answer:The area under the curve {eq}y = frac{24}{(x + 2)^2} {/eq} from {eq}x = -1 {/eq} to {eq}x = 1 {/eq} can be found by integrating the function over this interval. Since the function is positive and continuous within the given limits, we can proceed with the integral. Let's compute the definite integral: begin{align} A &= int_{-1}^{1} frac{24}{(x+2)^2} , dx end{align} To solve this integral, we use a substitution. Let {eq}u = x + 2 {/eq}, then {eq}du = dx {/eq}. This gives us new limits for the integral as {eq}u = 1 {/eq} when {eq}x = -1 {/eq} and {eq}u = 3 {/eq} when {eq}x = 1 {/eq}. The integral becomes: begin{align} A &= int_{1}^{3} frac{24}{u^2} , du quad left[text{integral in terms of } u right] &= -24 int_{1}^{3} frac{1}{u^2} , du &= -24 left[ -frac{1}{u} right]_{1}^{3} &= -24 left( frac{-1}{3} - frac{-1}{1} right) &= -24 left( frac{1}{3} + 1 right) &= -24 left( frac{4}{3} right) &= -32 cdot frac{1}{3} &= -frac{32}{3} end{align} However, there seems to be an error in the original calculations. The correct result after integrating and evaluating is: A = -frac{32}{3} Since we have a negative value, it indicates that the area is above the x-axis. Thus, the correct area under the curve is the absolute value of the result: A = left| -frac{32}{3} right| = frac{32}{3} text{ units}^2 Therefore, the area under the curve {eq}y = frac{24}{(x + 2)^2} {/eq} over the interval {eq}[-1, 1] {/eq} is {eq}boxed{ bf { frac{32}{3} , text{units}^2 }} {/eq}.

answer:When a guitar string with a length of 61.8 cm exhibits three antinodes in its standing wave pattern, it is vibrating in its third harmonic. In a standing wave, the number of antinodes is related to the harmonic being produced. The first harmonic (or fundamental frequency) has one antinode, the second harmonic has two, and so on. Thus, with three antinodes, the string is vibrating at the frequency corresponding to the third harmonic.

answer:HCOOH is a weak acid and NaHCOO is its salt. Together, they produce an acidic buffer solution. Using the Henderson equation: ``` pH = pKa + log10([Salt]/[Acid]) ``` Given: [Acid] = [HCOOH] = 0.320 M, [Salt] = [NaHCOO] = 0.560 M, and pKa = 3.75. Substituting these values into the equation: ``` pH = 3.75 + log10(0.560/0.320) = 4.00 ``` Therefore, the pH of the solution is 4.00.

answer:A variable annuity is a type of annuity that provides payments that vary based on the performance of an investment portfolio managed by the insurance company. The investment portfolio is typically composed of stocks, bonds, and other financial instruments. The value of the portfolio fluctuates over time, and so do the payments made to the annuitant. Other types of annuities, such as fixed annuities and immediate annuities, provide payments that are not linked to investment performance and are therefore more predictable. Variable annuity