First find the area of the sector, which includes the shaded region's area and the area of the triangle JKL. Let be the area of the sector. This area occurs in a fixed ratio with the area of the entire circle based on the measure of the central angle subtended by the arc LK:
To get the area of the shaded region, subtract from the area of the triangle.
The area of a triangle is 1/2 the base times the height. If we bisect the central angle with a line segment that meets the side KL at its midpoint, then we get a right triangle with hypotenuse 27 and one angle of measure 27º (half of the central angle). This triangle has base and height such that
So the right triangle has area approximately 1/2*12.258*24.057, or about 147.443. Triangle JKL is made up of two of these right triangles, so it has area of 294.887.
Subtracting this from gives an area of the shaded region of about 48.472, which we round up to 48.5.
Answer:
10
Step-by-step explanation:
<h2>—Math</h2>
½ x + 5 = 10
½ x = 10 – 5
x = 5 x 2
x = 10
<em> </em><em>follow </em><em>my </em><em>account</em><em> </em><em>^</em><em>^</em>
The and would be y=-1/3x-5 because to find slope you would use the equation y2-y1 divided by x2-x1 and find the slope of -4/12 which simplifies to -1/3 and the y-int is -5 so you would just plug that in.
Answer:
x=1/2y+2
Step-by-step explanation:
−4x+2y=−8
Step 1: Add -2y to both sides.
−4x+2y+−2y=−8+−2y
−4x=−2y−8
Step 2: Divide both sides by -4.
−4x
−4
=
−2y−8
−4