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