The unoccupied volume will be the volume of the can minus the volume of the three tennis balls.
V=HπR^2-3(4πr^3)/3
V=HπR^2-4πr^3 where H=height of can, R=radius of can, r=radius of ball
We are told H=7, R=1.5, and r=1.125 (2.25/2=d/2=r) so
V=7π1.5^2-4π1.125^3
V=15.75π-5.6953125π in^3
V≈31.58 in^3
If SU bisects TSV, then TSU = USV
4y + 11 = 6y + 5
6y - 4y = 11 - 5 = 6
y = 6/2 = 3
Therefore, m<TSU = 4(3) + 11 = 12 + 11 = 23
0.12 I think I’m not sure how to do this I ji at have an idea maybe I’m super wrong
Cos (π/2 - x) = sin x = 3/5
tan x = sin x / cos x = 3/5 / 4/5 = 3/4
csc x = 1/sin x = 1 / 3/5 = 5/3
sec x = 1/cos x = 1 / 4/5 = 5/4
cot x = 1/tan x = 1 / 3/4 = 4/3.
Answer:
Is there a link or image that shows us the mid segment?
Step-by-step explanation:
