answer:24,960

answer:Problem 1: We will use the time of the second car as the reference time t. That means that the first car starts accelerating at the time (t + 6) s. The displacement s for the initial value u and the acceleration a is given by the equation below, for any time t. s = ut + 1/2 at^2 Taking the acceleration of the first car as a1 = 3 m/s^2, the displacement of the car at any time t is: s1 = 1/2 a1 (t + 6)^2 Taking the acceleration of the second car as a2 = 5 m/s^2, the displacement of the car at any time t is: s2 = 1/2 a2 t^2 In the instance that the second car overtakes the first, the displacements are the same, i.e. s1 = s2 1/2 a1 (t + 6)^2 = 1/2 a2 t^2 a1 (t + 6)^2 = a2 t^2 3 (t + 6)^2 = 5 t^2 3t^2 + 36t + 108 = 5t^2 2t^2 - 36t - 108 = 0 t^2 - 18t - 54 = 0 t = (18 ± √(18^2 - 4(1)(-54))) / 2(1) t ≈ 20.8 s Therefore, it takes the second car approximately 20.8 seconds to overtake the first car. Problem 2: Taking the downwards direction as positive, we have the initial velocity u = 25 m/s, the gravitational acceleration a = g = 9.81 m/s^2, and the time of flight is t = 2 s. Using a suitable equation, the height s of the building is: s = ut + 1/2 gt^2 = (25 m/s) (2 s) + 1/2 (9.81 m/s^2) (2 s)^2 = 69.62 m Therefore, the building is 69.62 meters high.

answer:The correct representations are: x > -5/2 Step-by-step explanation: Starting with the given inequality -3(2x - 5) < 5(2 - x), let's solve it: 1. Expand the parentheses: -6x + 15 < 10 - 5x 2. Move all x terms to one side and constants to the other: -6x + 5x < 10 - 15 3. Simplify: -x < -5 4. Divide both sides by -1, remembering to reverse the inequality sign: x > 5/2 Therefore, the correct option is x > 5/2. Note that "x > 5" is incorrect, as it has been incorrectly derived.

answer:Yes, an alternative approach exists. Utilize the property that cosh(t) = frac{e^t + e^{-t}}2. Thus, we can rewrite f(t) as follows: f(t) = t^{10}left(frac{e^t + e^{-t}}2right) = frac{1}{2}t^{10}e^t + frac{1}{2}t^{10}e^{-t} Now apply the linearity of the Laplace transform and the specific Laplace transforms for functions of the form t^n e^{at}: mathcal{L}{f(t)} = frac{1}{2}mathcal{L}{t^{10}e^t} + frac{1}{2}mathcal{L}{t^{10}e^{-t}} Using the Laplace transform identities: mathcal{L}{t^n e^{at}} = frac{n!}{(s-a)^{n+1}} quad text{and} quad mathcal{L}{e^{ct} f(t)} = F(s-c) where n in mathbb{Z}^+ and F = mathcal{L}{f}, we find: mathcal{L}{f(t)} = frac{1}{2}cdotfrac{10!}{(s-1)^{11}} + frac{1}{2}cdotfrac{10!}{(s+1)^{11}} This approach simplifies the derivative calculation significantly, and other Laplace transform identities may also provide shortcuts.