answer:To sketch the graph of (y=3 cos (x+pi) -3), we can use the following steps: 1. **Identify the amplitude, period, and phase shift.** - The amplitude is the distance from the center line to the top or bottom of the wave. In this case, the amplitude is 3. - The period is the distance between two consecutive peaks or troughs of the wave. In this case, the period is (2pi). - The phase shift is the horizontal shift of the wave. In this case, the phase shift is (pi). 2. **Use the sum and difference formulas to simplify the equation.** - We can use the sum and difference formulas for cosine to rewrite the equation as follows: y = 3 (cos x cos pi - sin x sin pi) - 3 y = 3 (cos x (-1) - sin x (0)) - 3 y = -3 cos x - 3 3. **Sketch the graph of the simplified equation.** - The graph of (y = -3 cos x - 3) is a cosine wave with an amplitude of 3, a period of (2pi), and a phase shift of (pi). - The graph is shifted down 3 units from the graph of (y = cos x). - The graph is reflected over the x-axis. Here is a graph of (y = -3 cos x - 3): graph{y = -3cosx - 3 [-10, 10, -5, 5]}

answer:The inverse of the given matrix is: left( begin{array}{cc} -frac{2}{13} & frac{5}{13} frac{1}{13} & frac{4}{13} end{array} right)

answer:The slope of the short-run Phillips curve represents: e. The short-run trade-off between the unemployment rate and inflation The short-run Phillips curve shows the relationship between the unemployment rate and the inflation rate, assuming the natural unemployment rate and expected inflation rate remain constant. As depicted in the graph below, a higher inflation rate is associated with a lower unemployment rate, while a lower inflation rate corresponds to a higher unemployment rate. For instance, if inflation is 7%, unemployment is around 4%, but if inflation drops to 3%, unemployment increases to 8%. <img src="multimages/16/phillips6661858582167624453.png" alt="Short-run Phillips Curve Graph" />

answer:{eq}displaystyle int {sin (3x)cos (x)} dx {/eq} Using the product-to-sum identity {eq}displaystyle sin (A)cos (B) = frac{1}{2}[sin (A + B) + sin (A - B)] {/eq}, we have: {eq}displaystyle = frac{1}{2}int {[sin (3x + x) + sin (3x - x)]} dx displaystyle = frac{1}{2}int {[sin (4x) + sin (2x)]} dx {/eq} Integrating each term separately: {eq}displaystyle = frac{1}{2}left[ { - frac{{cos (4x)}}{4} - frac{{cos (2x)}}{2}} right] + C displaystyle = - frac{{cos (4x)}}{8} - frac{{cos (2x)}}{4} + C {/eq}