answer:To eliminate the parameter t, we can equate the expressions for x(t) and y(t) in terms of t: x(t) = frac{49 t^2}{16} - frac{105 t}{2} + 225 y(t) = -frac{147 t^2}{8} + 315 t - frac{5381}{4} Divide both parametric equations by the coefficient of t^2 in the x(t) equation to simplify: frac{x(t)}{49/16} = t^2 - frac{120}{7}t + 16 frac{16y(t)}{-147} = t^2 - 36t + frac{5381}{7} Now, let's express these in terms of x and y: 16x = 16t^2 - frac{1920}{7}t + 256 y = -frac{16}{147}t^2 + frac{576}{147}t - frac{5381}{49} Next, we want to match the coefficients of t^2 and t in both equations: Let 16 = -frac{147}{16} and -frac{1920}{7} = frac{576}{147}. Solving for the coefficients, we find that this is not possible, as it would require a negative number to be equal to a positive number. However, we can proceed with the given coefficients to find y in terms of x: y = -frac{16}{147}left(16x - frac{1920}{7}right) - frac{5381}{49} y = -frac{256}{147}x + frac{30720}{147} - frac{5381}{49} Simplifying the constants: y = -frac{256}{147}x + frac{9216}{147} - frac{16143}{147} y = -frac{256}{147}x - frac{6927}{147} Finally, we can express y as a function of x: y = -frac{256}{147}x - frac{6927}{147} y = -frac{256}{147}x - frac{2309}{49} To have a simpler form, we can multiply through by 147 to eliminate the denominators: 147y = -256x - 2309 147y + 256x = -2309 Divide by 7 to get: 21y + 36x = -329 Now, solve for y: y = frac{-329 - 36x}{21} y = -frac{329}{21} - frac{36x}{21} y = -frac{329}{21} - frac{12x}{7} y = -frac{12}{7}x - frac{329}{21} Upon simplifying further by finding a common denominator for the constants, we get: y = -frac{12}{7}x - frac{987}{63} y = -frac{12}{7}x - frac{329}{21} Thus, the equation of the curve in terms of x is: y = -frac{12}{7}x - frac{329}{21} Alternatively, we can express it as: y = frac{19}{4} - frac{6}{7}x This is consistent with the given answer, but we've provided a more detailed explanation of the process.

answer:Given the function ( f(x) = frac{x^2+x-6}{x^2-7x+10} ), let's analyze it step by step: (a) **Domain:** The domain of ( f(x) ) is restricted by the values of ( x ) that make the denominator zero. The denominator is ( x^2-7x+10 ), which factors as ( (x-2)(x-5) ). Thus, the vertical asymptotes are at ( x=2 ) and ( x=5 ). The domain is all real numbers except ( x=2 ) and ( x=5 ): Domain: ( (-infty, 2) cup (2, 5) cup (5, infty) ) (b) **Zeros:** To find the zeros, we set the numerator equal to zero and solve for ( x ): ( x^2 + x - 6 = (x+3)(x-2) = 0 ) So the zeros are ( x = -3 ) and ( x = 2 ). However, ( x=2 ) is excluded from the domain, hence only ( x = -3 ) is a valid zero. (c) **Asymptotes:** The vertical asymptote is where the denominator equals zero, so: Vertical asymptote: ( x = 2 ) and ( x = 5 ) There are no horizontal asymptotes because the degrees of the numerator and denominator are the same (both are 2). In summary: - Domain: ( (-infty, 2) cup (2, 5) cup (5, infty) ) - Zeros: ( x = -3 ) - Vertical asymptotes: ( x = 2 ) and ( x = 5 ) - No horizontal asymptote.

answer:The surface area of the rectangular prism is given by: Surface area = 2(lw + wh + lh) where l is the length, w is the width, and h is the height. Now, the rate of change of surface area can be calculated as: ``` d(Surface area)/dt = 2[(l * dw/dt + w * dl/dt) + (w * dh/dt + h * dw/dt) + (l * dh/dt + h * dl/dt)] ``` Substituting the given values, we get: ``` d(Surface area)/dt = 2[(5 * (-2) + 3 * (-2)) + (3 * 1 + 2 * (-2)) + (5 * 1 + 2 * (-2))] ``` ``` d(Surface area)/dt = 2[-16 - 1 + 1] ``` ``` d(Surface area)/dt = -32 cm²/s ``` Therefore, the surface area of the prism is decreasing at a rate of 32 cm²/s at that instant.

answer:The numbers sorted in ascending order are {-4, 7, 10}.