answer:1. Define the angles (alpha) and (beta): [ alpha = arccos left( frac{sqrt{6} + 1}{2sqrt{3}} right) quad text{and} quad beta = arccos left( sqrt{frac{2}{3}} right). ] Since (alpha, beta in left( 0, frac{pi}{2} right)), the difference ( A = alpha - beta ) will lie in the interval ( left( -frac{pi}{2}, frac{pi}{2} right) ). 2. We want to find (A) such that: [ A = arccos left( frac{sqrt{6} + 1}{2sqrt{3}} right) - arccos left( sqrt{frac{2}{3}} right). ] 3. Use the trigonometric identity for the difference of cosines: [ cos (alpha - beta) = cos alpha cos beta + sin alpha sin beta. ] However, since we have (arccos) terms, we use the sine function difference formula. Note that: [ sin A = sin (alpha - beta) = sin alpha cos beta - cos alpha sin beta. ] 4. Calculate (cos alpha) and (sin alpha): [ cos alpha = frac{sqrt{6} + 1}{2 sqrt{3}}. ] To find (sin alpha): [ sin^2 alpha + cos^2 alpha = 1 implies sin^2 alpha = 1 - left( frac{sqrt{6} + 1}{2 sqrt{3}} right)^2. ] Calculate (left( frac{sqrt{6} + 1}{2 sqrt{3}} right)^2): [ left( frac{sqrt{6} + 1}{2 sqrt{3}} right)^2 = frac{(sqrt{6} + 1)^2}{4 cdot 3} = frac{6 + 2 sqrt{6} + 1}{12} = frac{7 + 2 sqrt{6}}{12}. ] Therefore: [ sin^2 alpha = 1 - frac{7 + 2 sqrt{6}}{12} = frac{12 - 7 - 2 sqrt{6}}{12} = frac{5 - 2 sqrt{6}}{12}. ] And (sin alpha = sqrt{frac{5 - 2 sqrt{6}}{12}}) (noting that the value is positive in (left(0, frac{pi}{2} right))). 5. Calculate (cos beta) and (sin beta): [ cos beta = sqrt{frac{2}{3}}. ] [ sin^2 beta + cos^2 beta = 1 implies sin^2 beta = 1 - left( sqrt{frac{2}{3}} right)^2 = 1 - frac{2}{3} = frac{1}{3}. ] Therefore, (sin beta = sqrt{frac{1}{3}} = frac{1}{sqrt{3}}). 6. Use these values to calculate (sin A): [ sin A = sin alpha cos beta - cos alpha sin beta. ] Substituting the values we determined, we have: [ sin A = sqrt{frac{5 - 2 sqrt{6}}{12}} cdot sqrt{frac{2}{3}} - frac{sqrt{6} + 1}{2 sqrt{3}} cdot frac{1}{sqrt{3}}. ] After some simplification (omitted for brevity), this yields: [ sin A = -frac{1}{2}. ] 7. Determine (A) from (sin A = -frac{1}{2}): [ A = arcsin left( -frac{1}{2} right) = -frac{pi}{6}. ] 8. Finally, express the result in the form ( frac{a pi}{b} ): [ A = -frac{pi}{6} implies a = -1 quad text{and} quad b = 6. ] Thus: [ |a - b| = |-1 - 6| = 7. ] # Conclusion: [ boxed{7} ]

answer:First, let's find out how fast Pete can walk on his hands. We know that Pete walks on his hands at 2 miles per hour. Next, we are told that Pete walks backwards three times faster than Susan walks forwards. Since Pete walks backwards at 12 miles per hour, Susan walks forwards at 12 miles per hour / 3 = 4 miles per hour. We also know that Pete can walk on his hands at one quarter the speed that Tracy can do cartwheels. Since Pete walks on his hands at 2 miles per hour, Tracy can do cartwheels at 2 miles per hour * 4 = 8 miles per hour. Now we have Tracy's cartwheel speed (8 miles per hour) and Susan's forward walking speed (4 miles per hour). To find the ratio of Tracy's cartwheel speed to Susan's forward walking speed, we divide Tracy's speed by Susan's speed: Tracy's cartwheel speed / Susan's forward walking speed = 8 miles per hour / 4 miles per hour = 2. So, the ratio of Tracy's cartwheel speed to Susan's forward walking speed is boxed{2:1} .

answer:First, let's find the area of the square. Since the sides of the square are 6 centimeters, the area of the square (A_square) is: A_square = side_length * side_length A_square = 6 cm * 6 cm A_square = 36 cm² Now, let's find the length of the first rectangle. We know that the width of the first rectangle is 4 centimeters and its area is equal to the area of the square. Let x be the length of the first rectangle. The area of the first rectangle (A_rectangle1) is: A_rectangle1 = width * length 36 cm² = 4 cm * x x = 36 cm² / 4 cm x = 9 cm So, the length of the first rectangle is 9 centimeters. Next, let's find the length of the second rectangle. We know that the width of the second rectangle is 3 centimeters and its area is equal to half the area of the square. Let y be the length of the second rectangle. The area of the second rectangle (A_rectangle2) is: A_rectangle2 = width * length A_rectangle2 = (1/2) * A_square A_rectangle2 = (1/2) * 36 cm² A_rectangle2 = 18 cm² Now, we can find the length y: 18 cm² = 3 cm * y y = 18 cm² / 3 cm y = 6 cm So, the length of the second rectangle is 6 centimeters. Therefore, the lengths of both rectangles are (x, y) = boxed{(9} cm, 6 cm).

answer:To solve for the largest possible value of square + bigtriangleup given that square times bigtriangleup = 48, we should first list all pairs of whole numbers (square, bigtriangleup) whose product equals 48. Then, we will sum the values in each pair to find the maximum sum. 1. **Factorize 48**: We enumerate all possible pairs (whole number factors) of 48: - (1, 48) - (2, 24) - (3, 16) - (4, 12) - (6, 8) 2. **Calculate the sums for each pair**: - For (1, 48), the sum is 1 + 48 = 49 - For (2, 24), the sum is 2 + 24 = 26 - For (3, 16), the sum is 3 + 16 = 19 - For (4, 12), the sum is 4 + 12 = 16 - For (6, 8), the sum is 6 + 8 = 14 3. **Determine the maximum sum**: Comparing the sums: - 49, 26, 19, 16, 14 The largest sum from the available pairs is 49. Thus, the largest possible value of square + bigtriangleup is 49. The final answer is The correct choice is boxed{E}.