answer:Given f(x) = (x-1)^3 + 1, then f(-5) + f(7) = (-6)^3 + 1 + 6^3 + 1 = 2, f(-4) + f(6) = (-5)^3 + 1 + 5^3 + 1 = 2 ... f(-1) + f(3) = 2, f(0) + f(2) = 2, f(1) = 1 So, f(-5) + ldots + f(0) + ldots + f(7) = 2 times 6 + 1 = 13 Therefore, the correct choice is boxed{D}.

answer:Let the price of a pen be x and the price of a pencil be y, in cents. We form the following system of equations based on the new problem statement: [ 5x + 4y = 315, ] [ 3x + 6y = 243. ] We simplify the equations to make it easier to solve: 1. Multiply the first equation by 3 and the second equation by 5 to eliminate x: [ 15x + 12y = 945, ] [ 15x + 30y = 1215. ] 2. Subtract the first modified equation from the second: [ 15x + 30y - (15x + 12y) = 1215 - 945, ] [ 18y = 270. ] 3. Solve for y: [ y = frac{270}{18} = 15. ] Thus, the cost of one pencil is boxed{15} cents.

answer:Let's denote the number of tables as T. According to the first statement, 6.0 students were sitting at each table. This means that the total number of students is 6.0T. According to the second statement, if divided evenly, 5.666666667 (which is 5 + 2/3 or 17/3 when converted to a fraction) students should sit at each table. This means that the total number of students can also be represented as (17/3)T. Since the total number of students should be the same in both cases, we can set the two expressions equal to each other: 6.0T = (17/3)T To find the number of tables (T), we can divide both sides of the equation by 6.0: T = (17/3)T / 6.0 To simplify the right side, we can multiply by the reciprocal of 6.0, which is 1/6.0 or 1/6: T = (17/3) * (1/6) Now, we can simplify the fraction: T = 17 / (3 * 6) T = 17 / 18 Since the number of tables cannot be a fraction, we need to find a number of tables that, when multiplied by 5.666666667, gives us a whole number of students. We can do this by finding the least common multiple (LCM) of 3 and 6, which is 6. This means we need to find a multiple of 6 that, when divided by 17/3, gives us a whole number. The smallest such number is 6 itself, because: (17/3) * 6 = 17 * 2 = 34 So, there are boxed{6} tables in the lunchroom.

answer:# Problem: In a triangle (ABC) with side ratio (AB:AC=4:3), the angle bisector of (angle BAC) intersects side (BC) at point (L). Find the length of the segment (AL) if the magnitude of the vector (3 cdot overrightarrow{AB} + 4 cdot overrightarrow{AC}) is 2016. 1. **Understanding the angle bisector property:** By the Angle Bisector Theorem, the ratio in which (L) divides (BC) is the same as the ratio of (AB) to (AC). Therefore, [ frac{BL}{LC} = frac{AB}{AC} = frac{4}{3}. ] Hence, the vector from (B) to (L) can be written as: [ overrightarrow{BL} = frac{4}{4+3} cdot overrightarrow{BC} = frac{4}{7} cdot overrightarrow{BC}. ] 2. **Expressing vectors in terms of ( overrightarrow{A B} ) and ( overrightarrow{A C} ):** Since (overrightarrow{BC} = overrightarrow{AC} - overrightarrow{AB}), we substitute this into the expression for (overrightarrow{BL}): [ overrightarrow{BL} = frac{4}{7} cdot (overrightarrow{AC} - overrightarrow{AB}). ] 3. **Finding vector ( overrightarrow{A L} ):** The vector (overrightarrow{AL}) is given by: [ overrightarrow{AL} = overrightarrow{AB} + overrightarrow{BL}. ] Substitute the expression for (overrightarrow{BL}): [ overrightarrow{AL} = overrightarrow{AB} + frac{4}{7} (overrightarrow{AC} - overrightarrow{AB}) = overrightarrow{AB} + frac{4}{7} overrightarrow{AC} - frac{4}{7} overrightarrow{AB}. ] Simplify this expression: [ overrightarrow{AL} = left(1 - frac{4}{7}right) overrightarrow{AB} + frac{4}{7} overrightarrow{AC} = frac{3}{7} overrightarrow{AB} + frac{4}{7} overrightarrow{AC}. ] 4. **Relating to the given vector:** From the problem statement, ( 3 cdot overrightarrow{AB} + 4 cdot overrightarrow{AC} ) has a magnitude of 2016. Notice that: [ overrightarrow{AL} = frac{1}{7} (3 cdot overrightarrow{AB} + 4 cdot overrightarrow{AC}). ] 5. **Calculating the length of ( AL ):** Therefore, [ |overrightarrow{AL}| = left|frac{1}{7} (3 cdot overrightarrow{AB} + 4 cdot overrightarrow{AC})right| = frac{1}{7} cdot |3 cdot overrightarrow{AB} + 4 cdot overrightarrow{AC}|. ] Substituting the given magnitude: [ |overrightarrow{AL}| = frac{1}{7} cdot 2016 = 288. ] **Conclusion:** [ boxed{288} ]