answer:Part 1: Coloring the Cube 1. **Choosing Colors:** We are given 6 different colors, let's denote them by (C_1, C_2, C_3, C_4, C_5,) and (C_6). 2. **Fixing the First Face:** We can always rotate the cube such that (C_1) is on the top face. Therefore, the number of ways to choose the top face is (1). 3. **Choosing the Opposite Face:** After fixing (C_1) on the top face, there are 5 remaining colors. We have 5 choices for the color of the bottom face. 4. **Fixing the Orientation:** With (C_1) on the top and some color (say (C_2)) on the bottom, we can freely rotate the cube around the vertical axis passing through (C_1) and (C_2). To fix this rotational degree of freedom, we place the lowest remaining color (say (C_3)) on one of the vertical side faces (north position). 5. **Coloring the Remaining Faces:** With (C_1) on top, (C_2) on the bottom, and (C_3) in a fixed side position (north), we have 3 remaining faces to color with the remaining 3 colors. The number of ways to do this is (3! = 6). 6. **Total Number of Ways:** Therefore, the total number of ways to color the cube such that each face is a different color: [ 1 times 5 times 6 = 30 ] Conclusion: (boxed{30}) Part 2: Coloring the Octahedron 1. **Choosing Colors:** We are given 8 different colors, let's denote them by (C_1, C_2, C_3, C_4, C_5, C_6, C_7,) and (C_8). 2. **Fixing the First Face:** We can always rotate the octahedron such that (C_1) is on the top face. Therefore, the number of ways to choose the top face is (1). 3. **Choosing the Opposite Face:** With (C_1) on the top face, there are 7 remaining colors. We have 7 choices for the bottom face. 4. **Choosing Colors for the Adjacent Faces:** Now we need to color 3 of the faces that are adjacent to the top face. We select 3 out of the remaining 6 colors. The number of ways to choose 3 colors out of 6 is given by the binomial coefficient: [ binom{6}{3} = 20 ] 5. **Fixing the Orientation:** Among these 3 chosen colors, we fix the lowest one (say (C_2)) into a fixed position (north). There are (2!) ways to arrange the remaining two colors on the faces adjacent to the top face. 6. **Coloring the Remaining Faces:** We have 3 faces left and 3 remaining colors. There are (3! = 6) ways to assign these colors to the faces. 7. **Total Number of Ways:** Therefore, the total number of ways to color the octahedron such that each face is a different color: [ 1 times 7 times 20 times 2 times 6 = 1680 ] Conclusion: (boxed{1680})

answer:To approximate 0.30105 to the nearest hundredth using the rounding method, we look at the third decimal place which is 1. Since 1 is less than 5, we do not round up the second decimal place. Therefore, the approximation to the nearest hundredth is 0.30. For approximating 0.30105 to 3 significant figures, we consider the first three non-zero digits. The fourth digit after the decimal is 0, which does not affect the third significant figure. Thus, the approximation to 3 significant figures is 0.301. Therefore, the answers are: boxed{0.30} for the nearest hundredth, and boxed{0.301} for 3 significant figures.

answer:Let's denote the total number of quantities as ( n ), the sum of all the quantities as ( S ), the sum of the 3 quantities with an average of 4 as ( S_3 ), and the sum of the remaining 2 quantities with an average of 33 as ( S_2 ). From the information given, we can write the following equations: 1. The average of all quantities is 6: [ frac{S}{n} = 6 ] [ S = 6n ] (Equation 1) 2. The average of 3 of them is 4: [ frac{S_3}{3} = 4 ] [ S_3 = 4 times 3 ] [ S_3 = 12 ] (Equation 2) 3. The average of the remaining 2 numbers is 33: [ frac{S_2}{2} = 33 ] [ S_2 = 33 times 2 ] [ S_2 = 66 ] (Equation 3) Since the sum of all the quantities is the sum of the 3 quantities with an average of 4 plus the sum of the remaining 2 quantities with an average of 33, we can write: [ S = S_3 + S_2 ] Substituting the values from Equations 2 and 3 into the above equation, we get: [ S = 12 + 66 ] [ S = 78 ] Now, we substitute the value of ( S ) into Equation 1 to find ( n ): [ 6n = 78 ] [ n = frac{78}{6} ] [ n = 13 ] Therefore, there are boxed{13} quantities in total.

answer:**Analysis** Deductive reasoning is reasoning from general to specific, which is correct. The general pattern of deductive reasoning is the "syllogism" form, and the correctness of the conclusion of deductive reasoning depends on the major premise, minor premise, and form of reasoning. Therefore, the correct choice is boxed{C}.