mathrm{B} Gamma} comprising different digits Pi, mathrm{B}, and Gamma such that the following equality holds: [ overline{Pi mathrm{B} Gamma} = (Pi + mathrm{B} + Gamma) times (Pi + mathrm{B} + Gamma + 1). ] 2. Notice that: [ Pi + mathrm{B} + Gamma geq 3 quad text{and} quad Pi + mathrm{B} + Gamma leq 24. ] This is because Pi, mathrm{B}, and Gamma are distinct digits, ranging from 0 to 9. 3. Also, for the equation to hold, both sides of the equation need to yield the same number when divided by 9. This implies: [ overline{Pi mathrm{B} Gamma} equiv (Pi + mathrm{B} + Gamma) times (Pi + mathrm{B} + Gamma + 1) pmod{9}. ] 4. Simplifying, we notice that the products (Pi + mathrm{B} + Gamma) times (Pi + mathrm{B} + Gamma + 1) must be a two-digit number that is a product of two consecutive integers. We test values Pi + mathrm{B} + Gamma for possible results: [ 12 times 13 = 156 15 times 16 = 240 quad text{(Not a three-digit number)} ] Other higher values either yield products exceeding three digits or do not form digits between 100 to 999. 5. Among these values, we find that: [ overline{Pi mathrm{B} Gamma} = 156 = 12 times 13. ] For the equality to hold: [ (Pi + mathrm{B} + Gamma) = 12 quad text{and} quad (Pi + mathrm{B} + Gamma + 1) = 13. ] 6. To verify, we check if the digits sum to 12: [ 1 + 5 + 6 = 12. ] # Conclusion The only valid three-digit number overline{Pi mathrm{B} Gamma} satisfying the given condition is: [ boxed{156} ]

answer:The distance d from point C to line l satisfies d leqslant |CM|. When l perp CM, the distance from point C to line l is maximized. Therefore, k_{CM} cdot k_{l} = -1. Since k_{CM}= frac{2-0}{1-2}=-2, we have k_{l}= frac{1}{2}. Thus, the equation of line l is y-2= frac{1}{2}(x-1), which simplifies to x-2y+3=0. Therefore, the correct answer is boxed{D}. This problem involves identifying the condition for the maximum distance from point C to line l, calculating the precise slope, and then solving for the equation of the line. It tests the application of the relationship between a line and a circle, as well as computational skills.

answer:Firstly, recall the identity arccos x + arcsin x = frac{pi}{2} for x in [-1, 1]. Thus by substitution, if x = frac{1}{sqrt{2}}, then x^2 = frac{1}{2}. Using the identity, we have: [ arccosleft(frac{1}{2}right) + arcsinleft(frac{1}{2}right) = frac{pi}{2} ] [ arccosleft(frac{1}{2}right) = frac{pi}{3}, quad arcsinleft(frac{1}{2}right) = frac{pi}{6} ] These values are well-known angles in trigonometry. Now calculate gleft(frac{1}{sqrt{2}}right): [ gleft(frac{1}{sqrt{2}}right) = arccosleft(frac{1}{2}right) cdot arcsinleft(frac{1}{2}right) ] [ gleft(frac{1}{sqrt{2}}right) = frac{pi}{3} cdot frac{pi}{6} = frac{pi^2}{18} ] The solution has been calculated step by step, following the trigonometric identities and operations accurately. Conclusion: [ boxed{frac{pi^2}{18}} ]