By definition, the arc length is given by:
arc = R * theta * ((2 * pi) / 360)
Where,
theta: angle in degrees
R: radio
We have then:
(Arc) QPT if <QZT = 120:
theta = 360-120 = 240 degrees
R = 13.5 units
Substituting values we have:
(Arc) QPT = R * theta * ((2 * pi) / 360)
(Arc) QPT = (13.5) * (240) * ((2 * pi) / 360)
(Arc) QPT = 56.55 units
Answer:
(Arc) QPT = 56.55 units
Answer: area
Step-by-step explanation:
area of the wall
Plug in t = 0. This is the same as replacing every copy of 't' with '0'. After that you evaluate
h = -t^2 + 95
h = -(t)^2 + 95
h = -(0)^2 + 95
h = 0 + 95
h = 95
So when the time is t = 0 seconds, the height of the ball is 95 feet. A likely location is that the boy is on top of a very tall building (roughly 95 feet or so) when he threw this ball into the air.
Use combination to solve it
13C10 =
13C10 =
13C10 =
13C10 =
13C10 =
13C10 = 286
There are 286 ways
Answer:
6:1
Step-by-step explanation:
If it's 12:2, then divide it by 2 on each side to get 6:1 or 6
