answer:This problem involves the operations of complex numbers and the application of necessary and sufficient conditions for complex equality, covering the conjugate complex number. It is a basic problem. Let z = a + bi, where a and b are real numbers. Then, overline{z} = a - bi. Thus, 2z - overline{z} = a + 3bi = 1 + 6i. Equating real and imaginary parts, we get: a = 1 b = 2 Therefore, z = 1 + 2i. So the answer is boxed{1 + 2i}.

answer:To determine how many 1 times 10sqrt{2} rectangles can be cut from a 50 times 90 rectangle using cuts parallel to its edges, we can follow these steps: 1. **Calculate the area of the 50 times 90 rectangle:** [ text{Area of the large rectangle} = 50 times 90 = 4500 ] 2. **Calculate the area of one 1 times 10sqrt{2} rectangle:** [ text{Area of one small rectangle} = 1 times 10sqrt{2} = 10sqrt{2} ] 3. **Determine the number of 1 times 10sqrt{2} rectangles that can fit into the 50 times 90 rectangle:** [ text{Number of rectangles} = frac{text{Area of the large rectangle}}{text{Area of one small rectangle}} = frac{4500}{10sqrt{2}} ] 4. **Simplify the expression:** [ frac{4500}{10sqrt{2}} = frac{4500}{10} times frac{1}{sqrt{2}} = 450 times frac{1}{sqrt{2}} = 450 times frac{sqrt{2}}{2} = 450 times frac{sqrt{2}}{2} = 225sqrt{2} ] 5. **Approximate the value of 225sqrt{2}:** [ 225sqrt{2} approx 225 times 1.414 approx 318.15 ] Since the number of rectangles must be an integer, we take the floor value of the approximation: [ lfloor 318.15 rfloor = 318 ] Therefore, the maximum number of 1 times 10sqrt{2} rectangles that can be cut from a 50 times 90 rectangle is 318. The final answer is boxed{318}

answer:Since each circle has a radius of 3 inches, the diameter of one circle is (2 times 3 = 6) inches. Therefore, the side length of the square, housing four circles two by two without any gap, is (2) times the diameter of one circle, i.e., (2 times 6 = 12) inches. 1. Calculate the area of the square: [ text{Area of the square} = text{side length}^2 = 12^2 = 144 text{ square inches} ] 2. Calculate the area of one circle: [ text{Area of one circle} = pi times text{radius}^2 = pi times 3^2 = 9pi text{ square inches} ] 3. Since there are four such circles, the total area covered by the circles is: [ text{Total area covered by the circles} = 4 times 9pi = 36pi text{ square inches} ] The square has an area of 144 square inches and the total area covered by the circles is 36pi square inches. Conclusion: [ boxed{text{Area of the square} = 144 text{ square inches, Total area covered by the circles} = 36pi text{ square inches}} ]

answer:First, calculate the magnitude of the products separately: 1. For (18-5i)(14+6i): - The magnitude of 18-5i is sqrt{18^2 + 5^2} = sqrt{324 + 25} = sqrt{349}. - The magnitude of 14+6i is sqrt{14^2 + 6^2} = sqrt{196 + 36} = sqrt{232}. - The magnitude of the product is sqrt{349} cdot sqrt{232} = sqrt{349 times 232}. 2. For (3-12i)(4+9i): - The magnitude of 3-12i is sqrt{3^2 + 12^2} = sqrt{9 + 144} = sqrt{153}. - The magnitude of 4+9i is sqrt{4^2 + 9^2} = sqrt{16 + 81} = sqrt{97}. - The magnitude of the product is sqrt{153} cdot sqrt{97} = sqrt{153 times 97}. Now, calculate the magnitude of the difference: - First, we need to recognize that subtracting two complex numbers' products involves more calculation and cannot be simplified directly by the magnitudes. Hence, calculate the products: - (18-5i)(14+6i) = 18*14 + 18*6i - 5i*14 - 5i*6i = 252 + 108i - 70i + 30 = 282 + 38i - (3-12i)(4+9i) = 3*4 + 3*9i -12i*4 -12i*9i = 12 + 27i - 48i - 108 = -96 - 21i - The difference is (282 + 38i) - (-96 - 21i) = 378 + 59i. - The magnitude of 378 + 59i is sqrt{378^2 + 59^2} = sqrt{142884 + 3481} = sqrt{146365}. Finally, boxed{sqrt{146365}} is the result.