answer:Let's start by analyzing the sequences ( {x_n} ) and ( {y_n} ). 1. **Analyzing the sequence ( {x_n} ):** - Given: [ begin{array}{llll} x_0 = 1, & x_1 = 1, & x_{n+1} = x_n + 2x_{n-1} & (n=1,2, cdots) end{array} ] - Rearrange the recurrence relation: [ x_{n+1} - 2x_n = -(x_n - 2x_{n-1}) ] - The sequence ( {x_n - 2x_{n-1}} ) is a geometric progression with the first term computed as ( x_1 - 2x_0 = 1 - 2 = -1 ) and a common ratio of (-1). [ x_n - 2x_{n-1} = (-1)^n ] - Solving for ( x_n ): [ x_n - 2x_{n-1} = (-1)^n ] [ text{Let } z_n = x_n - 2x_{n-1} text{, we find form for } z_n text{ is} z_n = (-1)^n ] [ x_n = frac{2^{n+1} + (-1)^n}{3} ] 2. **Analyzing the sequence ( {y_n} ):** - Given: [ begin{array}{llll} y_0 = 1, & y_1 = 7, & y_{n+1} = 2 y_n + 3 y_{n-1} & (n=1,2, cdots) end{array} ] - Using the method for solving recurrence relations similarly: [ y_n = 2 cdot 3^n - (-1)^n ] 3. **Showing no common terms other than (1):** - Assume ( x_m = y_n ) for some ( m > 1 ) and ( n>1 ). - Equate the derived forms of ( x_n ) and ( y_n ): [ frac{2^{m+1} + (-1)^m}{3} = 2 cdot 3^n - (-1)^n ] - Simplify the equation: [ 2 cdot 3^{n+1} - 2^{m+1} = (-1)^m + 3(-1)^n ] [ 2 (3^{n+1} - 2^m) = (-1)^m + 3 (-1)^n ] - Determine the parity (odd/even nature) of ( m ) and ( n ). They must be different, because: 1. **(m) is even and (n) is odd:** [ m = 2k, quad n = 2t - 1 ] Substitute in: [ 2 (3^{2t} - 2^{2k}) = 1 + 3(-1) = -2 ] [ 3^{2t} + 1 = 2^{2k} = 4^k ] Since (3^{2t} + 1 equiv 2 pmod{4}), but (4^k equiv 0 pmod{4}), which cannot happen. Hence, no such (k) and (t) exist. 2. **(m) is odd and (n) is even:** [ m = 2k+1, quad n = 2t ] Substitute in: [ 2 (3^{2t+1} - 2^{2k+1}) = -1 + 3 cdot 1 = 2 ] [ 3^{2t+1} - 1 = 2^{2k+1} ] Since (3^{2t+1} - 1 equiv 2 pmod{4}), but (2^{2k+1} equiv 0 pmod{4}), which cannot happen. Hence, no such (k) and (t) exist. Combining the results, there can't exist integers (m > 1) and (n > 1) such that (x_m = y_n). 4. **Verification by modular arithmetic:** - Compute ( {x_n} mod 8 ) and ( {y_n} mod 8 ): [ begin{array}{lllllll} x: & mod 8 & 1, & 1, & 3, & 5, & 3, & 5, cdots y: & mod 8 & 1, & 7, & 1, & 7, & 1, & 7, cdots end{array} ] - period for ( { x_n } mod 8) is ( 3, 5 ) for subsequent elements, - period for ( { y_n } mod 8) is ( 1, 7 ). As ( {x_n} mod 8 ) never coincide with ( { y_n } mod 8) except for the first term, we conclude that: **Conclusion:** [ boxed{text{The two sequences } {x_n} text{ and } {y_n} text{ have no common terms except for 1.}} ]

answer:1. **Isolate the Absolute Value**: Start by subtracting 7 from both sides of the equation: [ |5y| + 7 = 47 ] [ |5y| = 47 - 7 ] [ |5y| = 40 ] 2. **Solve for Both Cases**: Since |5y| = 40, we have two cases: - When 5y = 40 - When 5y = -40 Solving these, we get: - For 5y = 40: [ y = frac{40}{5} = 8 ] - For 5y = -40: [ y = frac{-40}{5} = -8 ] 3. **Product of Solutions**: The product of y = 8 and y = -8 is: [ (8)(-8) = -64 ] Therefore, the product of the values of y is boxed{-64}.

answer:Let's denote the total value of the item as V. The import tax is only applied to the portion of V that exceeds 1,000. So, if we subtract 1,000 from the total value V, we get the amount that is subject to the 7% import tax. The import tax paid is 109.20, which is 7% of the amount over 1,000. Let's set up the equation: 0.07 * (V - 1000) = 109.20 Now, we solve for V: (V - 1000) = 109.20 / 0.07 (V - 1000) = 1560 Now, add 1,000 to both sides to find the total value V: V = 1560 + 1000 V = 2560 So, the total value of the item was boxed{2,560} .

answer:For the inequality x^2 - 8x + c > 0 to hold for some x, the quadratic expression must be positive for at least some values of x. Since the quadratic opens upwards (leading coefficient 1 > 0), it can either be positive everywhere or negative between its roots (if it has any). To determine when it can be positive, we first ensure it has roots (real roots imply it crosses the x-axis and thus can be positive away from the roots). We calculate the discriminant for real roots: [ b^2 - 4ac = (-8)^2 - 4 cdot 1 cdot c = 64 - 4c ] This discriminant must be positive for there to be real roots: [ 64 - 4c > 0 ] [ 64 > 4c ] [ 16 > c ] So, c must be less than 16. Yet, since we need c to be positive for meaningful real roots, [ 0 < c < 16 ] Conclusion with boxed answer: Thus, the interval for c such that x^2 - 8x + c > 0 has real solutions is boxed{(0, 16)}.