answer:1. **Calculate the distance from -4 to -1:** The distance on a number line is the absolute difference between the two points. Thus, the distance from -4 to -1 is: [ |-1 - (-4)| = |-1 + 4| = |3| = 3 text{ units} ] 2. **Calculate the distance from -1 to -8**:** The distance from -1 to -8 is: [ |-8 - (-1)| = |-8 + 1| = |-7| = 7 text{ units} ] 3. **Calculate the distance from -8 to 6**:** The distance from -8 to 6 is: [ |6 - (-8)| = |6 + 8| = |14| = 14 text{ units} ] 4. **Add the distances to find the total distance crawled:** The total distance the bug crawls is the sum of the distances calculated in steps 1, 2, and 3: [ 3 text{ units} + 7 text{ units} + 14 text{ units} = 24 text{ units} ] Thus, the total distance the bug crawls is 24 units. Conclusion: The problem statement is consistent, and the calculations add up correctly to the final answer. The final answer is boxed{textbf{(D)} 24}

answer:1. **Identify Given Information:** - Point ( M ) lies on the angle bisector of ( angle BAC ) in triangle ( ABC ). - Point ( N ) lies on the extension of side ( AB ) beyond point ( A ). - The lengths are given as ( AC = AM = 1 ). - The angles ( angle ANM ) and ( angle CNM ) are equal. 2. **Introduce Auxiliary Points:** - We place an auxiliary point ( K ) on the extension of segment ( NC ) beyond point ( C ). 3. **Claims:** - Since ( M ) lies on the angle bisectors of ( angle BAC ) and ( angle ANC ), it is equidistant from lines ( AC ) and ( NC ). This means ( M ) also lies on the angle bisector of ( angle ACK ). 4. **Key Observations:** - ( angle MAC ) and ( angle MCA ) are equal due to the isosceles triangle ( AMC ). - The equality of alternate interior angles ( angle CMA ) and ( angle MCK ) implies ( AM ) is parallel to ( NC ). 5. **Angle Calculations:** - Since ( AM parallel NC ), corresponding angles ( angle MAC ) and ( angle ACN ) are equal and both are ( frac{angle BAC}{2} ). - Thus, ( angle ANC = angle BAC - angle ACN = angle BAC - frac{angle BAC}{2} = frac{angle BAC}{2} ). 6. **Conclusion:** - Triangle ( NAC ) is isosceles with ( A N = AC = 1 ). By analyzing the triangle ( NAC ), we have therefore determined that: [ A N = 1 ] The length of segment ( AN ) is (boxed{1}).

answer:Since a_{n-1}=2a_{n} for n geqslant 2, n in mathbb{N}^+, we have frac{a_{n}}{a_{n-1}} = frac{1}{2}. This indicates that the sequence {a_n} is a geometric sequence with the first term a_1=1 and common ratio frac{1}{2}. The sum of the first 6 terms of a geometric sequence can be found using the formula S_n = frac{a_1 cdot (1 - r^n)}{1 - r}, where a_1 is the first term, r is the common ratio, and n is the number of terms. Substituting the given values, we get S_6 = frac{1 cdot (1 - frac{1}{2^6})}{1 - frac{1}{2}} = frac{63}{32}. Therefore, the correct answer is option C: boxed{frac{63}{32}}. This problem tests the understanding of geometric sequences and their sum, and requires reasoning and computational skills. It is of moderate difficulty.

answer:To find the total amount of sand that the four trucks have when they arrive at the construction yard, we need to subtract the amount of sand lost by each truck from the amount they initially carried. Then we'll sum up the remaining sand from all four trucks. For the first truck: Initial sand = 4.1 pounds Sand lost = 2.4 pounds Remaining sand = 4.1 pounds - 2.4 pounds = 1.7 pounds For the second truck: Initial sand = 5.7 pounds Sand lost = 3.6 pounds Remaining sand = 5.7 pounds - 3.6 pounds = 2.1 pounds For the third truck: Initial sand = 8.2 pounds Sand lost = 1.9 pounds Remaining sand = 8.2 pounds - 1.9 pounds = 6.3 pounds For the fourth truck: Initial sand = 10.5 pounds Sand lost = 2.1 pounds Remaining sand = 10.5 pounds - 2.1 pounds = 8.4 pounds Total remaining sand = 1.7 pounds + 2.1 pounds + 6.3 pounds + 8.4 pounds Total remaining sand = 18.5 pounds Now, let's calculate the percentage of sand lost by each truck: For the first truck: Percentage lost = (Sand lost / Initial sand) * 100 Percentage lost = (2.4 pounds / 4.1 pounds) * 100 Percentage lost = 58.54% For the second truck: Percentage lost = (3.6 pounds / 5.7 pounds) * 100 Percentage lost = 63.16% For the third truck: Percentage lost = (1.9 pounds / 8.2 pounds) * 100 Percentage lost = 23.17% For the fourth truck: Percentage lost = (2.1 pounds / 10.5 pounds) * 100 Percentage lost = 20% In conclusion, the four trucks have a total of 18.5 pounds of sand when they arrive at the construction yard. The percentage of sand lost by each truck during their respective journeys is boxed{58.54%} for the first truck, 63.16% for the second truck, 23.17% for the third truck, and 20% for the fourth truck.