Sign In
x≥2 & (I) y≥-3 & (II) x+y≤4 & (III) The boundary line of Inequality (I) is going to be a vertical line at x=2. The inequality is non-strict, so the boundary line will be solid. Since x-values greater than or equal to 2 lie to the right of 2 on a number line, the shaded region will be to the right of the boundary line.
The boundary line of Inequality (II), y≥-3, is going to be a horizontal line at y=-3. The inequality greater than or equal to is non-strict, so we will use a solid line, and the shaded region will be above the line.
For Inequality (III), we will need to isolate the y-variable in order to graph the boundary line. x+y=4 ⇔ y = - x + 4 In order to find which region to shade, we will use the test point (0,0). If substituting the point into the inequality results in a true statement, we will shade the region that contains the point. Otherwise, we will shade the region that does not contain the point.
The inequality is non-strict, so the boundary line will be solid. Since substituting the test point made the inequality true, we will shade the region that contains it.
The solution set of the system is the intersection of all the shaded regions.
The vertices are located at (2,2), (2,-3), and (7,-3).
A= 12bh.
Notice that the two sides are perpendicular. This means we can use one of them as the base and the other as the height of our triangle. Now that we have our base and height lengths, we can substitute their values into the formula.
b= 5, h= 5
Multiply
1/b* a = a/b
Calculate quotient
The area of the triangle is 12.5 square units.