answer:Consider a right triangle where the opposite side is 3 and the hypotenuse is 5. [ text{Let } theta = arcsin frac{3}{5}. ] By the Pythagorean theorem, the adjacent side can be calculated as: [ text{Adjacent} = sqrt{text{Hypotenuse}^2 - text{Opposite}^2} = sqrt{5^2 - 3^2} = sqrt{25 - 9} = sqrt{16} = 4. ] Now, compute cos theta: [ cos theta = frac{text{Adjacent}}{text{Hypotenuse}} = frac{4}{5}. ] Also, compute tan theta: [ tan theta = frac{text{Opposite}}{text{Adjacent}} = frac{3}{4}. ] Thus, the answers are: [ cos left( arcsin frac{3}{5} right) = boxed{frac{4}{5}} ] [ tan left( arcsin frac{3}{5} right) = boxed{frac{3}{4}} ]

answer:1. **Understanding the problem**: We are given two predictions for the positions of five athletes A, B, C, D, and E in a race. Each prediction is partially correct: - First prediction: A first, B second, C third, D fourth, E fifth. - Second prediction: C first, E second, A third, B fourth, D fifth. - The first prediction is correct for exactly three athletes. - The second prediction is correct for exactly two athletes. - We need to determine the actual finishing order. 2. **Identifying the incorrect predictions**: - In the first prediction, precisely two athletes are predicted incorrectly. - In the second prediction, precisely three athletes are predicted incorrectly. 3. **Comparing both predictions**: - If we assume the correct prediction positions from the first list match with the incorrects of the second, we can exchange two positions as per the condition. 4. **Analysis**: - We need to find a pair of athletes that, when swapped in the first prediction, fulfills both conditions for correctness in the first prediction and incorrectness in the second prediction. - Let's start testing possible pairs. - If we swap A and C in the first prediction, we change: - A from 1st to 3rd and C from 3rd to 1st. - Verifying this swap in the second prediction shows: - A remains third (same as the second prediction, so it's correct now). - C changes to 1st (correct as per second prediction). 5. **Confirming the results**: - After swapping A and C, the new order is: C first, B second, A third, D fourth, and E fifth in the first prediction. - This swap fulfills first prediction correctness for C, B, A maintaining the conditions: - Incorrectness for D and E in the first prediction. - Correctness for C and A as required in the second prediction. 6. **Final positions after verification**: - After ensuring all other permutations, the correct and unique solution is: - C finishes first, - B finishes second, - A finishes third, - D finishes fourth, - E finishes fifth. # Conclusion: [ boxed{C text{ first}, B text{ second}, A text{ third}, D text{ fourth}, E text{ fifth}} ]

answer:Alice must pay £12, which we can multiply by the conversion factor frac{1 text{USD}}{0.75 text{GBP}} to obtain the value in U.S. dollars. Carrying out the calculation, we find that Alice must use £12 cdot frac{1 text{USD}}{0.75 text{GBP}} = 12 div 0.75 = 16. Therefore, Alice must use boxed{16.00 text{USD}} for the book.

answer:Let's denote the initially calculated average height as ( A ). The sum of the heights of all 20 students based on the initially calculated average would be: [ text{Sum of heights based on initial average} = 20 times A ] Since it was later found that one student's height was incorrectly written as 151 cm instead of the actual 111 cm, the sum of the heights based on the initial average was 40 cm more than the actual sum of the heights (because 151 - 111 = 40). The actual sum of the heights of all 20 students is: [ text{Actual sum of heights} = 20 times 173 ] Therefore, the sum of the heights based on the initial average is: [ text{Sum of heights based on initial average} = text{Actual sum of heights} + 40 ] Now we can set up the equation: [ 20 times A = 20 times 173 + 40 ] Solving for ( A ): [ 20A = 20 times 173 + 40 ] [ 20A = 3460 + 40 ] [ 20A = 3500 ] [ A = frac{3500}{20} ] [ A = 175 ] The initially calculated average height was boxed{175} cm.