Calculate, correct to two decimal places, the area enclosed by the line 3x - 5y + 4 = 0 and the axes.

To find the area enclosed by the line 3x - 5y + 4 = 0 and the axes, we need to find the points where the line intersects with the x-axis and y-axis. First, let's find the x-intercept. To do this, we set y = 0 and solve for x: 3x - 5(0) + 4 = 0 3x + 4 = 0 3x = -4 x = -4/3 So the x-intercept is (-4/3, 0). Next, let's find the y-intercept. To do this, we set x = 0 and solve for y: 3(0) - 5y + 4 = 0 -5y + 4 = 0 -5y = -4 y = 4/5 So the y-intercept is (0, 4/5). Now we can draw a triangle with vertices at the origin (0,0), the x-intercept (-4/3, 0), and the y-intercept (0, 4/5). The base of the triangle is the x-intercept, which has a length of 4/3. The height of the triangle is the y-intercept, which has a length of 4/5. Therefore, the area of the triangle is: (1/2) * base * height = (1/2) * (4/3) * (4/5) = 8/15 = 0.53 (rounded to two decimal places) Therefore, the answer is option C: 0.53 square units.