answer:To find a relationship between (c) and (d) similar to the original problem, we start by manipulating the first equation: [ 4x - 3y = c ] Multiply this equation by 2 to make the coefficient of (x) align with the coefficient in the second equation: [ 8x - 6y = 2c ] Now, consider the second equation: [ 2y - 8x = d ] Rewriting it, we get: [ 8x - 2y = -d ] Equating the modified first equation and the rewritten second equation: [ 8x - 6y = 2c 8x - 2y = -d ] Subtract the second from the first: [ -4y = 2c + d ] [ -4y = 2c + d Rightarrow 2c + d = 0 Rightarrow 2c = -d ] This implies: [ frac{c}{d} = frac{-d}{2d} = boxed{-frac{1}{2}} ]

answer:First, consider the 6 Republicans sitting around the table. There are (6-1)! ways to arrange 6 Republicans in a circle, assuming one is fixed to avoid identical rotations. This equals 5! = 120. Next, we need to seat the 4 Democrats such that no two Democrats are next to each other. This can be done in the spaces between the Republicans. There are 6 gaps created by the 6 Republicans seated. We choose 4 out of these 6 gaps to seat the Democrats, which can be done in binom{6}{4} = 15 ways. Each set of 4 Democrats can be arranged in their chosen gaps in 4! = 24 ways. Thus, the total arrangement is the product of arranging the Republicans, choosing the gaps, and arranging the Democrats: 5! cdot binom{6}{4} cdot 4! = 120 cdot 15 cdot 24 = 43200. Consequently, the number of ways they can be seated around the table, under these conditions, is boxed{43,200}.

answer:First, ascertain the distance between the horizontal lines, which are y=-1 and y=3. The distance between these lines is 3 - (-1) = 4. This distance is the length of each side of the square. Next, determine the position of the fourth line which is vertical, parallel to x=2. - If x=b is to the left of x=2, it must also be 4 units away. Thus, b = 2 - 4 = -2. - If x=b is to the right of x=2, it also must be 4 units away, so b = 2 + 4 = 6. The possible values for b are -2 and 6, and their product is (boxed{-12}).

answer:First, note that prime numbers are not abundant as their only proper factor is 1, which is less than any prime number itself. We consider numbers from 1 to 49, checking the sum of their proper factors: - Directly, numbers like 12, 18, 20, 24 are already known to be abundant from previous problem. - Next numbers to check are 28, 30, 36, 40, 42, 48 which are potential candidates as non-prime numbers. - For 28 (1+2+4+7+14=28), the sum equals the number, hence not abundant. - For 30 (1+2+3+5+6+10+15=42 > 30), sum is greater than the number, so it is abundant. - For 36 (1+2+3+4+6+9+12+18=55 > 36), sum is greater than the number, so it is abundant. - For 40 (1+2+4+5+8+10+20=50 > 40), sum is greater than the number, so it is abundant. - For 42 (1+2+3+6+7+14+21=54 > 42), sum is greater than the number, so it is abundant. - For 48 (1+2+3+4+6+8+12+16+24=76 > 48), sum is greater than the number, so it is abundant. Thus, the numbers 12, 18, 20, 24, 30, 36, 40, 42, 48 are abundant. That gives us boxed{9} abundant numbers less than 50.