Distribute the -2 to get
-16x + 8 < 2x + 5
3 < 18x
1/6 < x
so x > 1/6
66/55 in simplest form is 6/5. Because...
66÷ 11= 6
55÷ 11= 5
Answer:
a polyhedron is a three dimensional shape with flat polygonal faces, straight edges and sharp corners or vertices.
a structure whose outer surfaces are triangular and converge to a single step at the top, making the shape roughly a pyramid in the geometric sense
polygon with three edges and three vertices. It is one of the basic shapes in geometry. A triangle with vertices A, B, and C is denoted.
solid geometry is the traditional name for the geometry of three-dimensional Euclidean space. Stereometry deals with the measurements of volumes of various solid figures including pyramids, prisms and other polyhedrons; cylinders; cones; truncated cones; and balls bounded by spheres
a solid geometric figure whose two end faces are similar, equal, and parallel rectilinear figures, and whose sides are parallelograms.
Step-by-step explanation:
Area inside the semi-circle and outside the triangle is (91.125π - 120) in²
Solution:
Base of the triangle = 10 in
Height of the triangle = 24 in
Area of the triangle = 

Area of the triangle = 120 in²
Using Pythagoras theorem,




Taking square root on both sides, we get
Hypotenuse = 23 inch = diameter
Radius = 23 ÷ 2 = 11.5 in
Area of the semi-circle = 

Area of the semi-circle = 91.125π in²
Area of the shaded portion = (91.125π - 120) in²
Area inside the semi-circle and outside the triangle is (91.125π - 120) in².
9 more keepers would need to be added i believe <span />