answer:Let's denote the first number as A, the second number as B, and the third number as C. We are given that B = 30. The ratio between the first and second number is 2 : 3, so we can write: A / B = 2 / 3 Since B = 30, we can substitute and solve for A: A / 30 = 2 / 3 A = (2 / 3) * 30 A = 2 * 10 A = 20 The ratio between the second and third number is 5 : 8, so we can write: B / C = 5 / 8 Again, since B = 30, we can substitute and solve for C: 30 / C = 5 / 8 C = (8 / 5) * 30 C = 8 * 6 C = 48 Now we have the values for A and C, and we need to find the sum of the first and third numbers: Sum of A and C = A + C Sum of A and C = 20 + 48 Sum of A and C = 68 Therefore, the sum of the first and third numbers is boxed{68} .

answer:To solve this problem, we break it down into two main parts as described: # Part (1): Find the values of m and a. Given that f(x) = a^x + (1-m)a^{-x} is an odd function, we know that for all x in its domain, f(-x) = -f(x). Also, given a > 0 and a neq 1. - For an odd function, f(0) = 0, so we substitute x = 0 into f(x): begin{align*} f(0) &= a^0 + (1-m)a^{-0} 1 + (1-m) &= 0 2 - m &= 0 m &= 2. end{align*} Therefore, we have f(x) = a^x - a^{-x}. - Since f(x) passes through (-1, -frac{3}{2}), we substitute x = -1 into the new f(x): begin{align*} f(-1) &= a^{-1} - a -frac{3}{2} &= frac{1}{a} - a. end{align*} Multiplying through by a and rearranging gives a^2 + frac{3}{2}a - 1 = 0, which simplifies further to solve for a. This quadratic equation yields a = 2 as the suitable solution (as a > 0 and a neq 1 by the problem's constraints). Thus, we have m = 2 and a = 2, so boxed{m = 2, , a = 2}. # Part (2): Determine if there exists a positive real number t such that g(x) leqslant 0. From part (1), f(x) = 2^x - 2^{-x}, so g(x) = log_{t}[2^{2x} + 2^{-2x} - t(2^x - 2^{-x})]. - For x in [1, log_2 3], we introduce k = 2^x - 2^{-x}, which means k in [frac{3}{2}, frac{8}{3}]. - We then define h(k) = k^2 - tk + 2. The requirement is g(x) leqslant 0 for all x in the given interval, which translates to needing h(k) geq 1. (i) For 0 < t < 1, h(k) needs to be an increasing function in the interval [frac{3}{2}, frac{8}{3}]. We then examine h(frac{3}{2}) geq 0, which simplifies to t leq frac{13}{6}. Thus, for 0 < t < 1, the condition is satisfied. (ii) For t > 1, we explore various configurations and conclude that no value of t satisfies the conditions within the interval [frac{3}{2}, frac{8}{3}] after solving the inequalities. In conclusion, the value of t exists and must be in the interval (0, 1) for the function g(x) leqslant 0 to hold for all x in [1, log_2 3]. Therefore, boxed{t in (0, 1)}.

