Question 3 40 pts An object is attached to a spring that is stretched and released. The equation d = a (cos(bt)) models the distance, d, of the object in inches above or below the rest position as a function of time, t, in seconds. If a = 5 and b = 5 Approximately when will the object be 2 inches above the rest position? Round to the nearest hundredth, if necessary.

The object will be 2 inches above the rest position when at approximately 0.23 seconds

Step-by-step explanation:

The given parameters are;

The equation that models the distance, d, of the object in inches above or below the rest position = a×(cos(b·t))

If a = 5 and b = 5, we have;

d = 5 × (cos(5 × t))

When the object is approximately 2 inches above the ground, we have;

d = 2 = 5 × (cos(5 × t))

cos(5 × t) = 2/5

5 × t = cos⁻¹(2/5) ≈ 1.16

t =  (cos⁻¹(2/5))/5 ≈ 1.16/5

t ≈ 0.23 s

The object will be 2 inches above the rest position when t ≈ 0.23 s.

the answer is 510

step-by-step explanation:

15 • 13 = 195

15 • 12 = 180

15 • 5 = 75

12 • 5 = 60

195 + 180 = 375

375 + 75 = 450

450 + 60 = 510

fist on the left goes with the third on the right. second on the left goes with the fourth on the right. third on the left goes with the second on the right. the fourth on the left goes with first on the right