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:
Bigger size / smaller size = 40
Step-by-step explanation:
Notice that we
36 / (9/10) = 30 / (3/4) = 40
Therefore the proportion model would be
Bigger size / smaller size = 40
x^2 = 100...by taking the square root of both sides, this eliminates the ^2
x = (+-) sqrt 100
x = (+-) 10
so the values of x are -10 and 10
Answer:
6+x with x equalling the amount of coins bought if that is what you were looking for
Step-by-step explanation:
if you want to make and equation out of this you don't know the amount of coins she bought so you use a variable (x) to replace that unknown number then since it says she bought more she is adding more to her collection which means it would be an addition problem, leaving you with 6+x for your final answer