answer:Let mathbf{v} = begin{pmatrix} x y end{pmatrix} be any vector and mathbf{r} be its reflection over begin{pmatrix} 6 4 end{pmatrix}. First, compute the projection mathbf{p} of mathbf{v} onto begin{pmatrix} 6 4 end{pmatrix}: From the projection formula, [ mathbf{p} = operatorname{proj}_{begin{pmatrix} 6 4 end{pmatrix}} begin{pmatrix} x y end{pmatrix} = frac{begin{pmatrix} x y end{pmatrix} cdot begin{pmatrix} 6 4 end{pmatrix}}{begin{pmatrix} 6 4 end{pmatrix} cdot begin{pmatrix} 6 4 end{pmatrix}} begin{pmatrix} 6 4 end{pmatrix}. ] Calculating the dot products, [ mathbf{p} = frac{6x + 4y}{52} begin{pmatrix} 6 4 end{pmatrix} = frac{6x + 4y}{52} begin{pmatrix} 6 4 end{pmatrix} = begin{pmatrix} frac{36x + 24y}{52} frac{24x + 16y}{52} end{pmatrix} = begin{pmatrix} frac{9x + 6y}{13} frac{6x + 4y}{13} end{pmatrix}. ] Since mathbf{p} is like the midpoint of mathbf{v} and mathbf{r}, [ mathbf{r} = 2 mathbf{p} - mathbf{v} = 2 begin{pmatrix} frac{9x + 6y}{13} frac{6x + 4y}{13} end{pmatrix} - begin{pmatrix} x y end{pmatrix} = begin{pmatrix} frac{5x + 12y}{13} frac{12x - 5y}{13} end{pmatrix}. ] Thus, the reflection matrix is: [ boxed{begin{pmatrix} 5/13 & 12/13 12/13 & -5/13 end{pmatrix}}. ] Conclusion: Each step results in boxed{begin{pmatrix} 5/13 & 12/13 12/13 & -5/13 end{pmatrix}} occurring identically as in the problem's provided solution, indicating that this reflection matrix remains the same if the original vector is scaled.

answer:To solve the given system of linear equations, we need to carefully analyze the structure of the system. The system is: [ begin{aligned} & x_{1}+2 x_{2}+3 x_{3}+ldots+9 x_{9}+10 x_{10}=55, & x_{2}+2 x_{3}+3 x_{4}+ldots+9 x_{10}+10 x_{1}=55, & vdots & x_{10}+2 x_{1}+3 x_{2}+ldots+9 x_{8}+10 x_{9}=55. end{aligned} ] Step 1: Sum all the equations Summing all ten equations, we get: [ begin{aligned} & (x_{1}+2 x_{2}+3 x_{3}+ldots+9 x_{9}+10 x_{10}) & +(x_{2}+2 x_{3}+3 x_{4}+ldots+9 x_{10}+10 x_{1}) & + ldots & +(x_{10}+2 x_{1}+3 x_{2}+ldots+9 x_{8}+10 x_{9}) = 10 times 55. end{aligned} ] Since each x_i appears in the coefficient of a corresponding linear term exactly 55 times in the summation of all the equations, we have: [ 55(x_{1} + x_{2} + x_{3} + ldots + x_{10}) = 550. ] From this, it follows that: [ x_{1} + x_{2} + x_{3} + ldots + x_{10} = 10. quad text{(Equation 1)} ] Step 2: Subtract one equation from the next Subtract the first equation from the second: [ begin{aligned} &(x_{2} + 2 x_{3} + 3 x_{4} + ldots + 9 x_{10} + 10 x_{1}) - (x_{1} + 2 x_{2} + 3 x_{3} + ldots + 9 x_{9} + 10 x_{10}) & = 55 - 55. end{aligned} ] Simplify the resulting equation: [ 9x_{1} - (x_{2} + x_{3} + x_{4} + ldots + x_{10}) = 0. ] Using text{Equation 1} (i.e., x_{1} + x_{2} + ldots + x_{10} = 10), we can rewrite the above equation as: [ 9x_{1} - (10 - x_{1}) = 0, ] which simplifies to: [ 9x_{1} - 10 + x_{1} = 0, ] [ 10x_{1} = 10, ] [ x_{1} = 1. ] Step 3: Extend to other variables By symmetry, the above result holds for each pair of subsequent equations. Therefore, all x_i must be equal because the system is invariant under cyclic permutations. Hence, we find: [ x_{2} = x_{3} = ldots = x_{10} = 1. ] Conclusion All variables are equal to 1: [ x_{1} = x_{2} = ldots = x_{10} = 1. ] Thus, the solution to the system of equations is: [ boxed{x_{1} = x_{2} = x_{3} = ldots = x_{10} = 1}. ]