Aurora

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question:John decides to start collecting art. He pays the same price for his first 3 pieces of art and the total price came to 45,000. The next piece of art was 50% more expensive than those. How much did all the art cost?

answer:To solve this problem, we break it down into steps: 1. **Calculate the cost of each of the first three pieces of art:** The total cost for the first three pieces is 45,000. Since he paid the same price for each, we divide the total by 3 to find the cost of one piece. [ frac{45,000}{3} = 15,000 ] 2. **Determine the cost of the next piece of art:** The next piece of art was 50% more expensive than the previous ones. To find out how much more expensive, we calculate 50% of 15,000. [ 15,000 times 0.5 = 7,500 ] Then, we add this to the original price of 15,000 to get the cost of the next piece. [ 15,000 + 7,500 = 22,500 ] 3. **Calculate the total cost of all the art:** Finally, we add the total cost of the first three pieces to the cost of the next piece to find the total cost. [ 45,000 + 22,500 = 67,500 ] Therefore, the total cost of all the art John collected is boxed{67,500}.

question:The perimeter of a triangle is some length and the inradius of the triangle is 1.5 cm. The area of the triangle is 29.25 cm². What is the perimeter of the triangle?

answer:The area of a triangle can be calculated using the formula: Area = inradius * semiperimeter where the semiperimeter is half of the perimeter of the triangle. Let's denote the perimeter of the triangle as P and the semiperimeter as s. Then we have: s = P/2 Given that the inradius (r) is 1.5 cm and the area (A) is 29.25 cm², we can set up the equation: A = r * s 29.25 = 1.5 * (P/2) Now we can solve for P: 29.25 = 1.5 * P/2 29.25 = 0.75 * P P = 29.25 / 0.75 P = 39 cm Therefore, the perimeter of the triangle is boxed{39} cm.

question:f an income of 10.5 yuan is represented as +10.5 yuan, then an expenditure of 6 yuan can be represented as ____ yuan.

answer:Since an income of 10.5 yuan is represented as +10.5 yuan, an expenditure of 6 yuan can be represented as -6 yuan. Therefore, the answer is boxed{-6}.

question:the adjoining figure, two circles with radii 10 and 8 are drawn with their centers 15 units apart. At point P, where the circles intersect, a line is drawn such that the chords QP and PR are of equal length. Find the square of the length of chord QP.

answer:Let the length of the chords QP and PR be x. Since the line is drawn such that QP equals PR, we need to consider the triangle formed by the centers of the two circles and the point P. By the Law of Cosines: [ c^2 = a^2 + b^2 - 2ab cos C ] where c is the distance between the centers, a and b are the radii of the circles, and C is the angle at P. 1. Using the given values: - a = 10 - b = 8 - c = 15 2. Incorporating the condition QP = PR = x, and since QP and PR subtend angles at respective circle centers: - cos^{-1}left(frac{x}{20}right) + cos^{-1}left(frac{x}{16}right) = 180^circ - cos^{-1}left(frac{15^2 - 10^2 - 8^2}{-2 times 10 times 8}right) - Simplify the right side: [ cos^{-1}left(frac{-11}{160}right) ] 3. Calculate the cosine of both sides, using cosine addition formula and simplifying: [ cos left(cos^{-1}left(frac{x}{20}right) + cos^{-1}left(frac{x}{16}right)right) = cosleft(180^circ - cos^{-1}left(frac{-11}{160}right)right) ] [ frac{x^2}{320} - frac{x^2}{320} cdot frac{-11}{160} = 1 - 2sin^2left(cos^{-1}left(frac{-11}{160}right)right) ] [ x^2 = 250 ] Thus, the square of the length of chord QP is boxed{250}.