Graph the feasible region for the following system of inequalities: 0.25x + 0.60y <= 75, 0.60x + 0.20y <= 90, x >= 0, y >= 0.

You're on a roll — I cannot draw the graph directly, but I can give you precise instructions and the key points you need to plot it accurately. The system of inequalities for Question 2 is: 1. 0.25x + 0.60y 75 2. 0.60x + 0.20y 90 3. x 0 4. y 0 Step 1: Draw the axes. Draw a coordinate plane with the x-axis representing the number of barrels of pesticide A (x) and the y-axis representing the number of barrels of pesticide B (y). Since x 0 and y 0, focus on the first quadrant. Step 2: Plot the boundary line for the first inequality (0.25x + 0.60y = 75). To find the x-intercept, set y=0: 0.25x = 75 x = (75)/(0.25) = 300. Plot (300, 0). To find the y-intercept, set x=0: 0.60y = 75 y = (75)/(0.60) = 125. Plot (0, 125). Draw a solid line connecting (300, 0) and (0, 125). Step 3: Plot the boundary line for the second inequality (0.60x + 0.20y = 90). To find the x-intercept, set y=0: 0.60x = 90 x = (90)/(0.60) = 150. Plot (150, 0). To find the y-intercept, set x=0: 0.20y = 90 y = (90)/(0.20) = 450. Plot (0, 450). Draw a solid line connecting (150, 0) and (0, 450). Step 4: Find the intersection point of the two lines. Solve the system of equations: 1) 0.25x + 0.60y = 75 2) 0.60x + 0.20y = 90 From previous calculations, the intersection point is approximately (125.81, 72.58). Plot this point. Step 5: Shade the feasible region. For 0.25x + 0.60y 75, shade the region below the line connecting (300,0) and (0,125). For 0.60x + 0.20y 90, shade the region below the line connecting (150,0) and (0,450). Since x 0 and y 0, the feasible region is restricted to the first quadrant. The feasible region is the area in the first quadrant that is below both lines. It is a polygon with the following vertices: (0,0) (150,0) ((3900)/(31), (2250)/(31)) ≈ (125.81, 72.58) (0,125) This shaded region represents all possible combinations of pesticide A and B production that satisfy the pollution laws. What's next?