answer:The volume of a box is calculated by multiplying its length, width, and height. Given the dimensions of the box as 5 units in length, 5 units in width, and 3 units in height, the volume can be found as follows: Volume = Length times Width times Height Volume = 5 times 5 times 3 Volume = 75 Therefore, the volume of the box is 75 cubic units.

answer:Jim's distance from the diving board when he lands can be explained using the principles of kinematics. We consider two dimensions: the horizontal (x-direction) and the vertical (y-direction). In the x-direction, Jim's motion is constant, while in the y-direction, he accelerates due to gravity. For the horizontal distance (x), Jim's displacement is given by his initial velocity (v) multiplied by the time (t) he spends in the air: {eq}displaystyle x = vt {/eq} In the vertical direction (y), Jim's displacement from the initial height (h) to the water is given by: {eq}displaystyle y = h - frac{1}{2}gt^2 {/eq} where {eq}g {/eq} is the acceleration due to gravity. To find the time Jim spends in the air before hitting the water (which is the same for both boards), we set the vertical displacement to 0 (since he lands at water level): {eq}displaystyle 0 = h - frac{1}{2}gt^2 {/eq} Solving for time gives: {eq}displaystyle t = sqrt{frac{2h}{g}} {/eq} Substituting this time into the horizontal displacement equation: {eq}displaystyle x = vsqrt{frac{2h}{g}} {/eq} This shows that the horizontal distance Jim travels depends on the height of the diving board (h). Since a higher board (h2) has a longer time (t2) before hitting the water than a lower one (h1), he lands farther from the end of the higher board: {eq}displaystyle x_{h2} > x_{h1} {/eq} Thus, Jim lands farther from the end of the diving board when running off a higher board because he spends more time in the air, allowing his horizontal velocity to carry him a greater distance.

answer:To prove the limit using the precise definition, let 0 < varepsilon < 1. We aim to find a delta > 0 such that |f(x) - L| < varepsilon whenever 0 < |x - a| < delta, where f(x) = frac{x^3 - 1}{x - 1}, L = 3, and a = 1. Observe that frac{x^3 - 1}{x - 1} = x^2 + x + 1 for x neq 1. Therefore, we need to show |x^2 + x + 1 - 3| = |x^2 + x - 2| < varepsilon. To handle the inequality |(x-1)(x+2)|<varepsilon, we can restrict x to the interval (0, 2) to ensure |x+2| leq 4. This leads to: |(x-1)(x+2)| leq 4|x-1| < 4delta For this to be less than varepsilon, we require delta leq frac{varepsilon}{4}. However, we also need to ensure delta leq 1 to maintain the interval restriction. Hence, choose delta = minleft(1, frac{varepsilon}{4}right). If 0 < |x - 1| < delta, we have: left|frac{x^3-1}{x-1} - 3right| = |(x-1)(x+2)| leq 4|x-1| < 4delta leq varepsilon This establishes the limit lim_{x to 1} frac{x^3-1}{x-1} = 3 using the precise definition of limits.

answer:The mean of a set of numbers is calculated by summing all the numbers and dividing by the count of numbers. Let's first simplify each term by multiplying by sqrt{3} to get rid of the fractions: 2 sqrt{3} cdot sqrt{3} = 2 cdot 3 = 6 -frac{7}{sqrt{3}} cdot sqrt{3} = -7 frac{1}{sqrt{3}} cdot sqrt{3} = 1 frac{11}{sqrt{3}} cdot sqrt{3} = 11 Now, compute the mean: Mean = frac{6 + (-7) + 1 + 11}{4} = frac{11}{4} So, the mean of the given set is frac{11}{4}.