An object is attached to a spring that is stretched and released. The equation d = -8 cos (pi/6t) models the distance, d, of the 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.

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7.9m and 4.3°

step-by-step explanation:

let east be +x and north be +y, x = distance * cos θ, y = distance * sin θ  

for 3.5 south, y = -3.5  

for 8.2, x = 8.2 * cos 30 and y = 8.2 * sin 30  

for 15 west, x = -15  

total x = 8.2 * cos 30 – 15 ≈ -7.9  

total y = -3.5 + 8.2 * sin 30 = 0.6  

magnitude = [(8.2 * cos 30 – 15)^2 + 0.6^2)]^0.5  

the magnitude is approximately 7.9 m  

tan θ = y/x = 0.6/-7.9  

the angle is approximately 4.3˚ north of west

the answer is: -22 also a

hope i

-lavira: )