An object is attached to a spring that is stretched and released. The equation d= -8*cos((pi/6)*t) models the distance, d, of th e object in inches above or below the rest position as a function of time, t, in seconds. Approximately when will the object be 6 inches above the rest position? Round to the nearest hundredth, if necessary.
2 answers:
Answer:
Please see the attached graph for answer.
The object will be at 6 inches above the rest position at approximately 4.62 seconds
Step-by-step explanation:
To easily solve this problem, we only need to graph the equation in a calculator and find the solution at d = 6 inches
We can see in the graph that at t = 4.62 the height is equal to d = 6.001 which is the approximation of the required value.
The graph is cyclic, because it is modeled as an harmonic oscillator without any damping.
Answer:
c
Step-by-step explanation:
4.62
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Answer:
4x-12
Step-by-step explanation:
2x-5[2x-7(5)}=2x+2x-7-5
4x-12
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Answer:
See this attachment...
Answer:
- 12
Step-by-step explanation:
Substitute both the values of x and y in the equation given
x=2, y=4
4(2)-5(4)= 8-20
= - 12
It would be B. She is 30 inches in height (y-axis) and she is 1 year old, or 12 months (x-axis). Hope this helped :)
The perimeter = 4 * length of a semicircle of radius 5 == 4* pi * 5 = 20 pi cm The area of one of the clear parts = 1/2 area of square - area semicircle = 1/2 * 10^2 - 1/2 pi *5^2 = 50 - 12.5pi So area of shaded parts = area of square - 4 (50 - 12.5pi) = 100 - 200 + 100pi = (100pi - 100) cm^2