answer:To find the minimum value of the expression x + frac{4}{x} given that x > 0, we can use the AM-GM inequality, which states that for any positive numbers a and b, the arithmetic mean (AM) is always greater than or equal to the geometric mean (GM). In this case, we can set a = x and b = frac{4}{x}. The arithmetic mean of a and b is frac{a+b}{2} = frac{x + frac{4}{x}}{2}, and the geometric mean of a and b is sqrt{ab} = sqrt{x cdot frac{4}{x}} = sqrt{4} = 2. By the AM-GM inequality, we have: [ frac{x + frac{4}{x}}{2} geqslant 2 ] Multiplying both sides by 2, we get: [ x + frac{4}{x} geqslant 4 ] The equality holds when x = frac{4}{x}, which implies x^2 = 4 and thus x = 2 (since x > 0). Therefore, the minimum value of x + frac{4}{x} when x > 0 is boxed{4}.

answer:Let's denote the distance from marker B to marker C as ( D ). Since this distance is half of the distance from marker A to marker B, we have ( D = frac{120}{2} = 60 ) miles. The total distance of the trip from marker A to marker C is ( 120 + 60 = 180 ) miles. The average speed for the entire trip is given as 50 miles per hour. The total time for the trip can be calculated using the formula: [ text{Total Time} = frac{text{Total Distance}}{text{Average Speed}} ] [ text{Total Time} = frac{180 text{ miles}}{50 text{ mph}} ] [ text{Total Time} = 3.6 text{ hours} ] The ride from marker A to marker B lasted 3 times as many hours as the rest of the ride. Let's denote the time from marker B to marker C as ( t ) hours. Then the time from marker A to marker B is ( 3t ) hours. The total time is the sum of these two times: [ 3t + t = 3.6 ] [ 4t = 3.6 ] [ t = frac{3.6}{4} ] [ t = 0.9 text{ hours} ] Now we can calculate the average speed from marker B to marker C using the formula: [ text{Average Speed} = frac{text{Distance}}{text{Time}} ] [ text{Average Speed} = frac{60 text{ miles}}{0.9 text{ hours}} ] [ text{Average Speed} = frac{60}{0.9} ] [ text{Average Speed} = 66.overline{6} text{ mph} ] So the average speed of the motorcyclist while driving from marker B to marker C was approximately boxed{66.7} miles per hour.

answer:# Problem: 1. Can the sum of the squares of two odd numbers be a perfect square of an integer? 2. Can the sum of the squares of three odd numbers be a perfect square of an integer? 1. Let's investigate whether the sum of the squares of two odd numbers can be a perfect square. 1. The residue of the square of an odd number when divided by 4 is always 1. [ text{If } n text{ is odd, then } n = 2k + 1 text{ for some integer } k. ] [ n^2 = (2k+1)^2 = 4k^2 + 4k + 1 implies n^2 equiv 1 pmod{4}. ] 2. Hence, given two odd numbers ( a ) and ( b ): [ a^2 equiv 1 pmod{4} quad text{and} quad b^2 equiv 1 pmod{4}. ] 3. Therefore, the sum of their squares is: [ a^2 + b^2 equiv 1 + 1 = 2 pmod{4}. ] 4. Since a perfect square modulo 4 can only be 0 or 1, the sum of ( a^2 + b^2 ) which is 2, cannot be a perfect square. **Conclusion:** [ boxed{text{No, the sum of the squares of two odd numbers cannot be a perfect square.}} ] 2. Now, let's examine whether the sum of the squares of three odd numbers can be a perfect square. 1. From the previous, we know the residue of the square of an odd number when divided by 4 is always 1. 2. Let ( a, b, ) and ( c ) be three odd numbers. Then: [ a^2 equiv 1 pmod{4}, quad b^2 equiv 1 pmod{4}, quad c^2 equiv 1 pmod{4}. ] 3. The sum of their squares would be: [ a^2 + b^2 + c^2 equiv 1 + 1 + 1 = 3 pmod{4}. ] 4. Since a perfect square modulo 4 can only be 0 or 1, the sum of ( a^2 + b^2 + c^2 ) which is 3, cannot be a perfect square. **Conclusion:** [ boxed{text{No, the sum of the squares of three odd numbers cannot be a perfect square.}} ]

answer:Let z be the image of 3 - i under the dilation centered at 1 + 2i with a scale factor of 4. Since the dilation transformation formula in complex numbers for a transformation centered at a + bi that transforms c + di to z with a scale factor k is given by: [ z - (a + bi) = k((c + di) - (a + bi)) ] Applying this formula with a + bi = 1 + 2i, c + di = 3 - i, and k = 4: [ z - (1 + 2i) = 4((3 - i) - (1 + 2i)) ] [ z - (1 + 2i) = 4((3 - 1) - (i - 2i)) ] [ z - (1 + 2i) = 4(2 - (-1)i) = 4(2 + i) ] [ z - (1 + 2i) = 8 + 4i ] Solving for z: [ z = (1 + 2i) + (8 + 4i) = 9 + 6i ] The dilated point z is boxed{9 + 6i}.