answer:1. **Understanding the Starting Condition:** The problem states that Catherine's gas tank is initially ( frac{1}{8} ) full. 2. **Adding Gasoline and Final Condition:** When 30 litres of gasoline are added, the tank becomes ( frac{3}{4} ) full. 3. **Calculating the Difference in Tank Capacity:** We need to find the difference between ( frac{3}{4} ) and ( frac{1}{8} ) in terms of tank capacity. [ frac{3}{4} - frac{1}{8} = frac{6}{8} - frac{1}{8} = frac{5}{8} ] Thus, the 30 litres of gas added corresponds to ( frac{5}{8} ) of the tank's capacity. 4. **Determining Total Tank Capacity:** Let ( T ) be the total tank capacity in litres. Since ( frac{5}{8} ) of the capacity is equal to 30 litres: [ frac{5}{8} T = 30 ] Solve for ( T ): [ T = 30 times frac{8}{5} = 30 times frac{8}{5} = 48 text{ litres} ] 5. **Finding the Remaining Capacity:** We need to fill the remaining ( frac{1}{4} ) of the tank's capacity: [ frac{1}{4} times 48 = 12 text{ litres} ] 6. **Calculating the Cost to Fill the Remaining Tank:** Given that the cost of gas is 1.38 per litre, the total cost to fill the remaining 12 litres is: [ 12 times 1.38 = 16.56 text{ dollars} ] # Conclusion: [ boxed{16.56} ]

answer:Let's calculate the number of fish Jason has each day, starting with the initial eight fish. Day 1: The number of fish triples, so 8 * 3 = 24 fish. Day 2: The number triples again, so 24 * 3 = 72 fish. Day 3: The number triples once more, so 72 * 3 = 216 fish. On the fourth day, he takes out two-fifths of the fish: Day 4: Two-fifths of 216 is (2/5) * 216 = 86.4, but since we can't have a fraction of a fish, we'll round down to 86 fish removed. So, 216 - 86 = 130 fish remain. The number of fish continues to triple on days 5 and 6: Day 5: The number triples, so 130 * 3 = 390 fish. Day 6: The number triples again, so 390 * 3 = 1170 fish. On the sixth day, he removes three-sevenths of the fish: Three-sevenths of 1170 is (3/7) * 1170 = 501 fish removed. So, 1170 - 501 = 669 fish remain. The number of fish continues to triple on days 7, 8, and 9: Day 7: The number triples, so 669 * 3 = 2007 fish. Day 8: The number triples again, so 2007 * 3 = 6021 fish. Day 9: The number triples once more, so 6021 * 3 = 18063 fish. On the ninth day, he adds 20 more fish: 18063 + 20 = 18083 fish. Therefore, on the ninth day, Jason has a total of boxed{18083} fish in his aquarium.

answer:1. Consider a convex n-gon (polygon with n sides). The sum of the interior angles of this n-gon can be calculated using the formula [ text{Sum of interior angles} = (n-2) cdot 180^circ ] 2. Since the polygon is convex, we can also find the sum of its exterior angles. By definition, the sum of the exterior angles of any polygon is always [ text{Sum of exterior angles} = 360^circ = 2 cdot 180^circ ] 3. In a convex polygon, each interior angle is less than 180^circ, and each exterior angle is less than 180^circ. Identifying non-acute angles (where the angles are geq90^circ) helps us determine the minimum number of such angles in the polygon. 4. The problem indicates that in a 1992-sided polygon (a 1992-gon), the interior angles that are not acute need to be calculated. In a convex polygon, there can be at most 3 obtuse exterior angles (angles greater than 90^circ), since the sum of any three such exterior angles will exceed 180^circ, and thus conflict with the total sum of 360^circ for exterior angles: [ 3 cdot 60^circ leq 180^circ Rightarrow text{obtuse angles less than or equal to three} ] 5. Thus, considering non-acute angles (angles geq 90^circ), the remaining interior angles will not be acute: [ n - 3 = 1992 - 3 = 1989 ] 6. Therefore, in a 1992-sided polygon, there are at least 1989 interior angles that are not acute. # Conclusion: boxed{text{B}}

answer:To solve (5.763 + 2.489), we align the numbers at the decimal point and add them digit by digit from right to left, making sure to carry over when a sum exceeds 9: [ begin{array}{@{}c@{;}c@{}c@{}c@{}c@{}c} & 5. & 7 & 6 & 3 + & 2. & 4 & 8 & 9 cline{1-6} & (5+2) & (7+4+1) & (6+8) & (3+9) & 8. & 1 & 4 & 2 & text{(Carry-over 2 from 8+6, and 1 from 3+9)} end{array} ] The sum of the hundreds place (0.003 position) is (3 + 9 = 12), carrying over 1 to the tens place (0.01 position); here, (6 + 8 + 1) (from the carry) = 15, and therefore we carry over another 1; and finally, in the units position (0.1), (7 + 4 + 1) (from carry) = 12, hence another carry over. After adjusting for these carried values: [ begin{array}{@{}c@{;}c@{}c@{}c@{}c@{}c} & 8. & 1 & 5 & 2 end{array} ] The answer is (boxed{8.152}).