The trigonometric function that models the distance (feet) of the rider from the camera as a function of time (seconds) is γ(t) = ωt, where ω is the angular velocity of merry-go-round.
Let, center of the merry-go-round is C and camera is placed at point A. B(t) define the position of the rider at any time t. The angle between these three-point A, C and B is y(t). Radius (r) of the merry-go-round is 3 feet and distance (d) of the rider from the camera is 6 and the angular velocity of the rider is ω.
Assume the rider is at the edge of the merry-go-round (as the position is not specified). So, the length of CB(t) is r. To solve this problem lets consider that angular velocity of merry-go-round is constant, ω = 0 and y(t) = 0.
Therefore, we have y(t) = ωt
So, the the distance (feet) of the rider from the camera is (from the triangle AB(t)C)
C(t) = √(r² + d² - 2rdcos(y(t)) = √(45 - 36cos(ωt) = 3√(5 - 4cos(ωt))
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Answer:
f ( -2 ) = 30
g ( -2 ) = 3
Step-by-step explanation:
when it says f(x) and then has f(-2) - you fill in all x's in the equation with -2.. like this:
f(x) = -7(-2-2) +2 g(x) = 2(-2)^2 -5
f(x) = -7(-4) +2 g(x) = 2(4) -5
f(x) = 28 + 2 g(x) = 8 - 5
f(x) = 30 g(x) = 3
hope this helps!!
Less than because if you subtract 11, 138 minus 9,405 you get the answer you get 1,733 and that is less than 2,000
Answer:
c) 0
Step-by-step explanation:
m= slope
m= (y2 -y1)/(x2-x1)
we have the points:
(12,6) and (-1,6)
x1= 12 y1= 6
x2= -1 y2= 6
so we have:
m= (6-6)/( -1 -12)
m= 0/ -13
m= 0
