Given the position function <em>s(t)</em>, you can get the acceleration function by differentiating <em>s</em> twice:
velocity = <em>s'(t)</em> = -5 sin(<em>t </em>) + 3 cos(3<em>t</em> )
acceleration = <em>s''(t)</em> = -5 cos(<em>t</em> ) - 9 sin(3<em>t</em> )
Then when <em>t</em> = <em>π</em>, the particle's acceleration is
<em>s''(π)</em> = -5 cos(<em>π</em>) - 9 sin(3<em>π</em>)
… = -5 • (-1) - 9 • 0 = 5
Answer:45
Step-by-step explanation:
Answer:
x = 48.
Step-by-step explanation:
If AE is 921, then EC is equivalent to that. First, you subtract 9 from 921, then divide by nineteen to get your answer: x = 48.
Use the identity
sec^2x = 1 + tan^2 x
- so sec x = sqrt(1 + tan^2 x) then:-
tan x + sqrt( 1 + tan^2 x) = 1
sqrt ( 1 + tan^2 x) = 1 - tan x
1 + tan^2 x = 1 + tan^2x - 2 tan x
0 = -2 tanx
tan x = 0
x = 0, π
π is an extraneous root because sec 180 = -1
So the answer is 0 degrees