answer:**Analysis** This problem examines the operations of the dot product, magnitude, and angle of vectors. The key to solving this problem is to transform the given conditions. **Solution** According to the problem, we have: [ begin{cases} |overrightarrow{a}|^2 - 4overrightarrow{a}cdotoverrightarrow{b} + 4|overrightarrow{b}|^2 = 1 quad (1) 4|overrightarrow{a}|^2 + 12overrightarrow{a}cdotoverrightarrow{b} + 9|overrightarrow{b}|^2 = 4 quad (2) end{cases} ] Solving these, we get 4overrightarrow{a}cdotoverrightarrow{b} = |overrightarrow{b}|^2. Substituting 4overrightarrow{a}cdotoverrightarrow{b} = |overrightarrow{b}|^2 into (1), we obtain |overrightarrow{a}|^2 + 3|overrightarrow{b}|^2 = 1. Therefore, (5overrightarrow{a}-3overrightarrow{b})cdot (overrightarrow{a}-9overrightarrow{b}) = 5|overrightarrow{a}|^2 - 48overrightarrow{a}cdotoverrightarrow{b} + 27|overrightarrow{b}|^2 = 5(|overrightarrow{a}|^2 + 3|overrightarrow{b}|^2) = 5. Hence, the answer is boxed{5}.

answer:The store started with 40.0 coloring books and sold 20.0 of them, leaving them with 20.0 coloring books. They then put 4.0 coupons in each of the remaining books. To find out how many coupons they used, we multiply the number of remaining coloring books by the number of coupons per book: 20.0 coloring books * 4.0 coupons/book = 80.0 coupons So, they used boxed{80.0} coupons in total.

answer:1. **Step 1: Understanding the problem and setting up the premise** We are given that the sum of the digits of a natural number ( n ) is 2006. We need to prove that ( n ) cannot be a perfect square of an integer. 2. **Step 2: Utilizing digit sum properties** The sum of the digits of a number ( n ), denoted by ( S(n) ), has a special property when taken modulo 3. Specifically, for any number ( n ): [ S(n) equiv n pmod{3} ] This means that the sum of the digits of ( n ) and ( n ) itself will have the same remainder when divided by 3. 3. **Step 3: Applying the remainder property** Given ( S(n) = 2006 ), we can write: [ 2006 equiv n pmod{3} ] 4. **Step 4: Calculating the remainder** Now, we calculate the remainder when 2006 is divided by 3: [ 2006 div 3 = 668 text{ remainder } 2 ] Therefore, [ 2006 equiv 2 pmod{3} ] which implies, [ n equiv 2 pmod{3} ] 5. **Step 5: Considering properties of perfect squares modulo 3** We need to consider the possible remainders when a perfect square is divided by 3. Let's examine the squares of all integers modulo 3: - ( 0^2 equiv 0 pmod{3} ) - ( 1^2 equiv 1 pmod{3} ) - ( 2^2 equiv 4 equiv 1 pmod{3} ) Hence, a perfect square modulo 3 can only be 0 or 1. 6. **Step 6: Contradiction realization** From steps 4 and 5, since ( n equiv 2 pmod{3} ) and ( 2 mod{3} ) does not correspond to any valid perfect square (as perfect squares can only be 0 or 1 modulo 3), we realize that ( n ) cannot be a perfect square because it yields the remainder 2 when divided by 3. # Conclusion Thus, if the sum of the digits of a natural number ( n ) is 2006, then ( n ) cannot be a perfect square of an integer. [ boxed{} ]

answer:1. Multiply both sides by ((x - 4)(x - 2)^2) to clear the denominators: [5x = A(x - 2)^2 + B(x - 4)(x - 2) + C(x - 4).] 2. Set (x = 4): [5 times 4 = A times 2^2 rightarrow 20 = 4A rightarrow A = 5.] 3. Set (x = 2): [5 times 2 = C times (-2) rightarrow 10 = -2C rightarrow C = -5.] 4. Substitute A = 5 and C = -5 back and simplify: [5x = 5(x - 2)^2 + B(x - 4)(x - 2) - 5(x - 4) rightarrow 5x = 5(x^2 - 4x + 4) + B(x^2 - 6x + 8) - 5x + 20.] Since coefficients of (x^2) and (x) on both sides must match, [0x^2 + 5x + 0 = (5 + B)x^2 + (-20 - 6B)x + (20 + 8B).] Equating the coefficients: - For (x^2): (5 + B = 0 rightarrow B = -5.] Therefore, the ordered triple (A, B, C) = boxed{(5, -5, -5)}.