Answer:
AFB, BDC, and FBD
Step-by-step explanation:
We have that
2r−9≤−6------> 2r ≤ -6+9-------> 2r ≤ 3-----> r ≤ 1.5
the solution is the interval (-∞, 1.5]
using a graph tool
see the attached figure
<span>The volume of the cylinder will be 2</span>π
As shown in figure, R is the base
radius of the cylinder.
H is the height of the cylinder.
Given: H = diameter of the base of
cylinder.
Now, diameter of the base = 2R
So, H = 2R
<span>Volume of cylinder = π×</span>×H
<span> = π×</span>×(2R)
<span> = 2 π </span> cubic units.